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+\section{Conjugate of Subgroup is Subgroup}
+Tags: Conjugacy, Subgroups, Conjugate of Subgroup is Subgroup
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]].
+Let $H \le G$ be a [[Definition:Subgroup|subgroup]] of $G$.
+Then the [[Definition:Conjugate of Group Subset|conjugate]] of $H$ by $a$ is a [[Definition:Subgroup|subgroup]] of $G$:
+:$\forall H \le G, a \in G: H^a \le G$
+\end{theorem}
+
+\begin{proof}
+Let $H \le G$.
+First, we show that $x, y \in H^a \implies x \circ y \in H^a$:
+{{begin-eqn}}
+{{eqn | l = x, y
+      | o = \in
+      | r = H^a
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = a x a^{-1}, a y a^{-1}
+      | o = \in
+      | r = H
+      | c = {{Defof|Conjugate of Group Subset}}
+}}
+{{eqn | ll= \leadsto
+      | l = \paren {a x a^{-1} } \paren {a y a^{-1} }
+      | o = \in
+      | r = H
+      | c = {{GroupAxiom|0}}
+}}
+{{eqn | ll= \leadsto
+      | l = a \paren {x y} a^{-1}
+      | o = \in
+      | r = H
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = x y
+      | o = \in
+      | r = H^a
+      | c = {{Defof|Conjugate of Group Subset}}
+}}
+{{end-eqn}}
+Next, we show that $x \in H^a \implies x^{-1} \in H^a$:
+{{begin-eqn}}
+{{eqn | l = x
+      | o = \in
+      | r = H^a
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = a x a^{-1}
+      | o = \in
+      | r = H
+      | c = {{Defof|Conjugate of Group Subset}}
+}}
+{{eqn | ll= \leadsto
+      | l = \paren {a x a^{-1} }^{-1} = a x^{-1} a^{-1}
+      | o = \in
+      | r = H
+      | c = [[Power of Conjugate equals Conjugate of Power]]
+}}
+{{eqn | ll= \leadsto
+      | l = x^{-1}
+      | o = \in
+      | r = H^a
+      | c = {{Defof|Conjugate of Group Subset}}
+}}
+{{end-eqn}}
+Thus by the [[Two-Step Subgroup Test]], $H^a \le G$.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let $*: G \times G / H \to G / H$ be the [[Definition:Group Action on Coset Space|group action on the (left) coset space]]:
+:$\forall g \in G, \forall g' H \in G / H: g * \paren {g' H} := \paren {g g'} H$
+It is established in [[Action of Group on Coset Space is Group Action]] that $*$ is a [[Definition:Group Action|group action]].
+Then from [[Stabilizer of Coset under Group Action on Coset Space]]:
+:$\Stab {a H} = a H a^{-1}$
+where $\Stab {a H}$ the [[Definition:Stabilizer|stabilizer]] of $a H$ under $*$.
+The result follows from [[Stabilizer is Subgroup]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Inner Automorphisms form Normal Subgroup of Automorphism Group}
+Tags: Normal Subgroups, Automorphism Groups, Inner Automorphisms
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]].
+Then the [[Definition:Set|set]] $\Inn G$ of all [[Definition:Inner Automorphism|inner automorphisms]] of $G$ is a [[Definition:Normal Subgroup|normal subgroup]] of the [[Definition:Automorphism Group of Group|automorphism group]] $\Aut G$ of $G$:
+:$\Inn G \lhd \Aut G$
+\end{theorem}
+
+\begin{proof}
+Let $G$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
+From [[Inner Automorphisms form Subgroup of Automorphism Group]], $\Inn G$ forms a [[Definition:Subgroup|subgroup]] of $\Aut G$.
+It remains to be shown that $\Inn G$ is [[Definition:Normal Subgroup|normal in $\Aut G$]].
+Let $\kappa_x: G \to G$ be the [[Definition:Inner Automorphism|inner automorphism]] defined as:
+:$\forall g \in G: \map {\kappa_x} g = x g x^{-1}$
+Let $\phi \in \Aut G$.
+If we can show that:
+:$\forall \phi \in \Aut G: \forall \kappa_x \in \Inn G: \phi \circ \kappa_x \circ \phi^{-1} \in \Inn G$
+then by the [[Normal Subgroup Test]]:
+:$\Inn G \lhd \Aut G$
+Fix $\kappa_x \in \Inn G$.
+We claim $\phi \circ \kappa_x \circ \phi^{-1} = \kappa_{\map \phi x}$.
+Since $\phi \in \Aut G$ then $\phi$ is, in particular, a [[Definition:Group Homomorphism|homomorphism]].
+Therefore:
+{{begin-eqn}}
+{{eqn | ll= \forall g \in G:
+      | l = \map {\paren {\phi \circ \kappa_x \circ \phi^{-1} } } g
+      | r = \map {\paren {\phi \circ \kappa_x} } {\map {\phi^{-1} } g}
+      | c = 
+}}
+{{eqn | r = \map \phi {x \phi^{-1} \paren g x^{-1} }
+      | c = 
+}}
+{{eqn | r = \map \phi x \map \phi {\map {\phi^{-1} } g} \map \phi {x^{-1} }
+      | c = 
+}}
+{{eqn | r = \map \phi x g \map \phi x^{-1}
+      | c = 
+}}
+{{eqn | r = \map {\kappa_{\map \phi x} } g
+      | c = 
+}}
+{{end-eqn}}
+Therefore:
+:$\phi \circ \kappa_g \circ \phi^{-1} = \kappa_{\map \phi g} \in \Inn G$
+Since $\kappa_x \in \Inn G$ and $\phi \in \Aut G$ were arbitrary:
+:$\Inn G \lhd \Aut G$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Subgroup of Abelian Group is Normal}
+Tags: Normal Subgroups, Abelian Groups
+
+\begin{theorem}
+Every [[Definition:Subgroup|subgroup]] of an [[Definition:Abelian Group|abelian group]] is [[Definition:Normal Subgroup|normal]].
+\end{theorem}
+
+\begin{proof}
+Let $H \le G$ where $G$ is [[Definition:Abelian Group|abelian]].
+Then:
+{{begin-eqn}}
+{{eqn | l = y
+      | o = \in
+      | r = H^a
+      | c = where $H^a$ is the [[Definition:Conjugate of Group Subset|conjugate of $H$ by $a$]]
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = a y a^{-1}
+      | o = \in
+      | r = H
+      | c = {{Defof|Conjugate of Group Subset}}
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = y
+      | o = \in
+      | r = H
+      | c = [[Conjugate of Commuting Elements|because $a y a^{-1} = y$]] as $G$ is [[Definition:Abelian Group|abelian]]
+}}
+{{end-eqn}}
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Subgroup is Normal iff it contains Product of Inverses}
+Tags: Normal Subgroups
+
+\begin{theorem}
+A [[Definition:Subgroup|subgroup]] $H$ of a [[Definition:Group|group]] $G$ is [[Definition:Normal Subgroup|normal]] {{iff}}:
+:$\forall a, b \in G: a b \in H \implies a^{-1} b^{-1} \in H$
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+Suppose $H$ is [[Definition:Normal Subgroup|normal]].
+Let $a b \in H$.
+{{begin-eqn}}
+{{eqn | l = a b
+      | o = \in
+      | r = H
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = a^{-1} \left({a b}\right) a
+      | o = \in
+      | r = H
+      | c = [[Subgroup is Normal iff Contains Conjugate Elements]]
+}}
+{{eqn | ll= \leadsto
+      | l = b a
+      | o = \in
+      | r = H
+      | c = [[Definition:Group|Group properties]]
+}}
+{{end-eqn}}
+As $H$ is a [[Definition:Group|group]], then $x \in H \iff x^{-1} \in H$.
+So $b a \in H \iff \left({b a}\right)^{-1} \in H \iff a^{-1} b^{-1} \in H$.
+Thus $\forall a, b \in G: a b \in H \implies a^{-1} b^{-1} \in H$.
+=== Sufficient Condition ===
+Suppose $a b \in H \implies a^{-1} b^{-1} \in H$.
+Then (from the above) we have $\forall a, b \in H: a b \in H \implies b a \in H$.
+Let $g \in G$, and let $h \in H$. Let $h = x g$ where $x \in G$.
+Now $g h g^{-1}  = g x g g^{-1} = g x$.
+By hypothesis, $x g \in H \implies g x \in H$.
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Intersection with Normal Subgroup is Normal}
+Tags: Normal Subgroups, Intersection with Normal Subgroup is Normal
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]].
+Let $H$ be a [[Definition:Subgroup|subgroup]] of $G$, and let $N$ be a [[Definition:Normal Subgroup|normal subgroup]] of $G$.
+Then $H \cap N$ is a [[Definition:Normal Subgroup|normal subgroup]] of $H$.
+\end{theorem}
+
+\begin{proof}
+By [[Intersection of Subgroups is Subgroup]], $H \cap N$ is a [[Definition:Subgroup|subgroup]] of $N$.
+It remains to be shown that $H$ is [[Definition:Normal Subgroup|normal]] in $H$.
+Because $N \lhd G$:
+:$\forall n \in N: \forall g \in G: g n g^{-1} \in N$
+Let $x \in H \cap N$.
+Because $H \le G$ and therefore [[Definition:Closed Algebraic Structure|closed]]:
+:$\forall x \in H \cap N: \forall g \in H: g x g^{-1} \in H$
+But since $x \in N$ and $N \lhd G$, $g x g^{-1} \in N$.
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Kernel is Normal Subgroup of Domain}
+Tags: Normal Subgroups, Kernels of Group Homomorphisms
+
+\begin{theorem}
+Let $\phi$ be a [[Definition:Group Homomorphism|group homomorphism]].
+Then the [[Definition:Kernel of Group Homomorphism|kernel]] of $\phi$ is a [[Definition:Normal Subgroup|normal subgroup]] of the [[Definition:Domain of Mapping|domain]] of $\phi$:
+:$\map \ker \phi \lhd \Dom \phi$
+\end{theorem}
+
+\begin{proof}
+Let $\phi: G_1 \to G_2$ be a [[Definition:Group Homomorphism|group homomorphism]], where the [[Definition:Identity Element|identities]] of $G_1$ and $G_2$ are $e_{G_1}$ and $e_{G_2}$ respectively.
+By [[Kernel of Group Homomorphism is Subgroup]]:
+:$\map \ker \phi \le \Dom \phi$
+Let $k \in \map \ker \phi, x \in G_1$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \map \phi {x k x^{-1} }
+      | r = \map \phi x \map \phi k \paren {\map \phi x}^{-1}
+      | c = [[Homomorphism to Group Preserves Inverses]]
+}}
+{{eqn | r = \map \phi x e_{G_2} \paren {\map \phi x}^{-1}
+      | c = {{Defof|Kernel of Group Homomorphism}}
+}}
+{{eqn | r = \map \phi x \paren {\map \phi x}^{-1}
+      | c = {{Defof|Identity Element}}
+}}
+{{eqn | r = e_{G_2}
+      | c = {{Defof|Inverse Element}}
+}}
+{{end-eqn}}
+So:
+:$x k x^{-1} \in \map \ker \phi$
+This is true for all $k \in \map \ker \phi$ and $x \in G_1$.
+From [[Subgroup is Normal iff Contains Conjugate Elements]], it follows that $\map \ker \phi$ is a [[Definition:Normal Subgroup|normal subgroup]] of $G_1$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Coset Product is Well-Defined}
+Tags: Coset Product, Coset Product is Well-Defined
+
+\begin{theorem}
+Let $\struct {G, \circ}$ be a [[Definition:Group|group]].
+Let $N$ be a [[Definition:Normal Subgroup|normal subgroup]] of $G$.
+Let $a, b \in G$.
+Then the [[Definition:Coset Product|coset product]]:
+:$\paren {a \circ N} \circ \paren {b \circ N} = \paren {a \circ b} \circ N$
+is [[Definition:Well-Defined Operation|well-defined]].
+\end{theorem}
+
+\begin{proof}
+Let $N \lhd G$ where $G$ is a [[Definition:Group|group]].
+Let $a, a', b, b' \in G: a \circ N = a' \circ N, b \circ N = b' \circ N$.
+To show that the coset product is [[Definition:Well-Defined Operation|well-defined]], we need to demonstrate that $\paren {a \circ b} \circ N = \paren {a' \circ b'} \circ N$.
+So:
+{{begin-eqn}}
+{{eqn | l = a \circ N
+      | r = a' \circ N
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = a^{-1} \circ a'
+      | o = \in
+      | r = N
+      | c = [[Cosets are Equal iff Product with Inverse in Subgroup]]
+}}
+{{eqn | ll= \leadsto
+      | l = b^{-1} \circ a^{-1} \circ a'
+      | o = \in
+      | r = b^{-1} \circ N
+      | c = {{Defof|Subset Product}}
+}}
+{{eqn | ll= \leadsto
+      | l = b^{-1} \circ a^{-1} \circ a'
+      | o = \in
+      | r = N \circ b^{-1}
+      | c = $N$ is a [[Definition:Normal Subgroup|normal subgroup]]
+}}
+{{eqn | ll= \leadsto
+      | l = \exists n \in N: b^{-1} \circ a^{-1} \circ a'
+      | r = n \circ b^{-1}
+      | c = {{Defof|Subset Product}}
+}}
+{{eqn | ll= \leadsto
+      | l = \paren {a \circ b}^{-1} \circ \paren {a' \circ b'}
+      | r = n \circ b^{-1} \circ b'
+      | c = [[Definition:Group|Group Properties]]
+}}
+{{eqn | ll= \leadsto
+      | l = \paren {a \circ b}^{-1} \circ \paren {a' \circ b'}
+      | o = \in
+      | r = N
+      | c = {{Defof|Subset Product}}
+}}
+{{end-eqn}}
+By [[Cosets are Equal iff Product with Inverse in Subgroup]]:
+:$\paren {a \circ b}^{-1} \circ \paren {a' \circ b'} \in N \implies \paren {a \circ b} \circ N = \paren {a' \circ b'} \circ N$
+and the proof is complete.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let $N \lhd G$ where $G$ is a [[Definition:Group|group]].
+Consider $\paren {a \circ N} \circ \paren {b \circ N}$ as a [[Definition:Subset Product|subset product]]:
+:$\paren {a \circ N} \circ \paren {b \circ N} = \set {a \circ n_1 \circ b \circ n_2: n_1, n_2 \in N}$
+This is justified by [[Coset Product of Normal Subgroup is Consistent with Subset Product Definition]].
+Since $N$ is [[Definition:Normal Subgroup|normal]], each [[Definition:Conjugate (Group Theory)|conjugate]] $b^{-1} \circ N \circ b$ is contained in $N$.
+So for each $n_1 \in N$ there is some $n_3 \in N$ such that $b^{-1} \circ n_1 \circ b = n_3$.
+So, if $a \circ n_1 \circ b \circ n_2 \in \paren {a \circ N} \circ \paren {b \circ N}$, it follows that:
+{{begin-eqn}}
+{{eqn | l = a \circ n_1 \circ b \circ n_2
+      | r = a \circ b \circ b^{-1} \circ n_1 \circ b \circ n_2
+      | c = 
+}}
+{{eqn | r = a \circ b \circ n_3 \circ n_2
+      | c = 
+}}
+{{eqn | o = \in
+      | r = \paren {a \circ b} \circ N
+      | c = {{Defof|Subset Product}}
+}}
+{{eqn | o = \in
+      | r = N \circ b^{-1}
+      | c = {{Defof|Normal Subgroup}}
+}}
+{{end-eqn}}
+That is:
+:$\paren {a \circ N} \circ \paren {b \circ N} \subseteq \paren {a \circ b} \circ N$
+Then:
+{{begin-eqn}}
+{{eqn | l = a \circ b \circ n
+      | o = \in
+      | r = \paren {a \circ b} \circ N
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = a \circ e \circ b \circ n
+      | o = \in
+      | r = \paren {a \circ N} \circ \paren {b \circ N}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \paren {a \circ b} \circ N
+      | o = \subseteq
+      | r = \paren {a \circ N} \circ \paren {b \circ N}
+      | c = 
+}}
+{{end-eqn}}
+So:
+:$\paren {a \circ N} \circ \paren {b \circ N} \subseteq \paren {a \circ b} \circ N$
+and
+:$\paren {a \circ b} \circ N \subseteq \paren {a \circ N} \circ \paren {b \circ N}$
+The result follows by definition of [[Definition:Set Equality/Definition 2|set equality]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let $N \lhd G$ where $G$ is a [[Definition:Group|group]].
+Let $a, a', b, b' \in G: N \circ a = N \circ a', N \circ b = N \circ b'$.
+We need to show that $N \circ a \circ b = N \circ a' \circ b'$.
+So:
+{{begin-eqn}}
+{{eqn | l = N \circ a \circ b
+      | r = N \circ a' \circ b
+      | c = as $N \circ a = N \circ a'$
+}}
+{{eqn | r = a' \circ N \circ b
+      | c = $N \circ a' = a' \circ N$ as $N$ is [[Definition:Normal Subgroup|normal]]
+}}
+{{eqn | r = a' \circ N \circ b'
+      | c = as $N \circ b = N \circ b'$
+}}
+{{eqn | r = N \circ a' \circ b'
+      | c = $N \circ a' = a' \circ N$ as $N$ is [[Definition:Normal Subgroup|normal]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let $N \lhd G$ where $G$ is a [[Definition:Group|group]].
+Let $a, b \in G$.
+We have:
+{{begin-eqn}}
+{{eqn | l = \paren {a \circ N} \circ \paren {b \circ N}
+      | r = a \circ N \circ b \circ N
+      | c = [[Subset Product within Semigroup is Associative]]
+}}
+{{eqn | r = a \circ b \circ N \circ N
+      | c = {{Defof|Normal Subgroup}}
+}}
+{{eqn | r = \paren {a \circ b} \circ N
+      | c = [[Product of Subgroup with Itself]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let $N \lhd G$ where $G$ is a [[Definition:Group|group]].
+Let $a, a', b, b' \in G: a \circ N = a' \circ N, b \circ N = b' \circ N$.
+To show that the coset product is [[Definition:Well-Defined Operation|well-defined]], we need to demonstrate that:
+:$\paren {a \circ b} \circ N = \paren {a' \circ b'} \circ N$
+So:
+{{begin-eqn}}
+{{eqn | l = a \circ N
+      | r = a' \circ N
+      | c =
+}}
+{{eqn | ll= \leadsto
+      | l = a
+      | o = \in
+      | r = a' \circ N
+      | c = {{Defof|Left Coset}}
+}}
+{{eqn | ll= \leadsto
+      | l = a
+      | r = a' \circ n_1
+      | c = for some $n_1 \in N$
+}}
+{{eqn-intertext|Similarly, $b' {{=}} b \circ n_2$ for some $n_2 \in N$.}}
+{{eqn | ll= \leadsto
+      | l = a' \circ b'
+      | r = a \circ n_1 \circ b \circ n_2
+      | c =
+}}
+{{eqn-intertext|But $N \circ b {{=}} b \circ N$, as $N$ is [[Definition:Normal Subgroup|normal]], and so:}}
+{{eqn | ll= \leadsto
+      | l = a' \circ b'
+      | r = a \circ b \circ n_3 \circ n_2
+      | c = as $n_1 \circ b = b \circ n_3$ for some $n_3 \in N$
+}}
+{{eqn | ll= \leadsto
+      | l = a' \circ b'
+      | o = \in
+      | r = a \circ b \circ N
+      | c = as $n_3 \circ n_2 \in N$
+}}
+{{eqn | ll= \leadsto
+      | l = a' \circ b' \circ N
+      | r = a \circ b \circ N
+      | c = {{Defof|Left Coset}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Quotient Group of Abelian Group is Abelian}
+Tags: Quotient Groups
+
+\begin{theorem}
+Let $G$ be an [[Definition:Abelian Group|abelian group]].
+Let $N \le G$.
+Then the [[Definition:Quotient Group|quotient group]] $G / N$ is [[Definition:Abelian Group|abelian]].
+\end{theorem}
+
+\begin{proof}
+First we note that because $G$ is [[Definition:Abelian Group|abelian]], from [[Subgroup of Abelian Group is Normal]] we have $N \lhd G$.
+Thus $G / N$ exists for ''all'' [[Definition:Subgroup|subgroups]] of $G$.
+Let $X = x N, Y = y N$ where $x, y \in G$.
+From the definition of [[Definition:Coset Product|coset product]]:
+{{begin-eqn}}
+{{eqn | l = X Y
+      | r = \paren {x N} \paren {y N}
+      | c = 
+}}
+{{eqn | r = \paren {x y N}
+      | c = 
+}}
+{{eqn | r = \paren {y x N}
+      | c = 
+}}
+{{eqn | r = \paren {y N} \paren {x N}
+      | c = 
+}}
+{{eqn | r = Y X
+      | c = 
+}}
+{{end-eqn}}
+Thus $G / N$ is [[Definition:Abelian Group|abelian]].
+{{qed}}
+[[Category:Quotient Groups]]
+meo3u1osmjyzur71i3fnrzfyink4s68
+\end{proof}<|endoftext|>
+\section{Quotient Theorem for Group Epimorphisms}
+Tags: Quotient Groups, Group Epimorphisms, Group Isomorphisms, Named Theorems
+
+\begin{theorem}
+Let $\struct {G, \oplus}$ and $\struct {H, \odot}$ be [[Definition:Group|groups]].
+Let $\phi: \struct {G, \oplus} \to \struct {H, \odot}$ be a [[Definition:Group Epimorphism|group epimorphism]].
+Let $e_G$ and $e_H$ be the [[Definition:Identity Element|identities]] of $G$ and $H$ respectively.
+Let $K = \map \ker \phi$ be the [[Definition:Kernel of Group Homomorphism|kernel]] of $\phi$.
+There is one and only one [[Definition:Group Isomorphism|group isomorphism]] $\psi: G / K \to H$ satisfying:
+:$\psi \circ q_K = \phi$
+where $q_K$ is the [[Definition:Quotient Epimorphism|quotient epimorphism]] from $G$ to $G / K$.
+\end{theorem}
+
+\begin{proof}
+Let $\mathcal R_\phi$ be the [[Definition:Equivalence Relation Induced by Mapping|equivalence on $G$ defined by $\phi$]].
+{{begin-eqn}}
+{{eqn | l = \forall x \in G: e_G
+      | o = \mathcal R_\phi
+      | r = x
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \map \phi x
+      | r = \map \phi {e_G}
+      | c = Definition of $\mathcal R_\phi$
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \map \phi x
+      | r = e_H
+      | c = [[Homomorphism to Group Preserves Identity]]
+}}
+{{end-eqn}}
+Thus:
+:$K = \eqclass {e_G} {\mathcal R_\phi}$
+From the [[Quotient Theorem for Epimorphisms]]:
+:$\mathcal R_\phi$ is compatible with $\oplus$
+Thus from [[Kernel is Normal Subgroup of Domain]]:
+:$K \lhd G$
+From [[Congruence Relation induces Normal Subgroup]], $\mathcal R_\phi$ is the [[Definition:Equivalence Relation Induced by Mapping|equivalence defined by $K$]].
+Thus, again by [[Quotient Theorem for Epimorphisms]], there is a unique [[Definition:Group Epimorphism|epimorphism]] $\psi: G / K \to H$ satisfying $\psi \circ q_K = \phi$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Congruence Relation induces Normal Subgroup}
+Tags: Normal Subgroups, Congruence Relations
+
+\begin{theorem}
+Let $\left({G, \circ}\right)$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
+Let $\mathcal R$ be a [[Definition:Congruence Relation|congruence relation]] for $\circ$.
+Let $H = \left[\!\left[{e}\right]\!\right]_\mathcal R$, where $\left[\!\left[{e}\right]\!\right]_\mathcal R$ is the [[Definition:Equivalence Class|equivalence class]] of $e$ under $\mathcal R$.
+Then:
+: $(1): \quad \left({H, \circ \restriction_H}\right)$ is a [[Definition:Normal Subgroup|normal subgroup]] of $G$
+: $(2): \quad \mathcal R$ is the [[Congruence Modulo Subgroup is Equivalence Relation|equivalence relation $\mathcal R_H$ defined by $H$]]
+: $(3): \quad \left({G / \mathcal R, \circ_\mathcal R}\right)$ is the [[Definition:Subgroup|subgroup]] $\left({G / H, \circ_H}\right)$ of the [[Definition:Semigroup|semigroup]] $\left({\mathcal P \left({G}\right), \circ_\mathcal P}\right)$.
+\end{theorem}
+
+\begin{proof}
+=== [[Congruence Relation on Group induces Normal Subgroup|Proof of Normal Subgroup]] ===
+{{:Congruence Relation on Group induces Normal Subgroup}}
+=== [[Normal Subgroup induced by Congruence Relation defines that Congruence|Proof of Equality of Relations]] ===
+{{:Normal Subgroup induced by Congruence Relation defines that Congruence}}
+=== [[Quotient Structure on Group defined by Congruence equals Quotient Group|Proof that Quotient Structure is Quotient Group]] ===
+{{:Quotient Structure on Group defined by Congruence equals Quotient Group}}
+\end{proof}<|endoftext|>
+\section{Preimage of Normal Subgroup of Quotient Group under Quotient Epimorphism is Normal}
+Tags: Group Isomorphisms, Group Epimorphisms, Quotient Groups
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]].
+Let $H \lhd G$ where $\lhd$ denotes that $H$ is a [[Definition:Normal Subgroup|normal subgroup]] of $G$.
+Let $K \lhd G/H$ and $L = q_H^{-1} \left[{K}\right]$, where:
+:$q_H: G \to G/H$ is the [[Definition:Quotient Group Epimorphism|quotient epimorphism]] from $G$ to the [[Definition:Quotient Group|quotient group]] $G/H$
+:$q_H^{-1} \left[{K}\right]$ is the [[Definition:Preimage of Subset under Mapping|preimage of $K$ under $q_H$]].
+Then:
+:$L \lhd G$
+\end{theorem}
+
+\begin{proof}
+By [[Quotient Mapping on Structure is Canonical Epimorphism]], both $q_K$ and $q_H$ are [[Definition:Group Epimorphism|epimorphisms]].
+From [[Composite of Group Epimorphisms is Epimorphism]] we have that $q_K \circ q_H: G \to \left({G / H}\right) / K$ is also an [[Definition:Group Epimorphism|epimorphism]].
+Now:
+:$\forall x \in G: x \in \ker \left({q_K \circ q_H}\right) \iff q_K \left({q_H \left({x}\right)}\right) = K = e_{G/H}$
+This means the same as:
+:$q_H \left({x}\right) \in \ker \left({q_K}\right) = K$
+But:
+:$q_H \left({x}\right) \in K \iff x \in q_H^{-1} \left({K}\right) = L$
+Thus:
+:$L = \ker \left({q_K \circ q_H}\right)$
+By [[Kernel is Normal Subgroup of Domain]]:
+: $L \lhd G$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Trivial Quotient Group is Quotient Group}
+Tags: Quotient Groups, Group Isomorphisms
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]].
+Then the [[Definition:Trivial Quotient Group|trivial quotient group]]:
+:$G / \set {e_G} \cong G$
+where:
+:$\cong$ denotes [[Definition:Group Isomorphism|group isomorphism]]
+:$e_G$ denotes the [[Definition:Identity Element|identity element]] of $G$
+is a [[Definition:Quotient Group|quotient group]].
+\end{theorem}
+
+\begin{proof}
+From [[Trivial Subgroup is Normal]]:
+:$\set {e_G} \lhd G$
+Let $x \in G$.
+Then:
+:$x \set {e_G} = \set {x e_G} = \set x$
+So each [[Definition:Left Coset|(left) coset]] of $G$ modulo $\set {e_G}$ has one [[Definition:Element|element]].
+Now we set up the [[Definition:Quotient Group Epimorphism|quotient epimorphism]] $\psi: G \to G / \set {e_G}$:
+:$\forall x \in G: \map \phi x = x \set {e_G}$
+which is of course a [[Definition:Surjection|surjection]].
+We now need to establish that it is an [[Definition:Injection|injection]].
+Let $p, q \in G$.
+{{begin-eqn}}
+{{eqn | l = \map \phi p
+      | r = \map \phi q
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = p \set {e_G}
+      | r = q \set {e_G}
+      | c = Definition of $\phi$
+}}
+{{eqn | ll= \leadsto
+      | l = \set p
+      | r = \set q
+      | c = from above
+}}
+{{eqn | ll= \leadsto
+      | l = p
+      | r = q
+      | c = {{Defof|Set Equality}}
+}}
+{{end-eqn}}
+So $\psi$ is a [[Definition:Group Isomorphism|group isomorphism]] and therefore:
+:$G / \set {e_G} \cong G$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Correspondence Theorem (Group Theory)}
+Tags: Normal Subgroups, Quotient Groups, Named Theorems
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]].
+Let $N \lhd G$ be a [[Definition:Normal Subgroup|normal subgroup]] of $G$.
+Then every [[Definition:Subgroup|subgroup]] of the [[Definition:Quotient Group|quotient group]] $G / N$ is of the form $H / N = \set {h N: h \in H}$, where $N \le H \le G$.
+Conversely, if $N \le H \le G$ then $H / N \le G / N$.
+The correspondence between [[Definition:Subgroup|subgroups]] of $G / N$ and [[Definition:Subgroup|subgroups]] of $G$ containing $N$ is a [[Definition:Bijection|bijection]].
+This [[Definition:Bijection|bijection]] maps [[Definition:Normal Subgroup|normal subgroups]] of $G / N$ onto [[Definition:Normal Subgroup|normal subgroups]] of $G$ which contain $N$.
+\end{theorem}
+
+\begin{proof}
+Let $H'$ be a [[Definition:Subgroup|subgroup]] of $G / N$, so that it consists of a certain [[Definition:Set|set]] $\set {h N}$ of [[Definition:Left Coset|left cosets]] of $N$ in $G$.
+Let us define the [[Definition:Subset|subset]] $\map \beta {H'} \subseteq G$:
+:$\map \beta {H'} = \set {g \in G: g N \in H'}$
+Then clearly:
+:$N \subseteq \map \beta {H'}$
+Also:
+:$e_G \in N$
+so:
+:$e_G \in \map \beta {H'}$
+Let $x, y \in \map \beta {H'}$. Then:
+{{begin-eqn}}
+{{eqn | l = x, y
+      | o = \in
+      | r = \map \beta {H'}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = x N, y N
+      | o = \in
+      | r = H'
+      | c = Definition of $\beta$
+}}
+{{eqn | ll= \leadsto
+      | l = \paren {x N} \paren {y N}
+      | r = x y N \in H'
+      | c = {{Defof|Quotient Group}}: $G / N$ and as $H'$ is a subgroup of $G / N$
+}}
+{{eqn | ll= \leadsto
+      | l = x y
+      | o = \in
+      | r = \map \beta {H'}
+      | c = Definition of $\beta$
+}}
+{{end-eqn}}
+We also have, from [[Quotient Group is Group]]:
+:$\paren {x N}^{-1} = x^{-1} N \implies x^{-1} \in \map \beta {H'}$
+Thus, by the [[Two-Step Subgroup Test]], $\map \beta {H'} \le G$ that contains $N$.
+Conversely, let $H$ be such that $N \le H \le G$.
+Let $\map \alpha H = \set {h N: h \in H} \subseteq G / N$.
+It is easily checked that $\map \alpha H \le G / N$.
+Now, let $X$ be the [[Definition:Set|set]] of [[Definition:Subgroup|subgroups]] of $G$ containing $N$ and $Y$ be the [[Definition:Set|set]] of all [[Definition:Subgroup|subgroups]] of $G / N$.
+We now need to show that $\alpha: X \to Y$ is a [[Definition:Bijection|bijection]].
+We do this by checking that $\beta: Y \to X$ is the inverse of $\alpha$.
+To do this, we show that $\alpha \circ \beta = I_Y$ and $\beta \circ \alpha = I_X$.
+Suppose $N \le H \le G$. Then:
+{{begin-eqn}}
+{{eqn | l = \map {\paren {\beta \circ \alpha} } H
+      | r = \map \beta {H / N}
+      | c = Definition of $\alpha$
+}}
+{{eqn | r = \set {g \in G: g N \in H / N}
+      | c = Definition of $\beta$
+}}
+{{eqn | r = H
+      | c = {{Defof|Coset}} $H / N$
+}}
+{{end-eqn}}
+Thus $\beta \circ \alpha = I_X$.
+Now let $H' \le G / N$. Then:
+{{begin-eqn}}
+{{eqn | l = \map {\paren {\alpha \circ \beta} } {H'}
+      | r = \map \alpha {\set {g \in G: g N \in H'} }
+      | c = Definition of $\beta$
+}}
+{{eqn | r = \set {g N \in H'}
+      | c = Definition of $\alpha$
+}}
+{{eqn | r = H'
+      | c = Definition of $H'$
+}}
+{{end-eqn}}
+Thus $\alpha \circ \beta = I_Y$.
+So, by [[Bijection iff Inverse is Bijection]], $\alpha$ is a [[Definition:Bijection|bijection]].
+Now let $H \lhd G$ such that $N \le H$.
+We show that $\map \alpha H = H / N \lhd G / N$.
+This follows by [[Definition:Normal Subgroup/Definition 3|definition 3 of Normal Subgroup]] because: for any $h \in H, g \in G$
+:$\paren {g N} \paren {h N} \paren {g N}^{-1} = g h g^{-1} N \in H / N$
+:$\paren {g N}^{-1} \paren {h N} \paren {g N} = g^{-1} h g N \in H / N$
+Conversely, let $H' \lhd G / N$.
+Recall:
+:$\map \beta {H'} = \set {g \in G : g N \in H'}$
+Hence, for any $x \in G$ we have:
+{{begin-eqn}}
+{{eqn | l = x \map \beta {H'} x^{-1}
+      | r = \set {x g x^{-1} \in G: g N \in H'}
+      | c = 
+}}
+{{eqn | r = \set {y \in G: x^{-1} y x N \in H'}
+      | c = 
+}}
+{{end-eqn}}
+Now for any $h' \in \map \beta {H'}$, we have:
+:$h'N \in H'$
+For all $x \in G$:
+:$\paren {x^{-1} N} \paren {h' N} \paren {x N} \in H'$
+From $H' \lhd G / N$:
+:$x^{-1} h' x N \in H'$
+This implies:
+:$h' \in x \map \beta {H'} x^{-1}$
+or:
+:$\map \beta {H'} \subseteq x \map \beta {H'} x^{-1}$
+Similarly, we can also show:
+:$\map \beta {H'} \subseteq x^{-1} \map \beta {H'} x$
+Hence by [[Definition:Normal Subgroup/Definition 4|definition 4 of Normal Subgroup]]:
+:$\map \beta {H'} \lhd G$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Centralizer of Group Element is Subgroup}
+Tags: Subgroups, Centralizers, Centralizer of Group Element is Subgroup
+
+\begin{theorem}
+Let $\struct {G, \circ}$ be a [[Definition:Group|group]] and let $a \in G$.
+Then $\map {C_G} a$, the [[Definition:Centralizer of Group Element|centralizer]] of $a$ in $G$, is a [[Definition:Subgroup|subgroup]] of $G$.
+\end{theorem}
+
+\begin{proof}
+Let $\struct {G, \circ}$ be a [[Definition:Group|group]].
+We have that:
+: $\forall a \in G: e \circ a = a \circ e \implies e \in C_G \paren a$
+Thus $C_G \paren a \ne \O$.
+Let $x, y \in C_G \paren a$.
+Then:
+{{begin-eqn}}
+{{eqn | l = x \circ a
+      | r = a \circ x
+      | c = 
+}}
+{{eqn | l = y \circ a
+      | r = a \circ y
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = x \circ y \circ a
+      | r = x \circ a \circ y
+      | c = 
+}}
+{{eqn | r = a \circ x \circ y
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = x \circ y
+      | o = \in
+      | r = C_G \paren a
+      | c = 
+}}
+{{end-eqn}}
+Thus $C_G \paren a$ is closed under $\circ$.
+Let $x \in C_G \paren a$.
+Then:
+{{begin-eqn}}
+{{eqn | l = x \circ a
+      | r = a \circ x
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = x^{-1} \circ x \circ a \circ x^{-1}
+      | r = x^{-1} \circ a \circ x \circ x^{-1}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = a \circ x^{-1}
+      | r = x^{-1} \circ a
+      | c = 
+}}
+{{end-eqn}}
+So:
+: $x \in C_G \paren a \implies x^{-1} \in C_G \paren a$
+Thus, by the [[Two-Step Subgroup Test]], the result follows.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let $\struct {G, \circ}$ be a [[Definition:Group|group]].
+We have that:
+:$\forall a \in G: e \circ a = a \circ e \implies e \in \map {C_G} a$
+Thus $\map {C_G} a \ne \O$.
+Let $x, y \in \map {C_G} a$.
+Then from [[Commutation with Group Elements implies Commuation with Product with Inverse]]:
+:$a \circ x \circ y^{-1} = x \circ y^{-1} \circ a$
+so:
+:$x \circ y^{-1} \in\map {C_G} a$
+The result follows by the [[One-Step Subgroup Test]], the result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Centralizer in Subgroup is Intersection}
+Tags: Subgroups, Centralizers, Set Intersection
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]].
+Let $H$ be a [[Definition:Subgroup|subgroup]] of $G$.
+Then:
+:$\forall x \in G: \map {C_H} x = \map {C_G} x \cap H$
+That is, the [[Definition:Centralizer of Group Element|centralizer]] of an [[Definition:Element|element]] in a [[Definition:Subgroup|subgroup]] is the [[Definition:Set Intersection|intersection]] of that [[Definition:Subgroup|subgroup]] with the [[Definition:Centralizer of Group Element|centralizer]] of the [[Definition:Element|element]] in the [[Definition:Group|group]].
+\end{theorem}
+
+\begin{proof}
+It is clear that:
+:$g \in \map {C_H} x \iff g \in \map {C_G} x \land g \in H$
+The result follows by definition of [[Definition:Set Intersection|set intersection]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Kernel of Inner Automorphism Group is Center}
+Tags: Inner Automorphisms, Centers of Groups
+
+\begin{theorem}
+Let the [[Definition:Mapping|mapping]] $\kappa: G \to \Inn G$ from a [[Definition:Group|group]] $G$ to its [[Definition:Inner Automorphism Group|inner automorphism group]] $\Inn G$ be defined as:
+:$\map \kappa a = \kappa_a$
+where $\kappa_a$ is the [[Definition:Inner Automorphism|inner automorphism]] of $G$ given by $a$.
+Then $\kappa$ is a [[Definition:Group Epimorphism|group epimorphism]], and its [[Definition:Kernel of Group Homomorphism|kernel]] is the [[Definition:Center of Group|center]] of $G$:
+:$\map \ker \kappa = \map Z G$
+\end{theorem}
+
+\begin{proof}
+Let $\kappa: G \to \Aut G$ be a [[Definition:Mapping|mapping]] defined by $\map \kappa x = \kappa_x$.
+It is clear that $\Img \kappa = \Inn G$.
+It is also clear that $\kappa$ is a [[Definition:Group Homomorphism|homomorphism]]:
+{{begin-eqn}}
+{{eqn | l = \map \kappa x \map \kappa y
+      | r = \kappa_x \circ \kappa_y
+      | c = 
+}}
+{{eqn | r = \kappa_{x y}
+      | c = [[Inner Automorphism is Automorphism]]
+}}
+{{eqn | r = \map \kappa {x y}
+      | c = 
+}}
+{{end-eqn}}
+Note that $\forall \kappa_x \in \Inn G: \exists x \in G: \map \kappa x = \kappa_x$.
+Thus $\kappa: G \to \Inn G$ is a [[Definition:Surjection|surjection]] and therefore an [[Definition:Group Epimorphism|group epimorphism]].
+Now we investigate the [[Definition:Kernel of Group Homomorphism|kernel]] of $\kappa$:
+{{begin-eqn}}
+{{eqn | l = \map \ker \kappa
+      | r = \set {x \in G: \kappa_x = I_G}
+      | c = {{Defof|Kernel of Group Homomorphism}}
+}}
+{{eqn | r = \set {x \in G: \forall g \in G: \map {\kappa_x} g = \map {I_G} g}
+      | c = [[Equality of Mappings]]
+}}
+{{eqn | r = \set {x \in G: \forall g \in G: x g x^{-1} = g}
+      | c = Definition of $\kappa_x$
+}}
+{{eqn | r = \set {x \in G: \forall g \in G: x g = g x}
+      | c = 
+}}
+{{eqn | r = \map Z G
+      | c = {{Defof|Center of Group}}
+}}
+{{end-eqn}}
+So the [[Definition:Kernel of Group Homomorphism|kernel]] of $\kappa$ is the [[Definition:Center of Group|center]] of $G$.
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Group equals Center iff Abelian}
+Tags: Abelian Groups, Centers of Groups
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]].
+Then $G$ is [[Definition:Abelian Group|abelian]] {{iff}} $\map Z G = G$, that is, {{iff}} $G$ equals its [[Definition:Center of Group|center]].
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+Let $G$ be [[Definition:Abelian Group|abelian]].
+Then:
+:$\forall a \in G: \forall x \in G: a x = x a$
+Thus:
+:$\forall a \in G: a \in \map Z G = G$
+{{qed|lemma}}
+=== Sufficient Condition ===
+Let $\map Z G = G$.
+Then by the definition of [[Definition:Center of Group|center]]:
+:$\forall a \in G: \forall x \in G: a x = x a$
+and thus $G$ is [[Definition:Abelian Group|abelian]] by definition.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Normalizer is Subgroup}
+Tags: Normalizers
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]].
+The [[Definition:Normalizer|normalizer]] of a [[Definition:Subset|subset]] $S \subseteq G$ is a [[Definition:Subgroup|subgroup]] of $G$.
+:$S \subseteq G \implies \map {N_G} S \le G$
+\end{theorem}
+
+\begin{proof}
+Let $a, b \in \map {N_G} S$.
+Then:
+{{begin-eqn}}
+{{eqn | l = S^{a b}
+      | r = \paren {S^b}^a
+      | c = [[Conjugate of Set by Group Product]]
+}}
+{{eqn | r = S^a
+      | c = {{Defof|Normal Subgroup}}
+}}
+{{eqn | r = S
+      | c = {{Defof|Normal Subgroup}}
+}}
+{{end-eqn}}
+Therefore $a b \in \map {N_G} S$.
+Now let $a \in \map {N_G} S$:
+:$a \in \map {N_G} S \implies S^{a^{-1} } = \paren {S^a}^{a^{-1} } = S^{a^{-1} a} = S$
+Therefore $a^{-1} \in \map {N_G} S$.
+Thus, by the [[Two-Step Subgroup Test]], $\map {N_G} S \le G$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Subgroup is Normal Subgroup of Normalizer}
+Tags: Normal Subgroups, Normalizers, Subgroups
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]].
+A [[Definition:Subgroup|subgroup]] $H \le G$ is a [[Definition:Normal Subgroup|normal subgroup]] of its [[Definition:Normalizer|normalizer]]:
+:$H \le G \implies H \lhd \map {N_G} H$
+\end{theorem}
+
+\begin{proof}
+From [[Subgroup is Subgroup of Normalizer]] we have that $H \le \map {N_G} H$.
+It remains to show that $H$ is [[Definition:Normal Subgroup|normal]] in $\map {N_G} H$.
+Let $a \in H$ and $b \in \map {N_G} H$.
+By the definition of [[Definition:Normalizer|normalizer]]:
+: $b a b^{-1} \in H$
+Thus $H$ is [[Definition:Normal Subgroup|normal]] in $\map {N_G} H$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Normalizer of Subgroup is Largest Subgroup containing that Subgroup as Normal Subgroup}
+Tags: Normalizers
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]].
+Let $H$ be a [[Definition:Subgroup|subgroup]] of $G$.
+Then $\map {N_G} H$, the [[Definition:Normalizer|normalizer]] of $H$ in $G$, is the largest [[Definition:Subgroup|subgroup]] of $G$ containing $H$ as a [[Definition:Normal Subgroup|normal subgroup]].
+\end{theorem}
+
+\begin{proof}
+From [[Subgroup is Subgroup of Normalizer]], we have that $H \le \map {N_G} H$.
+Now we need to show that $H \lhd \map {N_G} H$.
+For $a \in \map {N_G} H$, the [[Definition:Conjugate of Group Subset|conjugate]] of $H$ by $a$ in $\map {N_G} H$ is:
+{{begin-eqn}}
+{{eqn | l = H^a
+      | r = \set {x \in \map {N_G} H: a x a^{-1} \in H}
+      | c = {{Defof|Conjugate of Group Subset}}
+}}
+{{eqn | r = H^a \cap \map {N_G} H
+      | c = {{Defof|Set Intersection}}
+}}
+{{eqn | r = H \cap \map {N_G} H
+      | c = {{Defof|Normalizer}}
+}}
+{{eqn | r = H
+      | c = [[Intersection with Subset is Subset]]
+}}
+{{end-eqn}}
+so:
+:$\forall a \in \map {N_G} H: H^a = H$
+and so by [[Definition:Normal Subgroup|definition of normal subgroup]]:
+:$H \lhd \map {N_G} H$
+Now we need to show that $\map {N_G} H$ is the largest [[Definition:Subgroup|subgroup]] of $G$ containing $H$ such that $H \lhd \map {N_G} H$.
+That is, to show that any [[Definition:Subgroup|subgroup]] of $G$ in which $H$ is [[Definition:Normal Subgroup|normal]] is also a [[Definition:Subset|subset]] of $\map {N_G} H$.
+Take any $N$ such that $H \lhd N \le G$.
+In $N$, the [[Definition:Conjugate of Group Subset|conjugate]] of $H$ by $a \in N$ is $N \cap H^a = H$.
+Therefore:
+:$H \subseteq H^a$
+Similarly, $H \subseteq H^{a^{-1} }$, so:
+:$H^a \subseteq \paren {H^a}^{a^{-1} } = H$
+Thus:
+:$\forall a \in N: H^a = H, a \in \map {N_G} H$
+That is:
+:$N \subseteq \map {N_G} H$
+So what we have shown is that any [[Definition:Subgroup|subgroup]] of $G$ in which $H$ is [[Definition:Normal Subgroup|normal]] is a [[Definition:Subset|subset]] of $\map {N_G} H$, which is another way of saying that $\map {N_G} H$ is the largest such [[Definition:Subgroup|subgroup]].
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Normal Subgroup iff Normalizer is Group}
+Tags: Normal Subgroups, Normalizers
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]].
+Let $H$ be a [[Definition:Subgroup|subgroup]] of $G$.
+Then $H$ is [[Definition:Normal Subgroup|normal]] in $G$ {{iff}} the [[Definition:Normalizer|normalizer]] of $H$ is equal to $G$:
+:$H \lhd G \iff \map {N_G} H = G$
+\end{theorem}
+
+\begin{proof}
+=== Sufficient Condition ===
+Let $H$ be [[Definition:Normal Subgroup|normal]] in $G$.
+Then $G$ is trivially the largest [[Definition:Subgroup|subgroup]] of $G$ in which $H$ is [[Definition:Normal Subgroup|normal]].
+Thus from [[Normalizer of Subgroup is Largest Subgroup containing that Subgroup as Normal Subgroup]]:
+:$\map {N_G} H = G$
+{{qed|lemma}}
+=== Necessary Condition ===
+Let $\map {N_G} H = G$.
+From [[Subgroup is Normal Subgroup of Normalizer]], $H$ is [[Definition:Normal Subgroup|normal]] in $\map {N_G} H$.
+Hence $H$ is [[Definition:Normal Subgroup|normal]] in $G$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Normalizer of Conjugate is Conjugate of Normalizer}
+Tags: Normalizers, Conjugacy
+
+\begin{theorem}
+The [[Definition:Normalizer|normalizer]] of a [[Definition:Conjugate of Group Subset|conjugate]] is the [[Definition:Conjugate of Group Subset|conjugate]] of the [[Definition:Normalizer|normalizer]]:
+:$S \subseteq G \implies \map {N_G} {S^a} = \paren {\map {N_G} S}^a$
+\end{theorem}
+
+\begin{proof}
+From the definition of [[Definition:Conjugate of Group Subset|conjugate]]:
+:$S^a = \set {y \in G: \exists x \in S: y = a x a^{-1} } = a S a^{-1}$
+From the definition of [[Definition:Normalizer|normalizer]]:
+:$\map {N_G} S = \set {x \in G: S^x = S}$
+Thus:
+{{begin-eqn}}
+{{eqn | l = \map {N_G} {S^a}
+      | r = \set {x \in G: \paren {S^a}^x = S^a}
+      | c = 
+}}
+{{eqn | r = \set {x \in G: x a S a^{-1} x^{-1} = a S a^{-1} }
+      | c = 
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | l = \paren {\map {N_G} S}^a
+      | r = \set {x \in G: S^x = S}^a
+}}
+{{eqn | r = \set {x \in G: x S x^{-1} = S}^a
+}}
+{{eqn | r = \set {y \in G: \exists z \in \set {x \in G: x S x^{-1} = S}: y = a z a^{-1} }
+}}
+{{end-eqn}}
+Suppose that $x \in \map {N_G} S$.
+It is to be shown that:
+: $a x a^{-1} \in \map {N_G} {S^a} = \map {N_G} {a S a^{-1} }$
+To this end, compute:
+{{begin-eqn}}
+{{eqn | l = \paren {a x a^{-1} } a S a^{-1} \paren {a x a^{-1} }^{-1}
+      | r = \paren {a x a^{-1} } a S a^{-1} \paren {a x^{-1} a^{-1} }
+      | c = [[Power of Conjugate equals Conjugate of Power]]
+}}
+{{eqn | r = a x S x^{-1} a^{-1}
+}}
+{{eqn | r = a S a^{-1}
+      | c = as $x \in \map {N_G} S$
+}}
+{{end-eqn}}
+Hence $a x a^{-1} \in \map {N_G} {S^a}$, and it follows that:
+:$z \in \paren {\map {N_G} S}^a \implies z \in \map {N_G} {S^a}$
+Conversely, let $x \in \map {N_G} {S^a}$.
+That is, let $x \in G$ such that $x a S a^{-1} x^{-1} = a S a^{-1}$.
+Now if we can show that $a^{-1} x a \in \map {N_G} S$, then:
+:$x = a \left({a^{-1} x a}\right) a^{-1} \in \paren {\map {N_G} S}^a$
+establishing the remaining inclusion.
+Thus, we compute:
+{{begin-eqn}}
+{{eqn | l = \paren {a^{-1} x a} S \paren {a^{-1} x a}^{-1}
+      | r = a^{-1} x a S a^{-1} x^{-1} a
+      | c = [[Power of Conjugate equals Conjugate of Power]]
+}}
+{{eqn | r = a^{-1} a S a^{-1} a
+      | c = as $x \in \map {N_G} {S^a}$
+}}
+{{eqn | r = S
+}}
+{{end-eqn}}
+Combined with the above observation, this establishes that:
+:$z \in \map {N_G} {S^a} \implies z \in \paren {\map {N_G} S}^a$
+Hence the result, by definition of [[Definition:Set Equality/Definition 2|set equality]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Normalizer of Center is Group}
+Tags: Normalizers, Centers of Groups
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]].
+Let $\map Z G$ be the [[Definition:Center of Group|center]] of $G$.
+Let $x \in G$.
+Let $\map {N_G} x$ be the [[Definition:Normalizer|normalizer of $x$ in $G$]].
+Then:
+:$\map Z G = \set {x \in G: \map {N_G} x = G}$
+That is, the [[Definition:Center of Group|center]] of a [[Definition:Group|group]] $G$ is the [[Definition:Set|set]] of [[Definition:Element|elements]] $x$ of $G$ such that the [[Definition:Normalizer|normalizer]] of $x$ is the whole of $G$.
+Thus:
+:$x \in \map Z G \iff \map {N_G} x = G$
+and so:
+:$\index G {\map {N_G} x} = 1$
+where $\index G {\map {N_G} x}$ is the [[Definition:Index of Subgroup|index of $\map {N_G} x$ in $G$]].
+\end{theorem}
+
+\begin{proof}
+$\map {N_G} x$ is the [[Definition:Normalizer|normalizer]] of the [[Definition:Set|set]] $\set x$.
+Thus:
+{{begin-eqn}}
+{{eqn | l = \map {N_G} x
+      | r = G
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | lo= \forall a \in G:
+      | l = \set x^a
+      | r = \set x
+      | c = {{Defof|Normalizer}}
+}}
+{{eqn | ll= \leadstoandfrom
+      | lo= \forall a \in G:
+      | l = a x a^{-1}
+      | r = x
+      | c = {{Defof|Conjugate of Group Element}}
+}}
+{{eqn | ll= \leadstoandfrom
+      | lo= \forall a \in G:
+      | l = a x
+      | r = x a
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = x
+      | o = \in
+      | r = \map Z G
+      | c = {{Defof|Center of Group}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Quotient Group of Ideal is Coset Space}
+Tags: Ideal Theory, Quotient Groups, Cosets
+
+\begin{theorem}
+Let $\struct {R, +, \circ}$ be a [[Definition:Ring (Abstract Algebra)|ring]].
+Let $J$ be an [[Definition:Ideal of Ring|ideal]] of $R$.
+Let $\struct {R / J, +}$ be the [[Definition:Quotient Group|quotient group]] of $\struct {R, +}$ by $\struct {J, +}$.
+Then each element of $\struct {R / J, +}$ is a [[Definition:Coset|coset]] of $J$ in $R$, that is, is of the form $x + J = \set {x + j: j \in J}$ for some $x \in R$.
+The rule of addition of these [[Definition:Coset|cosets]] is: $\paren {x + J} + \paren {y + J} = \paren {x + y} + J$.
+The [[Definition:Identity Element|identity]] of $\struct {R / J, +}$ is $J$ and for each $x \in R$, the [[Definition:Inverse Element|inverse]] of $x + J$ is $\paren {-x} + J$.
+\end{theorem}
+
+\begin{proof}
+From [[Ideal is Additive Normal Subgroup]] that $J$ is a [[Definition:Normal Subgroup|normal subgroup]] of $R$ under $+$.
+From [[Quotient Ring Addition is Well-Defined]], $+$ is a [[Definition:Well-Defined Operation|well-defined operation]].
+The rest follows directly from the definition of [[Definition:Quotient Group|quotient group]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Quotient Ring is Ring/Quotient Ring Product is Well-Defined}
+Tags: Quotient Rings
+
+\begin{theorem}
+Let $\struct {R, +, \circ}$ be a [[Definition:Ring (Abstract Algebra)|ring]] whose [[Definition:Ring Zero|zero]] is $0_R$ and whose [[Definition:Unity of Ring|unity]] is $1_R$.
+Let $J$ be an [[Definition:Ideal of Ring|ideal]] of $R$.
+Let $\struct {R / J, +, \circ}$ be the [[Definition:Quotient Ring|quotient ring]] of $R$ by $J$.
+Then $\circ$ is [[Definition:Well-Defined Operation|well-defined]] on $R / J$, that is:
+:$x_1 + J = x_2 + J, y_1 + J = y_2 + J \implies x_1 \circ y_1 + J = x_2 \circ y_2 + J$
+\end{theorem}
+
+\begin{proof}
+From [[Left Cosets are Equal iff Product with Inverse in Subgroup]], we have:
+{{begin-eqn}}
+{{eqn | l = x_1 + J
+      | r = x_2 + J
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = x_1 + \paren {-x_2}
+      | o = \in
+      | r = J
+      | c = 
+}}
+{{end-eqn}}
+and:
+{{begin-eqn}}
+{{eqn | l = y_1 + J
+      | r = y_2 + J
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = y_1 + \paren {-y_2}
+      | o = \in
+      | r = J
+      | c = 
+}}
+{{end-eqn}}
+Hence from the [[Definition:Ideal of Ring|definition of ideal]]:
+{{begin-eqn}}
+{{eqn | l = \paren {x_1 + \paren {-x_2} } \circ y_1
+      | o = \in
+      | r = J
+}}
+{{eqn | l = x_2 \circ \paren {y_1 + \paren {-y_2} }
+      | o = \in
+      | r = J
+}}
+{{end-eqn}}
+Thus:
+{{begin-eqn}}
+{{eqn | l = \paren {x_1 + \paren {-x_2} } \circ y_1 + x_2 \circ \paren {y_1 + \paren {-y_2} }
+      | o = \in
+      | r = J
+      | c = [[Definition:Group|as $\struct {J, +}$ is a group]]
+}}
+{{eqn | ll= \leadsto
+      | l = x_1 \circ y_1 + \paren {-\paren {x_2 \circ y_2} }
+      | o = \in
+      | r = J
+      | c = [[Definition:Ring (Abstract Algebra)|Various ring properties]]
+}}
+{{eqn | ll= \leadsto
+      | l = x_1 \circ y_1 + J
+      | r = x_2 \circ y_2 + J
+      | c = [[Left Cosets are Equal iff Product with Inverse in Subgroup]]
+}}
+{{end-eqn}}
+<includeonly>{{qed|lemma}}</includeonly>
+\end{proof}<|endoftext|>
+\section{Quotient Ring of Commutative Ring is Commutative}
+Tags: Quotient Rings
+
+\begin{theorem}
+Let $\struct {R, +, \circ}$ be a [[Definition:Ring (Abstract Algebra)|ring]] whose [[Definition:Ring Zero|zero]] is $0_R$ and whose [[Definition:Unity of Ring|unity]] is $1_R$.
+Let $J$ be an [[Definition:Ideal of Ring|ideal]] of $R$.
+Let $\struct {R / J, +, \circ}$ be the [[Definition:Quotient Ring|quotient ring]] defined by $J$.
+If $\struct {R, +, \circ}$ is a [[Definition:Commutative Ring|commutative ring]], then so is $\struct {R / J, +, \circ}$.
+\end{theorem}
+
+\begin{proof}
+Let $\struct {R, +, \circ}$ be a [[Definition:Commutative Ring|commutative ring]]
+That means $\circ$ is [[Definition:Commutative Operation|commutative]] on $R$.
+Thus:
+{{begin-eqn}}
+{{eqn | ll= \forall x, y \in R:
+      | l = \paren {x + J} \circ \paren {y + J}
+      | r = x \circ y + J
+      | c = [[Definition:Quotient Ring|Definition of $\circ$]] in $R / J$
+}}
+{{eqn | r = y \circ x + J
+      | c = [[Definition:Commutative Operation|Commutativity]] of $\circ$
+}}
+{{eqn | r = \paren {y + J} \circ \paren {x + J}
+      | c = [[Definition:Quotient Ring|Definition of $\circ$]] in $R / J$
+}}
+{{end-eqn}}
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Quotient Ring of Ring with Unity is Ring with Unity}
+Tags: Quotient Rings, Rings with Unity
+
+\begin{theorem}
+Let $\struct {R, +, \circ}$ be a [[Definition:Ring with Unity|ring with unity]] whose [[Definition:Ring Zero|zero]] is $0_R$ and whose [[Definition:Unity of Ring|unity]] is $1_R$.
+Let $J$ be an [[Definition:Ideal of Ring|ideal]] of $R$.
+Let $\struct {R / J, +, \circ}$ be the [[Definition:Quotient Ring|quotient ring]] defined by $J$.
+Then $\struct {R / J, +, \circ}$ is a [[Definition:Ring with Unity|ring with unity]], and its [[Definition:Unity of Ring|unity]] is $1_R + J$.
+\end{theorem}
+
+\begin{proof}
+Let $\struct {R, +, \circ}$ be a [[Definition:Ring with Unity|ring with unity]].
+First, let $J \subsetneq R$.
+By [[Ideal of Unit is Whole Ring/Corollary|Ideal of Unit is Whole Ring: Corollary]]:
+:$1_R \in J \implies J = R$
+So $1_R \notin J$.
+Thus $1_R + J \ne J$, so $1_R + J \ne 0_{R/J}$.
+Now let $x \in R$.
+{{begin-eqn}}
+{{eqn | l = \paren {1_R + J} \circ \paren {x + J}
+      | r = 1_R \circ x + J
+      | c = [[Definition:Quotient Ring|Definition of $\circ$]] in $R / J$
+}}
+{{eqn | r = x + J
+      | c = {{Defof|Unity of Ring}}
+}}
+{{eqn | r = x \circ 1_R + J
+      | c = {{Defof|Unity of Ring}}
+}}
+{{eqn | r = \paren {x + J} \circ \paren {1_R + J}
+      | c = [[Definition:Quotient Ring|Definition of $\circ$]] in $R / J$
+}}
+{{end-eqn}}
+Thus $R / J$ has a [[Definition:Unity of Ring|unity]], and that [[Definition:Unity of Ring|unity]] is $1_R + J$.
+Now suppose $J = R$.
+Then $1_R + J = J$ and therefore $1_R = 0_R$.
+The only ring to have $1_R = 0_R$ is the [[Definition:Null Ring|null ring]].
+This is appropriate, because:
+:$R / J = R / R = \set {0_{R / R} }$
+which is the [[Definition:Null Ring|null ring]].
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Ring Epimorphism with Trivial Kernel is Isomorphism}
+Tags: Ring Epimorphisms, Ring Isomorphisms
+
+\begin{theorem}
+Let $\phi: \left({R_1, +_1, \circ_1}\right) \to \left({R_2, +_2, \circ_2}\right)$ be a [[Definition:Ring Epimorphism|ring epimorphism]].
+Then $\phi$ is a [[Definition:Ring Isomorphism|ring isomorphism]] iff $\ker \left({\phi}\right) = \left\{ {0_{R_1} }\right\}$.
+\end{theorem}
+
+\begin{proof}
+Follows directly from [[Kernel is Trivial iff Monomorphism]] and the definition of a [[Definition:Ring Epimorphism|ring epimorphism]].
+{{qed}}
+[[Category:Ring Epimorphisms]]
+[[Category:Ring Isomorphisms]]
+7eicjqz42mvwfnab4mh5x2fqlui7ox6
+\end{proof}<|endoftext|>
+\section{Quotient Epimorphism is Epimorphism/Ring}
+Tags: Quotient Rings
+
+\begin{theorem}
+Let $\struct {R, +, \circ}$ be a [[Definition:Ring (Abstract Algebra)|ring]] whose [[Definition:Ring Zero|zero]] is $0_R$ and whose [[Definition:Unity of Ring|unity]] is $1_R$.
+Let $J$ be an [[Definition:Ideal of Ring|ideal]] of $R$.
+Let $\struct {R / J, +, \circ}$ be the [[Definition:Quotient Ring|quotient ring]] defined by $J$.
+Let $\phi: R \to R / J$ be the [[Definition:Quotient Ring Epimorphism|quotient (ring) epimorphism]] from $R$ to $R / J$:
+:$x \in R: \map \phi x = x + J$
+Then $\phi$ is a [[Definition:Ring Epimorphism|ring epimorphism]] whose [[Definition:Kernel of Ring Homomorphism|kernel]] is $J$.
+\end{theorem}
+
+\begin{proof}
+Let $x, y \in R$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \map \phi {x + y}
+      | r = \paren {x + y} + J
+      | c = Definition of $\phi$
+}}
+{{eqn | r = \paren {x + J} + \paren {y + J}
+      | c = {{Defof|Quotient Ring Addition}}
+}}
+{{eqn | r = \map \phi x + \map \phi y
+      | c = 
+}}
+{{end-eqn}}
+and:
+{{begin-eqn}}
+{{eqn | l = \map \phi {x \circ y}
+      | r = \paren {x \circ y} + J
+      | c = Definition of $\phi$
+}}
+{{eqn | r = \paren {x + J} \circ \paren {y + J}
+      | c = {{Defof|Quotient Ring Product}}
+}}
+{{eqn | r = \map \phi x \, \map \phi y
+      | c = 
+}}
+{{end-eqn}}
+Thus $\phi$ is a [[Definition:Ring Homomorphism|homomorphism]].
+$\phi$ is [[Definition:Surjection|surjective]] because:
+:$\forall x + J \in R / J: x + J = \map \phi x$
+Therefore $\phi$ is an [[Definition:Ring Epimorphism|epimorphism]].
+Let $x \in \map \ker \phi$.
+Then:
+{{begin-eqn}}
+{{eqn | l = x
+      | o = \in
+      | r = \map \ker \phi
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \map \phi x
+      | r = 0_{R/J}
+      | c = {{Defof|Kernel of Ring Homomorphism}}
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = x + J
+      | r = J
+      | c = [[Definition:Quotient Ring|$J$ is the zero of $\struct {R / J, +, \circ}$]]
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = x
+      | o = \in
+      | r = J
+      | c = [[Left Coset Equals Subgroup iff Element in Subgroup]]
+}}
+{{end-eqn}}
+Thus:
+:$\map \ker \phi = J$
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Ring Homomorphism Preserves Subrings}
+Tags: Ring Homomorphisms, Subrings, Ring Homomorphism Preserves Subrings
+
+\begin{theorem}
+Let $\phi: \struct {R_1, +_1, \circ_1} \to \struct {R_2, +_2, \circ_2}$ be a [[Definition:Ring Homomorphism|ring homomorphism]].
+If $S$ is a [[Definition:Subring|subring]] of $R_1$, then $\phi \sqbrk S$ is a [[Definition:Subring|subring]] of $R_2$.
+\end{theorem}
+
+\begin{proof}
+Since $S \ne \O$, $\phi \sqbrk S \ne \O$.
+From [[Group Homomorphism Preserves Subgroups]], $\struct {\phi \sqbrk S, +_2}$ is a [[Definition:Subgroup|subgroup]] of $\struct {R_2, +_2}$.
+From [[Homomorphism Preserves Subsemigroups]], $\struct {\phi \sqbrk S, \circ_2}$ is a [[Definition:Subsemigroup|subsemigroup]] of $\struct {R_2, \circ_2}$.
+Thus, as $\struct {R_2, +_2}$ is a [[Definition:Group|group]] and $\struct {R_2, \circ_2}$ is a [[Definition:Semigroup|semigroup]], the result follows.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+From [[Morphism Property Preserves Closure]], $\phi \sqbrk {R_1}$ is a [[Definition:Closed Algebraic Structure|closed algebraic structure]].
+From [[Epimorphism Preserves Rings]], $\phi \sqbrk S$ is a [[Definition:Ring (Abstract Algebra)|ring]].
+Hence the result, from the definition of [[Definition:Subring|subring]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let $S$ be a [[Definition:Subring|subring]] of $R_1$.
+Since $S \ne \O$ it follows that $\phi \sqbrk S \ne \O$.
+Let $x, y \in \phi \sqbrk S$.
+Then:
+:$\exists s, t \in S: x = \map \phi s, y = \map \phi t$
+So:
+{{begin-eqn}}
+{{eqn | l = x +_2 \paren {-y}
+      | r = \map \phi s +_2 \paren {-\map \phi t}
+      | c = 
+}}
+{{eqn | r = \map \phi {s +_1 \paren {-t} }
+      | c = as $\phi$ is a [[Definition:Ring Homomorphism|homomorphism]]
+}}
+{{end-eqn}}
+As $S$ is a [[Definition:Subring|subring]] of $R_1$, it is [[Definition:Closed Algebraic Structure|closed]] under $+_1$ and the taking of [[Definition:Ring Negative|negatives]].
+Thus $s +_1 \paren {-t} \in S$ and so $x +_2 \paren {-y} \in \phi \sqbrk S$.
+Similarly:
+{{begin-eqn}}
+{{eqn | l = x \circ_2 y
+      | r = \map \phi s \circ_2 \map \phi t
+      | c = 
+}}
+{{eqn | r = \map \phi {s \circ_1 t}
+      | c = as $\phi$ is a [[Definition:Ring Homomorphism|homomorphism]]
+}}
+{{end-eqn}}
+Because $S$ is a [[Definition:Subring|subring]] of $R_1$, it is [[Definition:Closed Algebraic Structure|closed]] under $\circ_1$.
+Thus $s \circ_1 t \in S$ and so $x \circ_2 y \in \phi \sqbrk S$.
+The result follows from [[Subring Test]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Kernel of Ring Homomorphism is Subring}
+Tags: Ring Homomorphisms, Subrings
+
+\begin{theorem}
+Let $\phi: \struct {R_1, +_1, \circ_1} \to \struct {R_2, +_2, \circ_2}$ be a [[Definition:Ring Homomorphism|ring homomorphism]].
+Then the [[Definition:Kernel of Ring Homomorphism|kernel]] of $\phi$ is a [[Definition:Subring|subring]] of $R_1$.
+\end{theorem}
+
+\begin{proof}
+From [[Ring Homomorphism of Addition is Group Homomorphism]] and [[Kernel of Group Homomorphism is Subgroup]]:
+:$\struct {\map \ker \phi, +_1} \le \struct {R_1, +_1}$
+where $\le$ denotes [[Definition:Subgroup|subgroup]].
+Let $x, y \in \map \ker \phi$.
+{{begin-eqn}}
+{{eqn | l = \map \phi {x \circ_1 y}
+      | r = \map \phi x \circ_2 \map \phi y
+      | c = [[Definition:Morphism Property|Morphism Property]]
+}}
+{{eqn | r = 0_{R_2} \circ_2 0_{R_2}
+      | c = [[Definition:Kernel of Ring Homomorphism|Definition of Kernel]]
+}}
+{{eqn | r = 0_{R_2}
+      | c = 
+}}
+{{end-eqn}}
+Thus $x \circ_1 y \in \map \ker \phi$.
+Thus the conditions for [[Subring Test]] are fulfilled, and $\map \ker \phi$ is a [[Definition:Subring|subring]] of $R_1$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Kernel of Ring Homomorphism is Ideal}
+Tags: Ring Homomorphisms, Ideal Theory
+
+\begin{theorem}
+Let $\phi: \struct {R_1, +_1, \circ_1} \to \struct {R_2, +_2, \circ_2}$ be a [[Definition:Ring Homomorphism|ring homomorphism]].
+Then the [[Definition:Kernel of Ring Homomorphism|kernel]] of $\phi$ is an [[Definition:Ideal of Ring|ideal]] of $R_1$.
+\end{theorem}
+
+\begin{proof}
+By [[Kernel of Ring Homomorphism is Subring]], $\map \ker \phi$ is a [[Definition:Subring|subring]] of $R_1$.
+Let $s \in \map \ker \phi$, so $\map \phi s = 0_{R_2}$.
+Suppose $x \in R_1$. Then:
+{{begin-eqn}}
+{{eqn | l = \map \phi {x \circ_1 s}
+      | r = \map \phi x \circ_2 \map \phi s
+      | c = {{Defof|Morphism Property}}
+}}
+{{eqn | r = \map \phi x \circ_2 0_{R_2}
+      | c = as $s \in \map \ker \phi$
+}}
+{{eqn | r = 0_{R_2}
+      | c = Properties of $0_{R_2}$
+}}
+{{end-eqn}}
+and similarly for $\map \phi {s \circ_1 x}$.
+The result follows.
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Ideals of Division Ring}
+Tags: Ideal Theory
+
+\begin{theorem}
+Let $\struct {R, +, \circ}$ be a [[Definition:Division Ring|division ring]] whose [[Definition:Ring Zero|zero]] is $0_R$.
+The only [[Definition:Ideal of Ring|ideals]] of $\struct {R, +, \circ}$ are $\set {0_R}$ and $R$ itself.
+That is, $\struct {R, +, \circ}$ has no non-[[Definition:Null Ideal|null]] [[Definition:Proper Ideal|proper ideals]].
+\end{theorem}
+
+\begin{proof}
+From [[Null Ring is Ideal]], $\set {0_R}$ is an [[Definition:Ideal of Ring|ideal]] of $\struct {R, +, \circ}$, as $\struct {R, +, \circ}$, being a [[Definition:Division Ring|division ring]], is also a [[Definition:Ring (Abstract Algebra)|ring]].
+By definition, every non-[[Definition:Ring Zero|zero]] [[Definition:Element|element]] of a [[Definition:Division Ring|division ring]] is a [[Definition:Unit of Ring|unit]].
+So $S \ne \set {0_R} \implies \exists x \in S: x \ne 0_R$ such that $r$ is a [[Definition:Unit of Ring|unit]] of $R$.
+The result follows from [[Ideal of Unit is Whole Ring]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Quotient Ring of Kernel of Ring Epimorphism}
+Tags: Ring Epimorphisms, Ring Isomorphisms, Quotient Rings
+
+\begin{theorem}
+Let $\phi: \struct {R_1, +_1, \circ_1} \to \struct {R_2, +_2, \circ_2}$ be a [[Definition:Ring Epimorphism|ring epimorphism]].
+Let $K = \map \ker \phi$.
+Then there is a [[Definition:Unique|unique]] [[Definition:Ring Isomorphism|ring isomorphism]] $g: R_1 / K \to R_2$ such that:
+:$g \circ q_K = \phi$
+$\phi$ is an [[Definition:Ring Isomorphism|isomorphism]] {{iff}} $K = \set {0_{R_1} }$.
+\end{theorem}
+
+\begin{proof}
+From the [[Quotient Theorem for Epimorphisms]], there is [[Definition:Unique|one and only one]] [[Definition:Isomorphism (Abstract Algebra)|isomorphism]] that satisfies the conditions for each of the operations on $R_1$.
+So the first statement follows directly.
+From [[Kernel is Trivial iff Monomorphism]], $\phi$ is a [[Definition:Ring Monomorphism|ring monomorphism]] {{iff}} $K = \set {0_{R_1} }$.
+As $\phi$ is also an [[Definition:Ring Epimorphism|epimorphism]], the result follows.
+\end{proof}<|endoftext|>
+\section{Ring Epimorphism Preserves Ideals}
+Tags: Ideal Theory, Ring Epimorphisms
+
+\begin{theorem}
+Let $\phi: \struct {R_1, +_1, \circ_1} \to \struct {R_2, +_2, \circ_2}$ be a [[Definition:Ring Epimorphism|ring epimorphism]].
+Let $J$ be an [[Definition:Ideal of Ring|ideal]] of $R_1$.
+Then $\phi \sqbrk J$ is an [[Definition:Ideal of Ring|ideal]] of $R_2$.
+\end{theorem}
+
+\begin{proof}
+$J$ is an [[Definition:Ideal of Ring|ideal]] of $R_1$, so it is also a [[Definition:Subring|subring]] of $R_1$.
+From [[Ring Homomorphism Preserves Subrings]], it follows that $\phi \sqbrk J$ is a [[Definition:Subring|subring]] of $R_2$.
+Now suppose $u \in \phi \sqbrk J$.
+Let $v \in R_2$.
+Then $\exists x \in J, y \in R_1$ such that $\map \phi x = u, \map \phi y = v$.
+Thus, by the [[Definition:Morphism Property|morphism property]]:
+:$u \circ_2 v = \map \phi x \circ_2 \map \phi y = \map \phi {x \circ_1 y}$
+So $u \circ_2 v \in \phi \sqbrk J$ because $x \circ_1 y \in J$.
+Similarly $v \circ_2 u \in \phi \sqbrk J$ also.
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Preimage of Image of Subring under Ring Homomorphism}
+Tags: Ring Homomorphisms, Subrings
+
+\begin{theorem}
+Let $\phi: \left({R_1, +_1, \circ_1}\right) \to \left({R_2, +_2, \circ_2}\right)$ be a [[Definition:Ring Homomorphism|ring homomorphism]].
+Let $K = \ker \left({\phi}\right)$, where $\ker \left({\phi}\right)$ is the [[Definition:Kernel of Ring Homomorphism|kernel]] of $\phi$. 
+Let $J$ be a [[Definition:Subring|subring]] of $R_1$.
+Then:
+:$\phi^{-1} \left[{\phi \left[{J}\right]}\right] = J + K$
+\end{theorem}
+
+\begin{proof}
+Let $x \in \phi^{-1} \left[{\phi \left[{J}\right]}\right]$.
+Then:
+{{begin-eqn}}
+{{eqn | l=x
+      | o=\in
+      | r=\phi^{-1} \left[{\phi \left[{J}\right]}\right]
+      | c=
+}}
+{{eqn | ll=\implies
+      | l=\phi \left({x}\right)
+      | o=\in
+      | r=\phi \left[{J}\right]
+      | c=Definition of [[Definition:Preimage of Element under Mapping|preimage]]
+}}
+{{eqn | ll=\implies
+      | l=\exists b \in J: \phi \left({x}\right)
+      | r=\phi \left({b}\right)
+      | c=Definition of [[Definition:Image of Element under Mapping|image]]
+}}
+{{eqn | ll=\implies
+      | l=\phi \left({x}\right) + \left({- \phi \left({b}\right)}\right)
+      | r=0_{R_2}
+      | c=Definition of [[Definition:Ring Negative|ring negative]] on $R_2$
+}}
+{{eqn | ll=\implies
+      | l=\phi \left({x + \left({-b}\right)}\right)
+      | r=0_{R_2}
+      | c=as $\phi$ is a [[Definition:Ring Homomorphism|homomorphism]]
+}}
+{{eqn | ll=\implies
+      | l=x + \left({-b}\right)
+      | o=\in
+      | r=K
+      | c=Definition of [[Definition:Kernel of Ring Homomorphism|kernel]]
+}}
+{{eqn | ll=\implies
+      | l=x
+      | o=\in
+      | r=J + K
+      | c=as $x = b + \left({x + \left({-b}\right)}\right)$
+}}
+{{end-eqn}}
+So we have shown that:
+:$\phi^{-1} \left[{\phi \left[{J}\right]}\right] \subseteq J + K$
+Now suppose that $x \in J + K$.
+Then:
+{{begin-eqn}}
+{{eqn | l=x
+      | o=\in
+      | r=J + K
+      | c=
+}}
+{{eqn | ll=\implies
+      | l=\exists b \in J, a \in K: x
+      | r=b + a
+      | c=
+}}
+{{eqn | ll=\implies
+      | l=\phi \left({x}\right)
+      | r=\phi \left({b}\right) + \phi \left({a}\right)
+      | c=
+}}
+{{eqn | r=\phi \left({b}\right) + 0_{R_2}
+      | c=as $a \in K$
+}}
+{{eqn | r=\phi \left({b}\right)
+      | c=
+}}
+{{eqn | ll=\implies
+      | l=\phi \left({x}\right)
+      | o=\in
+      | r=\phi \left[{J}\right]
+      | c=as $b \in J$
+}}
+{{eqn | ll=\implies
+      | l=x
+      | o=\in
+      | r=\phi^{-1} \left[{\phi \left[{J}\right]}\right]
+      | c=Definition of [[Definition:Preimage of Subset under Mapping|preimage]]
+}}
+{{end-eqn}}
+So we have shown that:
+:$J + K \subseteq \phi^{-1} \left[{\phi \left[{J}\right]}\right]$
+Hence the result by definition of [[Definition:Set Equality/Definition 2|set equality]].
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Preimage of Subring under Ring Homomorphism is Subring}
+Tags: Ring Homomorphisms, Subrings
+
+\begin{theorem}
+Let $\phi: \left({R_1, +_1, \circ_1}\right) \to \left({R_2, +_2, \circ_2}\right)$ be a [[Definition:Ring Homomorphism|ring homomorphism]].
+Let $S_2$ be a [[Definition:Subring|subring]] of $R_2$.
+Then $S_1 = \phi^{-1} \left[{S_2}\right]$ is a [[Definition:Subring|subring]] of $R_1$ such that $\ker \left({\phi}\right) \subseteq S_1$.
+\end{theorem}
+
+\begin{proof}
+Let $K = \ker \left({\phi}\right)$ be the [[Definition:Kernel of Ring Homomorphism|kernel]] of $R_1$.
+We have that $0_{R_2} \in S_2$ and so $\left\{{0_{R_2}}\right\} \subseteq S_2$.
+From [[Subset Maps to Subset]]:
+:$\phi^{-1} \left[{\left\{{0_{R_2}}\right\}}\right] \subseteq \phi^{-1} \left[{S_2}\right] = S_1$
+But by definition, $K = \phi^{-1} \left[{\left\{{0_{R_2}}\right\}}\right]$
+and so $S_1$ is a [[Definition:Subset|subset]] of $R_1$ containing $K$, that is:
+:$K \subseteq S_1 \subseteq R_1$
+Now we need to show that $S_1$ is a [[Definition:Subring|subring]] of $R_1$.
+Let $r, r' \in S_1$.
+Then $\phi \left({r}\right), \phi \left({r'}\right) \in S_2$.
+Hence:
+{{begin-eqn}}
+{{eqn | l=\phi \left({r +_1 r'}\right)
+      | r=\phi \left({r}\right) +_2 \phi \left({r'}\right)
+      | c=as $\phi$ is a [[Definition:Ring Homomorphism|ring homomorphism]]
+}}
+{{eqn | o=\in
+      | r=S_2
+      | c=because $S_2$ is a [[Definition:Subring|subring]]
+}}
+{{end-eqn}}
+So:
+:$r + r' \in S_1$
+Then:
+{{begin-eqn}}
+{{eqn | l=\phi \left({-r}\right)
+      | r=-\phi \left({r}\right)
+      | c=[[Group Homomorphism Preserves Inverses]]
+}}
+{{eqn | o=\in
+      | r=S_2
+      | c=because $S_2$ is a [[Definition:Subring|subring]]
+}}
+{{end-eqn}}
+So:
+:$-r \in \phi^{-1} S_1$
+Finally:
+{{begin-eqn}}
+{{eqn | l=\phi \left({r \circ_1 r'}\right)
+      | r=\phi \left({r}\right) \circ_2 \phi \left({r'}\right)
+      | c=as $\phi$ is a [[Definition:Ring Homomorphism|ring homomorphism]]
+}}
+{{eqn | o=\in
+      | r=S_2
+      | c=because $S_2$ is a [[Definition:Subring|subring]]
+}}
+{{end-eqn}}
+So:
+:$r \circ_1 r' \in S_1$
+So from the [[Subring Test]] we have that $\phi^{-1} \left[{S_2}\right]$ is a [[Definition:Subring|subring]] of $R$ containing $K$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Preimage of Ideal under Ring Homomorphism is Ideal}
+Tags: Ring Homomorphisms, Ideal Theory
+
+\begin{theorem}
+Let $\phi: \left({R_1, +_1, \circ_1}\right) \to \left({R_2, +_2, \circ_2}\right)$ be a [[Definition:Ring Homomorphism|ring homomorphism]].
+Let $S_2$ be an [[Definition:Ideal of Ring|ideal]] of $R_2$.
+Then $S_1 = \phi^{-1} \left[{S_2}\right]$ is an [[Definition:Ideal of Ring|ideal]] of $R_1$ such that $\ker \left({\phi}\right) \subseteq S_1$.
+\end{theorem}
+
+\begin{proof}
+From [[Preimage of Subring under Ring Homomorphism is Subring]] we have that $S_1 = \phi^{-1} \left[{S_2}\right]$ is a [[Definition:subring|subring]] of $R_1$ such that $\ker \left({\phi}\right) \subseteq S_1$.
+We now need to show that $S_1$ is an [[Definition:Ideal of Ring|ideal]] of $R_1$.
+Let $s_1 \in S_1, r_1 \in R_1$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \phi \left({r_1 \circ_1 s_1}\right)
+      | r = \phi \left({r_1}\right) \circ_2 \phi \left({s_1}\right)
+      | c = as $\phi$ is a [[Definition:Ring Homomorphism|homomorphism]]
+}}
+{{eqn | r = S_2
+      | c = as $S_2$ is an [[Definition:Ideal of Ring|ideal]] of $R_2$
+}}
+{{end-eqn}}
+Thus:
+:$r_1 \circ_1 s_1 \in \phi^{-1} \left[{S_2}\right]= S_1$
+Similarly for $s_1 \circ_1 r_1$.
+So $S_1$ is an [[Definition:Ideal of Ring|ideal]] of $R_1$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Image of Preimage of Subring under Ring Epimorphism}
+Tags: Ring Epimorphisms, Subrings
+
+\begin{theorem}
+Let $\phi: \struct {R_1, +_1, \circ_1} \to \struct {R_2, +_2, \circ_2}$ be a [[Definition:Ring Epimorphism|ring epimorphism]].
+Let $S_2$ be a [[Definition:Subring|subring]] of $R_2$.
+Then:
+:$\phi \sqbrk {\phi^{-1} \sqbrk {S_2} } = S_2$
+\end{theorem}
+
+\begin{proof}
+As $\phi$ is an [[Definition:Ring Epimorphism|epimorphism]], it is a [[Definition:Surjection|surjection]], and so $\Img \phi = R_2$.
+So $S_2 \subseteq \Img {R_1}$.
+The result then follows from [[Image of Preimage under Mapping]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Diagonal Relation is Universally Congruent}
+Tags: Congruence Relations
+
+\begin{theorem}
+The [[Definition:Diagonal Relation|diagonal relation]] $\Delta_S$ on a [[Definition:Set|set]] $S$ is [[Definition:Universally Congruent|universally congruent]] with every [[Definition:Binary Operation|operation]] on $S$.
+\end{theorem}
+
+\begin{proof}
+We have that the [[Diagonal Relation is Equivalence|diagonal relation is an equivalence relation]].
+Let $\left({S, \circ}\right)$ be any [[Definition:Algebraic Structure|algebraic structure]].
+{{begin-eqn}}
+{{eqn | o=
+      | r=x_1 \Delta_S x_2 \land y_1 \Delta_S y_2
+      | c=
+}}
+{{eqn | o=\implies
+      | r=x_1 = x_2 \land y_1 = y_2
+      | c=Definition of [[Definition:Diagonal Relation|Diagonal Relation]]
+}}
+{{eqn | o=\implies
+      | r=x_1 \circ y_1 = x_2 \circ y_2
+      | c=(consequence of equality)
+}}
+{{eqn | o=\implies
+      | r=\left({x_1 \circ y_1}\right) \Delta_S \left({x_2 \circ y_2}\right)
+      | c=Definition of [[Definition:Diagonal Relation|Diagonal Relation]]
+}}
+{{end-eqn}}
+{{qed}}
+It can therefore be described as [[Definition:Universally Congruent|universally congruent]].
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Properties of Ordered Ring}
+Tags: Ordered Rings
+
+\begin{theorem}
+Let $\struct {R, +, \circ, \le}$ be an [[Definition:Ordered Ring|ordered ring]] whose [[Definition:Ring Zero|zero]] is $0_R$ and whose [[Definition:Unity of Ring|unity]] is $1_R$.
+Let $U_R$ be the [[Definition:Group of Units of Ring|group of units]] of $R$.
+Let $x, y, z \in \struct {R, +, \circ, \le}$.
+Then the following properties hold:
+: $(1): \quad x < y \iff x + z < y + z$. Hence $x \le y \iff x + z \le y + z$ (because $\struct {R, +, \le}$ is an [[Definition:Ordered_Group|ordered group]]).
+: $(2): \quad x < y \iff 0 < y + \paren {-x}$. Hence $x \le y \iff 0 \le y + \paren {-x}$
+: $(3): \quad 0 < x \iff \paren {-x} < 0$. Hence $0 \le x \iff \paren {-x} \le 0$
+: $(4): \quad x < 0 \iff 0 < \paren {-x}$. Hence $x \le 0 \iff 0 \le \paren {-x}$
+: $(5): \quad \forall n \in \Z_{>0}: x > 0 \implies n \cdot x > 0$
+: $(6): \quad x \le y, 0 \le z: x \circ z \le y \circ z, z \circ x \le z \circ y$
+: $(7): \quad x \le y, z \le 0: y \circ z \le x \circ z, z \circ y \le z \circ x$
+\end{theorem}<|endoftext|>
+\section{Positive Elements of Ordered Ring}
+Tags: Ordered Rings
+
+\begin{theorem}
+Let $\struct {R, +, \circ, \le}$ be an [[Definition:Ordered Ring|ordered ring]] [[Definition:Ring with Unity|with unity]] whose [[Definition:Ring Zero|zero]] is $0_R$ and whose [[Definition:Unity of Ring|unity]] is $1_R$.
+Let $P$ be the set of [[Definition:Positive|positive]] elements of $R$ , that is, $P = R_{\ge 0}$.
+Then:
+: $(1): \quad P + P \subseteq P$
+: $(2): \quad P \cap \paren {-P} = \set {0_R}$
+: $(3): \quad P \circ P \subseteq P$
+If $\le$ is a [[Definition:Total Ordering|total ordering]], that is, if $\struct {R, +, \circ, \le}$ is a [[Definition:Totally Ordered Ring|totally ordered ring]], then:
+: $(4): \quad P \cup \paren {-P} = R$
+The converse is also true:
+Let $\struct {R, +, \circ}$ be a [[Definition:Ring (Abstract Algebra)|ring]].
+Let $P \subseteq R$ such that $(1)$, $(2)$ and $(3)$ are satisfied.
+Then there is one and only one [[Definition:Ordering|ordering]] $\le$ [[Definition:Ordering Compatible with Ring Structure|compatible with the ring structure]] of $R$ such that $P = R_{\ge 0}$.
+Also, if $(4)$ is also satisfied, then $\le$ is a [[Definition:Total Ordering|total ordering]].
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+First, suppose that $\le$ is [[Definition:Ordering Compatible with Ring Structure|compatible with the ring structure]] of $R$.
+Hence:
+:$(OR1): \quad \le$ is [[Definition:Relation Compatible with Operation|compatible]] with $+$
+:$(OR2): \quad \forall x, y \in R: 0_R \le x, 0_R \le y \implies 0_R \le x \circ y$.
+$(1)$: Let $x, y \in R: 0_R \le x, 0_R \le y$.
+Then $0_R + 0_R \le x + y$ by the fact that $\le$ is [[Definition:Relation Compatible with Operation|compatible]] with $+$.
+Thus $0_R \le x + y$ and thus $x + y \in P$.
+$(2)$: By [[Properties of Ordered Ring]] item 4, $-P = \set {x \in R: x \le 0_R}$.
+Let $x \in P \cap \paren {-P}$.
+Then $x \le 0_R$ and $0_R \le x$.
+So from the [[Definition:Ordering|antisymmetric nature]] of $\le$, $x = 0_R$.
+$(3)$: This is equivalent to $0_R \le x, 0_R \le y \implies 0_R \le x \circ y$ which is one of the properties of being [[Definition:Ordering Compatible with Ring Structure|compatible with the ring structure]] of $R$.
+$(4)$: Now if $\le$ is a [[Definition:Total Ordering|total ordering]], then $\forall x \in R$, either $x \le 0_R$ or $0_R \le x$, and the result follows.
+=== Sufficient Condition ===
+Let $P \subseteq R$ such that $(1)$, $(2)$ and $(3)$ are satisfied.
+By item $(OR2)$ of [[Properties of Ordered Ring]], we have:
+:$x \le y \iff 0 \le y + \paren {-x}$
+so there is at most one ordering on $R$ compatible with the ring structure of $R$ such that $P = R_{\ge 0}$, namely, the one that satisfies:
+:$x \le y \iff y + \left({-x}\right) \in P$
+Now we need to show that $\le$ thus defined has the required properties.
+;[[Definition:Reflexive Relation|Reflexivity]]:
+By $(2)$:
+:$\forall x \in R: x \le x$ because $x + \paren {-x} = 0_R \in P$
+;[[Definition:Antisymmetric Relation|Antisymmetry]]:
+Let $x \le y$ and $y \le x$.
+Then $y + \paren {-x} \in P$ and $-\paren {y + \paren {-x} } = x + \paren {-y} \in P$.
+Thus by $(2)$, $y + \paren {-x} = 0_R$ and thus $y = x$.
+;[[Definition:Transitive Relation|Transitivity]]:
+If $x \le y$ and $y \le z$, then $y + \paren {-x} \in P$ and $z + \paren {-y} \in P$.
+But as $z + \paren {-x} = z + \paren {-y} + y + \paren {-x}$, we have that $z + \paren {-x} \in P$ from $(1)$.
+Hence $x \le z$.
+Let $x \le y$.
+Then $z + x \le z + y$ since $\paren {z + y} + \paren {-\paren {x + x} } = y + \paren {-x} \in P$.
+Finally, [[Definition:Ordering Compatible with Ring Structure|Ordering Compatible with Ring]] holds because of $(3)$.
+If $(4)$ holds, then $\forall x, y \in R$, either $y + \paren {-x} \in P$ or $x + \paren {-y} = -\paren {y + \paren {-x} }  \in P$, that is, either $x \le y$ or $y \le x$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Symmetric Difference with Intersection forms Ring}
+Tags: Commutative Algebra, Set Intersection, Symmetric Difference, Power Set, Symmetric Difference with Intersection forms Ring, Examples of Commutative and Unitary Rings
+
+\begin{theorem}
+Let $S$ be a [[Definition:Set|set]].
+Then $\struct {\powerset S, *, \cap}$ is a [[Definition:Commutative and Unitary Ring|commutative ring with unity]], in which the [[Definition:Unity of Ring|unity]] is $S$.
+This ring is not an [[Definition:Integral Domain|integral domain]].
+\end{theorem}
+
+\begin{proof}
+From [[Symmetric Difference on Power Set forms Abelian Group]], $\struct {\powerset S, *}$ is an [[Definition:Abelian Group|abelian group]], where $\O$ is the [[Definition:Identity Element|identity]] and each [[Definition:Element|element]] is [[Definition:Self-Inverse Element|self-inverse]].
+From [[Power Set with Intersection is Monoid]], $\struct {\powerset S, \cap}$ is a [[Definition:Commutative Monoid|commutative monoid]] whose [[Definition:Identity Element|identity]] is $S$.
+Also [[Intersection Distributes over Symmetric Difference]].
+Thus $\struct {\powerset S, \cap}$ is a [[Definition:Commutative and Unitary Ring|commutative ring with a unity]] which is $S$.
+From [[Intersection with Empty Set]]:
+:$\forall A \in \powerset S: A \cap \O = \O = \O \cap A$
+Thus $\O$ is indeed the [[Definition:Ring Zero|zero]].
+However, from [[Set Intersection Not Cancellable]], it follows that $\struct {\powerset S, *, \cap}$ is not an [[Definition:Integral Domain|integral domain]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+From [[Power Set is Closed under Symmetric Difference]] and [[Power Set is Closed under Intersection]], we have that both $\struct {\powerset S, *}$ and $\struct {\powerset S, \cap}$ are [[Definition:Closed Algebraic Structure|closed]].
+Hence $\powerset S$ is a [[Definition:Ring of Sets|ring of sets]], and hence a [[Ring of Sets is Commutative Ring|commutative ring]].
+From [[Intersection with Subset is Subset]], we have $A \subseteq S \iff A \cap S = A$. Thus we see that $S$ is the [[Definition:Unity of Ring|unity]].
+Also during the proof of [[Power Set with Intersection is Monoid]], it was established that $S$ is the [[Definition:Identity Element|identity]] of $\struct {\powerset S, \cap}$.
+We also note that [[Set Intersection Not Cancellable|set intersection is not cancellable]], so $\struct {\powerset S, *, \cap}$ is not an [[Definition:Integral Domain|integral domain]].
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Subfield Test}
+Tags: Subfields
+
+\begin{theorem}
+Let $\struct {F, +, \circ}$ be a [[Definition:Field (Abstract Algebra)|field]] whose [[Definition:Field Zero|zero]] is $0_F$.
+Let $K$ be a [[Definition:Subset|subset]] of $F$.
+Then $\struct {K, +, \circ}$ is a [[Definition:Subfield|subfield]] of $\struct {F, +, \circ}$ {{iff}} these all hold:
+:$(1): \quad K^* \ne \O$
+:$(2): \quad \forall x, y \in K: x + \paren {-y} \in K$
+:$(3): \quad \forall x, y \in K: x \circ y \in K$
+:$(4): \quad x \in K^* \implies x^{-1} \in K^*$
+where $K^*$ denotes $K \setminus \set {0_F}$.
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+Let $\struct {K, +, \circ}$ be a [[Definition:Subfield|subfield]] of $\struct {F, +, \circ}$.
+Then the conditions $(1)$ to $(4)$ all hold by virtue of the [[Definition:Field Axioms|field axioms]].
+{{qed|lemma}}
+=== Sufficient Condition ===
+Suppose the conditions $(1)$ to $(4)$ hold.
+From the [[Division Subring Test]], it follows that $\struct {K, +, \circ}$ is a [[Definition:Division Ring|division ring]].
+As $\struct {F, +, \circ}$ is a [[Definition:Field (Abstract Algebra)|field]], then $\circ$ is [[Definition:Commutative Operation|commutative]] on all of $F$.
+Therefore $\circ$ is [[Definition:Commutative Operation|commutative]] also on $K$ by [[Restriction of Commutative Operation is Commutative]].
+Thus $\struct {K, +, \circ}$ is a [[Definition:Field (Abstract Algebra)|field]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Field of Quotients of Subdomain}
+Tags: Fields of Quotients
+
+\begin{theorem}
+Let $\struct {F, +, \circ}$ be a [[Definition:Field (Abstract Algebra)|field]] whose [[Definition:Unity of Ring|unity]] is $1_F$.
+Let $\struct {D, +, \circ}$ be a [[Definition:Subdomain|subdomain]] of $\struct {F, +, \circ}$ whose [[Definition:Unity of Ring|unity]] is $1_D$.
+Let:
+:$K = \set {\dfrac x y: x \in D, y \in D^*}$
+where $\dfrac x y$ is the [[Definition:Division Product|division product]] of $x$ by $y$.
+Then $\struct {K, +, \circ}$ is a [[Definition:Field of Quotients|field of quotients]] of $\struct {D, +, \circ}$.
+\end{theorem}
+
+\begin{proof}
+$1_D = 1_F$ by [[Subdomain Test]].
+The sum and product of two elements of $K$ are also in $K$ by [[Addition of Division Products]] and [[Product of Division Products]].
+The additive and product inverses of $K$ are also in $K$ by [[Negative of Division Product]] and [[Inverse of Division Product]].
+Thus by [[Subfield Test]], $\struct {K, +, \circ}$ is a [[Definition:Subfield|subfield]] of $\struct {F, +, \circ}$ which clearly contains $\struct {D, +, \circ}$.
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Existence of Field of Quotients}
+Tags: Fields of Quotients, Integral Domains
+
+\begin{theorem}
+Let $\struct {D, +, \circ}$ be an [[Definition:Integral Domain|integral domain]].
+Then there exists a [[Definition:Field of Quotients|field of quotients]] of $\struct {D, +, \circ}$.
+\end{theorem}
+
+\begin{proof}
+Let $\struct {D, +, \circ}$ be an [[Definition:Integral Domain|integral domain]] whose [[Definition:Ring Zero|zero]] is $0_D$ and whose [[Definition:Unity of Ring|unity]] is $1_D$.
+=== Inverse Completion is an Abelian Group ===
+By [[Inverse Completion of Integral Domain Exists]], we can define the [[Definition:Inverse Completion|inverse completion]] $\struct {K, \circ}$ of $\struct {D, \circ}$.
+Thus $\struct {K, \circ}$ is a [[Definition:Commutative Semigroup|commutative semigroup]] such that:
+:$(1): \quad$ The [[Definition:Identity Element|identity]] of $\struct {K, \circ}$ is $1_D$
+:$(2): \quad$ Every [[Definition:Element|element]] $x$ of $\struct {D^*, \circ}$ has an [[Definition:Inverse Element|inverse]] $\dfrac {1_D} x$ in $\struct {K, \circ}$
+:$(3): \quad$ Every [[Definition:Element|element]] of $\struct {K, \circ}$ is of the form $x \circ y^{-1}$ (which from the definition of [[Definition:Division Product|division product]], we can also denote $x / y$), where $x \in D, y \in D^*$.
+It can also be noted that from [[Inverse Completion Less Zero of Integral Domain is Closed]], $\struct {K^*, \circ}$ is [[Definition:Closed Algebraic Structure|closed]].
+Hence $\struct {K^*, \circ}$ is an [[Definition:Abelian Group|abelian group]].
+=== Additive Operation on $K$ ===
+In what follows, we take for granted the rules of [[Definition:Associative Operation|associativity]], [[Definition:Commutative Operation|commutativity]] and [[Definition:Distributive Operation|distributivity]] of $+$ and $\circ$ in $D$.
+We require to extend the [[Definition:Binary Operation|operation]] $+$ on $D$ to an [[Definition:Binary Operation|operation]] $+'$ on $K$, so that $\struct {K, +', \circ}$ is a [[Definition:Field (Abstract Algebra)|field]].
+By [[Addition of Division Products]], we define $+'$ as:
+:$\forall a, c \in D, \forall b, d \in D^*: \dfrac a b +' \dfrac c d = \dfrac {a \circ d + b \circ c} {b \circ d}$
+where we have defined $\dfrac x y = x \circ y^{-1} = y^{-1} \circ x$ as [[Definition:Division Product|$x$ divided by $y$]].
+Next, we see that:
+:$\forall a, b \in D: a +' b = \dfrac {a \circ 1_D + b \circ 1_D} {1_D \circ 1_D} = a + b$
+So $+$ [[Definition:Operation Induced by Restriction|induces the given operation $+$]] on its substructure $D$, and we are justified in using $+$ for both operations.
+{{qed|lemma}}
+=== Addition on $K$ makes an Abelian Group ===
+Now we verify that $\struct {K, +}$ is an [[Definition:Abelian Group|abelian group]]
+Taking the [[Definition:Group Axioms|group axioms]] in turn:
+==== $\text G 0$: Closure ====
+Let $\dfrac a b, \dfrac c d \in K$.
+Then $a, c \in D$ and $b, d \in D^*$, and $\dfrac a b + \dfrac c d = \dfrac {a \circ d + b \circ c} {b \circ d}$.
+As $b, d \in D^*$ it follows that $b \circ d \in D^*$ because $D$ is an [[Definition:Integral Domain|integral domain]].
+By the fact of [[Definition:Closed Algebraic Structure|closure]] of $+$ and $\circ$ in $D$, $a \circ d + b \circ c \in D$.
+Hence $\dfrac a b + \dfrac c d \in K$ and $+$ is [[Definition:Closed Algebraic Structure|closed]].
+{{qed|lemma}}
+==== $\text G 1$: Associativity ====
+{{begin-eqn}}
+{{eqn | l = \paren {\frac a b + \frac c d} + \frac e f
+      | r = \frac {a \circ d + b \circ c} {b \circ d} + \frac e f
+      | c = 
+}}
+{{eqn | r = \frac {\paren {a \circ d + b \circ c} \circ f + b \circ d \circ e} {b \circ d \circ f}
+      | c = 
+}}
+{{eqn | r = \frac {a \circ d \circ f + b \circ c \circ f + b \circ d \circ e} {b \circ d \circ f}
+      | c = 
+}}
+{{eqn | r = \frac {a \circ d \circ f + b \circ \paren {c \circ f + d \circ e} } {b \circ d \circ f}
+      | c = 
+}}
+{{eqn | r = \frac a b + \frac {c \circ f + d \circ e} {d \circ f}
+      | c = 
+}}
+{{eqn | r = \frac a b + \paren {\frac c d + \frac e f}
+      | c = 
+}}
+{{end-eqn}}
+Hence $\dfrac a b + \dfrac c d \in K$ and $+$ is [[Definition:Associative Operation|associative]].
+{{qed|lemma}}
+==== $\text G 2$: Identity ====
+The [[Definition:Identity Element|identity]] for $+$ is $\dfrac 0 k$ where $k \in D^*$:
+{{begin-eqn}}
+{{eqn | l = \frac a b + \frac 0 k
+      | r = \frac {a \circ k + b \circ 0} {b \circ k}
+      | c = 
+}}
+{{eqn | r = \frac {a \circ k} {b \circ k}
+      | c = 
+}}
+{{eqn | r = \frac a b
+      | c = 
+}}
+{{end-eqn}}
+Similarly for $\dfrac 0 k + \dfrac a b$.
+{{qed|lemma}}
+==== $\text G 3$: Inverses ====
+The [[Definition:Inverse Element|inverse]] of $\dfrac a b$ for $+$ is $\dfrac {-a} b$:
+{{begin-eqn}}
+{{eqn | l = \frac a b + \frac {-a} b
+      | r = \frac {a \circ b + b \circ \paren {-a} } {b \circ b}
+      | c = 
+}}
+{{eqn | r = \frac {b \circ \paren {a + \paren {-a} } } {b \circ b}
+      | c = 
+}}
+{{eqn | r = \frac {b \circ 0} {b \circ b}
+      | c = 
+}}
+{{eqn | r = \frac 0 {b \circ b}
+      | c = 
+}}
+{{end-eqn}}
+From [[Existence of Field of Quotients#G2: Identity|above]], this is the [[Definition:Identity Element|identity]] for $+$.
+Similarly, $\dfrac {-a} b + \dfrac a b = \dfrac 0 {b \circ b}$.
+Hence $\dfrac {-a} b$ is the [[Definition:Inverse Element|inverse]] of $\dfrac a b$ for $+$.
+{{qed|lemma}}
+==== $\text C$: Commutativity ====
+{{begin-eqn}}
+{{eqn | l = \frac a b + \frac c d
+      | r = \frac {a \circ d + b \circ c} {b \circ d}
+      | c = 
+}}
+{{eqn | r = \frac {c \circ b + d \circ a} {d \circ b}
+      | c = 
+}}
+{{eqn | r = \frac c d + \frac a b
+      | c = 
+}}
+{{end-eqn}}
+Therefore, $\struct {K, +, \circ}$ is a [[Definition:Commutative and Unitary Ring|commutative ring with unity]].
+{{qed|lemma}}
+=== Product Distributes over Addition ===
+From [[Extension Theorem for Distributive Operations]], it follows directly that $\circ$ [[Definition:Distributive Operation|distributes]] over $+$.
+{{qed|lemma}}
+=== Product Inverses in $K$ ===
+From [[Ring Product with Zero]], we note that:
+:$\forall x \in D, y \in D^*: \dfrac x y \ne 0_D \implies x \ne 0_D$
+From [[Inverse of Division Product]]:
+:$\forall x, y \in D^*: \paren {\dfrac x y}^{-1} = \dfrac y x$
+Thus $\dfrac x y \in K$ has the [[Definition:Ring Product Inverse|ring product inverse]] $\dfrac y x \in K$.
+{{qed|lemma}}
+=== Inverse Completion is a Field ===
+We have that:
+:the [[Definition:Algebraic Structure|algebraic structure]] $\struct {K, +}$ is an [[Definition:Abelian Group|abelian group]]
+:the [[Definition:Algebraic Structure|algebraic structure]] $\struct {K^*, \circ}$ is an [[Definition:Abelian Group|abelian group]]
+:the operation $\circ$ [[Definition:Distributive Operation|distributes]] over $+$.
+Hence $\struct {K, +, \circ}$ is a [[Definition:Field (Abstract Algebra)|field]].
+We also have that $\struct {K, +, \circ}$ [[Definition:Submagma|contains $\struct {D, +, \circ}$ algebraically]] such that:
+:$\forall x \in K: \exists z \in D, y \in D^*: z = \dfrac x y$
+Thus $\struct {K, +, \circ}$ is a [[Definition:Field of Quotients|field of quotients]] of $\struct {D, +, \circ}$.
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Quotient Theorem for Monomorphisms}
+Tags: Fields of Quotients, Integral Domains, Ring Monomorphisms, Named Theorems
+
+\begin{theorem}
+Let $K, L$ be [[Definition:Field of Quotients|fields of quotients]] of [[Definition:Integral Domain|integral domains]] $\struct {R, +_R, \circ_R}, \struct {S, +_S, \circ_S}$ respectively.
+Let $\phi: R \to S$ be a [[Definition:Ring Monomorphism|monomorphism]].
+Then there is one and only one [[Definition:Ring Monomorphism|monomorphism]] $\psi: K \to L$ extending $\phi$, and:
+:$\forall x \in R, y \in R^*: \map \psi {\dfrac x y} = \dfrac {\map \phi x} {\map \phi y}$
+Also, if $\phi$ is a [[Definition:Ring Isomorphism|ring isomorphism]], then so is $\psi$.
+\end{theorem}
+
+\begin{proof}
+By definition, $\struct {K, \circ_R}$ and $\struct {L, \circ_S}$ are [[Definition:Inverse Completion|inverse completions]] of $\struct {R, \circ_R}$ and $\struct {S, \circ_S}$ respectively.
+{{Questionable|this is not the definition}}
+So by the [[Extension Theorem for Homomorphisms]], there is one and only one [[Definition:Ring Monomorphism|monomorphism]] $\psi: \struct {K, \circ_R} \to \struct {L, \circ_S$ extending $\phi$.
+Thus:
+:$\forall x \in R, y \in R^*: \map \psi {\dfrac x y} = \dfrac {\map \phi x} {\map \phi y}$
+By the [[Extension Theorem for Isomorphisms]], $\psi$ is an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]] if $\phi$ is.
+Thus, $\forall x, y \in R, z, w \in R^*$:
+{{begin-eqn}}
+{{eqn | l = \map \psi {\frac x z +_R \frac y w}
+      | r = \map \psi {\frac {\paren {x \circ_R w} +_R \paren {y \circ_R z} } {z \circ_R w} }
+      | c = [[Addition of Division Products]]
+}}
+{{eqn | r = \frac {\map \phi {\paren {x \circ_R w} +_R \paren {y \circ_R z} } } {\map \phi {z \circ_R w} }
+      | c = Definition of $\psi$
+}}
+{{eqn | r = \frac {\paren {\map \phi x \circ_S \map \phi w} +_S \paren {\map \phi y \circ_S \map \phi z} } {\map \phi z \circ_S \map \phi w}
+      | c = [[Definition:Morphism Property|Morphism Property]]
+}}
+{{eqn | r = \frac {\map \phi x} {\map \phi z} +_S \frac {\map \phi y} {\map \phi w}
+      | c = [[Addition of Division Products]]
+}}
+{{eqn | r = \map \psi {\frac x z} +_S \map \psi {\frac y w}
+      | c = Definition of $\psi$
+}}
+{{end-eqn}}
+Thus $\psi: K \to L$ is a [[Definition:Ring Monomorphism|monomorphism]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Field of Quotients is Unique}
+Tags: Fields of Quotients, Field Isomorphisms, Integral Domains
+
+\begin{theorem}
+Let $\struct {D, +, \circ}$ be an [[Definition:Integral Domain|integral domain]].
+Let $K, L$ be [[Definition:Field of Quotients|field of quotients]] of $\struct {D, +, \circ}$.
+Then there is [[Definition:Exactly One|one and only one]] [[Definition:Field Isomorphism|(field) isomorphism]] $\phi: K \to L$ satisfying:
+:$\forall x \in D: \map \phi x = x$
+\end{theorem}
+
+\begin{proof}
+Follows directly from the [[Quotient Theorem for Monomorphisms]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Divided by Positive Element of Field of Quotients}
+Tags: Fields of Quotients, Integral Domains
+
+\begin{theorem}
+Let $\struct {K, +, \circ}$ be the [[Definition:Field of Quotients|field of quotients]] of a [[Definition:Totally Ordered Ring|totally ordered]] [[Definition:Integral Domain|integral domain]] $\struct {D, +, \circ, \le}$.
+Then:
+:$\forall z \in K: \exists x, y \in D: z = \dfrac x y, y \in D_{>0}$
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Field of Quotients|field of quotients]]:
+:$\forall z \in K: \exists x, y \in D: z = \dfrac x y, y \in D_{\ne 0}$
+Suppose $z = x' / y'$ such that $y' \notin D_{>0}$.
+Then $y' < 0$ as $D$ is [[Definition:Totally Ordered Ring|totally ordered]].
+Then:
+{{begin-eqn}}
+{{eqn | l = x' / y'
+      | r = x' \circ \paren {y'}^{-1}
+      | c = {{Defof|Division}}
+}}
+{{eqn | r = \paren {-x'} \circ \paren {-y'}^{-1}
+      | c = [[Product of Ring Negatives]]
+}}
+{{eqn | r = \paren {-x'} / \paren {-y'}
+      | c = {Defof|Division}}
+}}
+{{end-eqn}}
+{{explain|first to second step above assumes more than is stated: $-\paren {y'}^{-1} {{=}} \paren {-y'}^{-1}$ - add that step and link to that result}}
+If $y' < 0$, then $\paren {-y'} > 0$ from [[Properties of Ordered Ring]] $(4)$.
+So all we need to do is set $x = -x', y = -y'$ and the result follows.
+{{qed}}
+[[Category:Fields of Quotients]]
+[[Category:Integral Domains]]
+1i02g1dcijv3u3rs5199dhvjn1utptf
+\end{proof}<|endoftext|>
+\section{Total Ordering on Field of Quotients is Unique}
+Tags: Fields of Quotients, Integral Domains
+
+\begin{theorem}
+Let $\struct {K, +, \circ}$ be a [[Definition:Field of Quotients|field of quotients]] of an [[Definition:Ordered Integral Domain|ordered integral domain]] $\struct {D, +, \circ, \le}$.
+Then there is one and only one [[Definition:Total Ordering|total ordering]] $\le'$ on $K$ which is [[Definition:Ordering Compatible with Ring Structure|compatible with its ring structure]] and induces on $D$ its given [[Definition:Total Ordering|total ordering]] $\le$.
+That ordering is the one defined by:
+:$P = \set {\dfrac x y \in K: x \in D_+, y \in D_+^*}$
+\end{theorem}
+
+\begin{proof}
+First, note that from [[Divided by Positive Element of Field of Quotients]]:
+:$\forall z \in K: \exists x, y \in R: z = \dfrac x y, y \in R_+^*$
+Now we show that $P$ satistfies conditions $(1)$ to $(4)$ of [[Positive Elements of Ordered Ring]].
+From [[Addition of Division Products]] and [[Product of Division Products]], it is clear that $P$ satisfies $(1)$ and $(3)$.
+{{qed|lemma}}
+Next we establish $(2)$.
+Let $z \in P \cap \paren {-P}$.
+Then $z \in P$ and $z \in \paren {-P}$. Thus:
+:$\exists x_1, x_2 \in D_+, y_1, y_2 \in D_+^*: z = x_1 / y_1, -z = x_2 / y_2$
+So:
+:$x_1 / y_1 = -\paren {x_2 / y_2} \implies x_1 \circ y_2 = -\paren {x_2 \circ y_1}$
+But $0 \le x_1 \circ y_2$ and $-\paren {x_2 \circ y_1} \le 0$.
+So $x_1 \circ y_2 = 0$.
+As $0 < y_2$, this means $x = 0$ and therefore $z = 0$.
+Thus $(2)$ has been established.
+{{qed|lemma}}
+Now to show that $(4)$ holds.
+Let $z = x / y$ where $x \in D, y \in D_+$.
+Suppose $0 \le x$. Then $z \in P$.
+However, suppose $x < 0$. Then $0 < \paren {-x}$ so $-z = \paren {-x} / y \in P$.
+Thus:
+:$z = -\paren {-z} \in -P$
+So:
+:$P \cup \paren {-P} = K$
+So by [[Positive Elements of Ordered Ring]], the relation $\le'$ on $K$ defined by $P$ is a [[Definition:Total Ordering|total ordering]] on $K$ [[Definition:Ordering Compatible with Ring Structure|compatible with its ring structure]].
+{{qed|lemma}}
+Now we need to show that the ordering induced on $D$ by $\le'$ is indeed $\le$.
+Let $z \in D_+$. Then $z = z / 1_D \in P$, as $1_D \in D_+^*$.
+Thus $D_+ \subseteq P$ and $D_+ \subseteq D$ so $D_+ \subseteq D \cap P$ from [[Intersection is Largest Subset]].
+Conversely, let $z \in D \cap P$. Then:
+:$\exists x \in D_+, y \in D_+^*: z = x / y$
+If $x = 0$ then $z = 0$, and if $0 < x$ then as $z \circ y = x$ and $0 < y$, it follows that $0 < z$ by item 1 of [[Properties of Ordered Ring#Total Ordering|Properties of a Totally Ordered Ring]].
+So $\forall z \in D: 0 \le z \iff z \in P$.
+Thus it follows that $z \in D \cap P \implies z \in D_+$, i.e. $D \cap P \subseteq D_+$.
+Thus $D_+ = D \cap P$.
+By item $(2)$ of [[Properties of Ordered Ring]], we have:
+:$x \le y \iff 0 \le y + \left({-x}\right)$
+and thus it follows that the ordering induced on $D$ by $\le'$ is $\le$.
+{{qed|lemma}}
+Now we need to show uniqueness.
+Let $\preceq$ be a total ordering on $K$ which is compatible with its ring structure and induces on $D$ the ordering $\le$.
+Let $Q = \set {z \in K: 0 \preceq z}$.
+We now show that $Q = P$.
+Let $x \in D_+, y \in D_+^*$.
+Then $0 \preceq x$ and $0 \prec 1 / y$ from item 4 of [[Properties of Ordered Ring#Total Ordering|Properties of a Totally Ordered Ring]].
+Thus by [[Definition:Ordering Compatible with Ring Structure|compatibility with ring structure]], $0 \preceq x / y$.
+Hence $P \subseteq Q$.
+Conversely, let $z \in Q$.
+Let $z = x / y$ where $x \in D, y \in D_+^*$.
+Then $x = z \circ y$ and by [[Definition:Ordering Compatible with Ring Structure|compatibility with ring structure]], $0 \preceq x$.
+Thus $0 \le x$ and hence $z \in P$, and so $Q \subseteq P$.
+So $Q = P$.
+Therefore, by item $(2)$ of [[Properties of Ordered Ring]], it follows that $\preceq$ is the same as $\le$.
+{{Qed}}
+{{improve|Put this in the context of Ordered Integral Domain.}}
+\end{proof}<|endoftext|>
+\section{Order Embedding between Quotient Fields is Unique}
+Tags: Ordered Rings, Fields of Quotients, Total Orderings, Order Embeddings
+
+\begin{theorem}
+Let $\struct {R_1, +_1, \circ_1, \le_1}$ and $\struct {S, +_2, \circ_2, \le_2}$ be [[Definition:Totally Ordered Ring|totally ordered]] [[Definition:Integral Domain|integral domains]].
+Let $K, L$ be [[Definition:Totally Ordered Ring|totally ordered]] [[Definition:Field of Quotients|fields of quotients]] of $\struct {R_1, +_1, \circ_1, \le_1}$ and $\struct {S, +_2, \circ_2, \le_2}$ respectively.
+Let $\phi: R \to S$ be a [[Definition:Order Embedding|order embedding]].
+Then there is one and only one [[Definition:Order Embedding|order embedding]] $\psi: K \to L$ extending $\phi$.
+Also:
+:$\forall x \in R, y \in R_{\ne 0}: \map \psi {\dfrac x y} = \dfrac {\map \phi x} {\map \phi y}$
+If $\phi: R \to S$ is an [[Definition:Order Isomorphism|order isomorphism]], then so is $\psi$.
+\end{theorem}
+
+\begin{proof}
+By [[Field of Quotients is Unique]], all we need to show is:
+:$\forall x_1, x_2 \in R, y_1, y_2 \in R_{> 0}: \dfrac {x_1} {y_1} \le \dfrac {x_2} {y_2} \iff \dfrac {\map \phi {x_1}} {\map \phi {y_1} } \le \dfrac {\map \phi {x_2} } {\map \phi {y_2} }$
+Let $x_1 / y_1 \le x_2 / y_2$, where $y_1, y_2 \in R_{> 0}$.
+As $y_1, y_2 \in R_{> 0}$, it follows that $0 < y_1 \circ_1 y_2$ and $0 < 1 / \paren {y_1 \circ_1 y_2}$.
+We also have:
+:$0 < \map \phi {y_1} \circ_2 \map \phi {y_2} = \map \phi {y_1 \circ_1 y_2}$
+Therefore:
+:$x_1 \circ_1 y_2 = \frac {x_1} {y_1} \circ_1 \paren {y_1 \circ y_2} \le \dfrac {x_2} {y_2} \circ_1 \paren {y_1 \circ_1 y_2} = x_2 \circ_1 y_1$
+Conversely, let $x_1 \circ_1 y_2 \le x_2 \circ_1 y_1$.
+Then:
+:$\dfrac {x_1} {y_1} = x_1 \circ_1 y_2 \circ_1 \paren {\dfrac 1 {y_1 \circ_1 y_2} } \le x_2 \circ_1 y_1 \circ_1 \paren {\dfrac 1 {y_1 \circ_1 y_2} } = \dfrac {x_2} {y_2}$
+That is, we have:
+:$\dfrac {x_1} {y_1} \le \dfrac {x_2} {y_2} \iff x_1 \circ_1 y_2 \le x_2 \circ_1 y_1$
+Similarly:
+:$\dfrac {\map \phi {x_1} } {\map \phi {y_1} } \le \dfrac {\map \phi {x_2} } {\map \phi {y_2} } \iff \map \phi {x_1} \circ_2 \map \phi {y_2} \le \map \phi {x_2} \circ_2 \map \phi {y_1}$
+Now $\phi: R \to S$ is an [[Definition:Order Embedding|order embedding]].
+Therefore:
+:$x_1 \circ_1 y_2 \le x_2 \circ_1 y_1 \iff \map \phi {x_1 \circ_1 y_2} \le \map \phi {x_2 \circ_1 y_1}$
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Strict Ordering Preserved under Product with Cancellable Element}
+Tags: Ordered Semigroups
+
+\begin{theorem}
+Let $\left({S, \circ, \preceq}\right)$ be an [[Definition:Ordered Semigroup|ordered semigroup]].
+Let $x, y, z \in S$ be such that:
+:$(1): \quad z$ is [[Definition:Cancellable Element|cancellable]] for $\circ$
+:$(2): \quad x \prec y$
+Then:
+: $x \circ z \prec y \circ z$
+: $z \circ x \prec z \circ y$
+\end{theorem}
+
+\begin{proof}
+Let $z$ be [[Definition:Cancellable Element|cancellable]] and $x \prec y$.
+Then by the definition of [[Definition:Ordered Semigroup|ordered semigroup]]:
+: $x \circ z \preceq y \circ z$
+From the fact that $z$ is [[Definition:Cancellable Element|cancellable]]:
+: $x \circ z = y \circ z \iff x = y$
+Thus as $x \circ z \ne y \circ z$ it follows from [[Strictly Precedes is Strict Ordering]] that:
+: $x \circ z \prec y \circ z$
+Similarly, $z \circ x \prec z \circ y$ follows from $z \circ x \preceq z \circ y$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Ordering of Inverses in Ordered Monoid}
+Tags: Order Theory, Monoids
+
+\begin{theorem}
+Let $\left({S, \circ, \preceq}\right)$ be an [[Definition:Ordered Monoid|ordered monoid]] whose [[Definition:Identity Element|identity]] is $e$.
+Let $x, y \in S$ be [[Definition:Invertible Element|invertible]].
+Then:
+:$x \prec y \iff y^{-1} \prec x^{-1}$
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+{{begin-eqn}}
+{{eqn | l = x
+      | o = \prec
+      | r = y
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = e
+      | r = x^{-1} \circ x \prec x^{-1} \circ y
+      | c = [[Strict Ordering Preserved under Product with Cancellable Element]]
+}}
+{{eqn | ll= \implies
+      | l = y^{-1}
+      | r = e \circ y^{-1} \prec x^{-1} \circ y \circ y^{-1} = x^{-1}
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = y^{-1}
+      | o = \prec
+      | r = x^{-1}
+      | c = 
+}}
+{{end-eqn}}
+{{qed|lemma}}
+=== Sufficient Condition ===
+{{begin-eqn}}
+{{eqn | l = y^{-1}
+      | o = \prec
+      | r = x^{-1}
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = x
+      | r = \left({x^{-1} }\right)^{-1} \prec \left({y^{-1} }\right)^{-1} = y
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = x
+      | o = \prec
+      | r = y
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Ordered Group Equivalences}
+Tags: Ordered Groups
+
+\begin{theorem}
+Let $\left({S, \circ, \preceq}\right)$ be an [[Definition:Ordered Group|ordered group]] whose [[Definition:Identity Element|identity]] is $e$.
+Let $x, y, z, \in S$.
+Then the following are all equivalent:
+: $(1): \quad x \prec y$
+: $(2): \quad x \circ z \prec y \circ z$
+: $(3): \quad z \circ x \prec z \circ y$
+: $(4): \quad y^{-1} \prec x^{-1}$
+: $(5): \quad e \prec y \circ x^{-1}$
+: $(6): \quad e \prec x^{-1} \circ y$
+\end{theorem}
+
+\begin{proof}
+Since $\left({S, \circ, \preceq}\right)$ is an [[Definition:Ordered Group|ordered group]], we have, for all $x,y,z \in S$:
+:$(R): \quad x \prec y \implies x \circ z \prec y \circ z$
+:$(L): \quad x \prec y \implies z \circ x \prec z \circ y$
+Using $(R)$ and $(L)$, let us demonstrate the equivalences.
+=== $(1) \iff (2)$ and $(1) \iff (5)$ ===
+We have by applying $(R)$ twice:
+{{begin-eqn}}
+{{eqn|l = x
+     |o = \prec
+     |r = y
+}}
+{{eqn|ll= \implies
+     |l = x \circ z
+     |o = \prec
+     |r = y \circ z
+}}
+{{eqn|ll= \implies
+     |l = x \circ z \circ z^{-1}
+     |o = \prec
+     |r = y \circ z \circ z^{-1}
+}}
+{{end-eqn}}
+This last identity comes down to $x \prec y$ by definition of [[Definition:Inverse (Abstract Algebra)|inverse]].
+Selecting $z = x^{-1}$ proves $(1) \iff (5)$ as well.
+{{qed|lemma}}
+=== $(1) \iff (3)$ and $(1) \iff (6)$ ===
+We have by applying $(L)$ twice:
+{{begin-eqn}}
+{{eqn|l = x
+     |o = \prec
+     |r = y
+}}
+{{eqn|ll= \implies
+     |l = z \circ x
+     |o = \prec
+     |r = z \circ y
+}}
+{{eqn|ll= \implies
+     |l = z^{-1} \circ z \circ x
+     |o = \prec
+     |r = z^{-1} \circ z \circ x
+}}
+{{end-eqn}}
+This last identity comes down to $x \prec y$ by definition of [[Definition:Inverse (Abstract Algebra)|inverse]].
+Selecting $z = x^{-1}$ proves $(1) \iff (6)$ as well.
+{{qed|lemma}}
+=== $(4) \iff (5)$ ===
+Applying that we know $(1) \iff (2)$, we obtain:
+{{begin-eqn}}
+{{eqn|l = y^{-1}
+     |o = \prec
+     |r = x^{-1}
+}}
+{{eqn|ll= \iff
+     |l = e
+     |o = \prec
+     |r = y \circ x^{-1}
+}}
+{{end-eqn}}
+since $y \circ y^{-1} = e$.
+{{qed|lemma}}
+The remaining equivalences follow from [[Material Equivalence is Equivalence]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Monomorphism from Total Ordering}
+Tags: Total Orderings, Monomorphisms
+
+\begin{theorem}
+Let the following conditions hold:
+: $(1): \quad$ Let $\left({S, \circ, \preceq}\right)$ and $\left({T, *, \preccurlyeq}\right)$ be [[Definition:Ordered Semigroup|ordered semigroups]].
+: $(2): \quad$ Let $\phi: S \to T$ be a [[Definition:Mapping|mapping]].
+: $(3): \quad$ Let $\preceq$ be a [[Definition:Total Ordering|total ordering]] on $S$.
+Then $\phi \left({S, \circ, \preceq}\right) \to \left({T, *, \preccurlyeq}\right)$ is a [[Definition:Ordered Structure Monomorphism|(structure) monomorphism]] iff:
+: $(1): \quad \phi$ is [[Definition:Strictly Increasing|strictly increasing]] from $\left({S, \preceq}\right)$ into $\left({T, \preccurlyeq}\right)$;
+: $(2): \quad \phi$ is a [[Definition:Homomorphism (Abstract Algebra)|homomorphism]] from $\left({S, \circ}\right)$ into $\left({T, *}\right)$.
+\end{theorem}
+
+\begin{proof}
+This follows:
+: $(1): \quad$ As a direct consequence of [[Mapping from Totally Ordered Set is Order Embedding iff Strictly Increasing]]
+: $(2): \quad$ From the definition of [[Definition:Monomorphism (Abstract Algebra)|monomorphism]] as a [[Definition:Homomorphism (Abstract Algebra)|homomorphism]] which is an [[Definition:Injection|injection]].
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Extension Theorem for Total Orderings}
+Tags: Total Orderings, Semigroups, Named Theorems
+
+\begin{theorem}
+Let the following conditions be fulfilled:
+:$(1):\quad$ Let $\struct {S, \circ, \preceq}$ be a [[Definition:Totally Ordered Commutative Semigroup|totally ordered commutative semigroup]]
+:$(2):\quad$ Let all the [[Definition:Element|elements]] of $\struct {S, \circ, \preceq}$ be [[Definition:Cancellable Element|cancellable]]
+:$(3):\quad$ Let $\struct {T, \circ}$ be an [[Definition:Inverse Completion|inverse completion]] of $\struct {S, \circ}$.
+Then:
+:$(1):\quad$ The [[Definition:Relation|relation]] $\preceq'$ on $T$ satisfying $\forall x_1, x_2, y_1, y_2 \in S: x_1 \circ \paren {y_1}^{-1} \preceq' x_2 \circ \paren {y_2}^{-1} \iff x_1 \circ y_2 \preceq x_2 \circ y_1$ is a [[Definition:Well-Defined Relation|well-defined relation]]
+:$(2):\quad$ $\preceq'$ is the only [[Definition:Total Ordering|total ordering]] on $T$ [[Definition:Relation Compatible with Operation|compatible with $\circ$]]
+:$(3):\quad$ $\preceq'$ is the only [[Definition:Total Ordering|total ordering]] on $T$ that induces the given [[Definition:Total Ordering|ordering]] $\preceq$ on $S$.
+\end{theorem}
+
+\begin{proof}
+By [[Inverse Completion is Commutative Semigroup]]:
+:every [[Definition:Element|element]] of $T$ is of the form $x \circ y^{-1}$ where $x, y \in S$.
+=== Proof that Relation is Well-Defined ===
+First we need to show that $\preceq'$ is [[Definition:Well-Defined Relation|well-defined]].
+So we need to show that if $x_1, x_2, y_1, y_2, z_1, z_2, w_1, w_2 \in S$ satisfy:
+{{begin-eqn}}
+{{eqn | l = x_1 \circ \paren {y_1}^{-1}
+      | r = x_2 \circ \paren {y_2}^{-1}
+      | c = 
+}}
+{{eqn | l = z_1 \circ \paren {w_1}^{-1}
+      | r = z_2 \circ \paren {w_2}^{-1}
+      | c = 
+}}
+{{eqn | l = x_1 \circ w_1
+      | o = \preceq
+      | r = y_1 \circ z_1
+      | c = 
+}}
+{{end-eqn}}
+then:
+:$x_2 \circ \paren {w_2}^{-1} = y_2 \circ \paren {z_2}^{-1}$
+We have:
+{{begin-eqn}}
+{{eqn | l = x_1 \circ \paren {y_1}^{-1}
+      | r = x_2 \circ \paren {y_2}^{-1}
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = x_1 \circ y_2
+      | r = x_2 \circ y_1
+      | c = 
+}}
+{{end-eqn}}
+and:
+{{begin-eqn}}
+{{eqn | l = z_1 \circ \paren {w_1}^{-1}
+      | r = z_2 \circ \paren {w_2}^{-1}
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = z_1 \circ w_2
+      | r = z_2 \circ w_1
+      | c = 
+}}
+{{end-eqn}}
+So:
+{{begin-eqn}}
+{{eqn | l = x_2 \circ w_2 \circ y_1 \circ z_1
+      | r = x_1 \circ w_1 \circ y_2 \circ z_2
+      | c = 
+}}
+{{eqn | o = \preceq
+      | r = y_1 \circ z_1 \circ y_2 \circ z_2
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = x_2 \circ w_2
+      | o = \preceq
+      | r = y_2 \circ z_2
+      | c = [[Strict Ordering Preserved under Product with Cancellable Element]]
+}}
+{{end-eqn}}
+Thus $\preceq'$ is a [[Definition:Well-Defined Relation|well-defined relation]] on $T$.
+{{qed|lemma}}
+=== Proof that Relation is Transitive ===
+To show that $\preceq'$ is [[Definition:Transitive Relation|transitive]]:
+{{begin-eqn}}
+{{eqn | ll= \forall x, y, z, w, u, v \in S:
+      | l = x \circ y^{-1}
+      | o = \preceq'
+      | r = z \circ w^{-1}
+      | c = 
+}}
+{{eqn | l = z \circ w^{-1}
+      | o = \preceq'
+      | r = u \circ v^{-1}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = x \circ w \circ v
+      | o = \preceq
+      | r = y \circ z \circ v
+      | c = 
+}}
+{{eqn | o = \preceq
+      | r = y \circ w \circ u
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = x \circ v
+      | o = \preceq
+      | r = y \circ u
+      | c = [[Strict Ordering Preserved under Product with Cancellable Element]]
+}}
+{{eqn | ll= \leadsto
+      | l = x \circ y^{-1}
+      | o = \preceq'
+      | r = u \circ v^{-1}
+      | c = 
+}}
+{{end-eqn}}
+Thus $\preceq'$ is [[Definition:Transitive Relation|transitive]].
+{{qed|lemma}}
+=== Proof that Relation is Total Ordering ===
+To show that $\preceq'$ is a [[Definition:Total Ordering|total ordering]] on $T$ [[Definition:Relation Compatible with Operation|compatible with $\circ$]]:
+Let $z_1, z_2 \in T$.
+Then:
+:$\exists x_1, x_2, y_1, y_2 \in S: z_1 = x_1 \circ y_1^{-1}, z_2 = x_2 \circ y_2^{-1}$
+Then:
+:$z_1 \circ y_1 = x_1$
+:$z_2 \circ y_2 = x_2$
+As $\preceq$ is a total ordering on $S$, either:
+:$z_1 \circ y_1 \circ y_2 \preceq z_2 \circ y_2 \circ y_1$
+or:
+:$z_2 \circ y_2 \circ y_1 \preceq z_1 \circ y_1 \circ y_2$
+{{WLOG}}, suppose that:
+:$z_1 \circ y_1 \circ y_2 \preceq z_2 \circ y_2 \circ y_1$
+So:
+{{begin-eqn}}
+{{eqn | l = z_1 \circ y_1 \circ y_2
+      | o = \preceq
+      | r = z_2 \circ y_2 \circ y_1
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = x_1 \circ y_2
+      | o = \preceq
+      | r = x_2 \circ y_1
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = x_1 \circ y_1^{-1}
+      | o = \preceq'
+      | r = x_2 \circ y_2^{-1}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = z_1
+      | o = \preceq'
+      | r = z_2
+      | c = 
+}}
+{{end-eqn}}
+and it is seen that $\preceq'$ is a [[Definition:Total Ordering|total ordering]] on $T$.
+{{qed|lemma}}
+=== Proof that Relation is Compatible with Operation ===
+Let $x_1, x_2, y_1, y_2 \in T$ such that $x_1 \preceq' x_2, y_1 \preceq' y_2$.
+We need to show that $x_1 \circ y_1 \preceq' x_2 \circ y_2$.
+Let:
+: $x_1 = r_1 \circ s_1^{-1}, x_2 = r_2 \circ s_2^{-1}$
+: $y_1 = u_1 \circ v_1^{-1}, y_2 = u_2 \circ v_2^{-1}$
+We have:
+{{begin-eqn}}
+{{eqn | l = x_1
+      | o = \preceq'
+      | r = x_2
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = r_1 \circ s_1^{-1}
+      | o = \preceq'
+      | r = r_2 \circ s_2^{-1}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = r_1 \circ s_2
+      | o = \preceq
+      | r = r_2 \circ s_1
+      | c = 
+}}
+{{end-eqn}}
+and
+{{begin-eqn}}
+{{eqn | l = y_1
+      | o = \preceq'
+      | r = y_2
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = u_1 \circ v_1^{-1}
+      | o = \preceq'
+      | r = u_2 \circ v_2^{-1}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = u_1 \circ v_2
+      | o = \preceq
+      | r = u_2 \circ v_1
+      | c = 
+}}
+{{end-eqn}}
+Because of the [[Definition:Relation Compatible with Operation|compatibility of $\preceq$ with $\circ$]] on $S$:
+{{begin-eqn}}
+{{eqn | l = \paren {r_1 \circ s_2} \circ \paren {u_1 \circ v_2}
+      | o = \preceq
+      | r = \paren {r_2 \circ s_1} \circ \paren {u_2 \circ v_1}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \paren {r_1 \circ u_1} \circ \paren {s_2 \circ v_2}
+      | o = \preceq'
+      | r = \paren {r_2 \circ u_2} \circ \paren {s_1 \circ v_1}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \paren {r_1 \circ u_1} \circ \paren {s_1 \circ v_1}^{-1}
+      | o = \preceq'
+      | r = \paren {r_2 \circ u_2} \circ \paren {s_2 \circ v_2}^{-1}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \paren {r_1 \circ s_1^{-1} } \circ \paren {u_1 \circ v_1^{-1} }
+      | o = \preceq'
+      | r = \paren {r_2 \circ s_2^{-1} } \circ \paren {u_2 \circ v_2^{-1} }
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = x_1 \circ y_1
+      | o = \preceq'
+      | r = x_2 \circ y_2
+      | c = 
+}}
+{{end-eqn}}
+Thus [[Definition:Relation Compatible with Operation|compatibility]] is proved.
+{{qed|lemma}}
+=== Proof about Restriction of Relation ===
+To show that the restriction of $\preceq'$ to $S$ is $\preceq$:
+{{begin-eqn}}
+{{eqn | l = \forall x, y \in S: x
+      | o = \preceq
+      | r = y
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \forall a \in S: \paren {x \circ a} \circ a
+      | o = \preceq
+      | r = \paren {y \circ a} \circ a
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = x
+      | r = \paren {x \circ a} \circ a^{-1}
+      | c = 
+}}
+{{eqn | o = \preceq
+      | r = \paren {y \circ a} \circ a^{-1}
+      | c = 
+}}
+{{eqn | r = y
+      | c = 
+}}
+{{end-eqn}}
+Conversely:
+{{begin-eqn}}
+{{eqn | l = \forall x, y \in S: x
+      | o = \preceq'
+      | r = y
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \exists u, v, z, w \in S: x
+      | r = u \circ v^{-1}
+      | c = 
+}}
+{{eqn | l = y
+      | r = z \circ w^{-1}
+      | c = 
+}}
+{{eqn | l = u \circ w
+      | o = \preceq
+      | r = z \circ v
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = x \circ v \circ w
+      | r = u \circ w
+      | c = 
+}}
+{{eqn | o = \preceq
+      | r = z \circ v
+      | c = 
+}}
+{{eqn | r = y \circ v \circ w
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = x
+      | o = \preceq
+      | r = y
+      | c = [[Strict Ordering Preserved under Product with Cancellable Element]]
+}}
+{{end-eqn}}
+{{qed|lemma}}
+=== Proof that Relation is Unique ===
+To show that $\preceq'$ is [[Definition:Unique|unique]]:
+Let $\preceq^*$ be any [[Definition:Ordering|ordering]] on $T$ [[Definition:Relation Compatible with Operation|compatible with $\circ$]] that induces $\preceq$ on $S$.
+Then:
+{{begin-eqn}}
+{{eqn | ll= \forall x, y, z, w \in S:
+      | l = x \circ y^{-1}
+      | o = \preceq^*
+      | r = z \circ w^{-1}
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = x \circ w
+      | o = \preceq
+      | r = y \circ z
+      | c = 
+}}
+{{end-eqn}}
+Then:
+{{begin-eqn}}
+{{eqn | l = x \circ y^{-1}
+      | o = \preceq^*
+      | r = z \circ w^{-1}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = x \circ w
+      | r = \paren {x \circ y^{-1} } \circ \paren {y \circ w}
+      | c = 
+}}
+{{eqn | o = \preceq
+      | r = \paren {z \circ w^{-1} } \circ \paren {y \circ w}
+      | c = 
+}}
+{{eqn | r = y \circ z
+      | c = 
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | l = x \circ w
+      | o = \preceq
+      | r = y \circ z
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = x \circ y^{-1}
+      | r = \paren {x \circ w} \circ w^{-1} \circ y^{-1}
+      | c = 
+}}
+{{eqn | o = \preceq^*
+      | r = \paren {y \circ z} \circ w^{-1} \circ y^{-1}
+      | c = 
+}}
+{{eqn | r = z \circ w^{-1}
+      | c = 
+}}
+{{end-eqn}}
+So:
+{{begin-eqn}}
+{{eqn | ll= \forall x, y, z, w \in S:
+      | l = x \circ y^{-1}
+      | o = \preceq^*
+      | r = z \circ w^{-1}
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = x \circ w
+      | o = \preceq
+      | r = y \circ z
+      | c = 
+}}
+{{end-eqn}}
+Hence as every [[Definition:Element|element]] of $T$ is of the form $x \circ y^{-1}$ where $x, y \in S$, the [[Definition:Total Ordering|orderings]] $\preceq^*$ and $\preceq'$ are identical.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Cancellability in Naturally Ordered Semigroup}
+Tags: Naturally Ordered Semigroup
+
+\begin{theorem}
+Let $\left({S, \circ, \preceq}\right)$ be a [[Definition:Naturally Ordered Semigroup|naturally ordered semigroup]].
+Then:
+\end{theorem}<|endoftext|>
+\section{Strict Lower Closure of Sum with One}
+Tags: Naturally Ordered Semigroup
+
+\begin{theorem}
+Let $\left({S, \circ, \preceq}\right)$ be a [[Definition:Naturally Ordered Semigroup|naturally ordered semigroup]].
+Then:
+:$\forall n \in \left({S, \circ, \preceq}\right): \left({n \circ 1}\right)^\prec = n^\prec \cup \left\{{n}\right\}$
+where $n^\prec$ is defined as the [[Definition:Strict Lower Closure of Element|strict lower closure]] of $n$, that is, the set of elements [[Definition:Strictly Precede|strictly preceding]] $n$.
+\end{theorem}
+
+\begin{proof}
+First note that as $\left({S, \circ, \preceq}\right)$ is [[Definition:Naturally Ordered Semigroup|well-ordered]] and hence [[Definition:Totally Ordered Set|totally ordered]], the [[Trichotomy Law (Ordering)|Trichotomy Law]] applies.
+Thus:
+{{begin-eqn}}
+{{eqn | l = \forall m \in S:
+      | o = 
+      | r = m \notin n^\prec
+      | c = 
+}}
+{{eqn | o = \iff
+      | r = \neg \ m \prec n
+      | c = Definition of [[Definition:Strict Lower Closure of Element|Strict Lower Closure]]
+}}
+{{eqn | o = \iff
+      | r = m = n \lor n \prec m
+      | c = [[Trichotomy Law (Ordering)|Trichotomy Law]]
+}}
+{{eqn | o = \iff
+      | r = n \preceq m
+      | c = Definition of [[Definition:Strictly Precede|Strictly Precede]]
+}}
+{{end-eqn}}
+So:
+{{begin-eqn}}
+{{eqn | l = \forall p \in S:
+      | o = 
+      | r = p \notin \left({n \circ 1}\right)^\prec
+      | c = 
+}}
+{{eqn | o = \iff
+      | r = n \circ 1 \preceq p
+      | c = from the above
+}}
+{{eqn | o = \iff
+      | r = n \prec p
+      | c = [[Sum with One is Immediate Successor in Naturally Ordered Semigroup]]
+}}
+{{end-eqn}}
+Similarly:
+{{begin-eqn}}
+{{eqn | l = \forall p \in S:
+      | o = 
+      | r = p \notin n^\prec \cup \left\{ {n}\right\}
+      | c = 
+}}
+{{eqn | o = \iff
+      | r = n \preceq p \land n \ne p
+      | c = from the above
+}}
+{{eqn | o = \iff
+      | r = n \prec p
+      | c = Definition of [[Definition:Strictly Precede|Strictly Precede]]
+}}
+{{end-eqn}}
+So:
+:$p \notin n^\prec \cup \left\{{n}\right\} \iff p \notin \left({n \circ 1}\right)^\prec$
+Thus:
+:$\complement_S \left({\left({n \circ 1}\right)^\prec}\right) = \complement_S \left({n^\prec \cup \left\{{n}\right\}}\right)$
+from the [[Definition:Relative Complement|definition of relative complement]].
+So:
+{{begin-eqn}}
+{{eqn | l = \left({n \circ 1}\right)^\prec
+      | r = \complement_S \left({\complement_S \left({\left({n \circ 1}\right)^\prec}\right)}\right)
+      | c = [[Relative Complement of Relative Complement]]
+}}
+{{eqn | r = \complement_S \left({\complement_S \left({n^\prec \cup \left\{ {n}\right\} }\right)}\right)
+      | c = from above
+}}
+{{eqn | r = n^\prec \cup \left\{ {n}\right\}
+      | c = [[Relative Complement of Relative Complement]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Closed Interval of Naturally Ordered Semigroup with Successor equals Union with Successor}
+Tags: Naturally Ordered Semigroup
+
+\begin{theorem}
+Let $\left({S, \circ, \preceq}\right)$ be a [[Definition:Naturally Ordered Semigroup|naturally ordered semigroup]].
+Then:
+:$\forall m, n \in \left({S, \circ, \preceq}\right): m \preceq n \implies \left[{m \,.\,.\, n \circ 1}\right] = \left[{m \,.\,.\, n}\right] \cup \left\{{n \circ 1}\right\}$
+where $\left[{m \,.\,.\, n}\right]$ is the [[Definition:Closed Interval|closed interval between $m$ and $n$]].
+\end{theorem}
+
+\begin{proof}
+Let $m \preceq n$. Then:
+{{begin-eqn}}
+{{eqn | o = 
+      | r = x \in \left[{m \,.\,.\, n \circ 1}\right]
+      | c = 
+}}
+{{eqn | o = \iff
+      | r = m \preceq x \land x \preceq \left({n \circ 1}\right)
+      | c = Definition of [[Definition:Closed Interval|Closed Interval]]
+}}
+{{eqn | o = \iff
+      | r = m \preceq x \land \left({x \prec n \circ 1 \lor x = n \circ 1}\right)
+      | c = Definition of [[Definition:Strictly Precede|Strictly Precedes]]
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | o = 
+      | r = x \in \left[{m \,.\,.\, n}\right] \cup \left\{ {n \circ 1}\right\}
+      | c = 
+}}
+{{eqn | o = \iff
+      | r = m \preceq x \land \left({x \preceq n \lor x = n \circ 1}\right)
+      | c = Definitions of [[Definition:Closed Interval|Closed Interval]] and [[Definition:Set Union|Union]]
+}}
+{{eqn | o = \iff
+      | r = m \preceq x \land \left({x \prec n \lor x = n \lor x = n \circ 1}\right)
+      | c = Definition of [[Definition:Strictly Precede|Strictly Precedes]]
+}}
+{{eqn | o = \iff
+      | r = m \preceq x \land \left({x \prec n \circ 1 \lor x = n \circ 1}\right)
+      | c = Definition of [[Definition:Strictly Precede|Strictly Precedes]]
+}}
+{{end-eqn}}
+Thus:
+:$\left[{m \,.\,.\, n \circ 1}\right] = \left[{m \,.\,.\, n}\right] \cup \left\{{n \circ 1}\right\}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Principle of Induction applied to Interval of Naturally Ordered Semigroup}
+Tags: Naturally Ordered Semigroup
+
+\begin{theorem}
+Let $\left({S, \circ, \preceq}\right)$ be a [[Definition:Naturally Ordered Semigroup|naturally ordered semigroup]].
+Let $\left[{p \,.\,.\, q}\right]$ be a [[Definition:Closed Interval|closed interval]] of $\left({S, \circ, \preceq}\right)$.
+Let $T \subseteq \left[{p \,.\,.\, q}\right]$ such that the [[Definition:Minimal Element|minimal element]] of $\left[{p \,.\,.\, q}\right]$ is in $T$.
+Let:
+: $x \in T: x \prec q \implies x \circ 1 \in T$
+Then:
+: $T = \left[{p \,.\,.\, q}\right]$
+\end{theorem}
+
+\begin{proof}
+Let $T' = T \cup \left\{{x \in S: q \prec x}\right\}$.
+Then $T'$ satisfies the conditions of the [[Principle of Mathematical Induction for Naturally Ordered Semigroup|Principle of Mathematical Induction]].
+Therefore:
+:$T' = \left\{{x \in S: p \preceq x}\right\}$
+Therefore:
+:$T = \left[{p \,.\,.\, q}\right]$
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Homomorphism of Powers}
+Tags: Homomorphisms
+
+\begin{theorem}
+Let $\struct {T_1, \odot}$ and $\struct {T_2, \oplus}$ be [[Definition:Semigroup|semigroups]].
+Let $\phi: \struct {T_1, \odot} \to \struct {T_2, \oplus}$ be a [[Definition:Semigroup Homomorphism|(semigroup) homomorphism]].
+\end{theorem}<|endoftext|>
+\section{Naturally Ordered Semigroup is Unique}
+Tags: Naturally Ordered Semigroup, Isomorphisms
+
+\begin{theorem}
+Let $\struct {S, \circ, \preceq}$ and $\struct {S', \circ', \preceq'}$ be [[Definition:Naturally Ordered Semigroup|naturally ordered semigroups]].
+Let:
+:$0'$ be the [[Definition:Smallest Element|smallest element]] of $S'$
+:$1'$ be the [[Definition:Smallest Element|smallest element]] of $S' \setminus \set {0'} = S'^*$.
+Then the [[Definition:Mapping|mapping]] $g: S \to S'$ defined as:
+:$\forall a \in S: \map g a = \circ'^a 1'$
+is an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]] from $\struct {S, \circ, \preceq}$ to $\struct {S', \circ', \preceq'}$.
+This [[Definition:Isomorphism (Abstract Algebra)|isomorphism]] is unique.
+Thus, up to [[Definition:Isomorphism (Abstract Algebra)|isomorphism]], there is only one [[Definition:Naturally Ordered Semigroup|naturally ordered semigroup]].
+\end{theorem}
+
+\begin{proof}
+=== Proof that Mapping is Isomorphism ===
+Let $T' = \Cdm g$, that is, the [[Definition:Codomain of Mapping|codomain]] of $g$.
+By [[Zero is Identity in Naturally Ordered Semigroup]], $0'$ is the [[Definition:Identity Element|identity]] for $\circ'$.
+Thus:
+:$\map g 0 = \circ'^0 1' = 0'$
+and so $0' \in T'$
+Suppose $x' \in T'$.
+Then:
+:$\map g n = x'$
+and so:
+{{begin-eqn}}
+{{eqn | l = x' \circ' 1'
+      | r = \circ'^n 1' \circ' 1'
+      | c = 
+}}
+{{eqn | r = \paren {\circ'^n 1'} \circ' \paren {\circ'^1 1'}
+      | c = 
+}}
+{{eqn | r = \circ'^{\paren {n \circ 1} } 1'
+      | c = 
+}}
+{{eqn | r = \map g {n \circ 1}
+      | c = 
+}}
+{{end-eqn}}
+So:
+:$x' \in T' \implies x' \circ' 1' \in T'$
+Thus, by the [[Principle of Mathematical Induction for Naturally Ordered Semigroup|Principle of Mathematical Induction]] applied to $S'$:
+:$T' = S'$
+So:
+:$\forall a' \in S': \exists a \in S: \map g a = a'$
+and so $g$ is [[Definition:Surjection|surjective]] by definition.
+From [[Index Laws for Semigroup/Sum of Indices|Index Laws for Semigroup: Sum of Indices]]:
+:$\map g {a \circ b} = \map g a \circ' \map g b$
+and therefore $g$ is a [[Definition:Homomorphism (Abstract Algebra)|homomorphism]] from $\struct {S, \circ}$ to $\struct {S', \circ'}$.
+Now:
+{{begin-eqn}}
+{{eqn | ll= \forall p \in S:
+      | l = \circ'^p 1'
+      | o = \prec'
+      | r = \circ'^p 1' \circ' 1'
+      | c = [[Sum with One is Immediate Successor in Naturally Ordered Semigroup]]
+}}
+{{eqn | r = \paren {\circ'^p 1'} \circ' \paren {\circ'^1 1'}
+      | c = 
+}}
+{{eqn | r = \circ'^{\paren {p \circ 1} } 1'
+      | c = [[Index Laws for Semigroup/Sum of Indices|Index Laws for Semigroup: Sum of Indices]]
+}}
+{{eqn | ll= \leadsto
+      | l = \circ'^p 1'
+      | o = \prec'
+      | r = \circ'^{\paren {p \circ 1} } 1'
+      | c = 
+}}
+{{end-eqn}}
+For $p \in S$, let $S_p$ be the [[Definition:Initial Segment|initial segment]] of $S$:
+:$S_p = \set {x \in S: x \prec p}$
+Let:
+:$T = \set {p \in S: \forall a \in S_p: \circ'^a 1' \prec' \circ'^p 1'}$
+Now $S_0 = \O \implies 0 \in T$.
+Suppose $p \in T$.
+Then:
+:$a \prec p \circ 1 \implies a \preceq p$
+By [[Strict Lower Closure of Sum with One]], either of these is the case:
+:$(1): \quad a \prec p: p \in T \implies \circ'^a 1' \prec' \circ'^p 1' \prec' \circ'^{\paren {p \circ 1} } 1'$
+:$(2): \quad a = p: \circ'^a 1' = \circ'^p 1' \prec' \circ'^{\paren {p \circ 1} } 1'$
+In either case, we have:
+:$p \in T \implies p \circ 1 \in T$, and by the[[Principle of Mathematical Induction for Naturally Ordered Semigroup|Principle of Mathematical Induction]]:
+:$T = S$
+So $n \prec p \implies \circ'^n 1' \prec' \circ'^p 1'$.
+Thus $g$ is a [[Definition:Surjection|surjective]] [[Definition:Monomorphism (Abstract Algebra)|monomorphism]] and therefore is an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]] from $\struct {S, \circ, \preceq}$ to $\struct {S', \circ', \preceq'}$.
+{{Qed}}
+=== Proof that Isomorphism is Unique ===
+Now we need to show that the [[Definition:Isomorphism (Abstract Algebra)|isomorphism]] $g$ is unique.
+Let $f: S \to S'$ be another [[Definition:Isomorphism (Abstract Algebra)|isomorphism]] different from $g$.
+{{AimForCont}} $\map f 1 \ne 1'$.
+We show by [[Principle of Mathematical Induction|induction]] that $1' \notin \Cdm f$.
+{{finish}}
+... Thus $1' \notin \Cdm f$ which is a contradiction.
+Thus $\map f 1 = 1$ and it follows
+{{finish}}
+that $f = g$.
+Thus the [[Definition:Isomorphism (Abstract Algebra)|isomorphism]] $g$ is unique.
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Consecutive Subsets of N}
+Tags: Natural Numbers
+
+\begin{theorem}
+Let $\N_k$ denote the [[Definition:Initial Segment of Natural Numbers|initial segment of the natural numbers]] determined by $k$:
+:$\N_k = \left\{{0, 1, 2, 3, \ldots, k - 1}\right\}$
+Then:
+:$\N_k = \N_{k + 1} \setminus \left\{{k}\right\}$
+In particular:
+:$\N_{k - 1} = \N_k \setminus \left\{{k - 1}\right\}$
+\end{theorem}
+
+\begin{proof}
+The result follows as a direct application of [[Strict Lower Closure of Sum with One]].
+{{qed}}
+[[Category:Natural Numbers]]
+ebnlj9h3zjmkiqhic6nj6grj7lm6oxj
+\end{proof}<|endoftext|>
+\section{Well-Ordering Principle}
+Tags: Number Theory, Natural Numbers, Named Theorems, Well-Orderings, Well-Ordering Principle, Ordering on Natural Numbers
+
+\begin{theorem}
+Every [[Definition:Non-Empty Set|non-empty]] [[Definition:Subset|subset]] of $\N$ has a [[Definition:Smallest Element|smallest (or '''first''') element]].
+This is called the '''well-ordering principle'''.
+The '''well-ordering principle''' also holds for $\N_{\ne 0}$.
+\end{theorem}
+
+\begin{proof}
+Consider the [[Definition:Natural Numbers|natural numbers]] $\N$ defined as the [[Definition:Naturally Ordered Semigroup|naturally ordered semigroup]] $\struct {S, \circ, \preceq}$.
+From its definition, $\struct {S, \circ, \preceq}$ is [[Definition:Well-Ordered Set|well-ordered]] by $\preceq$.
+The result follows.
+As $\N_{\ne 0} = \N \setminus \set 0$, by [[Set Difference is Subset]] $\N_{\ne 0} \subseteq \N$.
+As $\N$ is [[Definition:Well-Ordered Set|well-ordered]], by definition, every subset of $\N$ has a [[Definition:Smallest Element|smallest element]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let $S$ be a [[Definition:Non-Empty Set|non-empty]] [[Definition:Subset|subset]] of the [[Definition:Natural Numbers|set of natural numbers]] $\N$.
+We take as [[Definition:Axiom|axiomatic]] that $\N$ is itself a [[Definition:Subset|subset]] of the [[Definition:Real Number|set of real numbers]] $\R$.
+Thus $S \subseteq \R$.
+By definition:
+:$\forall n \in \N: n \ge 0$
+and so:
+:$\forall n \in S: n \ge 0$
+Hence $0$ is a [[Definition:Lower Bound of Subset of Real Numbers|lower bound]] of $S$.
+This establishes the fact that $S$ is [[Definition:Bounded Below Subset of Real Numbers|bounded below]].
+By the [[Continuum Property]], we have that $S$ admits an [[Definition:Infimum of Subset of Real Numbers|infimum]].
+Hence let $b = \inf S$.
+Because $b$ is the [[Definition:Infimum of Subset of Real Numbers|infimum]] of $S$, it follows that $b + 1$ is not a [[Definition:Lower Bound of Subset of Real Numbers|lower bound]] of $S$.
+So, for some $n \in S$:
+:$n < b + 1$
+{{AimForCont}} $n$ is not the [[Definition:Smallest Element|smallest element]] of $S$.
+Then there exists $m \in S$ such that:
+:$b \le m < n < b + 1$
+from which it follows that:
+:$0 < n - m < 1$
+But there exist no [[Definition:Natural Number|natural numbers]] $k$ such that $0 < k < 1$.
+Hence $n = b$ is the [[Definition:Smallest Element|smallest element]] of $S$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Equality of Natural Numbers}
+Tags: Natural Numbers
+
+\begin{theorem}
+Let $m, n \in \N$.
+Then:
+: $\N_m \sim \N_n \iff m = n$
+where $\sim$ denotes [[Definition:Set Equivalence|set equivalence]] and $\N_n$ denotes the [[Definition:Initial Segment of Natural Numbers|set of all natural numbers less than $n$]].
+\end{theorem}
+
+\begin{proof}
+By [[Set Equivalence is Equivalence Relation]], we have that:
+: $m = n \implies \N_m \sim \N_n$
+It remains to show that:
+: $m \ne n \implies \N_m \nsim \N_n$.
+Since the [[Definition:Natural Number|naturals]] are [[Definition:Totally Ordered Set|totally ordered]], it will be sufficient to show that:
+: $m \in \N_n \implies \N_m \nsim \N_n$
+Let $S = \left\{{n \in \N: \forall m \in \N_n: \N_m \nsim \N_n}\right\}$.
+That is, $S$ is the [[Definition:Set|set]] of all the [[Definition:Natural Numbers|natural numbers]] $n$ such that $\N_m \nsim \N_n$ for all $m \in \N_n$.
+We use [[Principle of Mathematical Induction|mathematical induction]] to prove that $S = \N$.
+From [[Initial Segment of Natural Numbers determined by Zero is Empty]]:
+:$\N_0 = \varnothing$
+Thus $0 \in S$.
+This is the [[Principle of Mathematical Induction#Basis for the Induction|basis for the induction]].
+Now, assume the [[Principle of Mathematical Induction#Induction Hypothesis|induction hypothesis]] that $n \in S$.
+We now complete the [[Principle of Mathematical Induction#Induction Step|induction step]], that is, to show that $n + 1 \in S$.
+Let $m \in \N_{n+1}$.
+If $m = 0$, then $\N_m \nsim \N_{n+1}$ because $\N_0 = \varnothing$ and $\N_{n+1} \ne \varnothing$.
+[[Proof by Contradiction|Aiming for a contradiction]], suppose that $m \ge 1$ and $\N_m \sim \N_{n+1}$.
+Then, by [[Set Equivalence Less One Element]], that means $\N_{m-1} \sim \N_n$.
+But then $m - 1 \in \N_n$ which [[Definition:Contradiction|contradicts]] the [[Principle of Mathematical Induction#Induction Hypothesis|induction hypothesis]] that $n \in S$.
+Thus $n + 1 \in S$.
+The result follows from the fact that [[Set Equivalence is Equivalence Relation]], in particular the [[Definition:Symmetric Relation|symmetry]] clause.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Principle of Counting}
+Tags: Natural Numbers, Named Theorems
+
+\begin{theorem}
+Let $T$ be a [[Definition:Set|set]] such that $T \sim \N_n$.
+Then:
+: $\forall m \in \N: n \ne m \implies T \nsim \N_m$
+\end{theorem}
+
+\begin{proof}
+The result follows directly from [[Equality of Natural Numbers]] and the fact that [[Set Equivalence is Equivalence Relation|set equivalence is transitive]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Cardinality Less One}
+Tags: Set Theory
+
+\begin{theorem}
+Let $S$ be a [[Definition:Finite Set|finite set]].
+Let:
+:$\left\lvert{S}\right\rvert = n + 1$
+where $\left\lvert{S}\right\rvert$ is the [[Definition:Cardinality|cardinality]] of $S$.
+Let $a \in S$.
+Then:
+:$\left\lvert{S \setminus \left\{ {a}\right\} }\right\rvert = n$
+where $\setminus$ denotes [[Definition:Set Difference|set difference]].
+\end{theorem}
+
+\begin{proof}
+This follows as an immediate consequence of [[Set Equivalence Less One Element]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Cardinality of Empty Set}
+Tags: Empty Set, Cardinality, Finite Sets
+
+\begin{theorem}
+:$\card S = 0 \iff S = \O$
+That is, the [[Definition:Empty Set|empty set]] is [[Definition:Finite Set|finite]], and has a [[Definition:Cardinality|cardinality]] of [[Definition:Zero (Number)|zero]].
+\end{theorem}
+
+\begin{proof}
+[[Definition:Zero (Number)|Zero]] is defined as the [[Definition:Cardinal|cardinal]] of the [[Definition:Empty Set|empty set]].
+The result follows from [[Finite Cardinals and Ordinals are Equivalent]].
+\end{proof}<|endoftext|>
+\section{Cardinality of Subset of Finite Set}
+Tags: Set Theory, Subsets, Cardinality
+
+\begin{theorem}
+Let $A$ and $B$ be [[Definition:Finite Set|finite sets]] such that $A \subseteq B$.
+Let
+:$\card B = n$
+where $\card {\, \cdot \,}$ denotes [[Definition:Cardinality|cardinality]].
+Then $\card A \le n$.
+\end{theorem}
+
+\begin{proof}
+Let $A \subseteq B$.
+There are two cases:
+$(1): \quad A \ne B$
+In this case:
+:$A \subsetneqq B$
+and from [[Cardinality of Proper Subset of Finite Set]]:
+:$\card A < n$
+$(2): \quad A = B$
+In this case:
+:$\card A = \card B$
+and so:
+:$\card A = n$
+In both cases:
+:$\card A \le n$
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Cardinality of Surjection}
+Tags: Surjections, Bijections, Cardinality
+
+\begin{theorem}
+Let $S$ be a [[Definition:Set|set]].
+Let:
+:$\card S = n$
+where $\card S$ denotes the [[Definition:Cardinality|cardinality]] of $S$.
+Let $f: S \to T$ be a [[Definition:Surjection|surjection]].
+Then $\card T \le n$.
+The equality:
+:$\card T = n$
+occurs {{iff}} $f$ is a [[Definition:Bijection|bijection]].
+\end{theorem}
+
+\begin{proof}
+We have that $\card S = n$.
+Then by definition of [[Definition:Cardinality|cardinality]]:
+:there is a [[Definition:Surjection|surjection]] from $S$ to $T$
+{{iff}}:
+:there is a [[Definition:Surjection|surjection]] from $\N_n$ to $T$.
+So we need consider the case only when $S = \N_n$.
+By definition of [[Definition:Surjection|surjection]]:
+:$\forall x \in T: f^{-1} \sqbrk {\set x} \ne \O$
+where $f^{-1} \sqbrk {\set x}$ denotes the [[Definition:Preimage of Subset under Mapping|preimage of $\set x$ under $f$]].
+From the [[Well-Ordering Principle]], $\N$ is [[Definition:Well-Ordered Set|well-ordered]].
+Therefore by [[Subset of Well-Ordered Set is Well-Ordered]], $S = \N_n$ is [[Definition:Well-Ordered Set|well-ordered]].
+We have that $f^{-1} \sqbrk {\set x} \subseteq S$.
+Therefore $f^{-1} \sqbrk {\set x}$ is also [[Definition:Well-Ordered Set|well-ordered]].
+Therefore, by definition of [[Definition:Well-Ordered Set|well-ordered set]], $f^{-1} \sqbrk {\set x}$ has a [[Definition:Smallest Element|smallest element]].
+Let $g: T \to S$ be the [[Definition:Mapping|mapping]] defined as:
+:$\forall x \in T: g \paren x =$ the [[Definition:Smallest Element|smallest element]] of $f^{-1} \sqbrk {\set x}$
+Let $x \in T$.
+Then:
+:$g \paren x \in f^{-1} \sqbrk {\set x}$
+so:
+:$f \paren {g \paren x} = x$
+Thus:
+:$f \circ g = I_T$
+and by [[Identity Mapping is Bijection]]:
+:$I_T: T \to g \sqbrk T$ is a [[Definition:Bijection|bijection]].
+By [[Cardinality of Subset of Finite Set]]:
+:$\card {g \sqbrk T} \le n$
+Let $\card {g \sqbrk T} = m$.
+By [[Set Equivalence is Equivalence Relation]]:
+:$\card T = m$
+[[Definition:Element|elements]].
+Suppose $m = n$.
+Then by [[Cardinality of Subset of Finite Set]]:
+:$g \sqbrk T = S$
+so $g: T \to S$ is a [[Definition:Bijection|bijection]].
+Therefore:
+{{begin-eqn}}
+{{eqn | l = f
+      | r = f \circ I_S
+      | c = 
+}}
+{{eqn | r = f \circ g \circ g^{-1}
+      | c = 
+}}
+{{eqn | r = I_S \circ g^{-1}
+      | c = 
+}}
+{{eqn | r = g^{-1}
+      | c = 
+}}
+{{end-eqn}}
+So $f: S \to T$ is a [[Definition:Bijection|bijection]].
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Equivalence of Mappings between Sets of Same Cardinality}
+Tags: Injections, Surjections, Bijections
+
+\begin{theorem}
+Let $S$ and $T$ be [[Definition:Finite|finite]] sets such that $\card S = \card T$.
+Let $f: S \to T$ be a [[Definition:Mapping|mapping]].
+Then the following statements are equivalent:
+: $(1): \quad f$ is [[Definition:Bijection|bijective]]
+: $(2): \quad f$ is [[Definition:Injection|injective]]
+: $(3): \quad f$ is [[Definition:Surjection|surjective]].
+\end{theorem}
+
+\begin{proof}
+$(2)$ implies $(3)$:
+Let $f$ be an [[Definition:Injection|injection]].
+Then by [[Cardinality of Image of Injection]]:
+:$\card S = \card {f \sqbrk S}$
+where $f \sqbrk S$ denotes the [[Definition:Image of Subset under Mapping|image]] of $S$ under $f$.
+Therefore the [[Definition:Subset|subset]] $f \sqbrk S$ of $T$ has the same number of [[Definition:Element|elements]] as $T$.
+Therefore:
+: $f \sqbrk S = T$
+and so $f$ is a [[Definition:Surjection|surjection]].
+{{qed|lemma}}
+$(3)$ implies $(1)$:
+Let $f$ be [[Definition:Surjection|surjective]].
+We have that $\card S = \card T$.
+By [[Cardinality of Surjection]], $f$ is [[Definition:Bijection|bijective]].
+{{qed|lemma}}
+$(1)$ implies $(2)$:
+Let $f$ be a [[Definition:Bijection|bijection]].
+By definition of [[Definition:Bijection|bijection]], $f$ is also an [[Definition:Injection|injection]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Natural Numbers are Infinite}
+Tags: Natural Numbers, Infinite Sets
+
+\begin{theorem}
+The [[Definition:Set|set]] $\N$ of [[Definition:Natural Numbers|natural numbers]] is [[Definition:Infinite Set|infinite]].
+\end{theorem}
+
+\begin{proof}
+Let the [[Definition:Mapping|mapping]] $s: \N \to \N$ be defined as:
+:$\forall n \in \N: \map s n = n + 1$
+$s$ is clearly an [[Definition:Injection|injection]].
+{{AimForCont}} $\N$ were [[Definition:Finite Set|finite]].
+By [[Equivalence of Mappings between Sets of Same Cardinality]] it follows that $s$ is a [[Definition:Surjection|surjection]].
+But:
+:$\forall n \in \N: \map s n \ge 0 + 1 > 0$
+So:
+:$0 \notin s \sqbrk \N$
+and $s$ is not a [[Definition:Surjection|surjection]].
+From this [[Definition:Contradiction|contradiction]] it is seen that $\N$ cannot be [[Definition:Finite Set|finite]].
+So, by definition, $\N$ is [[Definition:Infinite Set|infinite]].
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Finite Non-Empty Subset of Totally Ordered Set has Smallest and Greatest Elements}
+Tags: Total Orderings, Finite Non-Empty Subset of Totally Ordered Set has Smallest and Greatest Elements
+
+\begin{theorem}
+Let $\left({S, \preceq}\right)$ be a [[Definition:Totally Ordered Set|totally ordered set]].
+Then every [[Definition:Finite Set|finite]] $T$ such that $\varnothing \subset T \subseteq S$ has both a [[Definition:Smallest Element|smallest]] and a [[Definition:Greatest Element|greatest]] element.
+\end{theorem}
+
+\begin{proof}
+Let $A \subseteq \N_{>0}$ such that every $B \subseteq S$ such that $\left|{B}\right| = n$ has a [[Definition:Greatest Element|greatest]] and a [[Definition:Smallest Element|smallest]] element.
+Any $B \subseteq S$ such that $\left|{B}\right| = 1$ has $1$ element, $b \in B$ say.
+Then $b$ is both the [[Definition:Greatest Element|greatest]] and [[Definition:Smallest Element|smallest]] element of $B$.
+So $1 \in A$.
+Let $n \in A$.
+Let $B \subseteq S$ such that $\left|{B}\right| = n + 1$.
+Then $\exists b \in B$, and $\left|B \setminus \left\{{b}\right\}\right| = n$ elements by [[Cardinality Less One]].
+So, by the [[Principle of Mathematical Induction#Induction Hypothesis|induction hypothesis]], $B \setminus \left\{{b}\right\}$ has a [[Definition:Greatest Element|greatest]] element $c$ and a [[Definition:Smallest Element|smallest]] element $a$, as $n \in A$.
+Note that $b \ne c$ as $c \in B \setminus \left\{{b}\right\}$ but $b \notin B \setminus \left\{{b}\right\}$.
+So as $\left({S, \preceq}\right)$ is totally ordered, either $b \prec c$ or $c \prec b$ as $b \ne c$.
+If $b \prec c$ then $c$ is the [[Definition:Greatest Element|greatest]] element of $B$, otherwise it's $b$.
+Similarly, either $b \prec a$ or $a \prec b$, and thus either $a$ or $b$ is the [[Definition:Smallest Element|smallest]] element of $B$.
+Either way, $B$ has both a [[Definition:Greatest Element|greatest]] and a [[Definition:Smallest Element|smallest]] element, and therefore $n + 1 \in A$.
+Therefore, by the [[Principle of Mathematical Induction]], $A = \N_{>0}$ and the proof is complete.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+The result follows from:
+: [[Finite Non-Empty Subset of Ordered Set has Maximal and Minimal Elements]]
+: [[Minimal Element of Chain is Smallest Element]]
+: [[Maximal Element of Chain is Greatest Element]]
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Unique Isomorphism between Finite Totally Ordered Sets}
+Tags: Total Orderings, Order Isomorphisms
+
+\begin{theorem}
+Let $S$ and $T$ be [[Definition:Finite Set|finite sets]] such that:
+:$\card S = \card T$
+Let $\struct {S, \preceq}$ and $\struct {T, \preccurlyeq}$ be [[Definition:Totally Ordered Set|totally ordered sets]].
+Then there is exactly one [[Definition:Order Isomorphism|order isomorphism]] from $\struct {S, \preceq}$ to $\struct {T, \preccurlyeq}$.
+\end{theorem}
+
+\begin{proof}
+It is sufficient to consider the case where $\struct {T, \preccurlyeq}$ is $\struct {\N_n, \le}$ for some $n \in \N$.
+Let $A$ be the set of all $n \in \N$ such that if:
+:$(1): \quad S$ is any set such that $\card S = n$, and
+:$(2): \quad \preceq$ is any [[Definition:Total Ordering|total ordering]] on $S$
+then there is exactly one [[Definition:Order Isomorphism|isomorphism]] from $\struct {S, \preceq}$ to $\struct {\N_n, \le}$.
+$\O \in A$ from [[Empty Mapping is Mapping]] and [[Equivalence of Mappings between Sets of Same Cardinality]].
+Now let $n \in A$.
+Let $\struct {S, \preceq}$ be a [[Definition:Totally Ordered Set|totally ordered set]] with $n + 1$ elements.
+By [[Finite Non-Empty Subset of Totally Ordered Set has Smallest and Greatest Elements]], $S$ has a [[Definition:Greatest Element|greatest element]] $b$.
+Then $\card {S \setminus \set b} = n$ by [[Cardinality Less One]].
+The [[Definition:Total Ordering|total ordering]] on $S \setminus \set b$ is the one induced from that on $S$.
+So as $n \in A$, there exists a unique [[Definition:Order Isomorphism|order isomorphism]] $f: S \setminus \set b \to \struct {\N_n, \le}$.
+Let us define the mapping $g: S \to \N_{n + 1}$ as follows:
+:$\forall x \in S: \map g x = \begin{cases}
+\map f x: & x \in S \setminus \set b \\
+n: & x = b
+\end{cases}$
+This is the desired [[Definition:Order Isomorphism|order isomorphism]] from $\struct {S, \preceq}$ to $\struct {\N_{n + 1}, \le}$.
+Now let $h: \struct {S, \preceq}$ to $\struct {\N_{n + 1}, \le}$ be an [[Definition:Order Isomorphism|order isomorphism]].
+Then $\map h b = n$ (it has to be).
+So the [[Definition:Restriction of Mapping|restriction]] of $h$ to $S \setminus \set b$ is an [[Definition:Order Isomorphism|isomorphism]] from $S \setminus \set b$ to $\struct {\N_n, \le}$, and hence $h = f$ (as $n \in A$, any such [[Definition:Order Isomorphism|order isomorphism]] is unique).
+Thus:
+:$\forall x \in S \setminus \set b: \map h x = \map f x = \map g x$
+and:
+:$\map h b = n = \map g b$
+thus $h = g$.
+Therefore $n + 1 \in A$.
+The result follows from the [[Principle of Mathematical Induction]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Isomorphism to Closed Interval}
+Tags: Set Theory, Order Theory, Mapping Theory
+
+\begin{theorem}
+Let $m, n \in \N$ such that $m < n$.
+Then:
+: $\left|{\left[{m + 1 \,.\,.\, n}\right]}\right| = n - m$
+Let $h: \N_{n - m} \to \left[{m + 1 \,.\,.\, n}\right]$ be the [[Definition:Mapping|mapping]] defined as:
+:$\forall x \in \N_{n - m}: h \left({x}\right) = x + m + 1$
+Let the [[Definition:Ordering|orderings]] on $\left[{m + 1 \,.\,.\, n}\right]$ and $\N_{n - m}$ be those [[Definition:Restriction of Relation|induced]] by the [[Definition:Ordering|ordering]] of $\N$.
+Then $h$ a unique [[Definition:Order Isomorphism|order isomorphism]].
+\end{theorem}
+
+\begin{proof}
+From [[Unique Isomorphism between Finite Totally Ordered Sets]], it suffices to show that $h$ is an [[Definition:Order Isomorphism|order isomorphism]].
+To this end, remark that, for all $x, y \in \N_{n - m}$:
+{{begin-eqn}}
+{{eqn | l = h \left({x}\right)
+      | r = h \left({y}\right)
+}}
+{{eqn | ll= \iff
+      | l = x + m + 1
+      | r = y + m + 1
+}}
+{{eqn | ll = \iff
+      | l = x
+      | r = y
+      | c = [[Natural Number Addition is Cancellable]]
+}}
+{{end-eqn}}
+proving $h$ is an [[Definition:Injection|injection]], and so a [[Definition:Bijection|bijection]], from [[Equivalence of Mappings between Sets of Same Cardinality]].
+By [[Ordering on Natural Numbers is Compatible with Addition]] and [[Natural Number Addition is Cancellable for Ordering]], it follows that:
+:$x \le y \iff h \left({x}\right) \le h \left({y}\right)$
+so $h$ is an [[Definition:Order Isomorphism|order isomorphism]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Regular Representation wrt Cancellable Element on Finite Semigroup is Bijection}
+Tags: Semigroups, Regular Representations, Cancellability
+
+\begin{theorem}
+Let $\left({S, \circ}\right)$ be a [[Definition:Finite|finite]] [[Definition:Semigroup|semigroup]].
+Let $a \in S$ be [[Definition:Cancellable Element|cancellable]].
+Then:
+:the [[Definition:Left Regular Representation|left regular representation]] $\lambda_a$
+and:
+:the [[Definition:Right Regular Representation|right regular representation]] $\rho_a$
+of $\left({S, \circ}\right)$ with respect to $a$ are both [[Definition:Bijection|bijections]].
+\end{theorem}
+
+\begin{proof}
+By [[Cancellable iff Regular Representations Injective]], $\lambda_a$ and $\rho_a$ are [[Definition:Injection|injections]].
+We have that $S$ is [[Definition:Finite Set|finite]].
+From [[Injection from Finite Set to Itself is Surjection]], both $\lambda_a$ and $\rho_a$ are [[Definition:Surjection|surjections]].
+Thus $\lambda_a$ and $\rho_a$ are [[Definition:Injection|injective]] and [[Definition:Surjection|surjective]], and therefore [[Definition:Bijection|bijections]].
+{{qed}}
+[[Category:Semigroups]]
+[[Category:Regular Representations]]
+[[Category:Cancellability]]
+k62vgwldwv4np50anvdp0y1qep88fep
+\end{proof}<|endoftext|>
+\section{Power Set of Natural Numbers is not Countable}
+Tags: Power Set, Natural Numbers, Uncountable Sets
+
+\begin{theorem}
+The [[Definition:Power Set|power set]] $\powerset \N$ of the [[Definition:Natural Numbers|natural numbers]] $\N$ is not [[Definition:Countable Set|countable]].
+\end{theorem}
+
+\begin{proof}
+There is [[No Bijection from Set to its Power Set|no bijection from a set to its power set]].
+From [[Injection from Set to Power Set]], we have that there exists an [[Definition:Injection|injection]] $f: \N \to \powerset \N$.
+From the [[Cantor-Bernstein-Schröder Theorem]], there can be no [[Definition:Injection|injection]] $g: \powerset \N \to \N$.
+So, by definition, $\powerset \N$ is not [[Definition:Countable Set|countable]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Subset of Countably Infinite Set is Countable}
+Tags: Countable Sets, Subsets
+
+\begin{theorem}
+Every [[Definition:Subset|subset]] of a [[Definition:Countably Infinite Set|countably infinite set]] is [[Definition:Countable Set|countable]].
+\end{theorem}
+
+\begin{proof}
+Let $S = \set {a_0, a_1, a_2, \ldots}$ be [[Definition:Countably Infinite Set|countably infinite]].
+Let $T \subseteq S = \set {a_{n_0}, a_{n_1}, a_{n_2}, \ldots}$, where $a_{n_0}, a_{n_1}, a_{n_2}, \ldots$ are the [[Definition:Element|elements]] of $S$ also in $T$.
+If the set of numbers $\set {n_0, n_1, n_2, \ldots}$ has a largest number, then $T$ is [[Definition:Finite Set|finite]].
+Otherwise, consider the [[Definition:Bijection|bijection]] $i \leftrightarrow n_i$.
+This leads to the [[Definition:Bijection|bijection]] $i \leftrightarrow a_{n_i}$
+This latter [[Definition:Bijection|bijection]] is the required [[Definition:Bijection|one-to-one correspondence]] between the [[Definition:Element|elements]] of $T$ and those of $\N$, showing that $T$ is indeed [[Definition:Countable Set|countable]].
+{{finish|formal justification}}
+\end{proof}<|endoftext|>
+\section{Infinite Set has Countably Infinite Subset}
+Tags: Countable Sets, Infinite Sets, Subsets, Infinite Set has Countably Infinite Subset
+
+\begin{theorem}
+Every [[Definition:Infinite Set|infinite set]] has a [[Definition:Countably Infinite Set|countably infinite]] [[Definition:Subset|subset]].
+\end{theorem}
+
+\begin{proof}
+Let $S$ be an [[Definition:Infinite Set|infinite set]].
+We use [[Between Two Sets Exists Injection or Surjection]].
+Suppose that there exists an [[Definition:Injection|injection]] $\psi: \N \to S$.
+Let $T$ be the [[Definition:Image of Mapping|image]] of $\psi$.
+From [[Injection to Image is Bijection]], it follows that $\psi^{-1}: T \to \N$ is a [[Definition:Bijection|bijection]].
+Hence, $T$ is a [[Definition:Countably Infinite|countably infinite]] [[Definition:Subset|subset]] of $S$.
+Now, suppose that that there exists a [[Definition:Surjection|surjection]] $\phi: \N \to S$.
+From [[Surjection from Natural Numbers iff Countable]], it follows that $S$ is [[Definition:Countable Set|countably infinite]].
+So, from [[Set is Subset of Itself]], we have that $S$ is a [[Definition:Countable|countably infinite]] [[Definition:Subset|subset]] of $S$.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let $S$ be an [[Definition:Infinite Set|infinite set]].
+First an [[Definition:Injection|injection]] $f: \N \to S$ is constructed.
+Let $g$ be a [[Definition:Choice Function|choice function]] on $\powerset S \setminus \set \O$.
+Then define $f: \N \to S$ as follows:
+:$\forall n \in \N: f \left({n}\right) = \begin{cases}
+\map g S & : n = 0 \\
+\map g {S \setminus \set {\map f 0, \ldots, \map f {n - 1} } } & : n > 0
+\end{cases}$
+Since $S$ is [[Definition:Infinite Set|infinite]], $S \setminus \set {\map f 0, \ldots, \map f {n - 1} }$ is [[Definition:Non-Empty Set|non-empty]] for each $n \in \N$.
+Therefore $f \sqbrk \N$ is [[Definition:Infinite Set|infinite]].
+To show that $f$ is injective, let $m, n \in \N$, say $m < n$.
+Then:
+:$\map f m \in \set {\map f 0, \ldots, \map f {n - 1} }$
+but:
+:$\map f n \in S \setminus \set {\map f 0, \ldots, \map f {n - 1} }$
+Hence $\map f m \ne \map f n$.
+Since $f$ is [[Definition:Injection|injective]], it follows from [[Injection to Image is Bijection]] that $f$ is a [[Definition:Bijection|bijection]] from $\N$ to $f \sqbrk \N$.
+Thus $f \sqbrk \N$ is a [[Definition:Countable|countable]] subset of $S$.
+{{Qed}}
+\end{proof}
+
+\begin{proof}
+Let $S$ be an [[Definition:Infinite Set|infinite set]].
+First an [[Definition:Injection|injection]] $f: \N \to S$ is constructed.
+Let $f$ be a [[Definition:Choice Function|choice function]] on $\powerset S \setminus \set \O$.
+That is:
+:$\forall A \in \powerset S \setminus \set \O: \map f A \in A$
+This is justified only if the [[Axiom:Axiom of Choice|Axiom of Choice]] is accepted.
+Let $\CC$ be the [[Definition:Set of Sets|set]] of all [[Definition:Finite Set|finite subsets]] of $S$.
+Let $A \in \CC$.
+Since $S$ is [[Definition:Infinite Set|infinite]] it follows that $S \setminus A \ne \O$.
+So $S \setminus A \in \Dom f$.
+Let $g: \CC \to \CC$ be the [[Definition:Mapping|mapping]] defined as:
+:$\map g A = A \cup \set {\map f {S \setminus A} }$
+That is, $\map g A$ is constructed by joining $A$ with the element that $f$ chooses from $S \setminus A$.
+Consider the [[Recursion Theorem]] applied to $g$, starting with the set $\O$.
+We obtain a [[Definition:Mapping|mapping]] $U: \N \to \CC$ such that:
+:$\map U x = \begin{cases}
+\O & : x = 0 \\ \map U n \cup \set {\map f {S \setminus \map U n} } & : x = n^+
+\end{cases}$
+where here $\N$ is considered as the [[Definition:Set|set]] of [[Definition:Element|elements]] of the [[Definition:Von Neumann Construction of Natural Numbers|von Neumann construction of natural numbers]].
+Consider the mapping $v: \N \to S$, defined as:
+:$\forall n \in \N: \map v n = \map f {S \setminus \map U n}$
+We have that, by definition of $v$:
+:$(1): \quad \forall n \in \N: \map v n \notin \map U n$
+:$(2): \quad \forall n \in \N: \map v n \in \map U {n^+}$
+:$(3): \quad \forall m, n \in \N: n \le m \implies \map U n \subseteq \map U m$
+Then because $\map v n \in \map U m$ but $\map v m \notin \map U m$:
+:$(4): \quad \forall m, n \in \N: n < m \implies \map v n \ne \map v m$
+So $(4)$ implies that $v$ maps [[Definition:Distinct Elements|distinct elements]] of $\N$ onto [[Definition:Distinct Elements|distinct elements]] of $S$.
+Thus $v: \N \to S$ is an [[Definition:Injection|injection]].
+It follows from [[Injection to Image is Bijection]] that $v$ is a [[Definition:Bijection|bijection]] from $\N$ to $\map v \N$.
+Thus $\map v \N$ is the [[Definition:Countable|countable]] subset of $S$ that was required.
+{{Qed}}
+\end{proof}
+
+\begin{proof}
+Let $S$ be an [[Definition:Infinite Set|infinite set]].
+For all $n \in \N$, let:
+:$\mathcal F_n = \left\{{T \subseteq S : \left\vert{T}\right\vert = n}\right\}$
+where $\left\vert{T}\right\vert$ denotes the [[Definition:Cardinality|cardinality]] of $T$.
+From [[Set is Infinite iff exist Subsets of all Finite Cardinalities]]:
+:$\mathcal F_n$ is [[Definition:Non-Empty Set|non-empty]].
+Using the [[Axiom:Axiom of Countable Choice|axiom of countable choice]], we can obtain a [[Definition:Sequence|sequence]] $\left\langle{S_n}\right\rangle_{n \in \N}$ such that $S_n \in \mathcal F_n$ for all $n \in \N$.
+Define:
+:$\displaystyle T = \bigcup_{n \mathop \in \N} S_n \subseteq S$
+For all $n \in \N$, $S_n$ is a [[Definition:Subset|subset]] of $T$ whose [[Definition:Cardinality|cardinality]] is $n$.
+From [[Set is Infinite iff exist Subsets of all Finite Cardinalities]]:
+:$T$ is infinite.
+Because the [[Countable Union of Countable Sets is Countable|countable union of finite sets is countable]], $T$ is [[Definition:Countable Set|countable]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{No Bijection between Finite Set and Proper Subset}
+Tags: Subsets, Set Theory, No Bijection between Finite Set and Proper Subset
+
+\begin{theorem}
+A [[Definition:Finite Set|finite set]] can not be in [[Definition:Bijection|one-to-one correspondence]] with one of its [[Definition:Proper Subset|proper subsets]].
+That is, a [[Definition:Finite Set|finite set]] is not [[Definition:Dedekind-Infinite|Dedekind-infinite]].
+\end{theorem}
+
+\begin{proof}
+Let $S$ be a [[Definition:Finite Set|finite set]], and let $T$ be a [[Definition:Proper Subset|proper subset]] of $S$.
+Let $f: T \to S$ be an [[Definition:Injection|injection]].
+By [[Cardinality of Image of Injection]] and [[Cardinality of Subset of Finite Set]]:
+:$\size {\Img f} = \size T < \size S$
+Here, $\Img f$ denotes the [[Definition:Image of Mapping|image]] of $f$.
+Thus $\Img f \ne S$, and so $f$ is not a [[Definition:Bijection|bijection]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Proof by [[Principle of Mathematical Induction|induction]]:
+For all $n \in \N_{>0}$, let $P \left({n}\right)$ be the [[Definition:Proposition|proposition]]:
+:The set $\N_n = \left\{{0, 1, \ldots, n-1}\right\}$ of [[Definition:Initial Segment of Natural Numbers|natural numbers less than $n$]] is [[Definition:Set Equivalence|equivalent]] to none of its [[Definition:Proper Subset|proper subsets]].
+Here we use the definition of [[Definition:Set Equivalence|set equivalence]] to mean that $S$ is [[Definition:Set Equivalence|equivalent]] to $T$ if there exists a [[Definition:Bijection|bijection]] between them.
+=== Basis for the Induction ===
+$P \left({1}\right)$ is the case:
+:The set $\N_1 = \left\{{0}\right\}$ is [[Definition:Set Equivalence|equivalent]] to none of its [[Definition:Proper Subset|proper subsets]].
+The only [[Definition:Proper Subset|proper subset]] of $\left\{{0}\right\}$ is the [[Definition:Empty Set|empty set]] $\varnothing$, to which $\left\{{0}\right\}$ is trivially not [[Definition:Set Equivalence|equivalent]].
+This is our [[Principle of Mathematical Induction#Basis for the Induction|basis for the induction]].
+=== Induction Hypothesis ===
+Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 1$, then it logically follows that $P \left({k+1}\right)$ is true.
+So this is our [[Principle of Mathematical Induction#Induction Hypothesis|induction hypothesis]]:
+:The set $\N_k = \left\{{0, 1, \ldots, k-1}\right\}$ is [[Definition:Set Equivalence|equivalent]] to none of its [[Definition:Proper Subset|proper subsets]].
+Then we need to show:
+:The set $\N_{k+1} = \left\{{0, 1, \ldots, k}\right\}$ is [[Definition:Set Equivalence|equivalent]] to none of its [[Definition:Proper Subset|proper subsets]].
+=== Induction Step ===
+Suppose that $P \left({k+1}\right)$ is not true.
+Let $\phi: N_{k+1} \to S$ be a [[Definition:Bijection|bijection]] where $S \subsetneq N_{k+1}$.
+We aim to demonstrate that the existence of such a $\phi$ contradicts the [[No Bijection between Finite Set and Proper Subset/Proof 2#Induction Hypothesis|induction hypothesis]].
+We consider two cases:
+;Case 1.
+Let $S \subseteq N_k$.
+Then we define the [[Definition:Composition of Mappings|composite mapping]] $\phi \circ \phi$ defined as:
+:$\forall m \in \N_{k+1}: \phi \circ \phi \left({m}\right) = \phi \left({\phi \left({m}\right)}\right)$
+From [[Composite of Bijections is Bijection]], $\phi \circ \phi$ is a [[Definition:Bijection|bijection]].
+Now consider $T = \phi \circ \phi \left({\N_{k+1}}\right)$, the [[Definition:Image of Mapping|image set]] of $\N_{k+1}$ under $\phi \circ \phi$.
+Let $m, t \in \N_{k+1}$ such that:
+:$\phi \circ \phi \left({m}\right) = \phi \left({t}\right)$
+Since $\phi$ is an [[Definition:Injection|injection]] it follows that:
+:$\phi \left({m}\right) = t$
+As $S$ is a [[Definition:Proper Subset|proper subset]] of $\N_{k+1}$ it follows from [[Set Difference with Proper Subset]] that $\N_{k+1} \setminus S \ne \varnothing$.
+Let $x \in \N_{k+1} \setminus S$.
+Suppose $\phi \left({x}\right) \in T$.
+Then that would mean that $\exists m \in \N_{k+1}$ such that:
+: $\phi \circ \phi \left({m}\right) = \phi \left({x}\right)$
+which from above it would follow that:
+:$\phi \left({m}\right) = x$
+and so $x \in S$.
+But this can not happen as $S \cap \left({\N_{k+1} \setminus S}\right) = \varnothing$ from [[Set Difference Intersection with Second Set is Empty Set]].
+So it must be the case that $\phi \left({x}\right) \notin T$.
+Thus it can be concluded that $\phi \circ \phi$ is a [[Definition:Mapping|mapping]] from $\N_{k+1}$ onto a [[Definition:Proper Subset|proper subset]] $T$ of $S$.
+Now, consider the mapping $\psi: \N_k \to U$ where $U$ is a [[Definition:Proper Subset|proper subset]] of $\N_k$.
+We define $\psi$ as follows:
+:$\forall a \in \N_k: \psi \left({a}\right) = \begin{cases}
+a & : a \notin S = \phi \left({N_{k+1}}\right) \\
+\phi \left({a}\right) & : a \in S \end{cases}$
+The next step is to show that $\psi$ is an [[Definition:Injection|injection]].
+Consider $a, b \in \N_k$ such that $\psi \left({a}\right) = \psi \left({b}\right)$.
+Either $\psi \left({a}\right) = \psi \left({b}\right) \in \N_k \setminus S$, in which case:
+:$\psi \left({a}\right) = a = \psi \left({b}\right) = b$
+or $\psi \left({a}\right) = \psi \left({b}\right) \in S$, in which case:
+:$\psi \left({a}\right) = \phi \left({a}\right) = \psi \left({b}\right) = \phi \left({b}\right)$
+As $\phi$ is a [[Definition:Injection|injection]] it follows that $a = b$.
+So $\psi \left({a}\right) = \psi \left({b}\right) \implies a = b$ and so by definition $\psi$ is an [[Definition:Injection|injection]].
+Now consider $\psi \left({\N_k}\right)$.
+We have that $\psi \left({\N_k}\right) = \N_k \setminus \left({S \setminus T}\right)$ so that $\psi$ is an [[Definition:Injection|injection]] from $\N_k$ to a [[Definition:Proper Subset|proper subset]] of itself.
+This contradicts the [[No Bijection between Finite Set and Proper Subset/Proof 2#Induction Hypothesis|induction hypothesis]].
+;Case 2.
+Let $S \not \subseteq N_k$.
+This means that $k \in S$.
+There exists $x \in \N_{k+1}$ such that $\phi \left({x}\right) = k$, as $\phi: \N_{k+1} \to S$ is a [[Definition:Surjection|surjection]].
+There also exists $y \in \N_{k+1} \setminus S$ as $\N_{k+1} \setminus S \ne \varnothing$ from [[Set Difference with Proper Subset]].
+Let $\phi': \N_{k+1} \to S$ be defined as:
+:$\phi' \left({a}\right) = \begin{cases}
+\phi \left({a}\right) : & a \ne x \\
+y : & a = x \end{cases}$
+As $\phi$ is a [[Definition:Bijection|bijection]] it follows that $\phi': \N_{k+1} \to S\,'$ is also a [[Definition:Bijection|bijection]], where:
+:$S\,' = \left({S \setminus \left\{{k}\right\}}\right) \cup \left\{{y}\right\}$
+Proceeding as Case 1, we prove the same contradiction in the [[No Bijection between Finite Set and Proper Subset/Proof 2#Induction Hypothesis|induction hypothesis]].
+Thus by [[Proof by Contradiction]], the assumption that $\phi: N_{k+1} \to S$ can be a [[Definition:Bijection|bijection]] must be false.
+So the truth of $P \left({m}\right)$ for all $m \le k$ implies the proof of $P \left({k+1}\right)$.
+The result follows by the [[Second Principle of Mathematical Induction]].
+Therefore:
+:For all $n \in \N$, the set $\N_n = \left\{{0, 1, \ldots, n-1}\right\}$ of [[Definition:Initial Segment of Natural Numbers|natural numbers less than $n$]] is [[Definition:Set Equivalence|equivalent]] to none of its [[Definition:Proper Subset|proper subsets]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Infinite Set is Equivalent to Proper Subset}
+Tags: Infinite Set is Equivalent to Proper Subset, Set Equivalence, Infinite Sets, Subsets
+
+\begin{theorem}
+A [[Definition:Set|set]] is [[Definition:Infinite Set|infinite]] {{iff}} it is [[Definition:Set Equivalence|equivalent]] to one of its [[Definition:Proper Subset|proper subsets]].
+\end{theorem}
+
+\begin{proof}
+Let $T$ be an [[Definition:Infinite Set|infinite set]].
+By [[Infinite Set has Countably Infinite Subset]], it is possible to construct a [[Definition:Countable|countably infinite]] [[Definition:Subset|subset]] of $T$.
+Let $S = \set {a_1, a_2, a_3, \ldots}$ be such a [[Definition:Countable|countably infinite]] [[Definition:Subset|subset]] of $T$.
+Create a [[Definition:Partition (Set Theory)|Partition]] of $S$ into:
+:$S_1 = \set {a_1, a_3, a_5, \ldots}, S_2 = \set {a_2, a_4, a_6, \ldots}$
+Let a [[Definition:Bijection|bijection]] be established between $S$ and $S_1$, by letting $a_n \leftrightarrow a_{2 n - 1}$.
+This is extended to a [[Definition:Bijection|bijection]] between $S \cup \paren {T \setminus S} = T$ and $S_1 \cup \paren {T \setminus S} = T \setminus S_2$ by assigning each [[Definition:Element|element]] in $T \setminus S$ to itself.
+So a [[Definition:Bijection|bijection]] has been demonstrated between $T$ and one of its [[Definition:Proper Subset|proper subsets]] $T \setminus S_2$.
+That is, if $T$ is [[Definition:Infinite Set|infinite]], it is [[Definition:Set Equivalence|equivalent]] to one of its [[Definition:Proper Subset|proper subsets]].
+Now, let $T_0 \subsetneq T$ be a proper subset of $T$, and $f: T \to T_0$ be a [[Definition:Bijection|bijection]].
+It follows from [[No Bijection between Finite Set and Proper Subset]] that $T$ must be [[Definition:Infinite Set|infinite]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let $S$ be a [[Definition:Set|set]].
+Suppose $S$ is [[Definition:Finite Set|finite]].
+From [[No Bijection between Finite Set and Proper Subset]] we have that $S$ can not be [[Definition:Set Equivalence|equivalent]] to one of its [[Definition:Proper Subset|proper subsets]].
+Suppose $S$ is [[Definition:Infinite Set|infinite]].
+From [[Infinite Set has Countably Infinite Subset]], we can construct $v: \N \to S$ such that $v$ is an [[Definition:Injection|injection]].
+We now construct the [[Definition:Mapping|mapping]] $h: S \to S$ as follows.
+:$\map h x = \begin {cases}
+\map v {n + 1} & : \exists n \in \N: x = \map v n \\
+x & : x \notin \Img v
+\end {cases}$
+It is clear that $h$ is an [[Definition:Injection|injection]].
+But we have that $\map v 0 \notin \Img h$ and so $\Img h \subsetneq S$.
+The result follows from [[Injection to Image is Bijection]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let $X$ be a [[Definition:Set|set]] which has a [[Definition:Proper Subset|proper subset]] $Y$ such that:
+:$\card X = \card Y$
+where $\card X$ denotes the [[Definition:Cardinality|cardinality]] of $X$.
+Then:
+:$\exists \alpha \in \complement_X \paren Y$
+and
+:$Y \subsetneqq Y \cup \set \alpha \subseteq X$
+The [[Definition:Inclusion Mapping|inclusion mappings]]:
+:$i_Y: Y \to X: \forall y \in Y: i \paren y = y$
+:$i_{Y \cup \set \alpha}: Y \cup \set \alpha \to X: \forall y \in Y: i \paren y = y$
+give:
+:$\card X = \card Y \le \card Y + \mathbf 1 \le \card X $
+from which:
+:$\card X = \card Y + \mathbf 1 = \card X + \mathbf 1$
+So by definition $X$ is [[Definition:Infinite Set|infinite]].
+{{qed|lemma}}
+Now suppose $X$ is [[Definition:Infinite Set|infinite]].
+That is:
+:$\card X = \card X + \mathbf 1$
+Let $\alpha$ be any [[Definition:Object|object]] such that $\alpha \notin X$.
+Then there is a [[Definition:Bijection|bijection]] $f: X \cup \set \alpha \to X$.
+Let $f_{\restriction X}$ be the [[Definition:Restriction of Mapping|restriction]] of $f$ to $X$.
+Then from [[Injection to Image is Bijection]]:
+:$\image {f_{\restriction X} } = X \setminus \set {f \paren \alpha}$
+which is a [[Definition:Proper Subset|proper subset]] of $X$ which is [[Definition:Set Equivalence|equivalent]] to $X$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Cartesian Product of Countable Sets is Countable}
+Tags: Set Theory, Cartesian Product, Countable Sets, Cartesian Product of Countable Sets is Countable
+
+\begin{theorem}
+The [[Definition:Cartesian Product|cartesian product]] of two [[Definition:Countable Set|countable sets]] is [[Definition:Countable Set|countable]].
+\end{theorem}
+
+\begin{proof}
+Let $S = \set {s_0, s_1, s_2, \dotsc}$ and $T = \set {t_0, t_1, t_2, \dotsc}$ be [[Definition:Countable Set|countable sets]].
+If both $S$ and $T$ are [[Definition:Finite Set|finite]], the result follows immediately.
+Suppose either of $S$ or $T$ (or both) is [[Definition:Countably Infinite Set|countably infinite]].
+We can write the [[Definition:Element|elements]] of $S \times T$ in the form of an [[Definition:Infinite Set|infinite]] table:
+:$\begin{array} {*{4}c}
+ \tuple {s_0, t_0} & \tuple {s_0, t_1} & \tuple {s_0, t_2} & \cdots \\
+ \tuple {s_1, t_0} & \tuple {s_1, t_1} & \tuple {s_1, t_2} & \cdots \\
+ \tuple {s_2, t_0} & \tuple {s_2, t_1} & \tuple {s_2, t_2} & \cdots \\
+ \vdots  & \vdots  & \vdots & \ddots \\
+\end{array}$
+This table clearly contains all the [[Definition:Element|elements]] of $S \times T$.
+Now we can count the [[Definition:Element|elements]] of $S \times T$ by processing the table diagonally.
+First we pick $\tuple {s_0, t_0}$.
+Then we pick $\tuple {s_0, t_1}, \tuple {s_1, t_0}$. Then we pick $\tuple {s_0, t_2}, \tuple {s_1, t_1}, \tuple {s_2, t_0}$.
+We can see that all the [[Definition:Element|elements]] of $S \times T$ will (eventually) be listed, and there is a specific number ([[Definition:Element|element]] of $\N$) to index each of its [[Definition:Element|elements]] with.
+Thus we have the required [[Definition:Bijection|one-to-one correspondence]] between $S \times T$ and $\N$, and our assertion is proved.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let $S, T$ be [[Definition:Countable Set|countable sets]].
+From the definition of [[Definition:Countable Set|countable]], there exists a [[Definition:Injection|injection]] from $S$ to $\N$, and similarly one from $T$ to $\N$.
+Hence there exists an [[Definition:Injection|injection]] $g$ from $S \times T$ to $\N^2$.
+Now let us investigate the [[Definition:Cardinality|cardinality]] of $\N^2$.
+From the [[Fundamental Theorem of Arithmetic]], every [[Definition:Natural Number|natural number]] greater than $1$ has a unique [[Definition:Prime Decomposition|prime decomposition]].
+Thus, if a number can be written as $2^n 3^m$, it can be done thus in only one way.
+So, consider the [[Definition:Function|function]] $f: \N^2 \to \N$ defined by:
+:$\map f {n, m} = 2^n 3^m$.
+Now suppose $\exists m, n, r, s \in \N$ such that $\map f {n, m} = \map f {r, s}$.
+Then $2^n 3^m = 2^r 3^s$ so that $n = r$ and $m = s$.
+Thus $f$ is an [[Definition:Injection|injection]], whence $\N^2$ is [[Definition:Countably Infinite Set|countably infinite]].
+Now we see that as $g$ and $f$ are [[Definition:Injection|injective]], it follows from [[Composite of Injections is Injection]] that $f \circ g: S \times T \to \N$ is also [[Definition:Injection|injective]].
+Hence the result.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let $S, T$ be [[Definition:Countable Set|countable sets]].
+Let $S = \set {s_1, s_2, s_3, \dotsc}$ and $T = \set {t_1, t_2, t_3, \dotsc}$ be [[Definition:Countably Infinite Enumeration|enumerations]] of $S$ and $T$ respectively.
+Let $f: S \times T: \N$ be the [[Definition:Mapping|mapping]] defined as:
+:$\forall \tuple {s_k, t_l} \in S \times T: \map f {s_k, t_l} = \dfrac {\paren {k + l - 1} \paren {k + l - 2} } 2 + \dfrac {l + \paren {-1}^{k + 1} } 2 k + \dfrac {1 + \paren {-1}^{k + l - 1} } 2 l$
+Then $f$ gives an [[Definition:Countably Infinite Enumeration|enumeration]] of $S \times T$.
+This [[Definition:Countably Infinite Enumeration|enumeration]] can be depicted schematically as:
+:$\begin {array} {} \tuple {s_1, t_1} & & \tuple {s_1, t_2} & \to  & \tuple {s_1, t_3} & & \tuple {s_1, t_4} & \to & \dotsc \\
+\downarrow & \nearrow & & \swarrow & & \nearrow  & \dotsc \\
+\tuple {s_2, t_1} & & \tuple {s_2, t_2} & & \tuple {s_2, t_3} & \dotsc \\
+& \swarrow & & \nearrow & \dotsc \\
+\tuple {s_3, t_1} & & \tuple {s_2, t_3} & \dotsc \\
+\downarrow & \nearrow & \dotsc \\
+\tuple {s_3, t_1} & \dotsc \\
+\vdots \\
+\end{array}$
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Composition of Sequence with Mapping}
+Tags: Sequences
+
+\begin{theorem}
+Let $\left \langle {a_j} \right \rangle_{j \in B}$ be a [[Definition:Sequence|sequence]].
+Let $\sigma: A \to B$ be a [[Definition:Mapping|mapping]], where $A \subseteq \N$.
+Then $\left \langle {a_j} \right \rangle \circ \sigma$ is a [[Definition:Sequence|sequence]] whose value at each $k \in A$ is $a_{\sigma \left({k}\right)}$.
+Thus $\left \langle {a_j} \right \rangle \circ \sigma$ is denoted $\left \langle {a_{\sigma \left({k}\right)}} \right \rangle_{k \in A}$.
+\end{theorem}
+
+\begin{proof}
+By definition, a [[Definition:Sequence|sequence]] is a [[Definition:Mapping|mapping]] whose [[Definition:Domain of Mapping|domain]] is a [[Definition:Subset|subset]] of $\N$.
+Let the [[Definition:Range of Sequence|range]] of $\left \langle {a_j} \right \rangle_{j \in B}$ be $S$.
+Thus $\left \langle {a_j} \right \rangle_{j \in B}$ can be expressed using the [[Definition:Mapping|mapping]] $f: B \to S$ as:
+:$\forall j \in B: f \left({j}\right) = a_j$
+Let $k \in A$.
+Then $\sigma \left({k}\right) \in B$.
+By definition of [[Definition:Composition of Mappings|composition of mappings]]:
+{{begin-eqn}}
+{{eqn | l=f \circ \sigma \left({k}\right)
+      | r=f \left({\sigma \left({k}\right)}\right)
+      | c=
+}}
+{{eqn | ll=\implies
+      | l=\left \langle {a_j} \right \rangle_{j \in B}
+      | r=\left \langle {a_{\sigma \left({k}\right)} } \right \rangle_{\sigma \left({k}\right) \in B}
+      | c=
+}}
+{{eqn | r=\left \langle {a_{\sigma \left({k}\right)} } \right \rangle_{k \in A}
+      | c=as $\sigma \left({k}\right) \in B \implies k \in A$
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Sequences]]
+2wdusrdz15r73kktgzfr0g4a18i31vr
+\end{proof}<|endoftext|>
+\section{General Operation from Binary Operation}
+Tags: Abstract Algebra
+
+\begin{theorem}
+Let $\left({S, \oplus}\right)$ be a [[Definition:Magma|magma]].
+Then there a unique [[Definition:Sequence|sequence]] $\left \langle {\oplus_k} \right \rangle_{k \mathop \ge 1}$ such that:
+:$(1): \quad \forall n \in \N_{>0}: \oplus_n$ is an [[Definition:Operation on Set|$n$-ary operation]] on $S$ such that:
+:$(2): \quad \forall \left({a_1, \ldots, a_k}\right) \in S^k: \oplus_k \left({a_1, \ldots, a_k}\right) = \begin{cases}
+a: & k = 1 \\
+\oplus_n \left({a_1, \ldots, a_n}\right) \oplus a_{n+1}: & k = n + 1
+\end{cases}$
+In particular, $\oplus_2$ is the same as the given [[Definition:Binary Operation|binary operation]] $\oplus$.
+The $n$th [[Definition:Term of Sequence|term]] $\oplus_n$ of the [[Definition:Sequence|sequence]] $\left \langle {\oplus} \right \rangle$ is called the '''[[Definition:N-Ary Operation Defined by Binary Operation|$n$-ary operation defined by $\oplus$]]'''.
+\end{theorem}
+
+\begin{proof}
+Let $\Bbb S = \left\{{\odot:}\right.$ for some $n \in \N_{>0}$, $\odot$ is an [[Definition:Operation on Set|$n$-ary operation]] on $\left.{S}\right\}$.
+Let $s: \Bbb S \to \Bbb S$ be the [[Definition:Mapping|mapping]] defined as follows.
+Let $\odot$ be any [[Definition:Operation on Set|$n$-ary operation]] defined on $\Bbb S$.
+Then $s \left({\odot}\right)$ is the $\left({n+1}\right)$-ary operation defined by:
+:$\forall \left({a_1, \ldots, a_n, a_{n+1}}\right) \in S^{n+1}: s \left({\odot}\right) \left({a_1, \ldots, a_n, a_{n+1}}\right) = \odot \left({a_1, \ldots, a_n}\right) \oplus a_{n+1}$
+By the [[Principle of Recursive Definition]], there is a [[Definition:Unique|unique]] [[Definition:Sequence|sequence]] $\left \langle {\oplus_k} \right \rangle_{k \mathop \ge 1}$ such that:
+: $\oplus_1$ is the [[Definition:Unary Operation|unary operation]] defined as $\oplus_1 \left({a}\right) = a$
+and:
+: $\oplus_{n+1} = s \left({\oplus_n}\right)$ for each $n \in \N_{>0}$.
+Let $A$ be the [[Definition:Set|set]] defined as the set of all $n$ such that:
+:$\oplus_n$ is an [[Definition:Operation on Set|$n$-ary operation]] on $S$
+and:
+:by the definition of $s$, $(2)$ holds for every [[Definition:Ordered Tuple|ordered $n + 1$-tuple]] in $S^{n+1}$.
+It will be demonstrated by the [[Principle of Mathematical Induction]] that $A = \N$.
+=== Basis for the Induction ===
+We have that:
+:$\oplus_1$ is the unary operation defined as $\oplus_1 \left({a}\right) = a$
+By the [[Principle of Recursive Definition]]:
+:$\oplus_2 = s \left({\oplus_1}\right)$
+is the [[Definition:Unique|unique]] [[Definition:Binary Operation|operation]] on $S$ such that:
+{{begin-eqn}}
+{{eqn | lo= \forall \left({a_1, a_2}\right) \in S^2:
+      | l = \oplus_2 \left({a_1, a_2}\right)
+      | r = s \left({\oplus_1}\right) \left({a_1, a_2}\right)
+      | c = Definition of $s$
+}}
+{{eqn | r = \oplus_1 \left({a_1}\right) \oplus a_2
+      | c = 
+}}
+{{eqn | r = a_1 \oplus a_2
+      | c = 
+}}
+{{end-eqn}}
+By the definition of a [[Definition:Magma|magma]], $\oplus$ is a [[Definition:Binary Operation|operation]] defined such that:
+:$\forall \left({a_1, a_2}\right) \in S \times S: \oplus \left({a_1, a_2}\right) \in S$
+where $\oplus: S \times S \to S$ is a [[Definition:Mapping|mapping]].
+$a_1 \oplus a_2$ is defined as:
+:$\forall \left({a_1, a_2}\right) \in S \times S: a_1 \oplus a_2 := \oplus \left({a_1, a_2}\right)$
+So $1 \in A$.
+This is our [[Principle of Mathematical Induction#Basis for the Induction|basis for the induction]].
+=== Induction Hypothesis ===
+It is to be shown that, if $m \in A$ where $m \ge 1$, then it follows that $m + 1 \in A$.
+This is the [[Principle of Mathematical Induction#Induction Hypothesis|induction hypothesis]]:
+:$\oplus_m$ is an [[Definition:Operation on Set|$m$-ary operation]] on $S$
+and:
+:by the definition of $s$, $(2)$ holds for every [[Definition:Ordered Tuple|ordered $m + 1$-tuple]] in $S^{m+1}$.
+It is to be demonstrated that it follows that:
+:$\oplus_{m+1}$ is an [[Definition:Operation on Set|$m+1$-ary operation]] on $S$
+and:
+:by the definition of $s$, $(2)$ holds for every [[Definition:Ordered Tuple|ordered $m + 2$-tuple]] in $S^{m+2}$.
+=== Induction Step ===
+This is our [[Principle of Mathematical Induction#Induction Step|induction step]]:
+By the [[Principle of Recursive Definition]]:
+:$\oplus_{m+1} = s \left({\oplus_m}\right)$
+is the [[Definition:Unique|unique]] [[Definition:Binary Operation|operation]] on $S$ such that:
+{{begin-eqn}}
+{{eqn | lo= \forall \left({a_1, a_2, \ldots, a_{m+1} }\right) \in S^{m+1}:
+      | l = \oplus_{m+1} \left({a_1, a_2, \ldots, a_{m+1} }\right)
+      | r = s \left({\oplus_m}\right) \left({a_1, a_2, \ldots, a_{m+1} }\right)
+      | c = Definition of $s$
+}}
+{{eqn | r = \oplus_m \left({a_1, a_2, \ldots, a_m}\right) \oplus a_{m+1}
+      | c = 
+}}
+{{eqn | r = \left({a_1 \oplus a_2 \oplus \cdots \oplus a_m}\right) \oplus a_{m+1}
+      | c = 
+}}
+{{end-eqn}}
+By the definition of a [[Definition:Magma|magma]], $\oplus$ is a [[Definition:Binary Operation|operation]] defined such that:
+:$\forall \left({a_1, a_2}\right) \in S \times S: \oplus \left({a_1, a_2}\right) \in S$
+where $\oplus: S \times S \to S$ is a [[Definition:Mapping|mapping]].
+By the [[General Operation from Binary Operation#Induction Hypothesis|induction hypothesis]]:
+:$\oplus_m \left({a_1, a_2, \ldots, a_m}\right) \oplus a_{m+1} \in S$
+So by definition:
+$\oplus_{m+1} \left({a_1, a_2, \ldots, a_m, a_{m+1} }\right) \in S$
+Let $b := \oplus_{m+1} \left({a_1, a_2, \ldots, a_m, a_{m+1} }\right)$.
+But then by the [[General Operation from Binary Operation#Basis of the Induction|basis of the induction]]:
+:$b \oplus a_{m+2} \in S$
+That is:
+:$\oplus_{m+1} \left({a_1, a_2, \ldots, a_m, a_{m+1} }\right) \oplus a_{m+2} = \oplus_{m+2} \left({a_1, a_2, \ldots, a_m, a_{m+1}, a_{m+2} }\right) \in S$
+So $m \in A \implies m + 1 \in A$ and the result follows by the [[Principle of Mathematical Induction]]:
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Strictly Increasing Sequence induces Partition}
+Tags: Sequences
+
+\begin{theorem}
+Let $\left \langle {r_k} \right \rangle_{0 \mathop \le k \mathop \le n}$ be a [[Definition:Strictly Increasing Sequence|strictly increasing]] [[Definition:Finite Sequence|finite sequence]] of [[Definition:Natural Numbers|natural numbers]].
+Let:
+:$\forall k \in \left[{1 \,.\,.\, n}\right]: A_k := \left[{r_{k-1} + 1 \,.\,.\, r_k}\right]$
+Then:
+:$\left\{{A_k: k \in \left[{1 \,.\,.\, n}\right]}\right\}$ is a [[Definition:Partition (Set Theory)|partition]] of $\left[{r_0 + 1 \,.\,.\, r_n}\right]$.
+\end{theorem}
+
+\begin{proof}
+First we show that the [[Definition:Element|elements]] of $\left\{{A_k: k \in \left[{1 \,.\,.\, n}\right]}\right\}$ are [[Definition:Disjoint Sets|disjoint]].
+Let $j \in \left[{1 \,.\,.\, n}\right]$.
+We have that:
+: $\left \langle {r_k} \right \rangle_{0 \mathop \le k \mathop \le n}$ is [[Definition:Strictly Increasing Sequence|strictly increasing]]
+and:
+: $0 \le j - 1 < j \le n$
+Thus by [[Sum with One is Immediate Successor in Naturally Ordered Semigroup]]:
+: $r_0 \le r_{j-1} < r_j \le r_n \implies \left({r_0 + 1}\right) \le \left({r_{j-1} + 1}\right) \le r_j \le r_n$
+So:
+: $\varnothing \subset A_j \subseteq \left[{r_0 + 1 \,.\,.\, r_n}\right]$
+Because $\left \langle {r_k} \right \rangle_{0 \mathop \le k \mathop \le n}$ is [[Definition:Strictly Increasing Sequence|strictly increasing]]:
+:$1 \le j < k \le n \implies A_j \cap A_k = \varnothing$
+So the elements of $\left\{{A_k: k \in \left[{1 \,.\,.\, n}\right]}\right\}$ are [[Definition:Disjoint Sets|disjoint]], as we were to show.
+It remains to be established that:
+:if $m \in \left[{r_0 + 1 \,.\,.\, r_n}\right]$
+then:
+: $m \in A_k$ for some $k \in \left[{1\,.\,.\, n}\right]$.
+Let $m \in \left[{r_0 + 1 \,.\,.\, r_n}\right]$.
+Consider the set:
+:$J = \left\{{j \in \left[{0 \,.\,.\, n}\right]: m \le r_j}\right\}$
+Clearly $j \ne \varnothing$ as $n \in J$.
+Let $k$ be the [[Definition:Smallest Element|smallest]] element of $J$.
+Then because $r_0 < m$:
+: $k \ne 0$
+Thus:
+: $k \in \left[{1 \,.\,.\, n}\right]$
+By its definition:
+: $r_{k-1} < m \le r_k$
+Thus, by [[Sum with One is Immediate Successor in Naturally Ordered Semigroup]]:
+: $r_{k-1} + 1 \le m \le r_k$
+Therefore $m \in A_k$.
+{{qed}}
+{{improve|A better link instead of [[Sum with One is Immediate Successor in Naturally Ordered Semigroup]]}}
+\end{proof}<|endoftext|>
+\section{Fundamental Principle of Counting}
+Tags: Sequences, Fundamental Theorems, Set Theory, Combinatorics
+
+\begin{theorem}
+Let $A$ be a [[Definition:Finite Set|finite set]].
+Let $\sequence {B_n}$ be a [[Definition:Sequence of Distinct Terms|sequence of distinct]] [[Definition:Finite Set|finite]] [[Definition:Subset|subsets]] of $A$ which form a [[Definition:Set Partition|partition]] of $A$.
+Let $p_k = \size {B_k}$ for each $k \in \closedint 1 n$.
+Then:
+:$\displaystyle \size A = \sum_{k \mathop = 1}^n p_k$
+That is, the sum of the numbers of [[Definition:Element|elements]] in the [[Definition:Subset|subsets]] of a [[Definition:Set Partition|partition]] of a [[Definition:Set|set]] is equal to the total number of [[Definition:Element|elements]] in the [[Definition:Set|set]].
+\end{theorem}
+
+\begin{proof}
+Let $r_0 = 0$, and let:
+:$\displaystyle \forall k \in \closedint 1 n: r_k = \sum_{j \mathop = 1}^k {p_j}$
+Then:
+:$r_{k - 1} + p_k = r_k$
+so:
+:$r_{k - 1} < r_k$.
+Thus by [[Isomorphism to Closed Interval]], $\closedint {r_{k - 1} } {r_k}$ has $r_k - r_{k - 1} = p_k$ [[Definition:Element|elements]].
+As a consequence, there exists a [[Definition:Bijection|bijection]] $\sigma_k: B_k \to \closedint {r_{k - 1} } {r_k}$ for each $k \in \closedint 1 n$.
+Let $\sigma: A \to \N$ be the [[Definition:Mapping|mapping]] that satisfies:
+:$\forall x \in B_k: \forall k \in \N: \map \sigma x = \map {\sigma_k} x$
+By [[Strictly Increasing Sequence on Ordered Set]], $\sequence {r_k}_{0 \mathop \le k \mathop \le n}$ is a [[Definition:Strictly Increasing Sequence|strictly increasing sequence]] of [[Definition:Natural Numbers|natural numbers]].
+Thus by [[Strictly Increasing Sequence induces Partition]], $\sigma: A \to \closedint 1 {r_n}$ is a [[Definition:Bijection|bijection]].
+By [[Isomorphism to Closed Interval]], $\closedint 1 {r_n}$ has $r_n$ [[Definition:Element|elements]].
+Hence the result.
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Odd Number Theorem}
+Tags: Square Numbers, Proofs by Induction, Sums of Sequences, Named Theorems, Odd Number Theorem
+
+\begin{theorem}
+:$\displaystyle \sum_{j \mathop = 1}^n \paren {2 j - 1} = n^2$
+That is, the sum of the first $n$ [[Definition:Odd Integer|odd numbers]] is the $n$th [[Definition:Square Number|square number]].
+\end{theorem}
+
+\begin{proof}
+Proof by [[Principle of Mathematical Induction|induction]]:
+For all $n \in \N_{>0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
+:$\displaystyle n^2 = \sum_{j \mathop = 1}^n \paren {2 j - 1}$
+=== Basis for the Induction ===
+$\map P 1$ is true, as this just says $1^2 = 1$.
+This is our [[Definition:Basis for the Induction|basis for the induction]].
+=== Induction Hypothesis ===
+Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
+So this is our [[Definition:Induction Hypothesis|induction hypothesis]]:
+:$\displaystyle k^2 = \sum_{j \mathop = 1}^k \paren {2 j - 1}$
+Then we need to show:
+:$\displaystyle \paren {k + 1}^2 = \sum_{j \mathop = 1}^{k + 1} \paren {2 j - 1}$
+=== Induction Step ===
+This is our [[Definition:Induction Step|induction step]]:
+{{begin-eqn}}
+{{eqn | l = \paren {k + 1}^2
+      | r = k^2 + 2 k + 1
+      | c = 
+}}
+{{eqn | r = \sum_{j \mathop = 1}^k \paren {2 j - 1} + 2 k + 1
+      | c = [[Odd Number Theorem#Induction Hypothesis|Induction Hypothesis]]
+}}
+{{eqn | r = \sum_{j \mathop = 1}^k \paren {2 j - 1} + 2 \paren {k + 1} - 1
+      | c = 
+}}
+{{eqn | r = \sum_{j \mathop = 1}^{k + 1} \paren {2 j - 1}
+      | c = 
+}}
+{{end-eqn}}
+So $\map P k \implies \map P {k + 1}$ and the result follows by the [[Principle of Mathematical Induction]].
+Therefore:
+:$\displaystyle \forall n \in \N: n^2 = \sum_{j \mathop = 1}^n \paren {2 j - 1}$
+{{qed}}
+\end{proof}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \sum_{j \mathop = 1}^n \paren {2 j - 1}
+      | r = n^2
+      | c = [[Odd Number Theorem]]
+}}
+{{eqn | ll= \leadsto
+      | l = \sum_{j \mathop = 1}^n \paren {2 j - 1} + \sum_{j \mathop = 1}^n 1
+      | r = n^2 + n
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \sum_{j \mathop = 1}^n \paren {2 j}
+      | r = n \paren {n + 1}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \sum_{j \mathop = 1}^n j
+      | r = \frac {n \paren {n + 1} } 2
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{General Associativity Theorem}
+Tags: Associativity, Named Theorems, General Associativity Theorem
+
+\begin{theorem}
+If an [[Definition:Operation|operation]] is [[Definition:Associative Operation|associative]] on $3$ entities, then it is [[Definition:Associative Operation|associative]] on any number of them.
+\end{theorem}<|endoftext|>
+\section{General Commutativity Theorem}
+Tags: Named Theorems, Commutativity
+
+\begin{theorem}
+Let $\struct {S, \circ}$ be a [[Definition:Semigroup|semigroup]].
+Let $\family {a_k}_{1 \mathop \le k \mathop \le n}$ be a [[Definition:Finite Sequence|sequence]] of [[Definition:Element|elements]] of $S$.
+Suppose that:
+:$\forall i, j \in \closedint 1 n: a_i \circ a_j = a_j \circ a_i$
+Then for every [[Definition:Permutation|permutation]] $\sigma: \N_n \to \N_n$:
+:$a_{\map \sigma 1} \circ \cdots \circ a_{\map \sigma n} = a_1 \circ \cdots \circ a_n$
+where $\N_n$ is used here to denote the [[Definition:Initial Segment of One-Based Natural Numbers|initial segment]] of $\N_{>0}$:
+:$\N_n = \set {1, 2, \ldots, n}$
+\end{theorem}
+
+\begin{proof}
+The proof will proceed by the [[Principle of Mathematical Induction]] on $\N_{>0}$.
+Let $T$ be the [[Definition:Set|set]] of all $n \in \N_{>0}$ such that:
+:$a_{\map \sigma 1} \circ \cdots \circ a_{\map \sigma n} = a_1 \circ \cdots \circ a_n$
+holds for all [[Definition:Finite Sequence|sequences]] $\sequence {a_k}_{1 \mathop \le k \mathop \le n}$ of $n$ [[Definition:Element|elements]] of $S$ which satisfy:
+:$\forall i, j \in \closedint 1 n: a_i \circ a_j = a_j \circ a_i$
+for every [[Definition:Permutation|permutation]] $\sigma: \N_n \to \N_n$.
+=== Basis for the Induction ===
+We have that when $n = 1$:
+:$\map \sigma 1 = 1$
+and so:
+:$a_{\map \sigma 1} = a_1$
+for all $\sigma: \set 1 \to \set 1$.
+Thus $1 \in T$.
+This is our [[Definition:Basis for the Induction|basis for the induction]].
+=== Induction Hypothesis ===
+It is to be shown that, if $m \in T$ where $m \ge 1$, then it follows that $m + 1 \in T$.
+This is the [[Definition:Induction Hypothesis|induction hypothesis]]:
+:$a_{\map \sigma 1} \circ \cdots \circ a_{\map \sigma m} = a_1 \circ \cdots \circ a_m$
+holds for all [[Definition:Finite Sequence|sequences]] $\sequence {a_k}_{1 \mathop \le k \mathop \le m}$ of $m$ [[Definition:Element|elements]] of $S$ which satisfy:
+:$\forall i, j \in \closedint 1 m: a_i \circ a_j = a_j \circ a_i$
+for every [[Definition:Permutation|permutation]] $\sigma: \N^*_m \to \N^*_m$.
+It is to be demonstrated that it follows that:
+:$a_{\map \sigma 1} \circ \cdots \circ a_{\map \sigma {m + 1} } = a_1 \circ \cdots \circ a_{m + 1}$
+holds for all [[Definition:Finite Sequence|sequences]] $\sequence {a_k}_{1 \mathop \le k \mathop \le m + 1}$ of $m + 1$ [[Definition:Element|elements]] of $S$ which satisfy:
+:$\forall i, j \in \closedint 1 {m + 1}: a_i \circ a_j = a_j \circ a_i$
+for every [[Definition:Permutation|permutation]] $\sigma: \N_{m + 1} \to \N_{m + 1}$.
+=== Induction Step ===
+This is our [[Definition:Induction Step|induction step]]:
+Suppose $m \in T$.
+Let $\sequence {a_k}_{1 \le k \le m + 1}$ be a [[Definition:Finite Sequence|sequence]] of $m+1$ [[Definition:Element|elements]] of $S$ which satisfy:
+:$\forall i, j \in \closedint 1 {n + 1}: a_i \circ a_j = a_j \circ a_i$
+Let $\sigma: \N_{m + 1} \to \N_{m + 1}$ be a [[Definition:Permutation|permutation]] of $\closedint {1 {m + 1}$.
+There are three cases to consider:
+:$(1): \quad \map \sigma {m + 1} = m + 1$
+:$(2): \quad \map \sigma 1 = m + 1$
+:$(3): \quad \map \sigma r = m + 1$ for some $r \in \closedint 2 m$
+$(1): \quad$ Suppose $\map \sigma {m + 1} = m + 1$.
+Then the [[Definition:Restriction of Mapping|restriction]] of $\sigma$ to $\N_m$ is then a [[Definition:Permutation|permutation]] of $\N_m$.
+From the [[General Commutativity Theorem#Induction Hypothesis|induction hypothesis]]:
+:$m \in T$:
+Thus:
+:$a_{\map \sigma 1} \circ \cdots \circ a_{\map \sigma m} = a_1 \circ \cdots \circ a_m$
+from which:
+{{begin-eqn}}
+{{eqn | l = a_{\map \sigma 1} \circ \cdots \circ a_{\map \sigma {m + 1} }
+      | r = \paren {a_{\map \sigma 1} \circ \cdots \circ a_{\map \sigma m} } \circ a_{\map \sigma {m + 1} }
+      | c = 
+}}
+{{eqn | r = \paren {a_1 \circ \cdots \circ a_m} \circ a_{m + 1}
+      | c = 
+}}
+{{eqn | r = a_1 \circ \cdots \circ a_{m + 1}
+      | c = 
+}}
+{{end-eqn}}
+$(2): \quad$ Suppose $\map \sigma 1 = m + 1$.
+Let $\tau: \N^*_m \to \N^*_m$ be the mapping defined as:
+:$\forall k \in \closedint 1 m: \map \tau k = \map \sigma {k + 1}$
+From [[Closed Interval of Naturally Ordered Semigroup with Successor equals Union with Successor]]:
+:$\closedint 1 {m + 1} = \closedint 1 m \cup \set {m + 1}$
+Thus $\tau$ is clearly a [[Definition:Permutation|permutation]] on $\closedint 1 m$.
+Thus, as $m \in T$:
+:$a_{\map \tau 1} \circ \cdots \circ a_{\map \tau m} = a_1 \circ \cdots \circ a_m$
+So:
+{{begin-eqn}}
+{{eqn | l = a_{\map \sigma 1} \circ \cdots \circ a_{\map \sigma {m + 1} }
+      | r = a_{\map \sigma 1} \circ \paren {a_{\map \sigma 2} \circ \cdots \circ a_{\map \sigma {m + 1} } }
+      | c = 
+}}
+{{eqn | r = a_{m + 1} \circ \paren {a_{\map \tau 1} \circ \cdots \circ a_{\map \tau {m + 1} } }
+      | c = 
+}}
+{{eqn | r = a_{m + 1} \circ \paren {a_1 \circ \cdots \circ a_m}
+      | c = 
+}}
+{{eqn | r = \paren {a_1 \circ \cdots \circ a_m} \circ a_{m + 1}
+      | c = [[Element Commutes with Product of Commuting Elements]]
+}}
+{{eqn | r = a_1 \circ \cdots \circ a_{m + 1}
+      | c = 
+}}
+{{end-eqn}}
+$(3): \quad$ Suppose $\map \sigma r = m + 1$ for some $r \in \closedint 2 m$.
+Let $\tau: \N_{m + 1} \to \N_{m + 1}$ be a [[Definition:Mapping|mapping]] defined by:
+:$\forall k \in \N_{m + 1}: \map \tau k = \begin{cases}
+\map \sigma k & : k \in \closedint 1 {r - 1} \\
+\map \sigma {k + 1} & : k \in \closedint r m \\
+m + 1 & : k = m + 1 \end{cases}$
+It is seen that $\tau$ is a [[Definition:Permutation|permutation]] of $\N_{m + 1}$.
+So, by the first case:
+:$a_{\map \tau 1} \circ \cdots \circ a_{\map \tau {m + 1} } = a_1 \circ \cdots \circ a_{m + 1}$
+Thus:
+{{begin-eqn}}
+{{eqn | l = a_{\map \sigma 1} \circ \cdots \circ a_{\map \sigma {m + 1} }
+      | r = \paren {a_{\map \sigma 1} \circ \cdots \circ a_{\map \sigma {r - 1} } } \circ \paren {a_{\map \sigma r} \circ \paren {a_{\map \sigma {r + 1} } \circ \cdots \circ a_{\map \sigma {m + 1} } } }
+      | c = 
+}}
+{{eqn | r = \paren {a_{\map \tau 1} \circ \cdots \circ a_{\map \tau {r - 1} } } \circ \paren {a_{\map \tau {m + 1} } \circ \paren {a_{\map \tau r} \circ \cdots \circ a_{\map \tau m} } }
+      | c = 
+}}
+{{eqn | r = \paren {a_{\map \tau 1} \circ \cdots \circ a_{\map \tau {r - 1} } } \circ \paren {\paren {a_{\map \tau r} \circ \cdots \circ a_{\map \tau m} } \circ a_{\map \tau {m + 1} } }
+      | c = 
+}}
+{{eqn | r = a_{\map \tau 1} \circ \cdots \circ a_{\map \tau {m + 1} }
+      | c = 
+}}
+{{eqn | r = a_1 \circ \cdots \circ a_{m + 1}
+      | c = 
+}}
+{{end-eqn}}
+So in all cases, $m + 1 \in T$.
+Thus by the [[Principle of Mathematical Induction]]:
+:$T = \N_{>0}$
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{General Distributivity Theorem}
+Tags: Abstract Algebra, Summations, Named Theorems, Distributive Operations, General Distributivity Theorem
+
+\begin{theorem}
+:$\displaystyle \paren {\sum_{i \mathop = 1}^m a_i} * \paren {\sum_{j \mathop = 1}^n b_j} = \sum_{ {1 \mathop \le i \mathop \le m} \atop {1 \mathop \le j \mathop \le n} } \paren {a_i * b_j}$
+\end{theorem}
+
+\begin{proof}
+Proof by [[Principle of Mathematical Induction|induction]]:
+For all $n \in \Z_{> 0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
+:$\displaystyle \forall m \in \Z_{> 0}: \paren {\sum_{i \mathop = 1}^m a_i} * \paren {\sum_{j \mathop = 1}^n b_j} = \sum_{ {1 \mathop \le i \mathop \le m} \atop {1 \mathop \le j \mathop \le n} } \paren {a_i * b_j}$
+We have that $\struct {R, \circ, *}$ is a [[Definition:Ringoid (Abstract Algebra)|ringoid]], and so:
+:$\forall a, b, c \in S: \paren {a \circ b} * c = \paren {a * c} \circ \paren {b * c}$
+:$\forall a, b, c \in R: a * \paren {b \circ c} = \paren {a * b} \circ \paren {a * c}$
+=== Basis for the Induction ===
+$\map P 1$ is the case:
+:$\displaystyle \forall m \in \Z_{>0}: \paren {\sum_{i \mathop = 1}^m a_i} * b_1 = \sum_{1 \mathop \le i \mathop \le m} \paren {a_i * b_1}$
+This is demonstrated in [[General Distributivity Theorem/Lemma 1|Lemma 1]].
+This is our [[Definition:Basis for the Induction|basis for the induction]].
+=== Induction Hypothesis ===
+Now we need to show that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true.
+So this is our [[Definition:Induction Hypothesis|induction hypothesis]]:
+:$\displaystyle \forall m \in \Z_{> 0}: \paren {\sum_{i \mathop = 1}^m a_i} * \paren {\sum_{j \mathop = 1}^k b_j} = \sum_{ {1 \mathop \le i \mathop \le m} \atop {1 \mathop \le j \mathop \le k} } \paren {a_i * b_j}$
+Then we need to show:
+:$\displaystyle \forall m \in \Z_{> 0}: \paren {\sum_{i \mathop = 1}^m a_i} * \paren {\sum_{j \mathop = 1}^{k + 1} b_j} = \sum_{ {1 \mathop \le i \mathop \le m} \atop {1 \mathop \le j \mathop \le {k + 1} } } \paren {a_i * b_j}$
+=== Induction Step ===
+This is our [[Definition:Induction Step|induction step]]:
+{{begin-eqn}}
+{{eqn | l = \paren {\sum_{i \mathop = 1}^m a_i} * \paren {\sum_{j \mathop = 1}^{k + 1} b_j}
+      | r = \paren {\sum_{i \mathop = 1}^m a_i} * \paren {\paren {\sum_{j \mathop = 1}^k b_j} \circ b_{k + 1} }
+      | c = {{Defof|Summation}}
+}}
+{{eqn | r = \paren {\paren {\sum_{i \mathop = 1}^m a_i} * \paren {\sum_{j \mathop = 1}^k b_j} } \circ \paren {\paren {\sum_{i \mathop = 1}^m a_i} * b_{k + 1} }
+      | c = as $\struct {R, \circ, *}$ is a [[Definition:Ringoid (Abstract Algebra)|ringoid]]
+}}
+{{eqn | r = \paren {\paren {\sum_{i \mathop = 1}^m a_i} * \paren {\sum_{j \mathop = 1}^k b_j} } \circ \sum_{1 \mathop \le i \mathop \le m} \paren {a_i * b_{k + 1} }
+      | c = [[General Distributivity Theorem#Basis for the Induction|Basis for the Induction]]
+}}
+{{eqn | r = \sum_{ {1 \mathop \le i \mathop \le m} \atop {1 \mathop \le j \mathop \le k} } \paren {a_i * b_j} \circ \sum_{1 \mathop \le i \mathop \le m} \paren {a_i * b_{k + 1} }
+      | c = [[General Distributivity Theorem#Induction Hypothesis|Induction Hypothesis]]
+}}
+{{eqn | r = \sum_{ {1 \mathop \le i \mathop \le m} \atop {1 \mathop \le j \mathop \le {k + 1} } } \paren {a_i * b_j}
+      | c = [[Definition:Associative Operation|Associativity]] of $\circ$ in $\struct {R, \circ, *}$
+}}
+{{end-eqn}}
+So $\map P k \implies \map P {k + 1}$ and the result follows by the [[Principle of Mathematical Induction]].
+Therefore:
+:$\displaystyle \forall m, n \in \Z_{> 0}: \paren {\sum_{i \mathop = 1}^m a_i} * \paren {\sum_{j \mathop = 1}^n b_j} = \sum_{ {1 \mathop \le i \mathop \le m} \atop {1 \mathop \le j \mathop \le n} } \paren {a_i * b_j}$
+{{qed}}
+The same result can be obtained by fixing $n$ and using induction on $m$, which requires [[General Distributivity Theorem/Lemma 2|Lemma 2]] to be used for its [[Principle of Mathematical Induction#Basis for the Induction|base case]].
+\end{proof}<|endoftext|>
+\section{Associativity on Indexing Set}
+Tags: Indexed Families, Associativity
+
+\begin{theorem}
+Let $\left({S, \circ}\right)$ be a [[Definition:Commutative Semigroup|commutative semigroup]].
+Let $\left \langle {x_\alpha} \right \rangle_{\alpha \mathop \in A}$ be a [[Definition:Indexed Family|family of terms of $S$ indexed by]] a [[Definition:Finite Set|finite]] [[Definition:Non-Empty Set|non-empty set]] $A$.
+Let $\left \langle {B_k} \right \rangle_{1 \mathop \le k \mathop \le n}$ be a [[Definition:Family of Distinct Elements|family of distinct]] [[Definition:Subset|subsets]] of $A$ forming a [[Definition:Partition (Set Theory)|partition]] of $A$.
+Then:
+: $\displaystyle \prod_{k \mathop = 1}^n \left({\prod_{a \mathop \in B_k} x_\alpha}\right) = \prod_{\alpha \mathop \in A} x_\alpha$
+\end{theorem}
+
+\begin{proof}
+For each $k \in \N_{>0}$, let $\left|{B_k}\right| = p_k$.
+Let:
+:$r_0 = 0$
+:$\displaystyle \forall k \in \N_{>0}: r_k = \sum_{j \mathop = 1}^k {p_j}$
+and:
+:$p = r_n$
+Then:
+: $r_k - r_{k-1} = p_k$
+So, by [[Isomorphism to Closed Interval]], both $\left[{1 \,.\,.\, p_k}\right]$ and $\left[{r_{k - 1} + 1 \,.\,.\, r_k}\right]$ have $p_k$ elements.
+By [[Unique Isomorphism between Finite Totally Ordered Sets]], there is a unique [[Definition:Isomorphism (Abstract Algebra)|isomorphism]] $\tau_k: \left[{1 \,.\,.\, p_k}\right] \to \left[{r_{k - 1} + 1 \,.\,.\, r_k}\right]$ as both are [[Definition:Totally Ordered Set|totally ordered]].
+The [[Definition:Ordering|orderings]] on both of these are those induced by the ordering on $\N$.
+It is clear that $\tau_k$ is defined as:
+: $\forall j \in \left[{1 \,.\,.\, p_k}\right]: \tau_k \left({j}\right) = r_k + j$
+For each $k \in \left[{1 \,.\,.\, n}\right]$, let $\rho_k: \left[{1 \,.\,.\, p_k}\right] \to B_k$ be a [[Definition:Bijection|bijection]].
+By [[Strictly Increasing Sequence induces Partition]], the [[Definition:Mapping|mapping]] $\sigma: \left[{1 \,.\,.\, p}\right] \to A$ defined by:
+: $\displaystyle \forall j \in \left[{r_{k-1}+1 \,.\,.\, r_k}\right]: \forall k \in \left[{1 \,.\,.\, n}\right]: \sigma \left({j}\right) = \rho_k \left({\tau_k^{-1} \left({j}\right)}\right)$
+is a [[Definition:Bijection|bijection]].
+Let $\forall j \in \left[{1 \,.\,.\, p}\right]: y_j = x_{\sigma \left({j}\right)}$.
+By definition:
+: $\displaystyle \prod_{\alpha \mathop \in A} {x_\alpha} = \prod_{j \mathop = 1}^p {x_{\sigma \left({j}\right)}} = \prod_{j \mathop = 1}^p {y_j}$
+Also:
+: $\displaystyle \forall k \in \left[{1 \,.\,.\, n}\right]: \prod_{\alpha \mathop \in B_k} {x_\alpha} = \prod_{i \mathop = 1}^{p^k} {x_{\rho_k \left({i}\right)}}$
+Also by definition:
+: $\displaystyle \prod_{j \mathop = r_{k - 1} + 1}^{r_k} y_j = \prod_{i \mathop = 1}^{p_k} y_{\tau_k \left({i}\right)} = \prod_{i \mathop = 1}^{p_k} x_{\sigma \left({\tau_k \left({i}\right)}\right)} = \prod_{i \mathop = 1}^{p_k} x_{\rho_k \left({i}\right)}$
+So by the [[General Associativity Theorem]]:
+{{begin-eqn}}
+{{eqn | l = \prod_{a \mathop \in A} {x_a}
+      | r = \prod_{j \mathop = 1}^p y_j
+      | c = 
+}}
+{{eqn | r = \prod_{k \mathop = 1}^n \left({\prod_{j \mathop = r_{k - 1} \mathop + 1}^{r_k} {y_j} }\right)
+      | c = 
+}}
+{{eqn | r = \prod_{k \mathop = 1}^n \left({\prod_{i \mathop = 1}^{p_k} x_{\rho_k \left({i}\right)} }\right)
+      | c = 
+}}
+{{eqn | r = \prod_{k \mathop = 1}^n \left({\prod_{a \mathop \in B_k} x_\alpha}\right)
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{External Direct Product of Groups is Group}
+Tags: Group Theory, Direct Products, Group Direct Products
+
+\begin{theorem}
+Let $\struct {G_1, \circ_1}$ and $\struct {G_2, \circ_2}$ be [[Definition:Group|groups]] whose [[Definition:Identity Element|identity elements]] are $e_1$ and $e_2$ respectively.
+Let $\struct {G_1 \times G_2, \circ}$ be the [[Definition:Group Direct Product|external direct product]] of $G_1$ and $G_2$.
+Then $\struct {G_1 \times G_2, \circ}$ is a [[Definition:Group|group]] whose [[Definition:Identity Element|identity element]] is $\tuple {e_1, e_2}$.
+\end{theorem}
+
+\begin{proof}
+Taking the [[Definition:Group Axioms|group axioms]] in turn:
+=== G0: Closure ===
+From [[External Direct Product Closure]] it follows that $\struct {G_1 \times G_2, \circ}$ is [[Definition:Closed Algebraic Structure|closed]].
+{{qed|lemma}}
+=== G1: Associativity ===
+From [[External Direct Product Associativity]] it follows that $\circ$ is [[Definition:Associative Operation|associative]] on $G_1 \times G_2$.
+{{qed|lemma}}
+=== G2: Identity ===
+From [[External Direct Product Identity]] it follows that $\tuple {e_1, e_2}$ is the [[Definition:Identity Element|identity element]] of $\struct {G_1 \times G_2, \circ}$.
+{{qed|lemma}}
+=== G3: Inverses ===
+From [[External Direct Product Inverses]] it follows that $\tuple {g_1^{-1}, g_2^{-1} }$ is the [[Definition:Inverse Element|inverse element]] of $\tuple {g_1, g_2}$ in $\struct {G_1 \times G_2, \circ}$.
+{{qed|lemma}}
+All [[Definition:Group Axioms|group axioms]] are fulfilled, hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{External Direct Product of Projection with Canonical Injection}
+Tags: Abstract Algebra, Direct Products, Cartesian Product, Projections
+
+\begin{theorem}
+Let $\struct {S_1, \circ_1}$ and $\struct {S_2, \circ_2}$ be [[Definition:Algebraic Structure|algebraic structures]] with [[Definition:Identity Element|identity elements]] $e_1$ and $e_2$ respectively.
+Let $\struct {S_1 \times S_2, \circ}$ be the [[Definition:External Direct Product|external direct product]] of $\struct {S_1, \circ_1}$ and $\struct {S_2, \circ_2}$
+Let:
+:$\pr_1: \struct {S_1 \times S_2, \circ} \to \struct {S_1, \circ_1}$ be the [[Definition:First Projection|first projection]] from $\struct {S_1 \times S_2, \circ}$ to $\struct {S_1, \circ_1}$
+:$\pr_2: \struct {S_1 \times S_2, \circ} \to \struct {S_2, \circ_2}$ be the [[Definition:Second Projection|second projection]] from $\struct {S_1 \times S_2, \circ}$ to $\struct {S_2, \circ_2}$.
+Let:
+:$\inj_1: \struct {S_1, \circ_1} \to \struct {S_1 \times S_2, \circ}$ be the [[Definition:Canonical Injection (Abstract Algebra)|canonical injection]] from $\struct {S_1, \circ_1}$ to $\struct {S_1 \times S_2, \circ}$
+:$\inj_2: \struct {S_2, \circ_2} \to \struct {S_1 \times S_2, \circ}$ be the [[Definition:Canonical Injection (Abstract Algebra)|canonical injection]] from $\struct {S_2, \circ_2}$ to $\struct {S_1 \times S_2, \circ}$.
+Then:
+:$(1): \quad \pr_1 \circ \inj_1 = I_{S_1}$
+:$(2): \quad \pr_2 \circ \inj_2 = I_{S_2}$
+where $I_{S_1}$ and $I_{S_2}$ are the [[Definition:Identity Mapping|identity mappings]] on $S_1$ and $S_2$ respectively.
+\end{theorem}
+
+\begin{proof}
+Let $\tuple {s_1, s_2} \in S_1 \times S_2$.
+So, $s_1 \in S_1$ and $s_2 \in S_2$.
+From the definition of the [[Definition:Canonical Injection (Abstract Algebra)|canonical injection]], we have:
+:$\map {\inj_1} {s_1} = \tuple {s_1, e_2}$
+:$\map {\inj_2} {s_2} = \tuple {e_1, s_2}$
+From the definition of the [[Definition:First Projection|first projection]]:
+:$\map {\pr_1} {s_1, e_2} = s_1$
+and similarly from the definition of the [[Definition:Second Projection|second projection]]:
+:$\map {\pr_2} {e_1, s_2} = s_2$
+Thus:
+:$\map {\pr_1 \circ \inj_1} {s_1} = s_1$
+:$\map {\pr_2 \circ \inj_2} {s_2} = s_2$
+and the result follows from the definition of the [[Definition:Identity Mapping|identity mapping]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Inverse Completion of Natural Numbers}
+Tags: Inverse Completions, Natural Numbers
+
+\begin{theorem}
+There exists an [[Definition:Inverse Completion|inverse completion]] of the [[Definition:Natural Numbers|natural numbers]] under [[Definition:Natural Number Addition|addition]].
+\end{theorem}
+
+\begin{proof}
+The set of [[Definition:Natural Numbers|natural numbers]] under [[Definition:Natural Number Addition|addition]] can be denoted $\left ({\N, +}\right)$.
+From [[Natural Numbers under Addition form Commutative Monoid]], the [[Definition:Algebraic Structure|algebraic structure]] $\left ({\N, +}\right)$ is a [[Definition:Commutative Monoid|commutative monoid]].
+Therefore by definition of [[Definition:Commutative Monoid|commutative monoid]], $\left ({\N, +}\right)$ is a [[Definition:Commutative Semigroup|commutative semigroup]].
+From [[Natural Number Addition is Cancellable]], all of the elements of $\left ({\N, +}\right)$ are [[Definition:Cancellable Element|cancellable]].
+The result follows from the [[Inverse Completion Theorem]].
+{{qed}}
+[[Category:Inverse Completions]]
+[[Category:Natural Numbers]]
+q7fkui5qs8edlrc74tqw2t2cnac9r61
+\end{proof}<|endoftext|>
+\section{L'Hôpital's Rule}
+Tags: Differential Calculus, Limits of Functions, L'Hôpital's Rule
+
+\begin{theorem}
+Let $f$ and $g$ be [[Definition:Real Function|real functions]] which are [[Definition:Continuous on Interval|continuous]] on the [[Definition:Closed Real Interval|closed interval]] $\closedint a b$ and [[Definition:Differentiable on Interval|differentiable]] on the [[Definition:Open Real Interval|open interval]] $\openint a b$.
+Let:
+:$\forall x \in \openint a b: \map {g'} x \ne 0$
+where $g'$ denotes the [[Definition:Derivative|derivative]] of $g$ {{WRT|Differentiation}} $x$.
+Let:
+:$\map f a = \map g a = 0$
+Then:
+:$\displaystyle \lim_{x \mathop \to a^+} \frac {\map f x} {\map g x} = \lim_{x \mathop \to a^+} \frac {\map {f'} x} {\map {g'} x}$
+provided that the second limit exists.
+\end{theorem}
+
+\begin{proof}
+Let $l = \displaystyle \lim_{x \mathop \to a^+} \frac{f' \left({x}\right)}{g' \left({x}\right)}$.
+Let $\epsilon > 0$.
+By the definition of [[Definition:Limit of Real Function|limit]], we ought to find a $\delta > 0$ such that:
+:$\forall x: \left\vert{x - a}\right\vert < \delta \implies \left\vert{\dfrac {f \left({x}\right)} {g \left({x}\right)} - l}\right\vert < \epsilon$
+Fix $\delta$ such that:
+:$\forall x: \left\vert{x - a}\right\vert < \delta \implies \left\vert{\dfrac {f' \left({x}\right)} {g' \left({x}\right)} - l}\right\vert < \epsilon$
+which is possible by the definition of [[Definition:Limit of Real Function|limit]].
+Let $x$ be such that $\left\vert{x - a}\right\vert < \delta$.
+By the [[Cauchy Mean Value Theorem]] with $b = x$:
+: $\exists \xi \in \left({a \,.\,.\, x}\right): \dfrac {f' \left({\xi}\right)} {g' \left({\xi}\right)} = \dfrac {f \left({x}\right) - f \left({a}\right)} {g \left({x}\right) - g \left({a}\right)}$
+Since $f \left({a}\right) = g \left({a}\right) = 0$, we have:
+: $\exists \xi \in \left({a \,.\,.\, x}\right): \dfrac {f' \left({\xi}\right)} {g' \left({\xi}\right)} = \dfrac {f \left({x}\right)} {g \left({x}\right)}$
+Now, as $a < \xi < x$, it follows that $\left\vert{\xi - a}\right\vert < \delta$ as well.
+Therefore:
+:$\left\vert{\dfrac {f \left({x}\right)} {g \left({x}\right)} - l }\right\vert = \left\vert{ \dfrac {f' \left({\xi}\right)} {g' \left({\xi}\right)} - l}\right\vert < \epsilon$
+which leads us to the desired conclusion that:
+:$\displaystyle \lim_{x \mathop \to a^+} \frac {f \left({x}\right)} {g \left({x}\right)} = \lim_{x \mathop \to a^+} \frac {f' \left({x}\right)} {g' \left({x}\right)}$
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Take the [[Cauchy Mean Value Theorem]] with $b = x$:
+:$\exists \xi \in \openint a x: \dfrac {\map {f'} \xi} {\map {g'} \xi} = \dfrac {\map f x - \map f a} {\map g x - \map g a}$
+Then if $\map f a = \map g a = 0$ we have:
+:$\exists \xi \in \openint a x: \dfrac {\map {f'} \xi} {\map {g'} \xi} = \dfrac {\map f x} {\map g x}$
+Note that $\xi$ depends on $x$; that is, different values of $x$ may require different values of $\xi$ to make the above statement valid.
+It follows from [[Limit of Function in Interval]] that $\xi \to a$ as $x \to a$.
+Also, $\xi \ne a$ when $x > a$.
+So from '''Hypothesis $2$''' of [[Limit of Composite Function]], it follows that:
+:$\displaystyle \lim_{x \mathop \to a^+} \dfrac {\map {f'} \xi} {\map {g'} \xi} = \lim_{x \mathop \to a^+} \dfrac {\map {f'} x} {\map {g'} x}$
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Natural Numbers under Addition form Commutative Monoid}
+Tags: Natural Numbers, Examples of Monoids
+
+\begin{theorem}
+The [[Definition:Algebraic Structure|algebraic structure]] $\struct {\N, +}$ consisting of the [[Definition:Set|set]] of [[Definition:Natural Numbers|natural numbers]] $\N$ under [[Definition:Natural Number Addition|addition]] $+$ is a [[Definition:Commutative Monoid|commutative monoid]] whose [[Definition:Identity Element|identity]] is [[Definition:Zero (Number)|zero]].
+\end{theorem}
+
+\begin{proof}
+Consider the [[Definition:Natural Numbers|natural numbers]] $\N$ defined as the [[Definition:Naturally Ordered Semigroup|naturally ordered semigroup]].
+From the definition of the [[Definition:Naturally Ordered Semigroup|naturally ordered semigroup]], it follows that $\struct {\N, +}$ is a [[Definition:Commutative Semigroup|commutative semigroup]].
+From the definition of [[Definition:Zero (Number)|zero]], $\struct {\N, +}$ has $0 \in \N$ as the [[Definition:Identity Element|identity]], hence is a [[Definition:Monoid|monoid]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Natural Numbers form Commutative Semiring}
+Tags: Natural Numbers, Semirings
+
+\begin{theorem}
+The [[Definition:Semiring of Natural Numbers|semiring of natural numbers]] $\struct {\N, +, \times}$ forms a [[Definition:Commutative Semiring|commutative semiring]].
+\end{theorem}
+
+\begin{proof}
+The [[Definition:Algebraic Structure|algebraic structure]] $\struct {\N, +}$ is a [[Definition:Commutative Monoid|commutative monoid]] from [[Natural Numbers under Addition form Commutative Monoid]].
+Then we have:
+:$(1): \quad$ $\times$ is [[Definition:Closed Operation|closed]]
+:$(2): \quad$ [[Natural Number Multiplication is Associative]]
+:$(3): \quad$ [[Natural Number Multiplication is Commutative]]
+:$(4): \quad$ [[Natural Number Multiplication Distributes over Addition]]
+Thus $\struct {\N, \times}$ forms an [[Definition:Algebraic Structure|algebraic structure]] which is [[Definition:Closure (Abstract Algebra)|closed]] such that $\times$ is [[Definition:Associative Operation|associative]] and [[Definition:Commutative Operation|commutative]].
+So by definition, $\struct {\N, \times}$ is a [[Definition:Commutative Semigroup|commutative semigroup]].
+The result follows from definition of [[Definition:Commutative Semiring|commutative semiring]] and the [[Definition:Distributive Operation|distributivity]] of $\times$ over $+$.
+{{qed}}
+[[Category:Natural Numbers]]
+[[Category:Semirings]]
+3dw345e0hzrnc5uqio3ha4k7kqjsxnm
+\end{proof}<|endoftext|>
+\section{Integer Multiplication is Well-Defined}
+Tags: Integer Multiplication, Integer Multiplication is Well-Defined
+
+\begin{theorem}
+[[Definition:Integer Multiplication|Integer multiplication]] is [[Definition:Well-Defined Operation|well-defined]].
+\end{theorem}
+
+\begin{proof}
+Consider the [[Definition:Integer/Formal Definition|formal definition of the integers]]: $x = \eqclass {a, b} {}$ is an [[Definition:Equivalence Class|equivalence class]] of [[Definition:Ordered Pair|ordered pairs]] of [[Definition:Natural Numbers|natural numbers]].
+Consider the [[Definition:Mapping|mapping]] $\phi: \N_{>0} \to \Z_{>0}$ defined as:
+:$\forall u \in \N_{>0}: \map \phi u = u'$
+where $u' \in \Z$ be the [[Definition:Strictly Positive Integer|(strictly) positive integer]] $\eqclass {b + u, b} {}$.
+Let $v' = \eqclass {c + v, c} {}$.
+Then:
+{{begin-eqn}}
+{{eqn | l = u' v'
+      | r = \eqclass {b + u, b} {} \times \eqclass {c + v, c} {}
+      | c = 
+}}
+{{eqn | r = \eqclass {\paren {b + u} \paren {c + v} + b c, \paren {b + u} c + b \paren {c + v} } {}
+      | c = 
+}}
+{{eqn | r = \eqclass {b c + b v + c u + u v + b c, b c + u c + b c + b v} {}
+      | c = 
+}}
+{{eqn | r = \eqclass {b c + u v, b c} {}
+      | c = 
+}}
+{{eqn | r = \eqclass {b + u v, b} {}
+      | c = 
+}}
+{{eqn | r = \paren {u v}'
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Integer Multiplication is Closed}
+Tags: Integer Multiplication, Algebraic Closure
+
+\begin{theorem}
+The [[Definition:Set|set]] of [[Definition:Integer|integers]] is [[Definition:Closed Algebraic Structure|closed]] under [[Definition:Integer Multiplication|multiplication]]:
+:$\forall a, b \in \Z: a \times b \in \Z$
+\end{theorem}
+
+\begin{proof}
+Let us define $\eqclass {\tuple {a, b} } \boxminus$ as in the [[Definition:Integer/Formal Definition|formal definition of integers]].
+That is, $\eqclass {\tuple {a, b} } \boxminus$ is an [[Definition:Equivalence Class|equivalence class]] of [[Definition:Ordered Pair|ordered pairs]] of [[Definition:Natural Numbers|natural numbers]] under the [[Definition:Congruence Relation|congruence relation]] $\boxminus$.
+$\boxminus$ is the [[Definition:Congruence Relation|congruence relation]] defined on $\N \times \N$ by:
+:$\tuple {x_1, y_1} \boxminus \tuple {x_2, y_2} \iff x_1 + y_2 = x_2 + y_1$
+In order to streamline the notation, we will use $\eqclass {a, b} {}$ to mean $\eqclass {\tuple {a, b} } \boxminus$, [[Definition:Integer/Formal Definition/Notation|as suggested]].
+[[Definition:Integer Multiplication|Integer multiplication]] is defined as:
+:$\forall a, b, c, d \in \N: \eqclass {a, b} {} \times \eqclass {c, d} {} = \eqclass {a c + b d, a d + b c} {}$
+We have that:
+:$\forall a, b, c, d \in \N: \eqclass {a, b} {} \in \Z, \eqclass {c, d} {} \in \Z$
+Also:
+:$\forall a, b, c, d \in \N: \eqclass {a, b} {} \times \eqclass {c, d} {} = \eqclass {a c + b d, a d + b c} {}$
+But:
+:$a c + b d \in \N, a d + b c \in \N$
+So:
+:$\forall a, b, c, d \in \N: \eqclass {a c + b d, a d + b c} {} \in \Z$
+Therefore [[Definition:Integer Multiplication|integer multiplication]] is [[Definition:Closed Algebraic Structure|closed]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Integer Multiplication is Commutative}
+Tags: Integer Multiplication, Commutativity
+
+\begin{theorem}
+The operation of [[Definition:Integer Multiplication|multiplication]] on the [[Definition:Set|set]] of [[Definition:Integer|integers]] $\Z$ is [[Definition:Commutative Operation|commutative]]:
+:$\forall x, y \in \Z: x \times y = y \times x$
+\end{theorem}
+
+\begin{proof}
+From the [[Definition:Integer/Formal Definition|formal definition of integers]], $\eqclass {a, b} {}$ is an [[Definition:Equivalence Class|equivalence class]] of [[Definition:Ordered Pair|ordered pairs]] of [[Definition:Natural Numbers|natural numbers]].
+Let $x = \eqclass {a, b} {}$ and $y = \eqclass {c, d} {}$ for some $x, y \in \Z$.
+Then:
+{{begin-eqn}}
+{{eqn | l = x \times y
+      | r = \eqclass {a, b} {}\times \eqclass {c, d} {}
+      | c = {{Defof|Integer|subdef = Formal Definition}}
+}}
+{{eqn | r = \eqclass {a c + b d, a d + b c} {}
+      | c = {{Defof|Integer Multiplication}}
+}}
+{{eqn | r = \eqclass {c a + d b, d a + c b} {}
+      | c = [[Natural Number Multiplication is Commutative]]
+}}
+{{eqn | r = \eqclass {c a + d b, c b + d a} {}
+      | c = [[Natural Number Addition is Commutative]]
+}}
+{{eqn | r = \eqclass {c, d} {} \times \eqclass {a, b} {}
+      | c = {{Defof|Integer Multiplication}}
+}}
+{{eqn | r = y \times x
+      | c = {{Defof|Integer|subdef = Formal Definition}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Integer Multiplication is Associative}
+Tags: Integer Multiplication, Associativity
+
+\begin{theorem}
+The operation of [[Definition:Integer Multiplication|multiplication]] on the [[Definition:Set|set]] of [[Definition:Integer|integers]] $\Z$ is [[Definition:Associative Operation|associative]]:
+:$\forall x, y, z \in \Z: x \times \paren {y \times z} = \paren {x \times y} \times z$
+\end{theorem}
+
+\begin{proof}
+From the [[Definition:Integer/Formal Definition|formal definition of integers]], $\eqclass {a, b} {}$ is an [[Definition:Equivalence Class|equivalence class]] of [[Definition:Ordered Pair|ordered pairs]] of [[Definition:Natural Numbers|natural numbers]].
+Let $x = \eqclass {a, b} {}$, $y = \eqclass {c, d} {}$ and $z = \eqclass {e, f} {}$ for some $x, y, z \in \Z$.
+Then:
+{{begin-eqn}}
+{{eqn | l = x \times \paren {y \times z}
+      | r = \eqclass {a, b} {} \times \paren {\eqclass {c, d} {} \times \eqclass {e, f} {} }
+      | c = {{Defof|Integer|subdef = Formal Definition}}
+}}
+{{eqn | r = \eqclass {a, b} {} \times \eqclass {c e + d f, c f + d e} {}
+      | c = {{Defof|Integer Multiplication}}
+}}
+{{eqn | r = \eqclass {a \paren {c e + d f} + b \paren {c f + d e}, a \paren {c f + d e} + b \paren {c e + d f} } {}
+      | c = {{Defof|Integer Multiplication}}
+}}
+{{eqn | r = \eqclass {a c e + a d f + b c f + b d e, a c f + a d e + b c e + b d f} {}
+      | c = [[Natural Number Multiplication Distributes over Addition]]
+}}
+{{eqn | r = \eqclass {a c e + b d e + a d f + b c f, a c f + b d f + a d e + b c e} {}
+      | c = [[Natural Number Addition is Commutative]] and [[Natural Number Addition is Associative|Associative]]
+}}
+{{eqn | r = \eqclass {\paren {a c + b d} e + \paren {a d + b c} f, \paren {a c + b d} f + \paren {a d + b c} e} {}
+      | c = [[Natural Number Multiplication Distributes over Addition]]
+}}
+{{eqn | r = \eqclass {a c + b d, a d + b c} {} \times \eqclass {e, f} {}
+      | c = {{Defof|Integer Multiplication}}
+}}
+{{eqn | r = \paren {\eqclass {a, b} {} \times \eqclass {c, d} {} } \times \eqclass {e, f} {}
+      | c = {{Defof|Integer Multiplication}}
+}}
+{{eqn | r = \paren {x \times y} \times z
+      | c = {{Defof|Integer|subdef = Formal Definition}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Integer Multiplication Distributes over Addition}
+Tags: Integer Addition, Integer Multiplication, Distributive Operations
+
+\begin{theorem}
+The operation of [[Definition:Integer Multiplication|multiplication]] on the [[Definition:Set|set]] of [[Definition:Integer|integers]] $\Z$ is [[Definition:Distributive Operation|distributive]] over [[Definition:Integer Addition|addition]]:
+:$\forall x, y, z \in \Z: x \times \paren {y + z} = \paren {x \times y} + \paren {x \times z}$
+:$\forall x, y, z \in \Z: \paren {y + z} \times x = \paren {y \times x} + \paren {z \times x}$
+\end{theorem}
+
+\begin{proof}
+Let us define $\Z$ as in the [[Definition:Integer/Formal Definition|formal definition of integers]].
+That is, $\Z$ is an [[Definition:Inverse Completion|inverse completion]] of $\N$.
+From [[Natural Numbers form Commutative Semiring]], we have that:
+:All [[Definition:Element|elements]] of $\N$ are [[Definition:Cancellable Element|cancellable]] for [[Definition:Natural Number Addition|addition]]
+:[[Definition:Natural Number Addition|Addition]] and [[Definition:Natural Number Multiplication|multiplication]] are [[Definition:Commutative Operation|commutative]] and [[Definition:Associative Operation|associative]] on the [[Definition:Natural Numbers|natural numbers]] $\N$
+:[[Definition:Natural Number Multiplication|Natural number multiplication]] is [[Definition:Distributive Operation|distributive]] over [[Definition:Natural Number Addition|natural number addition]].
+The result follows from the [[Extension Theorem for Distributive Operations]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Construction of Inverse Completion}
+Tags: Inverse Completions
+
+\begin{theorem}
+This page consists of a series of linked theorems, each of which builds towards one result.
+To access the proofs for the individual theorems, click on the links which form the titles of each major section.
+\end{theorem}<|endoftext|>
+\section{Integer Multiplication Identity is One}
+Tags: Integer Multiplication
+
+\begin{theorem}
+The [[Definition:Identity Element|identity]] of [[Definition:Integer Multiplication|integer multiplication]] is $1$:
+:$\exists 1 \in \Z: \forall a \in \Z: a \times 1 = a = 1 \times a$
+\end{theorem}
+
+\begin{proof}
+Let us define $\eqclass {\tuple {a, b} } \boxtimes$ as in the [[Definition:Integer/Formal Definition|formal definition of integers]].
+That is, $\eqclass {\tuple {a, b} } \boxtimes$ is an [[Definition:Equivalence Class|equivalence class]] of [[Definition:Ordered Pair|ordered pairs]] of [[Definition:Natural Numbers|natural numbers]] under the [[Definition:Congruence Relation|congruence relation]] $\boxtimes$.
+$\boxtimes$ is the [[Definition:Congruence Relation|congruence relation]] defined on $\N \times \N$ by $\tuple {x_1, y_1} \boxtimes \tuple {x_2, y_2} \iff x_1 + y_2 = x_2 + y_1$.
+In order to streamline the notation, we will use $\eqclass {a, b} {}$ to mean $\eqclass {\tuple {a, b} } \boxtimes$, [[Definition:Integer/Formal Definition/Notation|as suggested]].
+From [[Construction of Inverse Completion|the method of construction]], $\eqclass {c + 1, c} {}$, where $c$ is any element of the [[Definition:Natural Numbers|natural numbers]] $\N$, is [[Definition:Isomorphic Copy|the isomorphic copy]] of $1 \in \N$.
+To ease the algebra, we will take $\eqclass {1, 0} {}$ as a canonical instance of this [[Definition:Equivalence Class|equivalence class]].
+Thus it is to be shown that:
+:$\forall a, b \in \N: \eqclass {a, b} {} \times \eqclass {1, 0} {} = \eqclass {a, b} {} = \eqclass {1, 0} {} \times \eqclass {a, b} {}$
+From [[Natural Numbers form Commutative Semiring]], we take it for granted that:
+:[[Definition:Natural Number Addition|addition]] and [[Definition:Natural Number Multiplication|multiplication]] are [[Definition:Commutative Operation|commutative]] and [[Definition:Associative Operation|associative]] on the [[Definition:Natural Numbers|natural numbers]] $\N$
+:[[Definition:Natural Number Multiplication|natural number multiplication]] is [[Definition:Distributive Operation|distributive]] over [[Definition:Natural Number Addition|natural number addition]].
+So:
+{{begin-eqn}}
+{{eqn | l = \eqclass {a, b} {} \times \eqclass {1, 0} {}
+      | r = \eqclass {1 \times a + 0 \times b, 0 \times a + 1 \times b} {}
+      | c = 
+}}
+{{eqn | r = \eqclass {a, b} {}
+      | c = from [[Construction of Inverse Completion/Equivalence Relation/Members of Equivalence Classes|Construction of Inverse Completion: Members of Equivalence Classes]]
+}}
+{{end-eqn}}
+So:
+:$\eqclass {a, b} {} \times \eqclass {1, 0} {} = \eqclass {a, b} {}$
+The identity $\eqclass {a, b} {} = \eqclass {1, 0} {} \times \eqclass {a, b} {}$ is demonstrated similarly.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Integer Multiplication has Zero}
+Tags: Integer Multiplication
+
+\begin{theorem}
+The [[Definition:Set|set]] of [[Definition:Integer|integers]] under [[Definition:Integer Multiplication|multiplication]] $\struct {\Z, \times}$ has a [[Definition:Zero Element|zero element]], which is $0$.
+\end{theorem}
+
+\begin{proof}
+Let us define $\eqclass {\tuple {a, b} } \boxminus$ as in the [[Definition:Integer/Formal Definition|formal definition of integers]].
+That is, $\eqclass {\tuple {a, b} } \boxminus$ is an [[Definition:Equivalence Class|equivalence class]] of [[Definition:Ordered Pair|ordered pairs]] of [[Definition:Natural Numbers|natural numbers]] under the [[Definition:Congruence Relation|congruence relation]] $\boxminus$.
+$\boxminus$ is the [[Definition:Congruence Relation|congruence relation]] defined on $\N \times \N$ by $\tuple {x_1, y_1} \boxminus \tuple {x_2, y_2} \iff x_1 + y_2 = x_2 + y_1$.
+In order to streamline the notation, we will use $\eqclass {a, b} {}$ to mean $\eqclass {\tuple {a, b} } \boxminus$, [[Definition:Integer/Formal Definition/Notation|as suggested]].
+From [[Construction of Inverse Completion/Identity of Quotient Structure|the method of construction]], $\eqclass {c, c} {}$, where $c$ is any element of the [[Definition:Natural Numbers|natural numbers]] $\N$, is the [[Definition:Identity (Abstract Algebra)|identity]] of $\struct {\Z, +}$.
+To ease the algebra, we will take $\eqclass {0, 0} {}$ as a canonical instance of this equivalence class.
+We need to show that:
+$\forall a, b, c \in \N: \eqclass {a, b} {} \times \eqclass {0, 0} {} = \eqclass {0, 0} {} = \eqclass {0, 0} {} \times \eqclass {a, b} {}$.
+From [[Natural Numbers form Commutative Semiring]], we can take it for granted that [[Definition:Natural Number Addition|addition]] and [[Definition:Natural Number Multiplication|multiplication]] are [[Definition:Commutative Operation|commutative]] on the [[Definition:Natural Numbers|natural numbers]] $\N$.
+{{begin-eqn}}
+{{eqn | l = \eqclass {a, b} {} \times \eqclass {0, 0} {}
+      | r = \eqclass {a \times 0 + b \times 0, a \times 0 + b \times 0} {}
+      | c = 
+}}
+{{eqn | r = \eqclass {0, 0} {}
+      | c = [[Construction of Inverse Completion/Equivalence Relation/Equivalence Class of Equal Elements|Construction of Inverse Completion: Equivalence Class of Equal Elements]]
+}}
+{{eqn | r = \eqclass {0 \times a + 0 \times b, 0 \times a + 0 \times b} {}
+      | c = 
+}}
+{{eqn | r = \eqclass {0, 0} {} \times \eqclass {a, b} {}
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Ring of Integers has no Zero Divisors}
+Tags: Integers
+
+\begin{theorem}
+The [[Definition:Integer|integers]] have no [[Definition:Zero Divisor of Ring|zero divisors]]:
+:$\forall x, y, \in \Z: x \times y = 0 \implies x = 0 \lor y = 0$
+\end{theorem}
+
+\begin{proof}
+Let us define $\eqclass {\tuple {a, b} } \boxminus$ as in the [[Definition:Integer/Formal Definition|formal definition of integers]].
+That is, $\eqclass {\tuple {a, b} } \boxminus$ is an [[Definition:Equivalence Class|equivalence class]] of [[Definition:Ordered Pair|ordered pairs]] of [[Definition:Natural Numbers|natural numbers]] under the [[Definition:Congruence Relation|congruence relation]] $\boxminus$.
+$\boxminus$ is the [[Definition:Congruence Relation|congruence relation]] defined on $\N \times \N$ by $\tuple {x_1, y_1} \boxminus \tuple {x_2, y_2} \iff x_1 + y_2 = x_2 + y_1$.
+In order to streamline the notation, we will use $\eqclass {a, b} {}$ to mean $\eqclass {\tuple {a, b} } \boxminus$, [[Definition:Integer/Formal Definition/Notation|as suggested]].
+From [[Construction of Inverse Completion/Identity of Quotient Structure|the method of construction]], $\eqclass {c, c} {}$, where $c$ is any element of the [[Definition:Natural Numbers|natural numbers]] $\N$, is the [[Definition:Identity Element|identity]] of $\struct {\Z, +}$.
+To ease the algebra, we will take $\eqclass {0, 0} {}$ as a canonical instance of this [[Definition:Equivalence Class|equivalence class]].
+We need to show that:
+:$\forall a, b, c, d \in \N: \eqclass {a, b} {} \times \eqclass {c, d} {} = \eqclass {0, 0} {} \implies \eqclass {a, b} {} = \eqclass {0, 0} {} \lor \eqclass {c, d} {} = \eqclass {0, 0} {}$
+From [[Natural Numbers form Commutative Semiring]], we can take it for granted that:
+:[[Definition:Natural Number Addition|addition]] and [[Definition:Natural Number Multiplication|multiplication]] are [[Definition:Commutative Operation|commutative]] and [[Definition:Associative Operation|associative]] on the [[Definition:Natural Numbers|natural numbers]] $\N$
+:[[Definition:Natural Number Multiplication|natural number multiplication]] is [[Definition:Distributive Operation|distributive]] over [[Definition:Natural Number Addition|natural number addition]].
+So:
+{{begin-eqn}}
+{{eqn | l = \eqclass {a, b} {} \times \eqclass {c, d} {}
+      | r = \eqclass {0, 0} {}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \eqclass {a c + b d, a d + b c} {}
+      | r = \eqclass {0, 0} {}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = a c + b d + 0
+      | r = a d + b c + 0
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = a c + b d
+      | r = a d + b c
+      | c = 
+}}
+{{end-eqn}}
+We have to be careful here, and bear in mind that $a, b, c, d$ are [[Definition:Natural Numbers|natural numbers]], and we have not defined (and, at this stage, will not define) subtraction on such entities.
+{{WLOG}}, suppose that $\eqclass {c, d} {} \ne \eqclass {0, 0} {}$.
+Then $c \ne d$.
+{{WLOG}}, suppose also that $c > d$.
+{{wtd|[[Ordering in terms of Addition]] should have a version for $\N$}}
+From [[Ordering in terms of Addition]], $\exists p \in \N: d + p = c$ where $p > 0$.
+Then:
+{{begin-eqn}}
+{{eqn | l = a c + b d
+      | r = a d + b c
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = a \paren {d + p} + b d
+      | r = a d + b \paren {d + p}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = a d + a p + b d
+      | r = a d + b d + b p
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = a p
+      | r = b p
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = a
+      | r = b
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \eqclass {a, b} {}
+      | r = \eqclass {0, 0} {}
+      | c = [[Construction of Inverse Completion/Equivalence Relation/Equivalence Class of Equal Elements|Construction of Inverse Completion: Equivalence Class of Equal Elements]]
+}}
+{{end-eqn}}
+Similarly for when $c < d$.
+Thus:
+:$\eqclass {c, d} {} \ne \eqclass {0, 0} {} \implies \eqclass {a, b} {} = \eqclass {0, 0} {}$
+A similar argument shows that:
+:$\eqclass {a, b} {} \ne \eqclass {0, 0} {} \implies \eqclass {c, d} {} = \eqclass {0, 0} {}$
+The equivalence between the two forms of the statement of this theorem follows from [[De Morgan's Laws (Logic)/Conjunction of Negations|De Morgan's Laws: Conjunction of Negations]] and the [[Rule of Transposition]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Integers form Integral Domain}
+Tags: Integers, Integral Domains
+
+\begin{theorem}
+The [[Definition:Integer|integers]] $\Z$ form an [[Definition:Integral Domain|integral domain]] under [[Definition:Integer Addition|addition]] and [[Definition:Integer Multiplication|multiplication]].
+\end{theorem}
+
+\begin{proof}
+First we note that the [[Integers form Commutative Ring with Unity|integers form a commutative ring with unity]] whose [[Integer Multiplication has Zero|zero is $0$]] and whose [[Integer Multiplication Identity is One|unity is $1$]].
+Next we see that the [[Ring of Integers has no Zero Divisors|$\struct {\Z, +, \times}$ has no divisors of zero]].
+So, by definition, the [[Definition:Algebraic Structure|algebraic structure]] $\struct {\Z, +, \times}$ is an [[Definition:Integral Domain|integral domain]] whose [[Definition:Ring Zero|zero]] is $0$ and whose [[Definition:Unity of Ring|unity]] is $1$.
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Positive Integer is Well-Defined}
+Tags: Integers
+
+\begin{theorem}
+"[[Definition:Positive Integer|Positive]]" as applied to an [[Definition:Integer|integer]] is [[Definition:Well-Defined Relation|well-defined]].
+\end{theorem}
+
+\begin{proof}
+Let us define $\eqclass {\tuple {a, b} } \boxminus$ as in the [[Definition:Integer/Formal Definition|formal definition of integers]].
+That is, $\eqclass {\tuple {a, b} } \boxminus$ is an [[Definition:Equivalence Class|equivalence class]] of [[Definition:Ordered Pair|ordered pairs]] of [[Definition:Natural Numbers|natural numbers]] under the [[Definition:Congruence Relation|congruence relation]] $\boxminus$.
+$\boxminus$ is the [[Definition:Congruence Relation|congruence relation]] defined on $\N \times \N$ by $\tuple {x_1, y_1} \boxminus \tuple {x_2, y_2} \iff x_1 + y_2 = x_2 + y_1$.
+In order to streamline the notation, we will use $\eqclass {a, b} {}$ to mean $\eqclass {\tuple {a, b} } \boxminus$, [[Definition:Integer/Formal Definition/Notation|as suggested]].
+Thus, what we are trying to prove is:
+:$\eqclass {a, b} {} = \eqclass {c, d} {} \land b < a \implies d < c$
+By definition:
+:$\eqclass {a, b} {} = \eqclass {c, d} {} \iff a + d = b + c$
+So:
+{{begin-eqn}}
+{{eqn | l = b
+      | o = <
+      | r = a
+}}
+{{eqn | ll= \leadsto
+      | l = \exists p \in \N: a
+      | r = b + p
+}}
+{{eqn | ll= \leadsto
+      | l = b + p + d
+      | r = b + c
+}}
+{{eqn | ll= \leadsto
+      | l = p + d
+      | r = c
+}}
+{{eqn | ll= \leadsto
+      | l = d
+      | o = <
+      | r = c
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Integers]]
+tfoqxrhyoxppc1d4o3kos1sbvkcow8n
+\end{proof}<|endoftext|>
+\section{Natural Numbers are Non-Negative Integers}
+Tags: Integers, Natural Numbers, Natural Numbers are Non-Negative Integers
+
+\begin{theorem}
+Let $m \in \Z$. Then:
+:$(1): \quad m \in \N \iff 0 \le m$
+:$(2): \quad m \in \N_{> 0} \iff 0 < m$
+:$(3): \quad m \notin \N \iff -m \in \N_{> 0}$
+That is, the [[Definition:Natural Numbers|natural numbers]] are precisely those [[Definition:Integer|integers]] which are greater than or equal to [[Definition:Zero (Algebra)|zero]].
+\end{theorem}
+
+\begin{proof}
+Consider the [[Definition:Integer/Formal Definition|formal definition of the integers]]: $x = \eqclass {a, b} {}$ is an [[Definition:Equivalence Class|equivalence class]] of [[Definition:Ordered Pair|ordered pairs]] of [[Definition:Natural Numbers|natural numbers]].
+Let $x \in \Z_{>0}$ be a [[Definition:Strictly Positive Integer|(strictly) positive integer]].
+Thus by definition:
+:$x > 0$
+This is equivalent to the condition that $a > b$.
+Hence:
+:$x = \eqclass {b + u, b} {}$
+for some $u \in \N_{>0}$.
+It is immediate that:
+:$\forall c \in \N: \eqclass {b + u, b} {} = \eqclass {c + u, c} {}$
+Consider the [[Definition:Mapping|mapping]] $\phi: \N_{>0} \to \Z_{>0}$ defined as:
+:$\forall u \in \N_{>0}: \map \phi u = u'$
+where $u' \in \Z$ be the [[Definition:Strictly Positive Integer|(strictly) positive integer]] $\eqclass {b + u, b} {}$.
+From the above we have that $\phi$ is [[Definition:Well-Defined Mapping|well-defined]].
+Let $\eqclass {b + u, b} {} = \eqclass {c + v, c} {}$.
+Then:
+:$b + u + c = b + c + v$
+and so:
+:$u = v$
+Hence $\phi$ is an [[Definition:Injection|injection]].
+Next note that all $u' \in \Z_{>0}$ can be expressed in the form:
+:$u' = \eqclass {b + u, b} {}$
+for arbitrary $b$.
+Hence $\phi$ is a [[Definition:Surjection|surjection]].
+Hence $\phi$ is an [[Definition:Isomorphism|isomorphism]].
+By defining $\map \phi 0 = \eqclass {b, b} {}$ we see that $\N$ and $\Z_{\ge 0}$ are [[Definition:Isomorphism|isomorphic]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Subtraction on Integers is Extension of Natural Numbers}
+Tags: Integers, Subtraction
+
+\begin{theorem}
+[[Definition:Integer Subtraction|Integer subtraction]] is an extension of the definition of [[Definition:Natural Number Subtraction|subtraction]] on the [[Definition:Natural Numbers|natural numbers]].
+\end{theorem}
+
+\begin{proof}
+Let $m, n \in \N: m \le n$.
+From [[Definition:Natural Number Subtraction|natural number subtraction]], $\exists p \in \N: m + p = n$ such that $n - m = p$.
+As $m, n, p \in \N$, it follows that $m, n, p \in \Z$ as well.
+However, as $\Z$ is the [[Definition:Inverse Completion|inverse completion]] of $\N$, it follows that $-m \in \Z$ as well, so it makes sense to express the following:
+{{begin-eqn}}
+{{eqn | l = \paren {n + \paren {-m} } + m
+      | r = n + \paren {\paren {-m} + m}
+      | c = 
+}}
+{{eqn | r = n
+      | c = 
+}}
+{{eqn | r = p + m
+      | c = 
+}}
+{{eqn | r = \paren {n - m} + m
+      | c = 
+}}
+{{end-eqn}}
+Thus, as all elements of $\Z$ are [[Definition:Cancellable Element|cancellable]], it follows that $n + \paren {-m} = n - m$.
+So:
+: $\forall m, n \in \Z, m \le n: n + \paren {-m} = n - m = n -_\N m$
+and the result follows.
+{{qed}}
+[[Category:Integers]]
+[[Category:Subtraction]]
+7wy2x7ny7d5h653ngmzazw9eqd8qcub
+\end{proof}<|endoftext|>
+\section{Non-Zero Integers are Cancellable for Multiplication}
+Tags: Integers, Non-Zero Integers are Cancellable for Multiplication
+
+\begin{theorem}
+Every non-zero [[Definition:Integer|integer]] is [[Definition:Cancellable Element|cancellable]] for [[Definition:Integer Multiplication|multiplication]].
+That is:
+:$\forall x, y, z \in \Z, x \ne 0: x y = x z \iff y = z$
+\end{theorem}
+
+\begin{proof}
+Let $x y = x z$.
+There are two cases to investigate: $x > 0$ and $x < 0$.
+Let $x > 0$.
+From [[Natural Numbers are Non-Negative Integers]], $x \in \N_{> 0}$.
+By the [[Extension Theorem for Distributive Operations]] and [[Ordering on Natural Numbers is Compatible with Multiplication]], $x$ is [[Definition:Cancellable Element|cancellable]] for multiplication.
+{{qed|lemma}
+Let $x < 0$.
+We know that the [[Integers form Integral Domain]] and are thus a [[Definition:Ring (Abstract Algebra)|ring]].
+Then $-x > 0$ and so:
+{{begin-eqn}}
+{{eqn | l = x y
+      | r = x z
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \paren {-\paren {-x} } y
+      | r = \paren {-\paren {-x} } z
+      | c = [[Negative of Ring Negative]]
+}}
+{{eqn | ll= \leadsto
+      | l = -\paren {\paren {-x} y}
+      | r = -\paren {\paren {-x} z}
+      | c = [[Product with Ring Negative]]
+}}
+{{eqn | ll= \leadsto
+      | l = -y
+      | r = -z
+      | c = as $-x$ is [[Definition:Strictly Positive Integer|(strictly) positive]], the above result holds
+}}
+{{eqn | ll= \leadsto
+      | l = y
+      | r = z
+      | c = $\struct {\Z, +}$ is a [[Definition:Group|group]]: {{GroupAxiom|3}}
+}}
+{{end-eqn}}
+{{qed|lemma}}
+So whatever non-zero value $x$ takes, it is [[Definition:Cancellable Element|cancellable]] for multiplication.
+{{Qed}}
+\end{proof}
+
+\begin{proof}
+Let $y, z \in \Z: y \ne z$.
+{{begin-eqn}}
+{{eqn | l = y
+      | o = \ne
+      | r = z
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = y - z
+      | o = \ne
+      | r = 0
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = x \paren {y - z}
+      | o = \ne
+      | r = 0
+      | c = [[Ring of Integers has no Zero Divisors]]
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = x y - x z
+      | o = \ne
+      | r = 0
+      | c = [[Integer Multiplication Distributes over Subtraction]]
+}}
+{{end-eqn}}
+The result follows by [[Rule of Transposition|transposition]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let $x y = x z$.
+There are two cases to investigate: $x > 0$ and $x < 0$.
+Let $x > 0$.
+From [[Natural Numbers are Non-Negative Integers]], $x \in \N_{> 0}$.
+By the [[Extension Theorem for Distributive Operations]] and [[Ordering on Natural Numbers is Compatible with Multiplication]], $x$ is [[Definition:Cancellable Element|cancellable]] for [[Definition:Integer Multiplication|multiplication]].
+{{qed|lemma}}
+Let $x < 0$.
+We know that the [[Integers form Integral Domain]] and are thus a [[Definition:Ring (Abstract Algebra)|ring]].
+Then $-x > 0$ and so:
+{{begin-eqn}}
+{{eqn | l = \paren {-x} y
+      | r = -\paren {x y}
+      | c = [[Product with Ring Negative]]
+}}
+{{eqn | r = -\paren {x z}
+      | c = $\struct {\Z, +}$ is a [[Definition:Group|group]]: {{GroupAxiom|3}}
+}}
+{{eqn | r = \paren {-x} z
+      | c = [[Product with Ring Negative]]
+}}
+{{eqn | ll= \leadsto
+      | l = y
+      | r = z
+      | c = from above: case where $x > 0$
+}}
+{{end-eqn}}
+{{qed|lemma}}
+So whatever non-zero value $x$ takes, it is [[Definition:Cancellable Element|cancellable]] for [[Definition:Integer Multiplication|multiplication]].
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Multiplicative Ordering on Integers}
+Tags: Integer Multiplication, Orderings on Integers
+
+\begin{theorem}
+Let $x, y, z \in \Z$ such that $z > 0$.
+Then:
+:$x < y \iff z x < z y$
+:$x \le y \iff z x \le z y$
+\end{theorem}
+
+\begin{proof}
+Let $z > 0$.
+Let $M_z: \Z \to \Z$ be the mapping defined as:
+:$\forall x \in \Z: \map {M_z} x = z x$
+All we need to do is show that $M_z$ is an [[Definition:Order Embedding|order embedding]] from $\struct {\Z, +, \le}$ to itself.
+By [[Monomorphism from Total Ordering]], we just need to show that:
+:$x < y \implies z x < z y$
+If $x < y$, then $0 < y - x$, so $z \in \N$ and $y - x \in \N$ by [[Natural Numbers are Non-Negative Integers]].
+Thus by [[Ordering on Natural Numbers is Compatible with Multiplication]]:
+:$z \paren {y - x} \in \N$
+Therefore
+:$0 < z \paren {y - x} = z y - z x$
+That is:
+:$z x < z y$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Invertible Integers under Multiplication}
+Tags: Integer Multiplication, Units of Rings
+
+\begin{theorem}
+The only [[Definition:Invertible Element|invertible elements]] of $\Z$ for [[Definition:Integer Multiplication|multiplication]] (that is, [[Definition:Unit of Ring|units of $\Z$]]) are $1$ and $-1$.
+\end{theorem}
+
+\begin{proof}
+Let $x > 0$ and $x y > 0$.
+Suppose $y \le 0$.
+Then from [[Multiplicative Ordering on Integers]] and [[Ring Product with Zero]]:
+:$x y \le x \, 0 = 0$
+From this contradiction we deduce that $y > 0$.
+Now, if we have $x > 0$ and $x y = 1$, then $y > 0$ and hence $y \in \N$ by [[Natural Numbers are Non-Negative Integers]].
+Hence by [[Invertible Elements under Natural Number Multiplication]] it follows that $x = 1$.
+Thus $1$ is the only element of $\N$ that is [[Definition:Invertible Element|invertible]] for [[Definition:Natural Number Multiplication|multiplication]].
+Therefore by [[Natural Numbers are Non-Negative Integers]] and [[Product with Ring Negative]], the result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Index Laws for Monoids}
+Tags: Monoids
+
+\begin{theorem}
+These results are an extension of the results in [[Index Laws for Semigroup]] in which the [[Definition:Domain of Mapping|domain]] of the indices is extended to include all [[Definition:Integer|integers]].
+Let $\struct {S, \circ}$ be a [[Definition:Monoid|monoid]] whose [[Definition:Identity Element|identity]] is $e$.
+Let $a \in S$ be [[Definition:Invertible Element|invertible]] for $\circ$.
+Let $n \in \N$.
+Let $a^n$ be the [[Definition:Power of Element/Monoid|$n$th power of $a$]]:
+:$a^n = \begin{cases}
+e : & n = 0 \\
+a^{n - 1} \circ a : & n > 0 \\
+\paren {a^{-n}}^{-1} : & n < 0
+\end{cases}$
+Then we have the following results:
+\end{theorem}<|endoftext|>
+\section{Totally Ordered Abelian Group Isomorphism}
+Tags: Isomorphisms, Abelian Groups
+
+\begin{theorem}
+Let $\left({\Z', +', \le'}\right)$ be a [[Definition:Totally Ordered Structure|totally ordered]] [[Definition:Abelian Group|abelian group]].
+Let $0'$ be the [[Definition:Identity Element|identity]] of $\left({\Z', +', \le'}\right)$.
+Let $\N' = \left\{{x \in \Z': x \ge' 0'}\right\}$.
+Let $\Z'$ contain at least two elements.
+Let $\N'$ be [[Definition:Well-Ordered Set|well-ordered]] for the ordering induced on $\N'$ by $\le'$.
+Then the [[Definition:Mapping|mapping]] $g: \Z \to \Z'$ defined by:
+:$\forall n \in \Z: g \left({n}\right) = \left({+'}\right)^n 1'$
+is an [[Definition:Group Isomorphism|isomorphism]] from $\left({\Z, +, \le}\right)$ onto $\left({\Z', +', \le'}\right)$, where $1'$ is the [[Definition:Smallest Element|smallest element]] of $\N' \setminus \left\{{0'}\right\}$.
+\end{theorem}
+
+\begin{proof}
+First we establish that $g$ is a [[Definition:Homomorphism (Abstract Algebra)|homomorphism]].
+Suppose $z \in \Z'$ such that $z \ne 0'$.
+Then by [[Ordering of Inverses in Ordered Monoid]], either $z >' 0'$ or $-z >' 0'$.
+Thus either:
+:$z \in \N' \setminus \left\{{0'}\right\}$
+or:
+:$-z \in \N' \setminus \left\{{0'}\right\}$
+and thus $\N' \setminus \left\{{0'}\right\}$ is not [[Definition:Empty Set|empty]].
+Therefore $\N' \setminus \left\{{0'}\right\}$ has a [[Definition:Minimal Element|minimal element]].
+Call this [[Definition:Minimal Element|minimal element]] $1'$.
+It is clear that $\N'$ is an ordered [[Definition:Semigroup|semigroup]] satisfying $(NO 1)$, $(NO 2)$ and $(NO 4)$ of the [[Definition:Naturally Ordered Semigroup Axioms|naturally ordered semigroup axioms]].
+Also:
+{{begin-eqn}}
+{{eqn | o = 
+      | r = 0' \le' x \le' y
+      | c = 
+}}
+{{eqn | o = \implies
+      | r = 0' \le' y - x
+      | c = 
+}}
+{{eqn | o = \implies
+      | r = y - x \in \N' \land x +' \left({y - x}\right) = y
+      | c = 
+}}
+{{end-eqn}}
+Thus $\N'$ also satisfies $(NO 3)$ of the [[Definition:Naturally Ordered Semigroup Axioms|naturally ordered semigroup axioms]].
+So $\left({\N', +', \le'}\right)$ is a [[Definition:Naturally Ordered Semigroup|naturally ordered semigroup]].
+So, by [[Naturally Ordered Semigroup is Unique]], the [[Definition:Restriction of Mapping|restriction]] to $\N$ of $g$ is an [[Definition:Group Isomorphism|isomorphism]] from $\left({\N, +, \le}\right)$ to $\left({\N', +', \le'}\right)$.
+By [[Index Laws for Monoids/Sum of Indices|Index Law for Sum of Indices]], $g$ is a [[Definition:Homomorphism (Abstract Algebra)|homomorphism]] from $\left({\Z, +}\right)$ into $\left({\Z', +'}\right)$.
+Next it is established that $g$ is [[Definition:Surjection|surjective]].
+Let $y \in \Z': y <' 0'$.
+{{begin-eqn}}
+{{eqn | o = 
+      | r = y <' 0'
+      | c = 
+}}
+{{eqn | o = \implies
+      | r = -y >' 0'
+      | c = 
+}}
+{{eqn | o = \implies
+      | r = \exists n \in \N: -y = g \left({n}\right)
+      | c = 
+}}
+{{eqn | o = \implies
+      | r = y = - g \left({n}\right) = g \left({-n}\right)
+      | c = [[Homomorphism with Identity Preserves Inverses]]
+}}
+{{end-eqn}}
+Therefore $g$ is a [[Definition:Surjection|surjection]].
+Now to show that $g$ is a [[Definition:Group Monomorphism|monomorphism]], that is, it is [[Definition:Injection|injective]].
+Let $n < m$.
+{{begin-eqn}}
+{{eqn | o = 
+      | r = n < m
+      | c = 
+}}
+{{eqn | o = \implies
+      | r = m - n \in \N_{>0}
+      | c = 
+}}
+{{eqn | o = \implies
+      | r = g \left({m}\right) - g \left({n}\right) - g \left({m - n}\right) \in \N' \setminus \left\{ {0'}\right\}
+      | c = 
+}}
+{{eqn | o = \implies
+      | r = g \left({n}\right) <' \left({g \left({m}\right) - g \left({n}\right)}\right) +' g \left({n}\right) = g \left({m}\right)
+      | c = [[Strict Ordering Preserved under Product with Cancellable Element]]
+}}
+{{end-eqn}}
+Therefore it can be seen that $g$ is [[Definition:Strictly Increasing|strictly increasing]].
+It follows from [[Monomorphism from Total Ordering]] that $g$ is a [[Definition:Group Monomorphism|monomorphism]] from $\left({\Z, +, \le}\right)$ to $\left({\Z', +', \le'}\right)$.
+A [[Definition:Surjection|surjective]] [[Definition:Group Monomorphism|monomorphism]] is an [[Definition:Group Isomorphism|isomorphism]], and the result follows.
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Integers under Addition form Totally Ordered Group}
+Tags: Integer Addition, Additive Group of Integers
+
+\begin{theorem}
+Then the [[Definition:Ordered Structure|ordered structure]] $\struct {\Z, +, \le}$ is a [[Definition:Totally Ordered Group|totally ordered group]].
+\end{theorem}
+
+\begin{proof}
+=== $\struct {\Z, +, \le}$ is an Ordered Structure ===
+==== $(1)$ ====
+By [[Integer Addition is Closed]], $\struct {\Z, +}$ is an [[Definition:Algebraic Structure|algebraic structure]].
+==== $(2)$ ====
+$\le$ is an [[Definition:Ordering|ordering]] on $\Z$.
+Thus, $\struct {\Z, \le}$ is an [[Definition:Ordered Set|ordered set]].
+==== $(3)$ ====
+By [[Ordering is Preserved on Integers by Addition]] and [[Integer Addition is Commutative]], $\le$ is [[Definition:Relation Compatible with Operation|compatible]] with $+$.
+Thus, $\struct {\Z, +, \le}$ is an [[Definition:Ordered Structure|ordered structure]].
+{{qed|lemma}}
+=== $\struct {\Z, +, \le}$ is a Totally Ordered Structure ===
+By definition, the [[Definition:Ordered Structure|ordered structure]] $\struct {\Z, +, \le}$ is a [[Definition:Totally Ordered Structure|totally ordered structure]] {{iff}} $\le$ is a [[Definition:Total Ordering|total ordering]].
+This follows from [[Ordering on Integers is Total Ordering]].
+{{qed|lemma}}
+=== $\struct {\Z, +, \le}$ is a Totally Ordered Group ===
+By definition, the [[Definition:Totally Ordered Structure|totally ordered structure]] $\struct {\Z, +, \le}$ is a [[Definition:Totally Ordered Group|totally ordered group]] {{iff}} $\struct {\Z, +}$ is a [[Definition:Group|group]].
+This follows from [[Integers under Addition form Abelian Group]].
+{{qed}}
+[[Category:Integer Addition]]
+[[Category:Additive Group of Integers]]
+15qysqr4r51uk7597b8j8vjkp15ng6w
+\end{proof}<|endoftext|>
+\section{Integers form Totally Ordered Ring}
+Tags: Integers, Totally Ordered Rings
+
+\begin{theorem}
+The [[Definition:Algebraic Structure|structure]] $\struct {\Z, +, \times, \le}$ is a [[Definition:Totally Ordered Ring|totally ordered ring]].
+\end{theorem}
+
+\begin{proof}
+From [[Integers form Commutative Ring]], $\struct {\Z, +, \times}$ is a [[Definition:Commutative Ring|commutative ring]].
+We need to show that $\le$ is a [[Definition:Ordering Compatible with Ring Structure|compatible ordering]] on $\Z$.
+That is, that:
+:$(1): \quad \le$ is [[Definition:Relation Compatible with Operation|compatible]] with $+$
+:$(2): \quad \forall x, y \in \Z: 0 \le x, 0 \le y \implies 0 \le x \times y$
+The first one follows from the fact that [[Integers under Addition form Totally Ordered Group]].
+Now, from [[Multiplicative Ordering on Integers]], we have:
+:$\forall z, x, y \in \Z, 0 < y: z \le x \iff y \times z \le y \times x$
+Let $z = 0, 0 \le x$.
+Then:
+:$0 = x \times z \le x \times y = y \times x$
+So if $y \ne 0$, the condition $0 \le x, 0 \le y \implies 0 \le x \times y$ holds.
+Now if $y = 0$, we have $x \times y = 0$ and thus $0 \le x \times y$.
+Thus $\le$ is [[Definition:Relation Compatible with Operation|compatible]] with $\Z$, and the result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Congruences on Rational Numbers}
+Tags: Congruence Relations, Rational Numbers
+
+\begin{theorem}
+There are only two [[Definition:Congruence Relation|congruence relations]] on the [[Definition:Field (Abstract Algebra)|field]] of [[Definition:Rational Number|rational numbers]] $\left({\Q, +, \times}\right)$:
+:$(1): \quad$ The [[Definition:Diagonal Relation|diagonal relation]] $\Delta_\Q$
+:$(2): \quad$ The [[Definition:Trivial Relation|trivial relation]] $\Q \times \Q$.
+\end{theorem}
+
+\begin{proof}
+From:
+:[[Diagonal Relation is Universally Compatible]] and
+:[[Trivial Relation is Universally Congruent]]
+we know that both these [[Definition:Relation|relations]] are [[Definition:Relation Compatible with Operation|compatible]] with both [[Definition:Rational Addition|addition]] and [[Definition:Rational Multiplication|multiplication]] on $\Q$.
+Now we need to show that these are the ''only'' such [[Definition:Relation|relations]].
+Let $\RR$ be a [[Definition:Congruence Relation|congruence]] on $\Q$, such that $\RR \ne \Delta_\Q$.
+{{begin-eqn}}
+{{eqn | l = \RR
+      | o = \ne
+      | r = \Delta_\Q
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \exists r, s \in \Q: r
+      | o = \ne
+      | r = s \land r \RR s
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \exists h \in \Q, h \ne 0: h
+      | r = r - s
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = h
+      | o = \RR
+      | r = 0
+      | c = as $\RR$ is a [[Definition:Congruence Relation|congruence relation]] compatible with $+$
+}}
+{{end-eqn}}
+Then:
+{{begin-eqn}}
+{{eqn | l = \forall x \in \Q: \paren {x / h}
+      | o = \RR
+      | r = \paren {x / h}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \paren {\paren {x / h} \times h}
+      | o = \RR
+      | r = \paren {\paren {x / h} \times 0}
+      | c = as $\RR$ is a [[Definition:Congruence Relation|congruence relation]] compatible with $\times$
+}}
+{{eqn | ll= \leadsto
+      | l = x
+      | o = \RR
+      | r = 0
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \eqclass 0 \RR
+      | r = \Q
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \RR
+      | r = \Q \times \Q
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Number of Elements in Partition}
+Tags: Combinatorics
+
+\begin{theorem}
+Let $S$ be a [[Definition:Set|set]].
+Let there be a [[Definition:Set Partition|partition]] on $S$ of $n$ [[Definition:Subset|subsets]], each of which has $m$ [[Definition:Element|elements]].
+Then:
+:$\card S = n m$
+\end{theorem}
+
+\begin{proof}
+Let the [[Definition:Set Partition|partition]] of $S$ be $S_1, S_2, \ldots, S_n$.
+Then:
+:$\forall k \in \set {1, 2, \ldots, n}: \card {S_k} = m$
+By definition of [[Definition:Integer Multiplication|multiplication]]:
+:$\displaystyle \sum_{k \mathop = 1}^n \card {S_k} = n m$
+and the result follows from the [[Fundamental Principle of Counting]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Cardinality of Complement}
+Tags: Combinatorics, Set Difference, Cardinality
+
+\begin{theorem}
+Let $T \subseteq S$ such that $\card S = n, \card T = m$.
+Then:
+:$\card {\relcomp S T} = \card {S \setminus T} = n - m$
+where:
+:$\relcomp S T$ denotes the [[Definition:Relative Complement|complement of $T$ relative to $S$]]
+:$S \setminus T$ denotes the [[Definition:Set Difference|difference between $S$ and $T$]].
+\end{theorem}
+
+\begin{proof}
+The result is obvious for $S = T$ or $T = \O$.
+Otherwise, $\set {T, S \setminus T}$ is a [[Definition:Set Partition|partition]] of $S$.
+Let $\card {S \setminus T} = p$.
+Then by the [[Fundamental Principle of Counting]]:
+: $m + p = n$
+and the result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Cardinality of Cartesian Product}
+Tags: Combinatorics, Cartesian Product, Cardinality, Cardinality of Cartesian Product
+
+\begin{theorem}
+Let $S \times T$ be the [[Definition:Cartesian Product|cartesian product]] of two [[Definition:Finite Set|finite sets]] $S$ and $T$.
+Then:
+:$\card {S \times T} = \card S \times \card T$
+where $\card S$ denotes [[Definition:Cardinality|cardinality]].
+\end{theorem}
+
+\begin{proof}
+Let $\card S = n$ and $\card T = m$.
+If either $n = 0$ or $m = 0$, then from [[Cartesian Product is Empty iff Factor is Empty]]:
+:$S \times T = \O$
+and the result holds, as $n m = 0 = \card \O$ from [[Cardinality of Empty Set]].
+So, we assume that $n > 0$ and $m > 0$.
+For each $a \in S$, we define the [[Definition:Mapping|mapping]] $g_a: T \to \set a \times T$ such that:
+:$\forall y \in T: \map {g_a} y = \tuple {a, y}$
+The mapping $g_a$ is a [[Definition:Bijection|bijection]], so:
+:$\card {\set a \times T} = m$
+Now let:
+:$\mathbb T = \set {\set a \times T: a \in S}$
+We define the [[Definition:Mapping|mapping]] $h: S \to \mathbb T$:
+:$\forall a \in S: \map h a = \set a \times T$
+The mapping $h$ is a [[Definition:Bijection|bijection]], so $\card {\mathbb T} = n$.
+Thus $\mathbb T$ is a [[Definition:Partition (Set Theory)|partition]] of $S \times T$ containing $n$ [[Definition:Set|sets]].
+Hence from [[Number of Elements in Partition]]:
+:$\card {S \times T} = n m$
+and the result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Cardinality of Set of All Mappings}
+Tags: Combinatorics, Mapping Theory, Cardinality, Cardinality of Set of All Mappings
+
+\begin{theorem}
+Let $S$ and $T$ be [[Definition:Set|sets]].
+The [[Definition:Cardinality|cardinality]] of the [[Definition:Set of All Mappings|set of all mappings]] from $S$ to $T$ (that is, the total number of [[Definition:Mapping|mappings]] from $S$ to $T$) is:
+:$\card {T^S} = \card T^{\card S}$
+\end{theorem}<|endoftext|>
+\section{Cardinality of Power Set of Finite Set}
+Tags: Power Set, Combinatorics, Cardinality, Cardinality of Power Set
+
+\begin{theorem}
+Let $S$ be a [[Definition:Set|set]] such that:
+:$\card S = n$
+where $\card S$ denotes the [[Definition:Cardinality|cardinality]] of $S$,
+Then:
+:$\card {\powerset S} = 2^n$
+where $\powerset S$ denotes the [[Definition:Power Set|power set]] of $S$.
+\end{theorem}
+
+\begin{proof}
+Let $T = \set {0, 1}$.
+For each $A \in \powerset S$, we consider the [[Definition:Characteristic Function of Set|characteristic function]] $\chi_A: S \to T$ defined as:
+:$\forall x \in S: \map {\chi_A} x = \begin{cases}
+1 & : x \in A \\
+0 & : x \notin A
+\end{cases}$
+Now consider the [[Definition:Mapping|mapping]] $f: \powerset S \to T^S$:
+:$\forall A \in \powerset S: \map f A = \chi_A$
+where $T^S$ is the [[Definition:Set of All Mappings|set of all mappings]] from $S$ to $T$.
+Also, consider the [[Definition:Mapping|mapping]] $g: T^S \to \powerset S$:
+:$\forall \phi \in T^S: \map g \phi = \phi^{-1} \sqbrk {\set 1}$
+where $\phi^{-1} \sqbrk {\set 1}$ is the [[Definition:Preimage of Subset under Mapping|preimage]] of $\set 1$ under $\phi$.
+Consider the [[Definition:Characteristic Function of Set|characteristic function]] of $\phi^{-1} \sqbrk {\set 1}$, denoted $\chi_{\phi^{-1} \sqbrk {\set 1} }$.
+We have:
+{{begin-eqn}}
+{{eqn | l = \forall x \in S: \map {\chi_{\phi^{-1} \sqbrk {\set 1} } } x
+      | r = \begin{cases} 1 & : x \in \phi^{-1} \sqbrk {\set 1} \\ 0 & : x \notin \phi^{-1} \sqbrk {\set 1} \end{cases}
+      | c =
+}}
+{{eqn | r = \begin{cases} 1 & : \map \phi x = 1 \\ 0 & : \map \phi x = 0 \end{cases}
+      | c =
+}}
+{{eqn | r = \map \phi x
+      | c =
+}}
+{{end-eqn}}
+So:
+{{begin-eqn}}
+{{eqn | l = \forall \phi \in T^S: \map {\paren {f \circ g} } \phi
+      | r = \map f {\phi^{-1} } {\sqbrk {\set 1} }
+      | c = 
+}}
+{{eqn | r = \chi_{\phi^{-1} \sqbrk {\set 1} }
+      | c = 
+}}
+{{eqn | r = \phi
+      | c = 
+}}
+{{end-eqn}}
+So $f \circ g = I_{T^S}$, that is, the [[Definition:Identity Mapping|identity mapping]] on $T^S$.
+So far so good. Now we consider the [[Definition:Preimage of Subset under Mapping|preimage]] of $\set 1$ under $\chi_A$:
+: $\chi_A^{-1} \sqbrk {\set 1} = A$
+from the definition of the [[Definition:Characteristic Function of Set|characteristic function]] $\chi_A$ above.
+Thus:
+{{begin-eqn}}
+{{eqn | l = \forall A \in \powerset S: \map {\paren {g \circ f} } A
+      | r = \map g {\map f A}
+      | c = 
+}}
+{{eqn | r = \map g {\chi_A}
+      | c = 
+}}
+{{eqn | r = \chi_A^{-1} \sqbrk {\set 1}
+      | c = 
+}}
+{{eqn | r = A
+      | c = 
+}}
+{{end-eqn}}
+So $g \circ f = I_{\powerset S}$, that is, the [[Definition:Identity Mapping|identity mapping]] on $\powerset S$.
+It follows from [[Bijection iff Left and Right Inverse]] that $f$ and $g$ are [[Definition:Bijection|bijections]].
+Thus by [[Cardinality of Set of All Mappings]] the result follows.
+{{Qed}}
+\end{proof}
+
+\begin{proof}
+Enumerating the [[Definition:Subset|subsets]] of $S$ is equivalent to counting all of the ways of selecting $k$ out of the $n$ elements of $S$ with $k = 0, 1, \ldots, n$.
+So, from [[Cardinality of Set of Subsets]], the number we are looking for is:
+:$\displaystyle \card {\powerset S} = \sum_{k \mathop = 0}^n \binom n k$
+But from the [[Binomial Theorem|binomial theorem]]:
+:$\displaystyle \paren {x + y}^n = \sum_{k \mathop = 0}^n \binom n k x^{n - k} y^k$
+It follows that:
+:$2^n = \displaystyle \paren {1 + 1}^n = \sum_{k \mathop = 0}^n \binom n k \paren 1^{n - k} \paren 1^k = \sum_{k \mathop = 0}^n \binom n k = \card {\powerset S}$
+{{Qed}}
+\end{proof}
+
+\begin{proof}
+Given an element $x$ of $S$, each [[Definition:Subset|subset]] of $S$ either includes $x$ or does not include $x$ (this follows directly from the [[Definition:Set|definition of a set]]), which gives us two possibilities.
+The same reasoning holds for any [[Definition:Element|element]] of $S$.
+One can intuitively see that this means that there are $\displaystyle \underbrace {2 \times 2 \times \ldots \times 2}_{\card S} = 2^{\card S}$ total possible combinations of [[Definition:Element|elements]] of $S$.
+This is exactly $\card {\powerset S}$.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Proof by [[Principle of Mathematical Induction|induction]]:
+For all $n \in \N$, let $P \left({n}\right)$ be the [[Definition:Proposition|proposition]]:
+:$\left|{S}\right| = n \implies \left|{\mathcal P \left({S}\right)}\right| = 2^n$
+Do not confuse $P \left({n}\right)$, which is a [[Definition:Propositional Function|propositional function]] on $\N$, with $\mathcal P \left({S}\right)$, the [[Definition:Power Set|power set]] of $S$.
+=== Basis for the Induction ===
+From [[Cardinality of Empty Set]], we have:
+: $S = \varnothing \iff \left|{S}\right| = 0$
+Then:
+: $\mathcal P \left({\varnothing}\right) = \left\{{\varnothing}\right\}$
+has one element, that is, $\varnothing$.
+So:
+: $\left|{\mathcal P \left({\varnothing}\right)}\right| = \left|{\left\{{\varnothing}\right\}}\right| = 1 = 2^0$
+thus confirming that $P \left({0}\right)$ holds.
+This is our [[Principle of Mathematical Induction#Basis for the Induction|basis for the induction]].
+=== Induction Hypothesis ===
+Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 2$, then it logically follows that $P \left({k+1}\right)$ is true.
+So this is our [[Principle of Mathematical Induction#Induction Hypothesis|induction hypothesis]]:
+:$\left|{S}\right| = k \implies \left|{\mathcal P \left({S}\right)}\right| = 2^k$
+Then we need to show:
+:$\left|{S}\right| = k + 1 \implies \left|{\mathcal P \left({S}\right)}\right| = 2^{k + 1}$
+=== Induction Step ===
+This is our [[Principle of Mathematical Induction#Induction Step|induction step]]:
+Let $\left|{S}\right| = k + 1$.
+Let $x \in S$.
+Consider the set $S' = S \setminus \left\{{x}\right\}$ where $x$ is any element of $S$.
+We see that $\left|{S'}\right| = k$.
+Now adjoin $x$ to all the [[Definition:Subset|subsets]] of $S'$.
+Counting the [[Definition:Subset|subsets]] of $S$, we have:
+: all the [[Definition:Subset|subsets]] of $S'$
+and:
+: all the [[Definition:Subset|subsets]] of $S'$ with $x$ adjoined to them.
+From the [[Cardinality of Power Set of Finite Set/Proof 3#Induction Hypothesis|induction hypothesis]], there are $2^k$ subsets of $S'$.
+Adding $x$ to each of these does not change their number, so there are another $2^k$ [[Definition:Subset|subsets]] of $S$ consisting of all the [[Definition:Subset|subsets]] of $S'$ with $x$ adjoined to them.
+In total that makes $2^k + 2^k = 2 \times 2^k = 2^{k + 1}$ [[Definition:Subset|subsets]] of $S$.
+So $P \left({k}\right) \implies P \left({k + 1}\right)$ and the result follows by the [[Principle of Mathematical Induction]].
+Therefore:
+:$\forall n \in \N: \left|{S}\right| = n \implies \left|{\mathcal P \left({S}\right)}\right| = 2^n$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Cardinality of Set of Injections}
+Tags: Injections, Combinatorics, Counting Arguments, Cardinality, Cardinality of Set of Injections
+
+\begin{theorem}
+Let $S$ and $T$ be [[Definition:Finite Set|finite sets]].
+The number of [[Definition:Injection|injections]] from $S$ to $T$, where $\card S = m, \card T = n$ is often denoted ${}^m P_n$, and is:
+:${}^m P_n = \begin{cases}
+\dfrac {n!} {\paren {n - m}!} & : m \le n \\
+0 & : m > n \end{cases}$
+\end{theorem}
+
+\begin{proof}
+Let $m > n$.
+By the [[Pigeonhole Principle]], there can be no [[Definition:Injection|injection]] from $S$ to $T$ when $\left|{S}\right| > \left|{T}\right|$.
+Once $\left|{T}\right|$ elements of $S$ have been used up, there is no element of $T$ left for the remaining elements of $S$ to be mapped to such that they all still map to different elements of $T$.
+Let $m = 0$.
+The only [[Definition:Injection|injection]] from $\varnothing \to T$ is $\varnothing \times T$ which is $\varnothing$.
+So if $m = 0$ there is $1 = n! / n!$ [[Definition:Injection|injection]].
+Let $0 < m \le n$.
+As in the proof of [[Cardinality of Set of All Mappings]], we can assume that $S = \N_m$ and $T = \N_n$.
+For each $k \in \left[{1 \,.\,.\, n}\right]$, let $\mathbb H \left({k, n}\right)$ be the set of all injections from $\N_k$ to $\N_n$.
+The proof now proceeds by [[Principle of Mathematical Induction|induction]].
+Let:
+:$\displaystyle \mathbb S = \left\{{k \in \left[{1 \,.\,.\, n}\right]: \left|{\mathbb H \left({k, n}\right)}\right| = \frac {n!} {\left({n - k}\right)!}}\right\}$
+=== Basis for the Induction ===
+Let $k = 1$.
+From [[Cardinality of Set of All Mappings]], there are $n^1 = n$ different mappings from $S$ to $T$.
+From [[Mapping from Singleton is Injection]], each one of these $n$ mappings is an [[Definition:Injection|injection]].
+Thus:
+:$\displaystyle \left|{\mathbb H \left({1, n}\right)}\right| = n = \frac {n!} {\left({n - 1}\right)!}$
+and so it follows that:
+:$1 \in \mathbb S$.
+This is the [[Principle of Mathematical Induction#Basis for the Induction|basis for the induction]].
+=== Induction Hypothesis ===
+We suppose that:
+:$\displaystyle \left|{\mathbb H \left({k, n}\right)}\right| = \frac {n!} {\left({n - k}\right)!}$.
+This is the [[Principle of Mathematical Induction#Induction Hypothesis|induction hypothesis]].
+We need to show that:
+:$\displaystyle \left|{\mathbb H \left({k+1, n}\right)}\right| = \frac {n!} {\left({n - \left({k+1}\right)}\right)!}$.
+=== Induction Step ===
+This is the [[Principle of Mathematical Induction#Induction Step|induction step]]:
+Let $k \in \mathbb S$ such that $k < n$.
+Let $\rho: \mathbb H \left({k + 1, n}\right) \to \mathbb H \left({k, n}\right)$ be the [[Definition:Mapping|mapping]] defined by:
+:$\forall f \in \mathbb H \left({k + 1, n}\right): \rho \left({f}\right) =$ the [[Definition:Restriction of Mapping|restriction]] of $f$ to $\N_k$
+Given that $g \in \mathbb H \left({k, n}\right)$ and $a \in \N_n - g \left({\N_k}\right)$, let $g_a: \N_{k + 1} \to \N_n$ be the [[Definition:Mapping|mapping]] defined as:
+:$g_a \left({x}\right) = \begin{cases}
+g \left({x}\right) & : x \in \N_k \\
+a & : x = k \end{cases}$
+Now $g$ is an [[Definition:Injection|injection]] as $g \in \mathbb H \left({k, n}\right)$, and as $g_a \left({a}\right) \notin g \left({\N_k}\right)$ it follows that $g_a$ is also an [[Definition:Injection|injection]].
+Hence:
+:$g_a \in \mathbb H \left({k + 1, n}\right)$
+It follows from the definition of $\rho$ that:
+:$\rho^{-1} \left({\left\{{g}\right\}}\right) = \left\{{g_a: a \in \N_n - g \left({\N_k}\right)}\right\}$
+Since $g$ is an [[Definition:Injection|injection]], $g \left({\N_k}\right)$ has $k$ elements.
+Therefore $\N_n - g \left({\N_k}\right)$ has $n - k$ elements by [[Cardinality of Complement]].
+As $G: a \to g_a$ is clearly a [[Definition:Bijection|bijection]] from $\N_n - g \left({\N_k}\right)$ onto $\rho^{-1} \left({\left\{{g}\right\}}\right)$, that [[Definition:Set|set]] has $n - k$ elements.
+Clearly:
+:$\left\{{\rho^{-1} \left({\left\{{g}\right\}}\right): g \in \mathbb H \left({k, n}\right)}\right\}$
+is a [[Definition:Partition (Set Theory)|partition]] of $\mathbb H \left({k + 1, n}\right)$.
+Therefore by [[Number of Elements in Partition]]:
+:$\left|{\mathbb H \left({k + 1, n}\right)}\right| = \left({n - k}\right) \dfrac {n!} {\left({n - k}\right)!} = \dfrac {n!} {\left({\left({n - k}\right) - 1}\right)!}$
+as $k \in \mathbb S$.
+But:
+:$\left({n - k}\right) - 1 = n - \left({k + 1}\right)$
+So:
+:$k + 1 \in \mathbb S$
+By [[Principle of Mathematical Induction|induction]]:
+:$\mathbb S = \left[{1 \,.\,.\, n}\right]$
+and in particular:
+:$m \in \mathbb S$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Cardinality of Set of Bijections}
+Tags: Combinatorics, Bijections, Cardinality, Cardinality of Set of Bijections
+
+\begin{theorem}
+Let $S$ and $T$ be sets such that $\size S = \size T = n$.
+Then there are $n!$ [[Definition:Bijection|bijections]] from $S$ to $T$.
+\end{theorem}
+
+\begin{proof}
+Follows directly from [[Cardinality of Set of Injections]] and [[Equivalence of Mappings between Sets of Same Cardinality]].
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Cardinality of Set of Subsets}
+Tags: Combinations, Combinatorics, Cardinality of Set of Subsets
+
+\begin{theorem}
+Let $S$ be a [[Definition:Set|set]] such that $\card S = n$.
+Let $m \le n$.
+Then the number of [[Definition:Subset|subsets]] $T$ of $S$ such that $\card T = m$ is:
+: ${}^m C_n = \dfrac {n!} {m! \paren {n - m}!}$
+\end{theorem}
+
+\begin{proof}
+For each $X \subseteq \N_n$ and $Y \subseteq S$, let $B \left({X, Y}\right)$ be the set of all [[Definition:Bijection|bijections]] from $X$ onto $Y$.
+Let $\Bbb S$ be the [[Definition:Set of Sets|set]] of all [[Definition:Subset|subsets]] of $S$ with $m$ [[Definition:Element|elements]].
+By [[Cardinality of Power Set of Finite Set]] and [[Cardinality of Subset of Finite Set]], $\Bbb S$ is [[Definition:Finite Set|finite]], so let $s = \left|{\Bbb S}\right|$.
+Let $\beta: B \left({\N_n, S}\right) \to \Bbb S$ be the [[Definition:Mapping|mapping]] defined as $\forall f \in B \left({\N_n, S}\right): \beta \left({f}\right) = f \left({\N_m}\right)$.
+For each $Y \in \Bbb S$, the [[Definition:Mapping|mapping]]:
+:$\Phi_Y: \beta^{-1} \left({Y}\right) \to B \left({\N_m, Y}\right) \times B \left({\N_n - \N_m, S - Y}\right)$
+defined as:
+:$\Phi_Y \left({f}\right) = \left({f_{\N_m}, f_{\N_n - \N_m}}\right)$
+is also (clearly) a bijection.
+By [[Cardinality of Set of Bijections]]:
+:$\left|{B \left({\N_m, Y}\right)}\right| = m!$
+and:
+:$\left|{B \left({\N_n - \N_m, S - Y}\right)}\right| = \left({n - m}\right)!$
+So by [[Cardinality of Cartesian Product]]:
+:$\left|{\beta^{-1} \left({Y}\right)}\right| = m! \left({n - m}\right)!$
+It is clear that $\left\{{\beta^{-1} \left({Y}\right): Y \in \Bbb S}\right\}$ is a [[Definition:Partition (Set Theory)|partition]] of $B \left({\N_n, S}\right)$.
+Therefore by [[Number of Elements in Partition]]:
+:$\left|{B \left({\N_n, S}\right)}\right| = m! \left({n - m}\right)! s$
+Consequently, as $\left|{B \left({\N_n, S}\right)}\right| = n!$ by [[Cardinality of Set of Bijections]], it follows that:
+:$m! \left({n - m}\right)! s = n!$
+and the result follows.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let ${}^m C_n$ be the number of [[Definition:Subset|subsets]] of $m$ [[Definition:Element|elements]] of $S$.
+From [[Number of Permutations]], the number of [[Definition:Permutation (Ordered Selection)|$m$-permutations of $S$]] is:
+:${}^m P_n = \dfrac {n!} {\paren {n - m}!}$
+Consider the way ${}^m P_n$ can be calculated.
+First one makes the selection of which $m$ elements of $S$ are to be arranged.
+This number is ${}^m C_n$.
+Then for each selection, the number of different arrangements of these is $m!$, from [[Number of Permutations]].
+So:
+{{begin-eqn}}
+{{eqn | l = m! \cdot {}^m C_n
+      | r = {}^m P_n
+      | c = [[Product Rule for Counting]]
+}}
+{{eqn | r = \frac {n!} {\paren {n - m}!}
+      | c = [[Number of Permutations]]
+}}
+{{eqn | ll= \leadsto
+      | l = {}^m C_n
+      | r = \frac {n!} {m! \paren {n - m}!}
+      | c = [[Product Rule for Counting]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let $\N_n$ denote the [[Definition:Set|set]] $\set {1, 2, \ldots, n}$.
+Let $\struct {S_n, \circ}$ denote the [[Definition:Symmetric Group on n Letters|symmetric group]] on $\N_n$.
+Let $r \in \N: 0 < r \le n$.
+Let $B_r$ denote the [[Definition:Set|set]] of all [[Definition:Subset|subsets]] of $\N_n$ of [[Definition:Cardinality|cardinality]] $r$:
+:$B_r  := \set {S \subseteq \N_n: \card S = r}$
+Let $*$ be the [[Definition:Mapping|mapping]] $*: S_n \times B_r \to B_r$ defined as:
+:$\forall \pi \in S_n, \forall S \in B_r: \pi * B_r = \pi \sqbrk S$
+where $\pi \sqbrk S$ denotes the [[Definition:Image of Subset under Mapping|image of $S$ under $\pi$]].
+From [[Group Action of Symmetric Group on Subset]] it is established that $*$ is a [[Definition:Group Action|group action]].
+The [[Definition:Stabilizer|stabilizer]] of any $U \in B_r$ is the [[Definition:Set|set]] of [[Definition:Permutation|permutations]] on $\N_n$ that fix $U$.
+Let $U = \set {1, 2, \ldots, r}$.
+So:
+:$\tuple {a_1, a_2, \ldots, a_r}$ can be any one of the $r!$ [[Definition:Permutation|permutations]] of $1, 2, \ldots, r$
+:$\tuple {a_{r + 1}, a_{r + 2}, \ldots, _n}$ can be any one of the $\paren {n - r}!$ [[Definition:Permutation|permutations]] of $r + 1, r + 2, \ldots, n$.
+Thus:
+:$\order {\Stab U} = r! \paren {n - r}!$
+From [[Group Action of Symmetric Group on Subset is Transitive]]:
+:$B_r = \Orb U$
+and so:
+:$\card {B_r} = \card {\Orb U}$
+From the [[Orbit-Stabilizer Theorem]]:
+:$\card {\Orb U} = \dfrac {\order {S_n} } {\order {\Stab U} } = \dfrac {n!} {r! \paren {n - r}!}$
+But $\card {B_r}$ is the number of [[Definition:Subset|subsets]] of $\N_n$ of [[Definition:Cardinality|cardinality]] $r$.
+Hence the result.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+The proof proceeds by [[Principle of Mathematical Induction|induction]].
+For all $n \in \Z_{\ge 0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
+:${}^m C_n = \dfrac {n!} {m! \paren {n - m}!}$
+=== Basis for the Induction ===
+$\map P 1$ is the case:
+:${}^m C_n = \dfrac {n!} {m! \paren {n - m}!}$
+{{begin-eqn}}
+{{eqn | l = {}^m C_1
+      | r = \begin {cases} 1 & : m = 0 \text { or } m = 1 \\ 0 & : \text {otherwise} \end {cases}
+      | c = 
+}}
+{{eqn | r = \dfrac {1!} {0! \paren {1 - 0}!}
+      | c = for $m = 0$
+}}
+{{eqn | r = \dfrac {1!} {1! \paren {1 - 1}!}
+      | c = for $m = 1$
+}}
+{{end-eqn}}
+Thus $\map P 1$ is seen to hold.
+This is the [[Principle of Mathematical Induction#Basis for the Induction|basis for the induction]].
+=== Induction Hypothesis ===
+Now it needs to be shown that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
+So this is the [[Principle of Mathematical Induction#Induction Hypothesis|induction hypothesis]]:
+:${}^m C_k = \dfrac {k!} {m! \paren {k - m}!}$
+from which it is to be shown that:
+:${}^m C_{k + 1} = \dfrac {\paren {k + 1}!} {m! \paren {k + 1 - m}!}$
+=== Induction Step ===
+This is the [[Principle of Mathematical Induction#Induction Step|induction step]]:
+The number of ways to choose $m$ [[Definition:Element|elements]] from $k + 1$ [[Definition:Element|elements]] is:
+:the number of ways to choose $m$ elements [[Definition:Element|elements]] from $k$ [[Definition:Element|elements]] (deciding not to select the $k + 1$th [[Definition:Element|element]])
+added to:
+:the number of ways to choose $m - 1$ elements [[Definition:Element|elements]] from $k$ [[Definition:Element|elements]] (after having selected the $k + 1$th [[Definition:Element|element]] for the $n$th selection)
+{{begin-eqn}}
+{{eqn | l = {}^m C_{k + 1}
+      | r = {}^m C_k + {}^{m - 1} C_k
+      | c = 
+}}
+{{eqn | r = \binom k m + \binom k {m - 1}
+      | c = [[Cardinality of Set of Subsets/Proof 4#Induction Hypothesis|Induction Hypothesis]]
+}}
+{{eqn | r = \binom {k + 1} m
+      | c = [[Pascal's Rule]]
+}}
+{{end-eqn}}
+So $\map P k \implies \map P {k + 1}$ and the result follows by the [[Principle of Mathematical Induction]].
+Therefore:
+:$\forall n \in \Z_{>0}: {}^m C_n = \dfrac {n!} {m! \paren {n - m}!}$
+\end{proof}<|endoftext|>
+\section{Set of Integers Bounded Below by Integer has Smallest Element}
+Tags: Set of Integers Bounded Below has Smallest Element
+
+\begin{theorem}
+Let $\Z$ be the [[Definition:Integer|set of integers]].
+Let $\le$ be the [[Definition:Ordering on Integers|ordering on the integers]].
+Let $\O \subset S \subseteq \Z$ such that $S$ is [[Definition:Bounded Below Set|bounded below]] in $\struct {\Z, \le}$.
+Then $S$ has a [[Definition:Smallest Element|smallest element]].
+\end{theorem}
+
+\begin{proof}
+We have that $S$ is [[Definition:Bounded Below Set|bounded below]] in $\Z$.
+So:
+: $\exists m \in \Z: \forall s \in S: m \le s$
+Hence:
+: $\forall s \in S: 0 \le s - m$
+Thus:
+:$T = \set {s - m: s \in S} \subseteq \N$
+The [[Well-Ordering Principle]] gives that $T$ has a [[Definition:Smallest Element|smallest element]], which we can call $b_T \in T$.
+Hence:
+:$\paren {\forall s \in S: b_T \le s - m} \land \paren {\exists b_S \in S: b_T = b_S - m}$
+So:
+{{begin-eqn}}
+{{eqn | o = 
+      | r = s \in S: b_S - m \le s - m
+      | c = 
+}}
+{{eqn | o = \leadsto
+      | r = \forall s \in S: b_S \le s
+      | c = Cancellability of elements of $\Z$
+}}
+{{eqn | o = \leadsto
+      | r = b_S \in S \land \paren {\forall s \in S: b_S \le s}
+      | c = {{Defof|Smallest Element}}
+}}
+{{end-eqn}}
+So $b_S$ is the [[Definition:Smallest Element|smallest element]] of $S$.
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Set of Integers Bounded Above by Integer has Greatest Element}
+Tags: Number Theory, Integers, Set of Integers Bounded Above has Greatest Element
+
+\begin{theorem}
+Let $\Z$ be the [[Definition:Integer|set of integers]].
+Let $\le$ be the [[Definition:Ordering on Integers|ordering on the integers]].
+Let $\O \subset S \subseteq \Z$ such that $S$ is [[Definition:Bounded Above Set|bounded above]] in $\struct {\Z, \le}$.
+Then $S$ has a [[Definition:Greatest Element|greatest element]].
+\end{theorem}
+
+\begin{proof}
+$S$ is [[Definition:Bounded Above Set|bounded above]], so $\exists M \in \Z: \forall s \in S: s \le M$.
+Hence $\forall s \in S: 0 \le M - s$.
+Thus the set $T = \set {M - s: s \in S} \subseteq \N$.
+The [[Well-Ordering Principle]] gives that $T$ has a [[Definition:Smallest Element|smallest element]], which we can call $b_T \in T$.
+Hence:
+:$\paren {\forall s \in S: b_T \le M - s} \land \paren {\exists g_S \in S: b_T = M - g_S}$
+So:
+{{begin-eqn}}
+{{eqn | o = 
+      | r = \forall s \in S: M - g_S \le M - s
+      | c = 
+}}
+{{eqn | o = \leadsto
+      | r = \forall s \in S: -g_S \le -s
+      | c = Cancellability of elements of $\Z$
+}}
+{{eqn | o = \leadsto
+      | r = \forall s \in S: g_S \ge s
+      | c = 
+}}
+{{eqn | o = \leadsto
+      | r = g_S \in S \land \paren {\forall s \in S: g_S \ge s}
+      | c = which is how the [[Definition:Greatest Element|greatest element]] is defined.
+}}
+{{end-eqn}}
+		 
+So $g_S$ is the [[Definition:Greatest Element|greatest element]] of $S$.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Since $S$ is [[Definition:Bounded Above Set|bounded above]], $\exists M \in \Z: \forall s \in S: s \le M$.
+Hence we can define the set $S' = \set {-s: s \in S}$.
+$S'$ is [[Definition:Bounded Below Set|bounded below]] by $-M$.
+So from [[Set of Integers Bounded Below by Integer has Smallest Element]], $S'$ has a [[Definition:Smallest Element|smallest element]], $-g_S$, say, where $\forall s \in S: -g_S \le -s$.
+Therefore $g_S \in S$ (by definition of $S'$) and $\forall s \in S: s \le g_S$.
+So $g_S$ is the [[Definition:Greatest Element|greatest element]] of $S$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Principle of Least Counterexample}
+Tags: Proof Techniques
+
+\begin{theorem}
+Suppose $P \paren n$ is a [[Definition:Propositional Function|condition]] on $n \in \set {x \in \Z: x \ge m \in \Z}$.
+Suppose next that: $\neg \paren {\forall n \ge m: P \paren n}$.
+(That is, not all $n \ge m$ satisfy $P \paren n$.)
+Then there is a '''least counterexample''', that is a [[Definition:Smallest Element|smallest]] [[Definition:Integer|integral]] value of $n$ for which $\neg P \paren n$.
+\end{theorem}
+
+\begin{proof}
+Let $S = \set {n \in \Z: n \ge m \in \Z: \neg P \paren n}$.
+That is, $S$ is the set of all elements in $\Z$ not less than $m$ for which the [[Definition:Propositional Function|condition]] is false.
+Since:
+:$\neg \paren {\forall n \ge m: P \paren n}$
+it follows that:
+:$S \ne \O$
+Also, $S \subseteq \Z$ and is [[Definition:Bounded Below Set|bounded below]] (by $m$).
+Therefore [[Set of Integers Bounded Below by Integer has Smallest Element|$S$ has a smallest element]], which proves the result.
+\end{proof}<|endoftext|>
+\section{Absolute Value is Bounded Below by Zero}
+Tags: Absolute Value Function
+
+\begin{theorem}
+Let $x \in \R$ be a [[Definition:Real Number|real number]].
+Then the [[Definition:Absolute Value|absolute value]] $\size x$ of $x$ is [[Definition:Bounded Below Mapping|bounded below]] by $0$.
+\end{theorem}
+
+\begin{proof}
+Let $x \ge 0$.
+Then $\size x = x \ge 0$.
+Let $x < 0$.
+Then $\size x = -x > 0$.
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Division Theorem}
+Tags: Division Theorem, Named Theorems, Divisors
+
+\begin{theorem}
+For every pair of [[Definition:Integer|integers]] $a, b$ where $b \ne 0$, there exist [[Definition:Unique|unique]] [[Definition:Integer|integers]] $q, r$ such that $a = q b + r$ and $0 \le r < \size b$:
+:$\forall a, b \in \Z, b \ne 0: \exists! q, r \in \Z: a = q b + r, 0 \le r < \size b$
+\end{theorem}
+
+\begin{proof}
+From [[Division Theorem/Positive Divisor|Division Theorem: Positive Divisor]]:
+:$\forall a, b \in \Z, b > 0: \exists! q, r \in \Z: a = q b + r, 0 \le r < b$
+That is, the result holds for [[Definition:Positive Integer|positive]] $b$.
+{{qed|lemma}}
+It remains to show that the result also holds for [[Definition:Negative Integer|negative]] values of $b$.
+Let $b < 0$.
+Consider:
+:$\size b = -b > 0$
+where $\size b$ denotes the [[Definition:Absolute Value|absolute value]] of $b$: by definition $\size b > 0$.
+From [[Division Theorem/Positive Divisor|Division Theorem: Positive Divisor]], we have the existence of $\tilde q, \tilde r \in \Z$ such that:
+:$a = \tilde q \size b + \tilde r, 0 \le \tilde r < \size b$
+Since $\size b = -b$:
+:$a = \tilde q \paren {-b} + \paren {\tilde r} = \paren {-\tilde q} b + \tilde r$
+Taking:
+:$q = -\tilde q, r = \tilde r$
+the existence has been proved of [[Definition:Integer|integers]] $q$ and $r$ that satisfy the requirements.
+The proof that they are [[Definition:Unique|unique]] is the same as that for the [[Division Theorem/Positive Divisor/Uniqueness|proof for positive $b$]], but with $\size b$ replacing $b$.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+=== Existence ===
+Consider the [[Definition:Arithmetic Sequence|arithmetic sequence]]:
+:$\ldots, a - 3 b, a - 2 b, a - b, a, a + b, a + 2 b, a + 3 b, \ldots$
+which extends in both directions.
+Then by the [[Well-Ordering Principle]], there must exist a smallest [[Definition:Positive Integer|non-negative]] element, denoted by $r$.
+So $r = a - q b$ for some $q \in \Z$.
+$r$ must be in the interval $\hointr 0 b$ because otherwise $r - b$ would be smaller than $r$ and a [[Definition:Positive Integer|non-negative]] element in the [[Definition:Arithmetic Sequence|sequence]].
+{{qed|lemma}}
+=== Uniqueness ===
+Suppose we have another pair $q_0$ and $r_0$ such that $a = b q_0 + r_0$, with $0 \le r_0 < b$.
+Then:
+:$b q + r = b q_0 + r_0$
+Factoring we see that:
+:$r - r_0 = b \paren {q_0 - q}$
+and so:
+:$b \divides \paren {r - r_0}$
+Since $0 \le r < b$ and $0 \le r_0 < b$, we have that:
+:$-b < r - r_0 < b$
+Hence:
+:$r - r_0 = 0 \implies r = r_0$
+So now:
+:$r - r_0 = 0 = b \paren {q_0 - q}$
+which implies that:
+:$q = q_0$
+Therefore the solution is [[Definition:Unique|unique]].
+{{Qed}}
+\end{proof}
+
+\begin{proof}
+From the equation $0_F = 0_F \circ d + 0_F$, the theorem is true for the trivial case $f = 0_F$.
+So, if there is a [[Definition:Counterexample|counterexample]] to be found, it will have a [[Definition:Degree of Polynomial over Field|degree]].
+{{AimForCont}} there exists at least one [[Definition:Counterexample|counterexample]].
+By a version of the [[Well-Ordering Principle]], we can assign a number $m$ to the lowest [[Definition:Degree of Polynomial over Field|degree]] possessed by any [[Definition:Counterexample|counterexample]].
+So, let $f$ denote a [[Definition:Counterexample|counterexample]] which has that minimum [[Definition:Degree of Polynomial over Field|degree]] $m$.
+If $m < n$, the equation $f = 0_F \circ d + f$ would show that $f$ was not a [[Definition:Counterexample|counterexample]].
+Therefore $m \ge n$.
+Suppose $d \divides f$ in $F \sqbrk X$.
+Then:
+:$\exists q \in F \sqbrk X: f = q \circ d + 0_F$
+and $f$ would not be a [[Definition:Counterexample|counterexample]].
+So $d \nmid f$ in $F \sqbrk X$.
+So, suppose that:
+{{begin-eqn}}
+{{eqn | l = f
+      | r = \sum_{k \mathop = 0}^m {a_k \circ X^k}
+}}
+{{eqn | l = d
+      | r = \sum_{k \mathop = 0}^n {b_k \circ X^k}
+}}
+{{eqn | l = m
+      | o = \ge
+      | r = n
+}}
+{{end-eqn}}
+We can create the [[Definition:Polynomial in Ring Element|polynomial]] $\paren {a_m \circ b_n^{-1} \circ X^{m - n} } \circ d$ which has the same [[Definition:Degree of Polynomial over Field|degree]] and [[Definition:Leading Coefficient of Polynomial|leading coefficient]] as $f$.
+Thus $f_1 = f - \paren {a_m \circ b_n^{-1} \circ X^{m - n} } \circ d$ is a [[Definition:Polynomial in Ring Element|polynomial]] of [[Definition:Degree of Polynomial over Field|degree]] less than $m$.
+Since $d \nmid f$, $f_1$ is a non-[[Definition:Null Polynomial over Ring|zero polynomial]].
+There is no [[Definition:Counterexample|counterexample]] of [[Definition:Degree of Polynomial over Field|degree]] less than $m$.
+Therefore:
+:$f_1 = q_1 \circ d + r$
+for some $q_1, r \in F \sqbrk X$, where either:
+:$r = 0_F$
+or:
+:$r$ is non-[[Definition:Null Polynomial over Ring|zero]] with [[Definition:Degree of Polynomial over Field|degree]] strictly less than $n$.
+Hence:
+{{begin-eqn}}
+{{eqn | l = f
+      | r = f_1 + \paren {a_m \circ b_n^{-1} \circ X^{m - n} } \circ d
+      | c = 
+}}
+{{eqn | r = \paren {q_1 + a_m \circ b_n^{-1} \circ X^{m - n} } \circ d + r
+      | c = 
+}}
+{{end-eqn}}
+Thus $f$ is not a [[Definition:Counterexample|counterexample]].
+From this [[Proof by Contradiction|contradiction]] follows the result.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Suppose $\map \deg f < \map \deg d$.
+Then we take $\map q X = 0$ and $\map r X = \map a X$ and the result holds.
+Otherwise, $\map \deg f \ge \map \deg d$.
+Let:
+:$\map f X = a_0 + a_1 X + a_2 x^2 + \cdots + a_m X^m$
+:$\map d X = b_0 + b_1 X + b_2 x^2 + \cdots + b_n X^n$
+We can subtract from $f$ a suitable multiple of $d$ so as to eliminate the highest term in $f$:
+:$\map f X - \map d X \cdot \dfrac {a_m} {b_n} x^{m - n} = \map p X$
+where $\map p X$ is some polynomial whose [[Definition:Degree of Polynomial over Field|degree]] is less than that of $f$.
+If $\map p X$ still has [[Definition:Degree of Polynomial over Field|degree]] higher than that of $d$, we do the same thing again.
+Eventually we reach:
+:$\map f X - \map d X \cdot \paren {\dfrac {a_m} {b_n} x^{m - n} + \dotsb} = \map r X$
+where either $r = 0_F$ or $r$ has [[Definition:Degree of Polynomial over Field|degree]] that is less than $n$.
+This approach can be formalised using the [[Principle of Complete Induction]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Proof by [[Principle of Mathematical Induction|induction]]:
+For all $n \in \N_{> 0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
+:$\forall f \in F \sqbrk X: \exists q, r \in F \sqbrk X: f = q \circ d + r$ provided that $\map \deg f < n$
+=== Basis for the Induction ===
+$\map P 0$ is the statement that $q$ and $r$ exist when $f = 0$.
+This is shown trivially to be true by taking $q = r = 0$.
+This is our [[Definition:Basis for the Induction|basis for the induction]].
+=== Induction Hypothesis ===
+Now we need to show that, if $\map P k$ is true, where $k \ge 0$, then it logically follows that $\map P {k + 1}$ is true.
+So this is our [[Definition:Induction Hypothesis|induction hypothesis]]:
+:$\forall f \in F \sqbrk X: \exists q, r \in F \sqbrk X: f = q \circ d + r$ provided that $\map \deg f < k$
+Then we need to show:
+:$\forall f \in F \sqbrk X: \exists q, r \in F \sqbrk X: f = q \circ d + r$ provided that $\map \deg f < k + 1$
+=== Induction Step ===
+This is our [[Definition:Induction Step|induction step]]:
+Let $f$ be such that $\map \deg f = n$.
+Let:
+:$f = a_0 + a_1 \circ x + a_2 \circ x^2 + \cdots + a_n \circ x^n$ where $a_n \ne 0$
+Let:
+:$d = b_0 + b_1 \circ x + b_2 \circ x^2 + \cdots + b_j \circ x^j$ where $b_j \ne 0$
+If $n < l$ then take $q = 0, r = f$.
+If $n \ge l$, consider:
+:$c := f - a_n b_j^{-1} x^{n - j} d$
+This has been carefully arranged so that the [[Definition:Polynomial Coefficient|coefficient]] of $x^n$ in $c$ is zero.
+Thus $\map \deg c < n$.
+Therefore, by the [[Division Theorem for Polynomial Forms over Field/Proof 3#Induction Hypothesis|induction hypothesis]]:
+:$c = d q_0 + r$
+where $\map \deg r < \map \deg d$.
+Therefore:
+{{begin-eqn}}
+{{eqn | l = f
+      | r = d \paren {q_0 + a_n b_j^{-1} x^{n - j} } + r
+      | c = 
+}}
+{{eqn | r = d q + r
+      | c = where $\map \deg r < \map \deg d$ and $q = q_0 + a_n b_j^{-1} x^{n - j}$
+}}
+{{end-eqn}}
+Thus the existence of $q$ and $r$ have been established.
+As for uniqueness, assume:
+:$d q + r = d q' + r'$
+with $\map \deg r < \map \deg d, \map \deg {r'} < \map \deg d$
+Then:
+:$d \paren {q - q'} = r' - r$
+By [[Degree of Sum of Polynomials]]:
+:$\map \deg {r' - r} \le \max \set {\map \deg {r'}, \map \deg r} < \map \deg d$
+and by [[Degree of Product of Polynomials over Integral Domain]]:
+:$\map \deg {d \paren {q - q'} } = \map \deg d + \map \deg {q - q'}$
+That is:
+:$\map \deg d < \map \deg d + \map \deg {q - q'}$
+and the only way for that to happen is for:
+:$\map \deg {q - q'} = -\infty$
+that is, for $q - q'$ to be the [[Definition:Null Polynomial|null polynomial]].
+That is, $q - q' = 0_F$ and by a similar argument $r' - r = 0_F$, demonstrating the uniqueness of $q$ and $r$.
+So $\map P k \implies \map P {k + 1}$ and the result follows by the [[Principle of Mathematical Induction]].
+Therefore:
+:$\forall f \in F \sqbrk X: \exists q, r \in F \sqbrk X: f = q \circ d + r$ such that either:
+::$(1): \quad r = 0_F$
+:or:
+::$(2): \quad r \ne 0_F$ and $r$ has [[Definition:Degree of Polynomial over Field|degree]] that is less than $n$.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Consider the set of [[Definition:Divisor of Integer|integer multiples]] $x \size b$ of $\size b$ less than or equal to $a$:
+:$M := \set {k \in \Z: \exists x \in \Z: k = x \size b, k \le a}$
+We have that:
+:$-\size a \size b \le -\size a \le a$
+and so $M \ne \O$.
+From [[Set of Integers Bounded Above by Integer has Greatest Element]], $M$ has a [[Definition:Greatest Element|greatest element]] $h \size b$.
+Then $h \size b \le a$ and so:
+:$a = h \size b + r$
+where $r \ge 0$.
+On the other hand:
+:$\paren {h + 1} \size b = h \size b + \size b > h \size b$
+So:
+:$\paren {h + 1} \size b > a$
+and:
+:$h \size b + \size b > h \size b + r$
+Thus:
+:$r \le b$
+Setting:
+:$q = h$ when $b > 0$
+:$q = -h$ when $b < 0$
+it follows that:
+:$h \size b = q b$
+and so:
+:$a = q b + r$
+as required.
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Odd Integer 2n + 1}
+Tags: Odd Integers
+
+\begin{theorem}
+Let $m$ be an [[Definition:Odd Integer|odd integer]].
+Then there exists [[Definition:Exactly One|exactly one]] [[Definition:Integer|integer]] $n$ such that $2 n + 1 = m$.
+\end{theorem}
+
+\begin{proof}
+Follows directly from the [[Division Theorem]].
+{{qed}}
+[[Category:Odd Integers]]
+mpi9q6vy7fvjpowo6jxu7u75i7r217j
+\end{proof}<|endoftext|>
+\section{Integer Divisor Results}
+Tags: Divisors, Integers, Integer Divisor Results
+
+\begin{theorem}
+Let $m, n \in \Z$ be [[Definition:Integer|integers]].
+Let $m \divides n$ denote that $m$ is a [[Definition:Divisor of Integer|divisor]] of $n$.
+The following results all hold:
+\end{theorem}<|endoftext|>
+\section{Zero Divides Zero}
+Tags: Divisors
+
+\begin{theorem}
+Let $n \in \Z$ be an [[Definition:Integer|integer]].
+Then:
+:$0 \divides n \implies n = 0$
+That is, [[Definition:Zero (Number)|zero]] is the only [[Definition:Integer|integer]] [[Definition:Divisor of Integer|divisible]] by [[Definition:Zero (Number)|zero]].
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = 0
+      | o = \divides
+      | r = n
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \exists q \in \Z: n
+      | r = q \times 0
+      | c = {{Defof|Divisor of Integer}}
+}}
+{{eqn | ll= \leadsto
+      | l = n
+      | r = 0
+      | c = Integers have no [[Definition:Zero Divisor of Ring|zero divisors]], as [[Integers form Integral Domain]].
+}}
+{{end-eqn}}
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Absolute Value of Integer is not less than Divisors}
+Tags: Number Theory, Divisors
+
+\begin{theorem}
+A (non-[[Definition:Zero (Number)|zero]]) [[Definition:Integer|integer]] is greater than or equal to its [[Definition:Divisor of Integer|divisors]] in [[Definition:Absolute Value|magnitude]]:
+:$\forall c \in \Z_{\ne 0}: a \divides c \implies a \le \size a \le \size c$
+\end{theorem}
+
+\begin{proof}
+Suppose $a \divides c$ for some $c \ne 0$.
+From [[Negative of Absolute Value]]:
+: $a \le \size a$
+Then:
+{{begin-eqn}}
+{{eqn | l = a
+      | o = \divides
+      | r = c
+      | c = 
+}}
+{{eqn | lll=\leadsto
+      | ll= \exists q \in \Z:
+      | l = c
+      | r = a q
+      | c = {{Defof|Divisor of Integer}}
+}}
+{{eqn | lll=\leadsto
+      | l = \size c
+      | r = \size a \size q
+      | c = 
+}}
+{{eqn | lll=\leadsto
+      | l = \size a \size q \ge \size a \times 1
+      | r = \size a
+      | c = 
+}}
+{{eqn | lll=\leadsto
+      | l = a \le \size a
+      | o = \le
+      | r = \size c
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Divisor Relation on Positive Integers is Partial Ordering}
+Tags: Divisors, Integers, Order Theory
+
+\begin{theorem}
+The [[Definition:Divisor of Integer|divisor relation]] is a [[Definition:Partial Ordering|partial ordering]] of $\Z_{>0}$.
+\end{theorem}
+
+\begin{proof}
+Checking in turn each of the criteria for an [[Definition:Ordering|ordering]]:
+=== [[Integer Divisor Results/Integer Divides Itself/Proof 1|Divisor Relation is Reflexive]] ===
+{{:Integer Divisor Results/Integer Divides Itself/Proof 1}}
+=== [[Divisor Relation is Transitive]] ===
+:$\forall x, y, z \in \Z: x \divides y \land y \divides z \implies x \divides z$:
+{{:Divisor Relation is Transitive/Proof 2}}
+=== [[Divisor Relation is Antisymmetric]] ===
+{{:Divisor Relation is Antisymmetric}}
+=== Divisor Ordering is Partial ===
+Let $a = 2$ and $b = 3$.
+Then neither $a \divides b$ nor $b \divides a$.
+Thus, while the [[Definition:Divisor of Integer|divisor relation]] is an [[Definition:Ordering|ordering]], it is specifically a [[Definition:Partial Ordering|partial ordering]]
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Common Divisor in Integral Domain Divides Linear Combination}
+Tags: Divisibility
+
+\begin{theorem}
+Let $\struct {D, +, \times}$ be an [[Definition:Integral Domain|integral domain]].
+Let $c$ be a [[Definition:Common Divisor of Ring Elements|common divisor]] of two elements $a$ and $b$ of $D$.
+That is:
+:$a, b, c \in D: c \divides a \land c \divides b$
+Then:
+:$\forall p, q \in D: c \divides \paren {p \times a + q \times b}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = c
+      | o = \divides
+      | r = a
+}}
+{{eqn | ll= \leadsto
+      | lo= \exists x \in D:
+      | l = a
+      | r = x \times c
+      | c = {{Defof|Divisor of Ring Element}}
+}}
+{{eqn | l = c
+      | o = \divides
+      | r = b
+}}
+{{eqn | ll= \leadsto
+      | lo= \exists y \in D:
+      | l = b
+      | r = y \times c
+      | c = {{Defof|Divisor of Ring Element}}
+}}
+{{eqn | ll= \leadsto
+      | lo= \forall p, q \in D:
+      | l = p \times a + q \times b
+      | r = p \times x \times c + q \times y \times c
+      | c = substituting for $a$ and $b$
+}}
+{{eqn | r = \paren {p \times x + q \times y} c
+      | c = as $\times$ is [[Definition:Distributive Operation|distributive]] over $+$
+}}
+{{eqn | ll= \leadsto
+      | lo= \exists z \in D:
+      | l = p \times a + q \times b
+      | r = z \times c
+      | c = where $z = p \times x + q \times y$
+}}
+{{eqn | ll= \leadsto
+      | l = c
+      | o = \divides
+      | r = \paren {p \times a + q \times b}
+      | c = {{Defof|Divisor of Ring Element}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Existence of Greatest Common Divisor}
+Tags: Greatest Common Divisor, Existence of Greatest Common Divisor
+
+\begin{theorem}
+Let $a, b \in \Z$ be [[Definition:Integer|integers]] such that $a \ne 0$ or $b \ne 0$.
+Then the [[Definition:Greatest Common Divisor of Integers|greatest common divisor]] of $a$ and $b$ exists.
+\end{theorem}
+
+\begin{proof}
+=== Proof of Existence ===
+We have that:
+:$\forall a, b \in \Z: 1 \divides a \land 1 \divides b$
+so $1$ is always a [[Definition:Common Divisor of Integers|common divisor]] of any two [[Definition:Integer|integers]].
+{{qed|lemma}}
+=== Proof of there being a Largest ===
+As the definition of $\gcd$ shows that it is symmetric, we can assume without loss of generality that $a \ne 0$.
+First we note that from [[Absolute Value of Integer is not less than Divisors]]:
+:$\forall c \in \Z: \forall a \in \Z_{>0}: c \divides a \implies c \le \size c \le \size a$
+The same applies for $c \divides b$.
+Now we have three different results depending on $a$ and $b$:
+{{begin-eqn}}
+{{eqn | l = a \ne 0 \land b \ne 0
+      | o = \implies
+      | r = \gcd \set {a, b} \le \min \set {\size a, \size b}
+      | c = 
+}}
+{{eqn | l = a = 0 \lor b = 0
+      | o = \implies
+      | r = \gcd \set {a, b} = \max \set {\size a, \size b}
+      | c = 
+}}
+{{eqn | l = a = b = 0
+      | o = \implies
+      | r = \forall x \in \Z: x \divides a \land x \divides b
+      | c = 
+}}
+{{end-eqn}}
+So if $a$ and $b$ are ''both'' [[Definition:Zero (Number)|zero]], then ''any'' $n \in \Z$ divides both, and there is no [[Definition:Greatest Common Divisor|greatest common divisor]].
+This is why the proviso that $a \ne 0 \lor b \ne 0$.
+So we have proved that [[Definition:Common Divisor|common divisors]] exist and are [[Definition:Bounded Above Set|bounded above]].
+Therefore, from [[Set of Integers Bounded Above by Integer has Greatest Element]] there is always a [[Definition:Greatest Common Divisor|'''greatest''' common divisor]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+By definition of [[Definition:Greatest Common Divisor|greatest common divisor]], we aim to show that there exists $c \in \Z_{>0}$ such that:
+{{begin-eqn}}
+{{eqn | l = c
+      | o = \divides
+      | r = a
+}}
+{{eqn | l = c
+      | o = \divides
+      | r = b
+}}
+{{end-eqn}}
+and:
+:$d \divides a, d \divides b \implies d \divides c$
+Consider the [[Definition:Set|set]] $S$:
+:$S = \set {s \in \Z_0: \exists x, y \in \Z: s = a x + b y}$
+$S$ is not [[Definition:Empty Set|empty]], because by setting $x = 1$ and $y = 0$ we have at least that $a \in S$.
+From the [[Well-Ordering Principle]], there exists a smallest $c \in S$.
+So, by definition, we have $c > 0$ is the smallest such that $c = a x + b y$ for some $x, y \in \Z$.
+Let $d$ be such that $d \divides a$ and $d \divides b$.
+Then from [[Common Divisor Divides Integer Combination]]:
+:$d \divides a x + b y$
+That is:
+:$d \divides c$
+We have that:
+{{begin-eqn}}
+{{eqn | lo= \exists t, u \in \Z:
+      | l = a
+      | r = c t + u:
+      | rr= 0 \le u < c
+      | c = [[Division Theorem]]
+}}
+{{eqn | ll= \leadsto
+      | l = a
+      | r = a x t + b y t + u
+      | c = Definition of $c$
+}}
+{{eqn | ll= \leadsto
+      | l = u
+      | r = r \paren {1 - x t} + b \paren {-y t} 
+      | c = rearranging
+}}
+{{eqn | ll= \leadsto
+      | l = u
+      | r = 0
+      | c = as $u < c$ and the definition of $c$
+}}
+{{eqn | ll= \leadsto
+      | l = c
+      | o = \divides
+      | r = a
+      | c = {{Defof|Divisor of Integer}}
+}}
+{{end-eqn}}
+[[Definition:Mutatis Mutandis|Mutatis mutandis]]:
+:$c \divides b$
+Now suppose $c'$ is such that:
+{{begin-eqn}}
+{{eqn | l = c'
+      | o = \divides
+      | r = a
+}}
+{{eqn | l = c'
+      | o = \divides
+      | r = b
+}}
+{{end-eqn}}
+and:
+:$d \divides a, d \divides b \implies d \divides c'$
+Then we have immediately that:
+:$c' \divides c$
+and by the same coin:
+$c \divides c'$
+and so:
+:$c = c'$
+demonstrating that $c$ is [[Definition:Unique|unique]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Greatest Common Divisor is at least 1}
+Tags: Greatest Common Divisor
+
+\begin{theorem}
+Let $a, b \in \Z$ be [[Definition:Integer|integers]].
+The [[Definition:Greatest Common Divisor of Integers|greatest common divisor]] of $a$ and $b$ is at least $1$:
+:$\forall a, b \in \Z_{\ne 0}: \gcd \set {a, b} \ge 1$
+\end{theorem}
+
+\begin{proof}
+From [[One Divides all Integers]]:
+:$\forall a, b \in \Z: 1 \divides a \land 1 \divides b$
+and so:
+:$1 \le \gcd \set {a, b}$
+as required.
+{{qed}}
+[[Category:Greatest Common Divisor]]
+2nrji97dz25wb80mslueng8xuvr7oza
+\end{proof}<|endoftext|>
+\section{GCD of Integer and Divisor}
+Tags: Greatest Common Divisor
+
+\begin{theorem}
+Let $a, b \in \Z_{>0}$, i.e. [[Definition:Integer|integers]] such that $a, b > 0$.
+Then:
+: $a \divides b \implies \gcd \set {a, b} = a$
+\end{theorem}
+
+\begin{proof}
+$a \divides b$ by hypothesis, $a \divides a$ from [[Integer Divides Itself]].
+Thus $a$ is a [[Definition:Common Divisor of Integers|common divisor]] of $a$ and $b$.
+Note that from [[Absolute Value of Integer is not less than Divisors]]:
+:$\forall x \in \Z: x \divides a \implies x \le \size a$.
+As $a$ and $b$ are both [[Definition:Positive Integer|positive]], the result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{GCD for Negative Integers}
+Tags: Greatest Common Divisor
+
+\begin{theorem}
+:$\gcd \set {a, b} = \gcd \set {\size a, b} = \gcd \set {a, \size b} = \gcd \set {\size a, \size b}$
+Alternatively, this can be put:
+:$\gcd \set {a, b} = \gcd \set {-a, b} = \gcd \set {a, -b} = \gcd \set {-a, -b}$
+ 
+which follows directly from the above.
+\end{theorem}
+
+\begin{proof}
+Note that $\size a = \pm a$.
+Suppose that:
+:$u \divides a$
+where $\divides$ denotes [[Definition:Divisor of Integer|divisibility]].
+Then:
+:$\exists q \in \Z: a = q u$
+Then:
+:$\size a = \pm q u = \paren {\pm q} u \implies u \divides \size a$
+So every [[Definition:Divisor of Integer|divisor]] of $a$ is a [[Definition:Divisor of Integer|divisor]] of $\size a$.
+Similarly, note that:
+:$a = \pm \size a$
+so every [[Definition:Divisor of Integer|divisor]] of $\size a$ is a [[Definition:Divisor of Integer|divisor]] of $a$.
+So it follows that the [[Definition:Common Divisor of Integers|common divisors]] of $a$ and $b$ are the same as those of $a$ and $\size b$, and so on.
+In particular:
+:$\gcd \set {a, b} = \gcd \set {a, \size b}$
+and so on.
+{{Qed}}
+[[Category:Greatest Common Divisor]]
+le9bxzi7pir8rzelwcgd7h3vqlvza11
+\end{proof}<|endoftext|>
+\section{GCD with Zero}
+Tags: Greatest Common Divisor
+
+\begin{theorem}
+Let $a \in \Z$ be an [[Definition:Integer|integer]] such that $a \ne 0$.
+Then:
+:$\gcd \left\{{a, 0}\right\} = \left\lvert{a}\right\rvert$
+where $\gcd$ denotes [[Definition:Greatest Common Divisor of Integers|greatest common divisor (GCD)]].
+\end{theorem}
+
+\begin{proof}
+Follows from:
+: [[Integer Divides Zero]]
+: [[GCD for Negative Integers]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Set of Integer Combinations equals Set of Multiples of GCD}
+Tags: Integer Combinations, Greatest Common Divisor
+
+\begin{theorem}
+The [[Definition:Set|set]] of all [[Definition:Integer Combination|integer combinations]] of $a$ and $b$ is precisely the set of all [[Definition:Divisor of Integer|integer multiples]] of the [[Definition:Greatest Common Divisor of Integers|GCD]] of $a$ and $b$:
+:$\gcd \set {a, b} \divides c \iff \exists x, y \in \Z: c = x a + y b$
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+Let $d = \gcd \set {a, b}$.
+Then:
+:$d \divides c \implies \exists m \in \Z: c = m d$
+So:
+{{begin-eqn}}
+{{eqn | l = \exists p, q \in \Z: d
+      | r = p a + q b
+      | c = [[Bézout's Lemma]]
+}}
+{{eqn | ll= \leadsto
+      | l = c
+      | r = m d
+      | c = 
+}}
+{{eqn | r = m p a + m q b
+      | c = 
+}}
+{{eqn | r = \paren {m p} a + \paren {m q} b
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \exists x, y \in \Z: c
+      | r = x a + y b
+      | c = 
+}}
+{{end-eqn}}
+Thus:
+:$\gcd \set {a, b} \divides c \implies \exists x, y \in \Z: c = x a + y b$
+{{qed|lemma}}
+=== Sufficient Condition ===
+Suppose $\exists x, y \in \Z: c = x a + y b$.
+From [[Common Divisor Divides Integer Combination]], we have:
+:$\gcd \set {a, b} \divides \paren {x a + y b}$
+It follows directly that $\gcd \set {a, b} \divides c$ and the proof is finished.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{GCD with Remainder}
+Tags: Greatest Common Divisor
+
+\begin{theorem}
+Let $a, b \in \Z$.
+Let $q, r \in \Z$ such that $a = q b + r$.
+Then:
+:$\gcd \set {a, b} = \gcd \set {b, r}$
+where $\gcd \set {a, b}$ is the [[Definition:Greatest Common Divisor of Integers|greatest common divisor]] of $a$ and $b$.
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | o = 
+      | r = \gcd \set {a, b} \divides a \land \gcd \set {a, b} \divides b
+      | c = {{Defof|Greatest Common Divisor of Integers}}
+}}
+{{eqn | ll= \leadsto
+      | o = 
+      | r = \gcd \set {a, b} \divides \paren {a - q b}
+      | c = [[Common Divisor Divides Integer Combination]]
+}}
+{{eqn | ll= \leadsto
+      | o = 
+      | r = \gcd \set {a, b} \divides r
+      | c = as $r = a - q b$
+}}
+{{eqn | ll= \leadsto
+      | o = 
+      | r = \gcd \set {a, b} \le \gcd \set {b, r}
+      | c = {{Defof|Greatest Common Divisor of Integers}}
+}}
+{{end-eqn}}
+The argument works the other way about:
+{{begin-eqn}}
+{{eqn | o = 
+      | r = \gcd \set {b, r} \divides b \land \gcd \set {b, r} \divides r
+      | c = {{Defof|Greatest Common Divisor of Integers}}
+}}
+{{eqn | ll= \leadsto
+      | o = 
+      | r = \gcd \set {b, r} \divides \paren {q b + r}
+      | c = [[Common Divisor Divides Integer Combination]]
+}}
+{{eqn | ll= \leadsto
+      | o = 
+      | r = \gcd \set {b, r} \divides a
+      | c = as $a = q b + r$
+}}
+{{eqn | ll= \leadsto
+      | o = 
+      | r = \gcd \set {b, r} \le \gcd \set {a, b}
+      | c = {{Defof|Greatest Common Divisor of Integers}}
+}}
+{{end-eqn}}
+Thus:
+:$\gcd \set {a, b} = \gcd \set {b, r}$
+{{qed}}
+[[Category:Greatest Common Divisor]]
+52xifv9uor3uhps3z23ov9u5wtec8hn
+\end{proof}<|endoftext|>
+\section{Integer Combination of Coprime Integers}
+Tags: Coprime Integers, Integer Combinations, Integer Combination of Coprime Integers
+
+\begin{theorem}
+Two [[Definition:Integer|integers]] are [[Definition:Coprime Integers|coprime]] {{iff}} there exists an [[Definition:Integer Combination|integer combination]] of them equal to $1$:
+:$\forall a, b \in \Z: a \perp b \iff \exists m, n \in \Z: m a + n b = 1$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = a
+      | o = \perp
+      | r = b
+}}
+{{eqn | ll= \leadsto
+      | l = \gcd \set {a, b}
+      | r = 1
+      | c = {{Defof|Coprime Integers}}
+}}
+{{eqn | ll= \leadsto
+      | l = \exists m, n \in \Z: m a + n b
+      | r = 1
+      | c = [[Bézout's Lemma]]
+}}
+{{end-eqn}}
+Then we have:
+{{begin-eqn}}
+{{eqn | l = \exists m, n \in \Z: m a + n b
+      | r = 1
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \gcd \set {a, b}
+      | o = \divides
+      | r = 1
+      | c = [[Set of Integer Combinations equals Set of Multiples of GCD]]
+}}
+{{eqn | ll= \leadsto
+      | l = \gcd \set {a, b}
+      | r = 1
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = a
+      | o = \perp
+      | r = b
+      | c = {{Defof|Coprime Integers}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+=== Sufficient Condition ===
+Let $a, b \in \Z$ be such that $\exists m, n \in \Z: m a + n b = 1$.
+Let $d$ be a [[Definition:Divisor of Integer|divisor]] of both $a$ and $b$.
+Then:
+:$d \divides m a + n b$
+and so:
+:$d \divides 1$
+Thus:
+:$d = \pm 1$
+and so:
+:$\gcd \set {a, b} = 1$
+Thus, by definition, $a$ and $b$ are [[Definition:Coprime Integers|coprime]].
+{{qed|lemma}}
+=== Necessary Condition ===
+Let $a \perp b$.
+Thus they are not both $0$.
+Let $S$ be defined as:
+:$S = \set {a m + b n: m, n \in \Z}$
+$S$ contains at least one [[Definition:Strictly Positive Integer|strictly positive integer]], because for example $a^2 + b^2 \in S$.
+By [[Set of Integers Bounded Below has Smallest Element]], let $d$ be the [[Definition:Smallest Element|smallest element]] of $S$ which is [[Definition:Strictly Positive Integer|strictly positive]].
+Let $d = a x + b y$.
+It remains to be shown that $d = 1$.
+By the [[Division Theorem]]:
+:$a = d q + r$ where $0 \le r < d$
+Then:
+{{begin-eqn}}
+{{eqn | l = r
+      | r = a - d q
+      | c = 
+}}
+{{eqn | r = a - \paren {a x + b y} q
+      | c = 
+}}
+{{eqn | r = a \paren {1 - x q} + b \paren {- y q}
+      | c = 
+}}
+{{eqn | o = \in
+      | r = S
+      | c = 
+}}
+{{end-eqn}}
+But we have that $0 \le r < d$.
+We have defined $d$ as the [[Definition:Smallest Element|smallest element]] of $S$ which is [[Definition:Strictly Positive Integer|strictly positive]]
+Hence it follows that $r$ cannot therefore be [[Definition:Strictly Positive Integer|strictly positive]] itself.
+Hence $r = 0$ and so $a = d q$.
+That is:
+:$d \divides a$
+By a similar argument:
+:$d \divides b$
+and so $d$ is a [[Definition:Common Divisor of Integers|common divisor]] of both $a$ and $b$.
+But the [[Definition:Greatest Common Divisor|GCD]] of $a$ and $b$ is $1$.
+Thus it follows that, as $d \in S$:
+:$\exists m, n \in \Z: m a + n b = 1$
+{{qed}}
+\end{proof}
+
+\begin{proof}
+=== Sufficient Condition ===
+Let $a, b \in \Z$ be such that $\exists m, n \in \Z: m a + n b = 1$.
+Let $d$ be a [[Definition:Divisor of Integer|divisor]] of both $a$ and $b$.
+Then:
+:$d \mathrel \backslash m a + n b$
+and so:
+:$d \mathrel \backslash 1$
+Thus:
+: $d = \pm 1$
+and so:
+: $\gcd \left\{ {a, b}\right\} = 1$
+Thus, by definition, $a$ and $b$ are [[Definition:Coprime Integers|coprime]].
+{{qed|lemma}}
+=== Necessary Condition ===
+Let $a \perp b$.
+Thus they are not both $0$.
+Let $S$ be defined as:
+:$S = \left\{ {\lambda a + \mu b: \lambda, \mu \in \Z}\right\}$
+$S$ contains at least one [[Definition:Strictly Positive Integer|strictly positive integer]], because for example:
+: $a \in S$ (setting $\lambda = 1$ and $\mu = 0$)
+: $b \in S$ (setting $\lambda = 0$ and $\mu = 1$)
+By [[Set of Integers Bounded Below has Smallest Element]], let $d$ be the [[Definition:Smallest Element|smallest element]] of $S$ which is [[Definition:Strictly Positive Integer|strictly positive]].
+Let $d = \alpha a + \beta b$.
+Let $c \in S$, such that $\lambda_0 a + \mu_0 b = c$ for some $\lambda_0, \mu_0 \in \Z$.
+By the [[Division Algorithm]]:
+:$\exists \gamma, \delta \in \Z: c = \gamma d + \delta$
+where $0 \le \delta < d$
+Then:
+{{begin-eqn}}
+{{eqn | l = \delta
+      | r = c - \gamma d
+      | c = 
+}}
+{{eqn | r = \left({\lambda_0 a + \mu_0 b}\right) - \gamma \left({\alpha a + \beta b}\right)
+      | c = 
+}}
+{{eqn | r = \left({\lambda_0 - \gamma \alpha}\right) a + \left({\mu_0 - \gamma \beta}\right) b
+      | c = 
+}}
+{{eqn | o = \in
+      | r = S
+      | c = 
+}}
+{{end-eqn}}
+But we have that $0 \le \delta < d$.
+We have defined $d$ as the [[Definition:Smallest Element|smallest element]] of $S$ which is [[Definition:Strictly Positive Integer|strictly positive]]
+Hence it follows that $\delta$ cannot therefore be [[Definition:Strictly Positive Integer|strictly positive]] itself.
+Hence $\delta = 0$ and so $c = \gamma d$.
+That is:
+:$d \mathrel \backslash c$
+and so the [[Definition:Smallest Element|smallest element]] of $S$ which is [[Definition:Strictly Positive Integer|strictly positive]] is a [[Definition:Divisor of Integer|divisor]] of both $a$ and $b$.
+But $a$ and $b$ are [[Definition:Coprime Integers|coprime]].
+Thus it follows that, as $d \mathrel \backslash 1$:
+:$d = 1$
+and so, by definition of $S$:
+:$\exists m, n \in \Z: m a + n b = 1$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Divisor of Sum of Coprime Integers}
+Tags: Number Theory, Divisors, Coprime Integers
+
+\begin{theorem}
+Let $a, b, c \in \Z_{>0}$ such that:
+:$a \perp b$ and $c \divides \paren {a + b}$.
+where:
+:$a \perp b$ denotes $a$ and $b$ are [[Definition:Coprime Integers|coprime]]
+:$c \divides \paren {a + b}$ denotes that $c$ is a [[Definition:Divisor of Integer|divisor]] of $a + b$.
+Then $a \perp c$ and $b \perp c$.
+That is, a [[Definition:Divisor of Integer|divisor]] of the sum of two [[Definition:Coprime Integers|coprime integers]] is [[Definition:Coprime Integers|coprime]] to both.
+\end{theorem}
+
+\begin{proof}
+Let $d \in \Z_{>0}: d \divides c \land d \divides a$.
+Then:
+{{begin-eqn}}
+{{eqn | l = d
+      | o = \divides
+      | r = \paren {a + b}
+      | c = as $c \divides \paren {a + b}$
+}}
+{{eqn | ll= \leadsto
+      | l = d
+      | o = \divides
+      | r = \paren {a + b - a}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = d
+      | o = \divides
+      | r = b
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = d
+      | r = 1
+      | c = as $d \divides a$ and $d \divides b$ which are [[Definition:Coprime Integers|coprime]]
+}}
+{{end-eqn}}
+A similar argument shows that if $d \divides c \land d \divides b$ then $d \divides a$.
+It follows that:
+: $\gcd \set {a, c} = \gcd \set {b, c} = 1$
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Integers Divided by GCD are Coprime}
+Tags: Greatest Common Divisor, Coprime Integers
+
+\begin{theorem}
+Any pair of [[Definition:Integer|integers]], not both zero, can be reduced to a pair of [[Definition:Coprime Integers|coprime]] ones by dividing them by their [[Definition:Greatest Common Divisor of Integers|GCD]]:
+:$\gcd \set {a, b} = d \iff \dfrac a d, \dfrac b d \in \Z \land \gcd \set {\dfrac a d, \dfrac b d} = 1$
+That is:
+:$\dfrac a {\gcd \set {a, b} } \perp \dfrac b {\gcd \set {a, b} }$
+\end{theorem}
+
+\begin{proof}
+Let $d = \gcd \set {a, b}$.
+We have:
+:$d \divides a \iff \exists s \in \Z: a = d s$
+:$d \divides b \iff \exists t \in \Z: b = d t$
+So:
+{{begin-eqn}}
+{{eqn | l = \exists m, n \in \Z: d
+      | r = m a + n b
+      | c = [[Bézout's Identity]]
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = d
+      | r = m d s + n d t
+      | c = Definition of $s$ and $t$
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = 1
+      | r = m s + n t
+      | c = dividing through by $d$
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \gcd \set {s, t}
+      | r = 1
+      | c = [[Bézout's Identity]]
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \gcd \set {\frac a d, \frac b d}
+      | r = 1
+      | c = Definition of $s$ and $t$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Product of Coprime Factors}
+Tags: Coprime Integers
+
+\begin{theorem}
+Let $a, b, c \in \Z$ such that $a$ and $b$ are [[Definition:Coprime Integers|coprime]].
+Let both $a$ and $b$ be [[Definition:Divisor of Integer|divisors]] of $c$.
+Then $a b$ is also a [[Definition:Divisor of Integer|divisor]] of $c$.
+That is:
+:$a \perp b \land a \divides c \land b \divides c \implies a b \divides c$
+\end{theorem}
+
+\begin{proof}
+We have:
+:$a \divides c \implies \exists r \in \Z: c = a r$
+:$b \divides c \implies \exists s \in \Z: c = b s$
+So:
+{{begin-eqn}}
+{{eqn | l = a
+      | o = \perp
+      | r = b
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \exists m, n \in \Z: m a + n b
+      | r = 1
+      | c = [[Integer Combination of Coprime Integers]]
+}}
+{{eqn | ll= \leadsto
+      | l = c m a + c n b
+      | r = c
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = b s m a + a r n b
+      | r = c
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = a b \paren {s m + r n}
+      | r = c
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = a b
+      | o = \divides
+      | r = c
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Existence of Lowest Common Multiple}
+Tags: Lowest Common Multiple, Existence of Lowest Common Multiple
+
+\begin{theorem}
+Let $a, b \in \Z: a b \ne 0$.
+The [[Definition:Lowest Common Multiple of Integers|lowest common multiple]] of $a$ and $b$, denoted $\lcm \set {a, b}$, always exists.
+\end{theorem}
+
+\begin{proof}
+We prove its existence thus:
+$a b \ne 0 \implies \size {a b} \ne 0$
+Also $\size {a b} = \pm a b = a \paren {\pm b} = \paren {\pm a} b$.
+So it definitely exists, and we can say that:
+:$0 < \lcm \set {a, b} \le \size {a b}$
+Now we prove it is the lowest. That is:
+:$a \divides n \land b \divides n \implies \lcm \set {a, b} \divides n$
+Let $a, b \in \Z: a b \ne 0, m = \lcm \set {a, b}$.
+Let $n \in \Z: a \divides n \land b \divides n$.
+We have:
+: $n = x_1 a = y_1 b$
+: $m = x_2 a = y_2 b$
+As $m > 0$, we have:
+{{begin-eqn}}
+{{eqn | l = n
+      | r = m q + r: 0 \le r < \size m = m
+      | c = 
+}}
+{{eqn | lll=\leadsto
+      | l = r
+      | r = n - m q
+      | c =
+}}
+{{eqn | r = 1 \times n + \paren {-q} \times m
+      | c =
+}}
+{{eqn | lll=\leadsto
+      | l = r
+      | r = x_1 a + \paren {-q} x_2 a
+      | c =
+}}
+{{eqn | r = y_1 b + \paren {-q} y_2 b
+      | c =
+}}
+{{eqn | lll=\leadsto
+      | l = a
+      | o = \divides
+      | r = r
+      | c =
+}}
+{{eqn | ll= \land
+      | l = b
+      | o = \divides
+      | r = r
+      | c = 
+}}
+{{end-eqn}}
+Since $r < m$, and $m$ is the smallest ''positive'' [[Definition:Common Multiple|common multiple]] of $a$ and $b$, it follows that $r = 0$.
+So:
+: $\forall n \in \Z: a \divides n \land b \divides n: \lcm \set {a, b} \divides n$
+That is, $\lcm \set {a, b}$ divides any [[Definition:Common Multiple|common multiple]] of $a$ and $b$.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Either $a$ and $b$ are [[Definition:Coprime Integers|coprime]] or they are not.
+Let:
+:$a \perp b$
+where $a \perp b$ denotes that $a$ and $b$ are [[Definition:Coprime Integers|coprime]].
+Let $a b = c$.
+Then:
+:$a \divides c, b \divides c$
+where $a \divides c$ denotes that $a$ is a [[Definition:Divisor of Integer|divisor]] of $c$.
+Suppose both $a \divides d, b \divides d$ for some $d \in \N_{> 0}: d < c$.
+Then:
+:$\exists e \in \N_{> 0}: a e = d$
+:$\exists f \in \N_{> 0}: b f = d$
+Therefore:
+:$a e = b f$
+and from {{EuclidPropLink|book = VII|prop = 19|title = Relation of Ratios to Products}}:
+:$a : b = f : e$
+But $a$ and $b$ are [[Definition:Coprime Integers|coprime]].
+From:
+:{{EuclidPropLink|prop = 21|title = Coprime Numbers form Fraction in Lowest Terms}}
+and:
+:{{EuclidPropLink|prop = 20|title = Ratios of Fractions in Lowest Terms}}
+it follows that $b \divides e$
+Since:
+:$a b = c$
+and:
+:$a e = d$
+it follows from {{EuclidPropLink|prop = 17|title = Multiples of Ratios of Numbers}} that:
+:$b : e = c : d$
+But $b \divides e$ and therefore:
+:$c \divides d$
+But $c > d$ which is impossible.
+Therefore $a$ and $b$ are both the [[Definition:Divisor of Integer|divisor]] of no number less than $c$.
+Now suppose $a$ and $b$ are not [[Definition:Coprime Integers|coprime]].
+Let $f$ and $e$ be the least numbers of those which have the same [[Definition:Ratio|ratio]] with $a$ and $b$.
+Then from {{EuclidPropLink|prop = 19|title = Relation of Ratios to Products}}:
+:$a e = b f$
+Let $a e = c$.
+Then $b f = c$.
+Hence:
+:$a \divides c$
+:$b \divides c$
+Suppose $a$ and $b$ are both the [[Definition:Divisor of Integer|divisor]] of some number $d$ which is less than $c$.
+Let:
+:$a g = d$
+and:
+:$b h = d$
+Therefore:
+:$a g = b h$
+and so by {{EuclidPropLink|prop = 19|title = Relation of Ratios to Products}}:
+:$a : b = f : e$
+Also:
+:$f : e = h : g$
+But $f, e$ are the least such.
+From {{EuclidPropLink|prop = 20|title = Ratios of Fractions in Lowest Terms}}:
+:$e \divides g$
+Since $a e = c$ and $a g = d$, from {{EuclidPropLink|prop = 17|title = Multiples of Ratios of Numbers}}:
+:$e : g = c : d$
+But:
+:$e \divides g$
+So $c \divides d$
+But $c > d$ which is impossible.
+Therefore $a$ and $b$ are both the [[Definition:Divisor of Integer|divisor]] of no number less than $c$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Product of GCD and LCM}
+Tags: Greatest Common Divisor, Lowest Common Multiple, Product of GCD and LCM
+
+\begin{theorem}
+:$\lcm \set {a, b} \times \gcd \set {a, b} = \size {a b}$
+where:
+:$\lcm \set {a, b}$ denotes the [[Definition:Lowest Common Multiple of Integers|lowest common multiple]] of $a$ and $b$
+:$\gcd \set {a, b}$ denotes the [[Definition:Greatest Common Divisor of Integers|greatest common divisor]] of $a$ and $b$.
+\end{theorem}
+
+\begin{proof}
+It is sufficient to prove that $\lcm \set {a, b} \times \gcd \set {a, b} = a b$, where $a, b \in \Z_{>0}$.
+{{begin-eqn}}
+{{eqn | l = d 
+      | r = \gcd \set {a, b}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = d
+      | o = \divides
+      | r = a b
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | lo= \exists n \in \Z_{>0}:
+      | l = a b
+      | r = d n
+      | c=
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | l = d \divides a
+      | o = \land
+      | r = d \divides b
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | lo= \exists u, v \in \Z:
+      | l = a = d u
+      | o = \land
+      | r = b = d v
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = d u b = d n
+      | o = \land
+      | r = a d v = d n
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = n = b u
+      | o = \land
+      | r = n = a v
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = a \divides n
+      | o = \land
+      | r = b \divides n
+      | c = 
+}}
+{{end-eqn}}
+Now we have $a \divides m \land b \divides m \implies m = a r = b s$.
+Also, by [[Bézout's Lemma]] we have $d = a x + b y$.
+So:
+{{begin-eqn}}
+{{eqn | l = m d
+      | r = a x m + b y m
+      | c = 
+}}
+{{eqn | r = b s a x + a r b y
+      | c = 
+}}
+{{eqn | r = a b \paren {s x + r y}
+      | c = 
+}}
+{{eqn | r = d n \paren {s x + r y}
+      | c = 
+}}
+{{end-eqn}}
+So:
+:$m = n \paren {s x + r y}$
+Thus:
+:$n \divides m \implies n \le \size m$
+while:
+:$a b = d n = \gcd \set {a, b} \times \lcm \set {a, b}$
+as required.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let $a = g m$ and $b = g n$, where $g = \gcd \set {a, b}$ and $m$ and $n$ are [[Definition:Coprime Integers|coprime]].
+The existence of $m$ and $n$ are proved by [[Integers Divided by GCD are Coprime]].
+Since $a = g m \divides g m n$ and $b = g n \divides g m n$, $g m n$ is the [[Definition:Lowest Common Multiple of Integers|LCM]] of $a$ and $b$.
+{{explain|why?}}
+Then it follows that:
+:$\lcm \set {a, b} \times \gcd \set {a, b} = g m n \times g = g m \times g n = \size {a b}$
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let $d := \gcd \set {a, b}$. 
+Then by definition of the [[Definition:Greatest Common Divisor of Integers|GCD]], there exist $j_1, j_2 \in \Z$ such that $a = d j_1$ and $b = d j_2$.
+Because $d$ [[Definition:Divisor of Integer|divides]] both $a$ and $b$, it must [[Definition:Divisor of Integer|divide]] their [[Definition:Integer Multiplication|product]]:
+:$\exists l \in \Z$ such that $a b = d l$
+Then we have: 
+{{begin-eqn}}
+{{eqn | l = d l
+      | m = \paren {d j_1} b
+      | mo= =
+      | r = a \paren {d j_2}
+}}
+{{eqn | ll= \leadsto
+      | l = l
+      | m = j_1 b
+      | mo= =
+      | r = a j_2
+}}
+{{end-eqn}}
+showing that $a \divides l$ and $b \divides l$.
+That is, $l$ is a [[Definition:Common Multiple|common multiple]] of $a$ and $b$.
+Now it must be shown that $l$ is the least such number.
+Let $m$ be any [[Definition:Common Multiple|common multiple]] of $a$ and $b$.
+Then there exist $k_1, k_2 \in \Z$ such that $m = a k_1 = b k_2$.
+By [[Bézout's Lemma]]:
+:$\exists x, y \in \Z: d = a x + b y$
+So:
+{{begin-eqn}}
+{{eqn | l = m d
+      | r = m a x + m b y
+}}
+{{eqn | r = \paren {b k_2} a x + \paren {a k_1} b y
+}}
+{{eqn | r = a b \paren {b k_2 + a k_1}
+}}
+{{eqn | r = d l \paren {b k_2 + a k_1}
+}}
+{{end-eqn}}
+Thus:
+:$m = l \paren {b k_2 + a k_1}$
+that is, $l \divides m$.
+Hence by definition of the [[Definition:Lowest Common Multiple of Integers|LCM]]:
+:$\lcm \set {a, b} = l$
+In conclusion:
+:$a b = d l = \gcd \set {a, b} \cdot \lcm \set {a, b}$
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let:
+{{begin-eqn}}
+{{eqn | l = m
+      | r = {p_1}^{k_1} {p_2}^{k_2} \dotsm {p_r}^{k_r}
+}}
+{{eqn | l = n
+      | r = {p_1}^{l_1} {p_2}^{l_2} \dotsm {p_r}r^{l_r}
+      | c = 
+}}
+{{end-eqn}}
+From [[LCM from Prime Decomposition]]:
+:$\lcm \set {m, n} = p_1^{\max \set {k_1, l_1} } p_2^{\max \set {k_2, l_2} } \dotsm p_r^{\max \set {k_r, l_r} }$
+From [[GCD from Prime Decomposition]]:
+:$\gcd \set {m, n} = p_1^{\min \set {k_1, l_1} } p_2^{\min \set {k_2, l_2} } \dotsm p_r^{\min \set {k_r, l_r} }$
+From [[Sum of Maximum and Minimum]], for all $i \in \set {1, 2, \ldots, r}$:
+:$\min \set {k_i, l_i} + \max \set {k_i, l_i} = k_i + l_i$
+Hence:
+{{begin-eqn}}
+{{eqn | l = \gcd \set {m, n} \times \lcm \set {m, n}
+      | r = p_1^{k_1 + l_1} p_2^{k_2 + l_2} \dotsm p_r^{k_r + l_r}
+      | c = 
+}}
+{{eqn | r = p_1^{k_1} p_1^{l_1} p_2^{k_2} p_2^{l_2} \dotsm p_r^{k_r} p_r^{l_r}
+      | c = 
+}}
+{{eqn | r = p_1^{k_1} p_2^{k_2} \dotsm p_r^{k_r} \times p_1^{l_1} p_2^{l_2} \dotsm p_r^{l_r}
+      | c = 
+}}
+{{eqn | r = m n
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Congruent to Zero if Modulo is Divisor}
+Tags: Modulo Arithmetic
+
+\begin{theorem}
+Let $a, z \in \R$.
+Then $a$ is [[Definition:Congruence (Number Theory)|congruent]] to $0$ modulo $z$ {{iff}} $a$ is an [[Definition:Integer Multiple|integer multiple]] of $z$.
+:$\exists k \in \Z: k z = a \iff a \equiv 0 \pmod z$
+If $z \in \Z$, then further:
+:$z \divides a \iff a \equiv 0 \pmod z$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \exists k \in \Z: a
+      | r = k z
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \exists k \in \Z: a
+      | r = 0 + k z
+      | c = 
+}}
+{{end-eqn}}
+Thus by definition of [[Definition:Congruence (Number Theory)|congruence]], $a \equiv 0 \pmod z$ and the result is proved.
+If $z$ is an [[Definition:Integer|integer]], then by definition of [[Definition:Divisor of Integer|divisor]]:
+:$z \divides a \iff \exists k \in \Z: a = k z$
+Hence the result for [[Definition:Integer|integer]] $z$.
+{{qed}}
+[[Category:Modulo Arithmetic]]
+87apiiualuplsor2u54hkqv8t4i4x69
+\end{proof}<|endoftext|>
+\section{Integer is Congruent Modulo Divisor to Remainder}
+Tags: Modulo Arithmetic
+
+\begin{theorem}
+Let $a \in \Z$.
+Let $a$ have a [[Definition:Remainder|remainder]] $r$ on [[Definition:Integer Division|division]] by $m$.
+Then:
+: $a \equiv r \pmod m$
+where the notation denotes that $a$ and $r$ are [[Definition:Congruence Modulo Integer|congruent modulo $m$]].
+\end{theorem}
+
+\begin{proof}
+Let $a$ have a [[Definition:Remainder|remainder]] $r$ on [[Definition:Integer Division|division]] by $m$.
+Then:
+: $\exists q \in \Z: a = q m + r$
+Hence by definition of [[Definition:Congruence Modulo Integer|congruence modulo $m$]]:
+:$a \equiv r \pmod m$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Integer is Congruent to Integer less than Modulus}
+Tags: Modulo Arithmetic
+
+\begin{theorem}
+Let $m \in \Z$.
+Then each [[Definition:Integer|integer]] is [[Definition:Congruence Modulo Integer|congruent (modulo $m$)]] to precisely one of the [[Definition:Integer|integers]] $0, 1, \ldots, m - 1$.
+\end{theorem}
+
+\begin{proof}
+=== Proof of Existence ===
+Let $a \in \Z$.
+Then from the [[Division Theorem]]: $\exists r \in \set {0, 1, \ldots, m - 1}: a \equiv r \pmod m$.
+=== Proof of Uniqueness ===
+Suppose that:
+:$\exists r_1, r_2 \in \set {0, 1, \ldots, m - 1}: a \equiv r_1 \pmod m \land a \equiv r_2 \pmod m$
+Then:
+:$\exists r_1, r_2 \in \Z: a = q_1 m + r_1 = q_2 m + r_2$
+This contradicts the uniqueness clause in the [[Division Theorem]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Modulo Addition is Well-Defined}
+Tags: Modulo Addition, Modulo Addition is Well-Defined
+
+\begin{theorem}
+Let $m \in \Z$ be an [[Definition:Integer|integer]].
+Let $\Z_m$ be the [[Definition:Integers Modulo m|set of integers modulo $m$]].
+The [[Definition:Modulo Addition|modulo addition]] operation on $\Z_m$, defined by the rule:
+:$\eqclass a m +_m \eqclass b m = \eqclass {a + b} m$
+is a [[Definition:Well-Defined Operation|well-defined operation]].
+That is:
+:If $a \equiv b \pmod m$ and $x \equiv y \pmod m$, then $a + x \equiv b + y \pmod m$.
+\end{theorem}
+
+\begin{proof}
+The equivalence class $\eqclass a m$ is defined as:
+:$\eqclass a m = \set {x \in \Z: x = a + k m: k \in \Z}$
+That is, the set of all [[Definition:Integer|integers]] which differ from $a$ by an [[Definition:Integer Multiple|integer multiple]] of $m$.
+Thus the notation for addition of two [[Definition:Integers Modulo m|set of integers modulo $m$]] is not usually $\eqclass a m +_m \eqclass b m$.
+What is more normally seen is $a + b \pmod m$.
+Using this notation:
+{{begin-eqn}}
+{{eqn | l = a
+      | o = \equiv
+      | r = b
+      | rr= \pmod m
+      | c =
+}}
+{{eqn | lo= \land
+      | l = c
+      | o = \equiv
+      | r = d
+      | rr= \pmod m
+      | c =
+}}
+{{eqn | ll= \leadsto
+      | l = a \bmod m
+      | r = b \bmod m
+      | c = {{Defof|Congruence Modulo Integer}}
+}}
+{{eqn | lo= \land
+      | l = c \bmod m
+      | r = d \bmod m
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = a
+      | r = b + k_1 m
+      | c = for some $k_1 \in \Z$
+}}
+{{eqn | lo= \land
+      | l = c
+      | r = d + k_2 m
+      | c = for some $k_2 \in \Z$
+}}
+{{eqn | ll= \leadsto
+      | l = a + c
+      | r = b + d + \paren {k_1 + k_2} m
+      | c = {{Defof|Integer Addition}}
+}}
+{{eqn | ll= \leadsto
+      | l = a + c
+      | o = \equiv
+      | r = b + d
+      | rr= \pmod m
+      | c = {{Defof|Modulo Addition}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+We need to show that if:
+:$\eqclass {x'} m = \eqclass x m$
+:$\eqclass {y'} m = \eqclass y m$
+then:
+:$\eqclass {x' + y'} m = \eqclass {x + y} m$
+Since:
+:$\eqclass {x'} m = \eqclass x m$
+and:
+:$\eqclass {y'} m = \eqclass y m$
+it follows from the definition of [[Definition:Integers Modulo m|set of integers modulo $m$]] that:
+:$x \equiv x' \pmod m$
+and:
+:$y \equiv y' \pmod m$
+By definition, we have:
+:$x \equiv x' \pmod m \implies \exists k_1 \in \Z: x = x' + k_1 m$
+:$y \equiv y' \pmod m \implies \exists k_2 \in \Z: y = y' + k_2 m$
+which gives us:
+:$x + y = x' + k_1 m + y' + k_2 m = x' + y' + \left({k_1 + k_2}\right) m$
+As $k_1 + k_2$ is an [[Definition:Integer|integer]], it follows that, by definition:
+:$x + y \equiv \left({x' + y'}\right) \pmod m$
+Therefore, by the definition of [[Definition:Integers Modulo m|integers modulo $m$]]:
+:$\eqclass {x' + y'} m = \eqclass {x + y} m$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Modulo Multiplication is Well-Defined}
+Tags: Modulo Multiplication, Modulo Multiplication is Well-Defined
+
+\begin{theorem}
+The [[Definition:Modulo Multiplication|multiplication modulo $m$]] operation on $\Z_m$, the set of [[Definition:Integers Modulo m|integers modulo $m$]], defined by the rule:
+:$\eqclass x m \times_m \eqclass y m = \eqclass {x y} m$
+is a [[Definition:Well-Defined Operation|well-defined operation]].
+That is:
+:If $a \equiv b \pmod m$ and $x \equiv y \pmod m$, then $a x \equiv b y \pmod m$.
+\end{theorem}
+
+\begin{proof}
+We need to show that if:
+:$\eqclass {x'} m = \eqclass x m$
+and:
+:$\eqclass {y'} m = \eqclass y m$
+then:
+:$\eqclass {x' y'} m = \eqclass {x y} m$
+We have that:
+:$\eqclass {x'} m = \eqclass x m$
+and:
+:$\eqclass {y'} m = \eqclass y m$
+It follows from the definition of [[Definition:Residue Class|residue class modulo $m$]] that:
+:$x \equiv x' \pmod m$
+and:
+:$y \equiv y' \pmod m$
+By definition, we have:
+:$x \equiv x' \pmod m \implies \exists k_1 \in \Z: x = x' + k_1 m$
+:$y \equiv y' \pmod m \implies \exists k_2 \in \Z: y = y' + k_2 m$
+which gives us:
+:$x y = \paren {x' + k_1 m} \paren {y' + k_2 m} = x' y' + \paren {x' k_2 + y' k_1} m + k_1 k_2 m^2$
+Thus by definition:
+:$x y \equiv \paren {x' y'} \pmod m$
+Therefore, by the definition of [[Definition:Residue Class|residue class modulo $m$]]:
+:$\eqclass {x' y'} m = \eqclass {x y} m$
+{{qed}}
+\end{proof}
+
+\begin{proof}
+The equivalence class $\eqclass a m$ is defined as:
+:$\eqclass a m = \set {x \in \Z: x = a + k m: k \in \Z}$
+that is, the set of all [[Definition:Integer|integers]] which differ from $a$ by an [[Definition:Integer Multiple|integer multiple]] of $m$.
+Thus the notation for addition of two [[Definition:Residue Class|residue classes modulo $z$]] is not usually $\eqclass a m +_m \eqclass b m$.
+What is more normally seen is:
+: $a b \pmod m$
+Using this notation:
+{{begin-eqn}}
+{{eqn | l = a
+      | o = \equiv
+      | r = b
+      | rr= \pmod m
+      | c = 
+}}
+{{eqn | lo= \land
+      | l = c
+      | o = \equiv
+      | r = d
+      | rr= \pmod m
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = a \bmod m
+      | r = b \bmod m
+      | c = {{Defof|Congruence Modulo Integer}}
+}}
+{{eqn | lo= \land
+      | l = c \bmod m
+      | r = d \bmod m
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = a
+      | r = b + k_1 m
+      | c = for some $k_1 \in \Z$
+}}
+{{eqn | lo= \land
+      | l = c
+      | r = d + k_2 m
+      | c = for some $k_2 \in \Z$
+}}
+{{eqn | ll= \leadsto
+      | l = a c
+      | r = \paren {b + k_1 m} \paren {d + k_2 m}
+      | c = {{Defof|Multiplication}}
+}}
+{{eqn | r = b d + b k_2 m + d k_1 m + k_1 k_2 m^2
+      | c = [[Integer Multiplication Distributes over Addition]]
+}}
+{{eqn | r = b d + \paren {b k_2 + d k_1 + k_1 k_2 m} m
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = a c
+      | o = \equiv
+      | r = b d
+      | rr= \pmod m
+      | c = {{Defof|Modulo Multiplication}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Congruence of Product}
+Tags: Modulo Arithmetic
+
+\begin{theorem}
+Let $a, b, z \in \R$.
+Let $a$ be [[Definition:Congruence (Number Theory)|congruent to $b$ modulo $z$]], that is:
+:$a \equiv b \pmod z$
+Then:
+:$\forall m \in \Z: m a \equiv m b \pmod z$
+\end{theorem}
+
+\begin{proof}
+Let $m \in \Z$ and $a \equiv b \pmod z$.
+Suppose $m = 0$.
+Then the {{RHS}} of the assertion degenerates to $0 \equiv 0 \pmod z$ which is trivially true.
+Otherwise, from [[Congruence by Product of Moduli]], we have:
+:$a \equiv b \iff m a \equiv m b \pmod z$
+As $m \in \Z$, it follows that $m z$ is an [[Definition:Integer Multiple|integer multiple]] of $z$.
+Hence from [[Congruence by Divisor of Modulus]], it follows that:
+:$m a \equiv m b \implies m a \equiv m b \pmod z$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Congruence of Powers}
+Tags: Modulo Arithmetic
+
+\begin{theorem}
+Let $a, b \in \R$ and $m \in \Z$.
+Let $a$ be [[Definition:Congruence Modulo Integer|congruent to $b$ modulo $m$]], that is:
+:$a \equiv b \pmod m$
+Then:
+:$\forall n \in \Z_{\ge 0}: a^n \equiv b^n \pmod m$
+\end{theorem}
+
+\begin{proof}
+Proof by [[Principle of Mathematical Induction|induction]]:
+For all $n \in \Z_{\ge 0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
+:$a \equiv b \implies a^k \equiv b^k \pmod m$
+$\map P 0$ is trivially true, as $a^0 = b^0 = 1$.
+$\map P 1$ is true, as this just says:
+:$a \equiv b \pmod m$
+=== Basis for the Induction ===
+$\map P 2$ is the case:
+:$a^2 \equiv b^2 \pmod m$
+which follows directly from the fact that [[Modulo Multiplication is Well-Defined]].
+This is our [[Principle of Mathematical Induction#Basis for the Induction|basis for the induction]].
+=== Induction Hypothesis ===
+Now we need to show that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true.
+So this is our [[Principle of Mathematical Induction#Induction Hypothesis|induction hypothesis]]:
+:$a \equiv b \implies a^k \equiv b^k \pmod m$
+Then we need to show:
+:$a \equiv b \implies a^{k + 1} \equiv b^{k + 1} \pmod m$
+=== Induction Step ===
+This is our [[Principle of Mathematical Induction#Induction Step|induction step]]:
+Suppose $a^k \equiv b^k \pmod m$.
+Then $a^k a \equiv b^k b \pmod m$ by definition of [[Definition:Modulo Multiplication|modulo multiplication]].
+Thus $a^{k + 1} \equiv b^{k + 1} \pmod m$.
+So $\map P k \implies \map P {k + 1}$ and the result follows by the [[Principle of Mathematical Induction]].
+Therefore:
+:$\forall n \in \Z_+: a \equiv b \implies a^n \equiv b^n \pmod m$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Modulo Addition is Associative}
+Tags: Modulo Addition
+
+\begin{theorem}
+[[Definition:Modulo Addition|Addition modulo $m$]] is [[Definition:Associative|associative]]:
+:$\forall \eqclass x m, \eqclass y m, \eqclass z m \in \Z_m: \paren {\eqclass x m +_m \eqclass y m} +_m \eqclass z m = \eqclass x m +_m \paren {\eqclass y m +_m \eqclass z m}$
+where $\Z_m$ is the [[Definition:Integers Modulo m|set of integers modulo $m$]].
+That is:
+:$\forall x, y, z \in \Z: \paren {x + y} + z \equiv x + \paren {y + z} \pmod m$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \paren {\eqclass x m +_m \eqclass y m} +_m \eqclass z m
+      | r = \eqclass {x + y} m +_m \eqclass z m
+      | c = {{Defof|Modulo Addition}}
+}}
+{{eqn | r = \eqclass {\paren {x + y} + z} m
+      | c = {{Defof|Modulo Addition}}
+}}
+{{eqn | r = \eqclass {x + \paren {y + z} } m
+      | c = [[Associative Law of Addition]]
+}}
+{{eqn | r = \eqclass x m +_m \eqclass {y + z} m
+      | c = {{Defof|Modulo Addition}}
+}}
+{{eqn | r = \eqclass x m +_m \paren {\eqclass y m +_m \eqclass z m}
+      | c = {{Defof|Modulo Addition}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Modulo Addition is Commutative}
+Tags: Modulo Addition
+
+\begin{theorem}
+[[Definition:Modulo Addition|Modulo addition]] is [[Definition:Commutative Operation|commutative]]:
+:$\forall x, y, z \in \Z: x + y \pmod m = y + x \pmod m$
+\end{theorem}
+
+\begin{proof}
+From the definition of [[Definition:Modulo Addition|modulo addition]], this is also written:
+:$\forall m \in \Z: \forall \eqclass x m, \eqclass y m \in \Z_m: \eqclass x m +_m \eqclass y m = \eqclass y m +_m \eqclass x m$
+Hence:
+{{begin-eqn}}
+{{eqn | l = \eqclass x m +_m \eqclass y m
+      | r = \eqclass {x + y} m
+      | c = {{Defof|Modulo Addition}}
+}}
+{{eqn | r = \eqclass {y + x} m
+      | c = [[Commutative Law of Addition]]
+}}
+{{eqn | r = \eqclass y m +_m \eqclass x m
+      | c = {{Defof|Modulo Addition}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Modulo Addition has Identity}
+Tags: Modulo Addition
+
+\begin{theorem}
+Let $m \in \Z$ be an [[Definition:Integer|integer]].
+Then [[Definition:Modulo Addition|addition modulo $m$]] has an [[Definition:Identity Element|identity]]:
+:$\forall \eqclass x m \in \Z_m: \eqclass x m +_m \eqclass 0 m = \eqclass x m = \eqclass 0 m +_m \eqclass x m$
+That is:
+:$\forall a \in \Z: a + 0 \equiv a \equiv 0 + a \pmod m$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \eqclass x m +_m \eqclass 0 m
+      | r = \eqclass {x + 0} m
+      | c = {{Defof|Modulo Addition}}
+}}
+{{eqn | r = \eqclass x m
+      | c = 
+}}
+{{eqn | r = \eqclass {0 + x} m
+      | c = 
+}}
+{{eqn | r = \eqclass 0 m +_m \eqclass x m
+      | c = {{Defof|Modulo Addition}}
+}}
+{{end-eqn}}
+Thus $\eqclass 0 m$ is the [[Definition:Identity Element|identity]] for [[Definition:Modulo Addition|addition modulo $m$]].
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Modulo Addition has Inverses}
+Tags: Modulo Addition
+
+\begin{theorem}
+Let $m \in \Z$ be an [[Definition:Integer|integer]].
+Then [[Definition:Modulo Addition|addition modulo $m$]] has [[Definition:Inverse Element|inverses]]:
+For each [[Definition:Element|element]] $\eqclass x m \in \Z_m$, there exists the element $\eqclass {-x} m \in \Z_m$ with the property:
+:$\eqclass x m +_m \eqclass {-x} m = \eqclass 0 m = \eqclass {-x} m +_m \eqclass x m$
+where $\Z_m$ is the [[Definition:Integers Modulo m|set of integers modulo $m$]].
+That is:
+:$\forall a \in \Z: a + \paren {-a} \equiv 0 \equiv \paren {-a} + a \pmod m$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \eqclass x m +_m \eqclass {-x} m
+      | r = \eqclass {x + \paren {-x} } m
+      | c = {{Defof|Modulo Addition}}
+}}
+{{eqn | r = \eqclass 0 m
+      | c = 
+}}
+{{eqn | r = \eqclass {\paren {-x} + x} m
+      | c = 
+}}
+{{eqn | r = \eqclass {-x} m +_m \eqclass x m
+      | c = {{Defof|Modulo Addition}}
+}}
+{{end-eqn}}
+As $-x$ is a perfectly good [[Definition:Integer|integer]], $\eqclass {-x} m \in \Z_m$, whatever $x$ may be.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Modulo Multiplication is Closed}
+Tags: Modulo Multiplication
+
+\begin{theorem}
+[[Definition:Modulo Multiplication|Multiplication modulo $m$]] is [[Definition:Closed Algebraic Structure|closed]] on the set of [[Definition:Integers Modulo m|integers modulo $m$]]:
+:$\forall \eqclass x m, \eqclass y m \in \Z_m: \eqclass x m \times_m \eqclass y m \in \Z_m$.
+\end{theorem}
+
+\begin{proof}
+From the definition of [[Definition:Modulo Multiplication|multiplication modulo $m$]], we have:
+:$\eqclass x m \times_m \eqclass y m = \eqclass {x y} m$
+By the [[Division Theorem]]:
+:$x y = q m + r$ where $0 \le r < m$
+Therefore for all $0 \le r < m$:
+:$\eqclass {x y} m = \eqclass r m$
+Therefore from the definition of [[Definition:Integers Modulo m|integers modulo $m$]]:
+:$\eqclass {x y} m \in \Z_m$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Modulo Multiplication is Associative}
+Tags: Modulo Multiplication, Modulo Multiplication is Associative
+
+\begin{theorem}
+[[Definition:Modulo Multiplication|Multiplication modulo $m$]] is [[Definition:Associative|associative]]:
+:$\forall \eqclass x m, \eqclass y m, \eqclass z m \in \Z_m: \paren {\eqclass x m \times_m \eqclass y m} \times_m \eqclass z m = \eqclass x m \times_m \paren {\eqclass y m \times_m \eqclass z m}$
+That is:
+:$\forall x, y, z \in \Z_m: \paren {x \cdot_m y} \cdot_m z = x \cdot_m \paren {y \cdot_m z}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \paren {\eqclass x m \times_m \eqclass y m} \times_m \eqclass z m
+      | r = \eqclass {x y} m \times_m \eqclass z m
+      | c = {{Defof|Modulo Multiplication}}
+}}
+{{eqn | r = \eqclass {\paren {x y} z} m
+      | c = {{Defof|Modulo Multiplication}}
+}}
+{{eqn | r = \eqclass {x \paren {y z} } m
+      | c = [[Integer Multiplication is Associative]]
+}}
+{{eqn | r = \eqclass x m \times_m \eqclass {y z} m
+      | c = {{Defof|Modulo Multiplication}}
+}}
+{{eqn | r = \eqclass x m \times_m \paren {\eqclass y m \times_m \eqclass z m}
+      | c = {{Defof|Modulo Multiplication}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let $j$ be the largest [[Definition:Integer|integer]] such that:
+:$j m \le x y$
+Let $p$ be the largest [[Definition:Integer|integer]] such that:
+:$p m \le y z$
+By definition of [[Definition:Modulo Multiplication|multiplication modulo $m$]]:
+:$x \cdot_m y = x y - j m$
+:$y \cdot_m z = y z - p m$
+Let $k$ be the largest [[Definition:Integer|integer]] such that:
+:$k m \le \paren {x y - j m} z$
+Let $q$ be the largest [[Definition:Integer|integer]] such that:
+:$q m \le x \paren {y z - p m}$
+Then:
+:$\paren {j z + k} m \le \paren {x y} z$
+:$\paren {q + x p} m \le x \paren {y z}$
+Thus:
+{{begin-eqn}}
+{{eqn | l = \paren {x \cdot_m y} \cdot_m z
+      | r = \paren {x y - j m} z - k m
+      | c = {{Defof|Modulo Multiplication}}
+}}
+{{eqn | l = x \cdot_m \paren {y \cdot_m z}
+      | r = x \left({y z - p m}\right) - q m
+      | c = {{Defof|Modulo Multiplication}}
+}}
+{{end-eqn}}
+But suppose that there exists an [[Definition:Integer|integer]] $s$ such that:
+:$s m \le \paren {x y} z$
+and:
+:$j z + k < s$
+Then:
+:$\paren {j z + k + 1} m \le \paren {x y} z$
+from which:
+:$\paren {k + 1} m \le \paren {x y - j m} z$
+But this contradicts the definition of $k$.
+Thus $j z + k$ is the largest of those [[Definition:Integer|integers]] $i$ such that $i m \le \paren {x y} z$.
+Similarly, $q + x p$ is the largest of those [[Definition:Integer|integers]] $i$ such that $i m \le x \paren {y z}$.
+From [[Integer Multiplication is Associative]]:
+:$\paren {x y} z = x \paren {y z}$
+Thus $j z + k = q + x p$ and so:
+{{begin-eqn}}
+{{eqn | l = \paren {x \cdot_m y} \cdot_m z
+      | r = \paren {x y - j m} z - k m
+      | c = {{Defof|Modulo Multiplication}}
+}}
+{{eqn | r = x y z - \paren {j z + k} m
+      | c = 
+}}
+{{eqn | r = x y z - \paren {q + x p} m
+      | c = 
+}}
+{{eqn | r = x \paren {y z - p m} - q m
+      | c = 
+}}
+{{eqn | r = x \cdot_m \paren {y \cdot_m z}
+      | c = {{Defof|Modulo Multiplication}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Modulo Multiplication is Commutative}
+Tags: Modulo Multiplication
+
+\begin{theorem}
+[[Definition:Modulo Multiplication|Multiplication modulo $m$]] is [[Definition:Commutative Operation|commutative]]:
+:$\forall \eqclass x m, \eqclass y m \in \Z_m: \eqclass x m \times_m \eqclass y m = \eqclass y m \times_m \eqclass x m$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \eqclass x m \times_m \eqclass y m
+      | r = \eqclass {x y} m
+      | c = {{Defof|Modulo Multiplication}}
+}}
+{{eqn | r = \eqclass {y x} m
+      | c = [[Integer Multiplication is Commutative]]
+}}
+{{eqn | r = \eqclass y m \times_m \eqclass x m
+      | c = {{Defof|Modulo Multiplication}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Modulo Multiplication has Identity}
+Tags: Modulo Multiplication
+
+\begin{theorem}
+[[Definition:Modulo Multiplication|Multiplication modulo $m$]] has an [[Definition:Identity Element|identity]]:
+:$\forall \eqclass x m \in \Z_m: \eqclass x m \times_m \eqclass 1 m = \eqclass x m = \eqclass 1 m \times_m \eqclass x m$
+\end{theorem}
+
+\begin{proof}
+Follows directly from the definition of [[Definition:Modulo Multiplication|multiplication modulo $m$]]:
+{{begin-eqn}}
+{{eqn | l = \eqclass x m \times_m \eqclass 1 m
+      | r = \eqclass {x \times 1} m
+      | c = 
+}}
+{{eqn | r = \eqclass x m
+      | c = 
+}}
+{{eqn | r = \eqclass {1 \times x} m
+      | c = 
+}}
+{{eqn | r = \eqclass 1 m \times_m \eqclass x m
+      | c = 
+}}
+{{end-eqn}}
+Thus $\eqclass 1 m$ is the [[Definition:Identity Element|identity]] for [[Definition:Modulo Multiplication|multiplication modulo $m$]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Modulo Multiplication Distributes over Modulo Addition}
+Tags: Modulo Multiplication, Modulo Addition, Distributive Operations
+
+\begin{theorem}
+[[Definition:Modulo Multiplication|Multiplication modulo $m$]] is [[Definition:Distributive Operation|distributive]] over [[Definition:Modulo Addition|addition modulo $m$]]:
+:$\forall \eqclass x m, \eqclass y m, \eqclass z m \in \Z_m$:
+:: $\eqclass x m \times_m \paren {\eqclass y m +_m \eqclass z m} = \paren {\eqclass x m \times_m \eqclass y m} +_m \paren {\eqclass x m \times_m \eqclass z m}$
+:: $\paren {\eqclass x m +_m \eqclass y m} \times_m \eqclass z m = \paren {\eqclass x m \times_m \eqclass z m} +_m \paren {\eqclass y m \times_m \eqclass z m}$
+where $\Z_m$ is the [[Definition:Integers Modulo m|set of integers modulo $m$]].
+That is, $\forall x, y, z, m \in \Z$:
+: $x \paren {y + z} \equiv x y + x z \pmod m$
+: $\paren {x + y} z \equiv x z + y z \pmod m$
+\end{theorem}
+
+\begin{proof}
+Follows directly from the definition of [[Definition:Modulo Multiplication|multiplication modulo $m$]] and [[Definition:Modulo Addition|addition modulo $m$]]:
+{{begin-eqn}}
+{{eqn | l = \eqclass x m \times_m \paren {\eqclass y m +_m \eqclass z m}
+      | r = \eqclass x m \times_m \eqclass {y + z} m
+      | c = 
+}}
+{{eqn | r = \eqclass {x \paren {y + z} } m
+      | c = 
+}}
+{{eqn | r = \eqclass {\paren {x y} + \paren {x z} } m
+      | c = 
+}}
+{{eqn | r = \eqclass {x y} m +_m \eqclass {x z} m
+      | c = 
+}}
+{{eqn | r = \paren {\eqclass x m \times_m \eqclass y m} +_m \paren {\eqclass x m \times_m \eqclass z m}
+      | c = 
+}}
+{{end-eqn}}
+And the second is like it, namely this:
+{{begin-eqn}}
+{{eqn | l = \paren {\eqclass x m +_m \eqclass y m} \times_m \eqclass z m
+      | r = \eqclass {x + y} m \times_m \eqclass z m
+      | c = 
+}}
+{{eqn | r = \eqclass {\paren {x + y} z} m
+      | c = 
+}}
+{{eqn | r = \eqclass {\paren {x z} + \paren {y z} } m
+      | c = 
+}}
+{{eqn | r = \eqclass {x z} m +_m \eqclass {y z} m
+      | c = 
+}}
+{{eqn | r = \paren {\eqclass x m \times_m \eqclass z m} +_m \paren {\eqclass y m \times_m \eqclass z m}
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Intersection of Congruence Classes}
+Tags: Modulo Arithmetic, Set Intersection, Lowest Common Multiple, Intersection of Congruence Classes
+
+\begin{theorem}
+Let $\mathcal R_m$ denote [[Definition:Congruence Modulo Integer|congruence modulo $m$]] on the [[Definition:Set|set]] of [[Definition:Integer|integers]] $\Z$.
+Then:
+:$\mathcal R_m \cap \mathcal R_n = \mathcal R_{\lcm \set {m, n} }$
+where $\lcm \set {m, n}$ is the [[Definition:Lowest Common Multiple of Integers|lowest common multiple]] of $m$ and $n$.
+In the language of [[Definition:Modulo Arithmetic|modulo arithmetic]], this is equivalent to:
+:$a \equiv b \pmod m, a \equiv b \pmod n \implies a \equiv b \pmod {\lcm \set {m, n} }$
+\end{theorem}
+
+\begin{proof}
+Let $\tuple {a, b} \in \mathcal R_m \cap \mathcal R_n$.
+That is, let $\tuple {a, b} \in \mathcal R_m$ and $\tuple {a, b} \in \mathcal R_n$.
+That means, by definition of [[Definition:Congruence Modulo Integer|congruence]]:
+:$a \equiv b \pmod m$
+:$a \equiv b \pmod n$
+Thus by definition of [[Definition:Congruence (Number Theory)/Integers/Integer Multiple|congruence]]:
+:$\exists r, s \in \Z: a - b = r m, a - b = s n$
+Let $d = \gcd \set {m, n}$ so that $m = d m', n = d n', m' \perp n'$.
+Substituting for $m$ and $n$:
+:$r d m' = s d n'$ and so $r m' = s n'$.
+So $n' \divides r m'$ and $m' \perp n'$ so by [[Euclid's Lemma]] $n' \divides r$.
+So we can put $r = k n'$ and get:
+:$a - b = r m = k m n' = k m \dfrac n d = k \dfrac {m n} d$
+But:
+:$\dfrac {m n} d = \dfrac {m n} {\gcd \set {m, n} }$
+So by [[Product of GCD and LCM]]:
+:$a - b = k \lcm \set {m, n}$
+So:
+:$a \equiv b \pmod {\lcm \set {m, n} }$
+and hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Mappings Between Residue Classes}
+Tags: Residue Classes
+
+\begin{theorem}
+Let $\eqclass a m$ be the [[Definition:Residue Class|residue class of $a$ (modulo $m$)]].
+Let $\phi: \Z_m \to \Z_n$ be a [[Definition:Mapping|mapping]] given by:
+:$\map \phi {\eqclass x m} = \eqclass x n$
+Then $\phi$ is [[Definition:Well-Defined Mapping|well defined]] {{iff}} $m$ is a [[Definition:Divisor of Integer|divisor]] of $n$.
+\end{theorem}
+
+\begin{proof}
+For $\phi$ to be [[Definition:Well-Defined Mapping|well defined]], we require that:
+:$\forall x, y \in \Z_m: \eqclass x m = \eqclass y m \implies \map \phi {\eqclass x m} = \map \phi {\eqclass y m}$
+Now:
+:$\eqclass x m = \eqclass y m \implies x - y \divides m$
+For $\map \phi {\eqclass x m} = \map \phi {\eqclass y m}$ we require that:
+:$\eqclass x n = \eqclass y n \implies x - y \divides n$
+Thus $\phi$ is [[Definition:Well-Defined Mapping|well defined]] {{iff}}:
+:$x - y \divides m \implies x - y \divides n$
+That is, {{iff}} $m \divides n$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{GCD from Congruence Modulo m}
+Tags: Modulo Arithmetic, Greatest Common Divisor
+
+\begin{theorem}
+Let $a, b \in \Z, m \in \N$.
+ 
+Let $a$ be [[Definition:Congruence Modulo Integer|congruent]] to $b$ modulo $m$.
+Then the [[Definition:Greatest Common Divisor of Integers|GCD]] of $a$ and $m$ is equal to the [[Definition:Greatest Common Divisor of Integers|GCD]] of $b$ and $m$.
+That is:
+:$a \equiv b \pmod m \implies \gcd \set {a, m} = \gcd \set {b, m}$
+\end{theorem}
+
+\begin{proof}
+We have:
+:$a \equiv b \pmod m \implies \exists k \in \Z: a = b + k m$
+Thus:
+:$a = b + k m$
+and the result follows directly from [[GCD with Remainder]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Prime Number has 4 Integral Divisors}
+Tags: Prime Numbers
+
+\begin{theorem}
+Let $p$ be an [[Definition:Integer|integer]].
+Then $p$ is a [[Definition:Prime Number|prime number]] {{iff}} $p$ has exactly four [[Definition:Integer|integral]] [[Definition:Divisor of Integer|divisors]]: $1, -1, p, -p$.
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+Let $p$ be a [[Definition:Prime Number/Definition 1|prime number]] from the definition that $p$ has exactly $2$ [[Definition:Divisor of Integer|divisors]] which are [[Definition:Positive Integer|positive integers]].
+From [[One Divides all Integers]] and [[Integer Divides Itself]] those [[Definition:Positive Integer|positive integers]] are $1$ and $p$.
+Also, we have $-1 \divides p$ and $-p \divides p$ from [[One Divides all Integers]] and [[Integer Divides its Negative]].
+{{AimForCont}}:
+: $\exists x < 0: x \divides p$
+where $x \ne -1$ and $x \ne -p$.
+Then:
+: $\size x \divides x \divides p$
+and so $\size x$ is therefore a [[Definition:Positive Integer|positive integer]] other than $1$ and $p$ that [[Definition:Divisor of Integer|divides]] $p$.
+This is a [[Proof by Contradiction|contradiction]] of the condition for $p$ to be [[Definition:Prime Number/Definition 1|prime]].
+So $-1$ and $-p$ are the only [[Definition:Negative Integer|negative integers]] that [[Definition:Divisor of Integer|divide]] $p$.
+It follows that $p$ has exactly those four [[Definition:Divisor of Integer|divisors]].
+{{qed|lemma}}
+=== Sufficient Condition ===
+Suppose $p$ has the [[Definition:Divisor of Integer|divisors]] $1, -1, p, -p$.
+It follows that $1$ and $p$ are the only [[Definition:Positive Integer|positive integers]] that [[Definition:Divisor of Integer|divide]] $p$.
+Thus $p$ has exactly two [[Definition:Divisor of Integer|divisors]] which are [[Definition:Positive Integer|positive integers]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Prime not Divisor implies Coprime}
+Tags: Prime Numbers, Coprime Integers, Prime not Divisor implies Coprime
+
+\begin{theorem}
+Let $p, a \in \Z$.
+If $p$ is a [[Definition:Prime Number|prime number]] then:
+: $p \nmid a \implies p \perp a$
+where:
+: $p \nmid a$ denotes that $p$ does not [[Definition:Divisor of Integer|divide]] $a$
+: $p \perp a$ denotes that $p$ and $a$ are [[Definition:Coprime Integers|coprime]].
+It follows directly that if $p$ and $q$ are [[Definition:Prime Number|primes]], then:
+: $p \divides q \implies p = q$
+: $p \ne q \implies p \perp q$.
+\end{theorem}
+
+\begin{proof}
+Let $p \in \Bbb P, p \nmid a$.
+We need to show that $\gcd \set {a, p} = 1$.
+Let $\gcd \set {a, p} = d$.
+As $d \divides p$, we must have $d = 1$ or $d = p$ by [[GCD with Prime]].
+But if $d = p$, then $p \divides a$ by definition of [[Definition:Greatest Common Divisor of Integers|greatest common divisor]].
+So $d \ne p$ and therefore $d = 1$.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let $p$ be a [[Definition:Prime Number|prime number]].
+Let $a \in \Z$ be such that $p$ is not a [[Definition:Divisor of Integer|divisor]] of $a$.
+{{AimForCont}} $p$ and $a$ are not [[Definition:Coprime Integers|coprime]].
+Then:
+:$\exists c \in \Z_{>1}: c \divides p, c \divides a$
+where $\divides$ denotes [[Definition:Divisor of Integer|divisibility]].
+But then by definition of [[Definition:Prime Number|prime]]:
+:$c = p$
+Thus:
+:$p \divides a$
+The result follows by [[Proof by Contradiction]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Composite Number has Two Divisors Less Than It}
+Tags: Divisors
+
+\begin{theorem}
+Let $n \in \Z_{> 1}$ such that $n \notin \mathbb P$.
+Then:
+:$\exists a, b \in \Z: 1 < a < n, 1 < b < n: n = a b$
+That is, a non-[[Definition:Prime Number|prime]] number greater than $1$ can be expressed as the product of two [[Definition:Positive Integer|positive integers]] strictly greater than $1$ and less than $n$.
+Note that these two numbers are not necessarily [[Definition:Distinct Elements|distinct]].
+\end{theorem}
+
+\begin{proof}
+Since $n \notin \mathbb P$, it has a [[Definition:Positive Integer|positive]] [[Definition:Divisor of Integer|factor]] $a$ such that $a \ne 1$ and $a \ne n$.
+Hence $\exists b \in \Z: n = a b$.
+Thus by definition of [[Definition:Divisor of Integer|factor]]:
+:$a \divides n$
+where $\divides$ denotes [[Definition:Divisor of Integer|divisibility]].
+From [[Divisor Relation on Positive Integers is Partial Ordering]]:
+:$a \le n$
+As $a \ne n$, it follows that $a < n$.
+From [[One Divides all Integers]]:
+:$1 \divides a$
+Thus from [[Divisor Relation on Positive Integers is Partial Ordering]]:
+:$1 \le a$
+Similarly, as $1 \ne a$ it follows that $1 < a$.
+Since $a \ne n$, it follows that $b \ne 1$.
+Similarly, since $a \ne 1$, it follows that $b \ne n$.
+Thus:
+:$b \divides n: 1 \ne b \ne n$
+Arguing as above, we show that $1 < b < n$ and the result follows.
+{{qed}}
+Note that we have not shown (and it is not necessarily the case) that $a \ne b$.
+[[Category:Divisors]]
+197jd22ea9wbyx8phcgp1vaw454jncw
+\end{proof}<|endoftext|>
+\section{Condition for Divisibility of Powers of Prime}
+Tags: Prime Numbers
+
+\begin{theorem}
+Let $p$ be a [[Definition:Prime Number|prime]].
+Let $k, l \in \Z_{>0}$.
+Then:
+:$p^k \divides p^l \iff k \le l$
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+Let $k \le l$.
+Then:
+:$l - k \ge 0$
+Thus $p^k, p^{l - k} \in \Z$ such that $p^l = p^k p^{l - k}$.
+Thus:
+:$p^k \divides p^l$
+{{qed|lemma}}
+=== Sufficient Condition ===
+Let $p^k \divides p^l$.
+Then:
+:$\exists b \in \Z_{>0}: p^l = p^k b$
+By the [[Fundamental Theorem of Arithmetic]], $b$ has a unique [[Definition:Prime Decomposition|prime decomposition]].
+Either $b = 1$ (in which case $k = l$) or $b$ has a [[Definition:Prime Decomposition|prime decomposition]] consisting entirely of $p$'s.
+In this case:
+:$\exists m \in \Z: b = p^m$
+Hence:
+:$p^{l - k} = p^m$
+Thus from the [[Fundamental Theorem of Arithmetic]]:
+:$l - k = m > 0$
+Thus:
+:$l > k$
+{{qed|lemma}}
+The result follows from combining the two results.
+{{qed}}
+[[Category:Prime Numbers]]
+tan97zsd81ac6y8j52yd6eemo9afhp9
+\end{proof}<|endoftext|>
+\section{Exponents of Primes in Prime Decomposition are Less iff Divisor}
+Tags: Prime Numbers
+
+\begin{theorem}
+Let $a, b \in \Z_{>0}$.
+Then $a \divides b$ {{iff}}:
+:$(1): \quad$ every [[Definition:Prime Number|prime]] $p_i$ in the [[Definition:Prime Decomposition|prime decomposition]] of $a$ appears in the [[Definition:Prime Decomposition|prime decomposition]] of $b$
+and:
+:$(2): \quad$ the [[Definition:Exponent|exponent]] of each $p_i$ in $a$ is less than or equal to its [[Definition:Exponent|exponent]] in $b$.
+\end{theorem}
+
+\begin{proof}
+Let $a, b \in \Z_{>0}$.
+Let their [[Definition:Prime Decomposition|prime decomposition]] be:
+:$a = p_1^{k_1} p_2^{k_2} \dotsm p_n^{k_n}$
+:$b = q_1^{l_1} q_2^{l_2} \dotsm q_n^{l_n}$
+=== Necessary Condition ===
+Let:
+:$(1): \quad$  [[Definition:Prime Number|prime]] in the [[Definition:Prime Decomposition|prime decomposition]] of $a$ appear in the [[Definition:Prime Decomposition|prime decomposition]] of $b$
+and:
+:$(2): \quad$ its [[Definition:Exponent|exponent]] in $a$ be less than or equal to its [[Definition:Exponent|exponent]] in $b$.
+Then:
+{{begin-eqn}}
+{{eqn | l = a
+      | r = p_1^{k_1} p_2^{k_2} \dotsm p_r^{k_r}
+      | c = 
+}}
+{{eqn | l = b
+      | r = p_1^{l_1} p_2^{l_2} \dotsm p_r^{l_r} \dotsm p_s^{l_s}
+      | c = 
+}}
+{{end-eqn}}
+where:
+:$k_1 \le l_1, k_2 \le l_2, \dotsc, k_r \le l_r, r \le s$
+Thus:
+:$d = p_1^{l_1 - k_1} p_2^{l_2 - k_2} \dotsm p_r^{l_r - k_r} \in \Z$
+and so:
+:$b = a d$
+So $a \divides b$.
+{{qed|lemma}}
+=== Sufficient Condition ===
+Let $a \divides b$.
+Let $a = p_1^{k_1} p_2^{k_2} \dotsm p_r^{k_r}$ be the [[Definition:Prime Decomposition|prime decomposition]] of $a$.
+Then:
+:$\forall i \in \N_r: p_i^{k_i} \divides a$
+Hence by [[Divisor Relation on Positive Integers is Partial Ordering]] each $p_i^{k_i}$ also [[Definition:Divisor of Integer|divides]] $b$.
+Thus:
+:$\exists c \in \Z: b = p_i^{k_i} c$
+The [[Definition:Prime Decomposition|prime decomposition]] of $b$ is therefore:
+:$b = p_i^{k_i} \times \paren {\text {prime decomposition of $c$} }$
+which may need to be rearranged.
+So $p_i$ must occur in the [[Definition:Prime Decomposition|prime decomposition]] of $b$ with an [[Definition:Exponent|exponent]] at least as big as $k_i$.
+The result follows.
+{{qed}}
+[[Category:Prime Numbers]]
+eovrd4xpmf3wv3snuvq7s6yuwm49psk
+\end{proof}<|endoftext|>
+\section{Not Coprime means Common Prime Factor}
+Tags: Coprime Integers, Prime Numbers
+
+\begin{theorem}
+Let $a, b \in \Z$.
+If $d \divides a$ and $d \divides b$ such that $d > 1$, then $a$ and $b$ have a [[Definition:Common Divisor of Integers|common divisor]] which is [[Definition:Prime Number|prime]].
+\end{theorem}
+
+\begin{proof}
+As $d > 1$, it has a [[Definition:Prime Decomposition|prime decomposition]].
+Thus there exists a prime $p$ such that $p \divides d$.
+From [[Divisor Relation on Positive Integers is Partial Ordering]], we have $p \divides d, d \divides a \implies p \divides a$, and similarly for $b$.
+The result follows.
+{{qed}}
+[[Category:Coprime Integers]]
+[[Category:Prime Numbers]]
+1dbh22tlukqincvbw8rkczuduc9vuvj
+\end{proof}<|endoftext|>
+\section{Set of Divisors of Integer}
+Tags: Number Theory
+
+\begin{theorem}
+Let $n \in \Z_{>1}$.
+Let $n$ be expressed in its [[Definition:Prime Decomposition|prime decomposition]]:
+:$n = p_1^{k_1} p_2^{k_2} \dotsm p_r^{k_r}$
+where $p_1 < p_2 < \dotsb < p_r$ are distinct [[Definition:Prime Number|primes]] and $k_1, k_2, \ldots, k_r$ are [[Definition:Positive Integer|positive integers]].
+The [[Definition:Set|set]] of [[Definition:Divisor of Integer|divisors]] of $n$ is:
+:$\set {p_1^{h_1} p_2^{h_2} \dotsm p_r^{h_r}: 0 \le h_i \le k_i, i = 1, 2, \ldots, r}$
+\end{theorem}
+
+\begin{proof}
+Each [[Definition:Integer|integer]] in the given set is a [[Definition:Divisor of Integer|divisor]] of $n$ because:
+:$(1): \quad \forall i: k_i - h_i \ge 0$
+:$(2): \quad n = \paren {p_1^{h_1} p_2^{h_2} \dotsm p_r^{h_r} } p_1^{k_1 - h_1} p_2^{k_2 - h_2} \ldots p_r^{k_r - h_r}$
+from [[Exponents of Primes in Prime Decomposition are Less iff Divisor]].
+By the [[Fundamental Theorem of Arithmetic]], these [[Definition:Integer|integers]] are distinct.
+It is necessary to show that the [[Definition:Integer|integers]] in this [[Definition:Set|set]] are the ''only'' [[Definition:Divisor of Integer|divisors]] of $n$.
+Let $d > 1$ and let $p \in \mathbb P: p \divides d$.
+{{begin-eqn}}
+{{eqn | o = 
+      | r = p \divides d \land d \divides n
+      | c = 
+}}
+{{eqn | o = \leadsto
+      | r = p \divides n
+      | c = [[Divisor Relation on Positive Integers is Partial Ordering]]
+}}
+{{eqn | o = \leadsto
+      | r = \exists i: p = p_i, 1 \le i \le r
+      | c = 
+}}
+{{eqn | o = \leadsto
+      | r = p \in \set {p_i: 1 \le i \le r}
+      | c = 
+}}
+{{eqn | o = \leadsto
+      | r = d = p_1^{h_1} p_2^{h_2} \dotsm p_r^{h_r}: 0 \le h_i
+      | c = 
+}}
+{{end-eqn}}
+It remains to be shown that:
+:$\forall i: h_1 \le k_i$
+First note that:
+:$d \divides n \implies \forall i: p_i^{k_i} \divides n$
+From above, all the primes $p_i$ are [[Definition:Distinct|distinct]].
+Therefore by [[Prime not Divisor implies Coprime]]:
+:$p_1 \nmid p_2^{k_2} p_3^{k_3} \dotsm p_r^{k_r} \implies \gcd \set {p_1, p_2^{k_2} p_3^{k_3} \ldots p_r^{k_r} } = 1$
+So:
+:$p_1^{h_1} \divides n \implies n = p_1^{k_1} \paren {p_2^{k_2} p_3^{k_3} \dotsm p_r^{k_r} }$
+ 
+By [[Euclid's Lemma]]:
+:$p_1^{h_1} \divides p_1^{k_1} \implies h_1 \le k_1$
+and the same argument applies to each of the other [[Definition:Prime Factor|prime factors]] of $n$.
+The result follows.
+{{qed}}
+[[Category:Number Theory]]
+rzbwqi92fhcc2ar5kuqb694dh6aozep
+\end{proof}<|endoftext|>
+\section{Sum Less Minimum is Maximum}
+Tags: Algebra
+
+\begin{theorem}
+For all [[Definition:Number|numbers]] $a, b$ where $a, b$ in $\N, \Z, \Q$ or $\R$:
+:$a + b - \min \left({a, b}\right) = \max \left({a, b}\right)$
+\end{theorem}
+
+\begin{proof}
+From [[Sum of Maximum and Minimum]] we have that $a + b = \max \left({a, b}\right) + \min \left({a, b}\right)$.
+Thus $a + b - \min \left({a, b}\right) = \max \left({a, b}\right)$ follows by subtracting $\min \left({a, b}\right)$ from both sides.
+It is clear that this result applies when $a, b$ in $\Z, \Q$ or $\R$, as [[Definition:Subtraction|subtraction]] is well-defined throughout those number sets.
+However, it can be seen to apply in $\N$ as well, despite the fact that $n - m$ is defined on $\N$ only when $n \ge m$.
+This is because the fact that $a + b \ge \min \left({a, b}\right)$ follows immediately again from $a + b = \max \left({a, b}\right) + \min \left({a, b}\right)$.
+{{qed}}
+[[Category:Algebra]]
+qfqfes3zbhckirbxe25ex8hvfroraqj
+\end{proof}<|endoftext|>
+\section{Sum Less Maximum is Minimum}
+Tags: Algebra
+
+\begin{theorem}
+For all [[Definition:Number|numbers]] $a, b$ where $a, b$ in $\N, \Z, \Q$ or $\R$:
+:$a + b - \max \left({a, b}\right) = \min \left({a, b}\right)$.
+\end{theorem}
+
+\begin{proof}
+From [[Sum of Maximum and Minimum]] we have that $a + b = \max \left({a, b}\right) + \min \left({a, b}\right)$.
+Thus $a + b - \max \left({a, b}\right) = \min \left({a, b}\right)$ follows by subtracting $\max \left({a, b}\right)$ from both sides.
+It is clear that this result applies when $a, b$ in $\Z, \Q$ or $\R$, as [[Definition:Subtraction|subtraction]] is well-defined throughout those number sets.
+However, it can be seen to apply in $\N$ as well, despite the fact that $n - m$ is defined on $\N$ only when $n \ge m$.
+This is because the fact that $a + b \ge \max \left({a, b}\right)$ follows immediately again from $a + b = \max \left({a, b}\right) + \min \left({a, b}\right)$.
+{{qed}}
+[[Category:Algebra]]
+ef4y60wxp42ltsv2eex2nzjibeles2j
+\end{proof}<|endoftext|>
+\section{GCD and LCM from Prime Decomposition}
+Tags: Greatest Common Divisor, Lowest Common Multiple, Prime Numbers
+
+\begin{theorem}
+Let $m, n \in \Z$.
+Let:
+:$m = p_1^{k_1} p_2^{k_2} \dotsm p_r^{k_r}$
+:$n = p_1^{l_1} p_2^{l_2} \dotsm p_r^{l_r}$
+:$p_i \divides m \lor p_i \divides n, 1 \le i \le r$.
+That is, the [[Definition:Prime Number|primes]] given in these [[Definition:Prime Decomposition|prime decompositions]] may be [[Definition:Divisor of Integer|divisors]] of ''either'' of the numbers $m$ or $n$.
+Note that if one of the [[Definition:Prime Number|primes]] $p_i$ does not appear in the [[Definition:Prime Decomposition|prime decompositions]] of either one of $m$ or $n$, then its corresponding index $k_i$ or $l_i$ will be zero.
+Then the following results apply:
+:$\gcd \set {m, n} = p_1^{\min \set {k_1, l_1} } p_2^{\min \set {k_2, l_2} } \ldots p_r^{\min \set {k_r, l_r} }$
+:$\lcm \set {m, n} = p_1^{\max \set {k_1, l_1} } p_2^{\max \set {k_2, l_2} } \ldots p_r^{\max \set {k_r, l_r} }$
+\end{theorem}
+
+\begin{proof}
+The proof of these results can be found in:
+:[[GCD from Prime Decomposition]]
+:[[LCM from Prime Decomposition]]
+{{Qed}}
+[[Category:Greatest Common Divisor]]
+[[Category:Lowest Common Multiple]]
+[[Category:Prime Numbers]]
+iur6iagonkv1hmri44ycxc6lbtsc4ux
+\end{proof}<|endoftext|>
+\section{GCD and LCM Distribute Over Each Other}
+Tags: Greatest Common Divisor, Lowest Common Multiple, Distributive Operations
+
+\begin{theorem}
+Let $a, b, c \in \Z$.
+Then:
+: $\lcm \set {a, \gcd \set {b, c} } = \gcd \set {\lcm \set {a, b}, \lcm \set {a, c} }$
+: $\gcd \set {a, \lcm \set {b, c} } = \lcm \set {\gcd \set {a, b}, \gcd \set {a, c} }$
+That is, [[Definition:Greatest Common Divisor of Integers|greatest common divisor]] and [[Definition:Lowest Common Multiple of Integers|lowest common multiple]] are [[Definition:Distributive Operation|distributive]] over each other.
+\end{theorem}
+
+\begin{proof}
+=== LCM Distributive over GCD ===
+Let $p_s$ be any of the prime divisors of $a, b$ or $c$, and let $s_a, s_b$ and $s_c$ be its exponent in each of those numbers.
+Let $x = \lcm \set {a, \gcd \set {b, c} }$.
+Then from [[GCD and LCM from Prime Decomposition]], the exponent of $p_s$ in $x$ is $\max \set {s_a, \min \set {s_b, s_c} }$.
+From [[Max and Min Operations are Distributive over Each Other]], $\max$ distributes over $\min$.
+Therefore:
+: $\max \set {s_a, \min \set {s_b, s_c} } = \min \set {\max \set {s_a, s_b}, \max \set {s_a, s_c} }$
+Hence it follows that [[Definition:Lowest Common Multiple of Integers|LCM]] is [[Definition:Distributive Operation|distributive]] over [[Definition:Greatest Common Divisor of Integers|GCD]].
+{{qed|lemma}}
+=== GCD Distributive over LCM ===
+Let $p_s$ be any of the prime divisors of $a, b$ or $c$, and let $s_a, s_b$ and $s_c$ be its exponent in each of those numbers.
+Let $x = \gcd \set {a, \lcm \set {b, c} }$.
+Then from [[GCD and LCM from Prime Decomposition]], the exponent of $p_s$ in $x$ is $\min \set {s_a, \max \set {s_b, s_c} }$.
+From [[Max and Min Operations are Distributive over Each Other]], $\min$ distributes over $\max$.
+Therefore:
+: $\min \set {s_a, \max \set {s_b, s_c} } = \max \set {\min \set {s_a, s_b}, \min \set {s_a, s_c} }$
+and the result follows.
+Hence it follows that [[Definition:Greatest Common Divisor of Integers|GCD]] is [[Definition:Distributive Operation|distributive]] over [[Definition:Lowest Common Multiple of Integers|LCM]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Tau Function from Prime Decomposition}
+Tags: Tau Function, Tau Function from Prime Decomposition
+
+\begin{theorem}
+Let $n$ be an [[Definition:Integer|integer]] such that $n \ge 2$.
+Let the [[Definition:Prime Decomposition|prime decomposition]] of $n$ be:
+:$n = p_1^{k_1} p_2^{k_2} \cdots p_r^{k_r}$
+Let $\map \tau n$ be the [[Definition:Tau Function|tau function]] of $n$.
+Then:
+:$\displaystyle \map \tau n = \prod_{j \mathop = 1}^r \paren {k_j + 1}$
+\end{theorem}
+
+\begin{proof}
+We have:
+:$d \divides n \implies \forall i: 1 \le i \le r: d = p_1^{l_1} p_2^{l_2} \ldots p_1^{l_1}, 0 \le l_i \le k_i$
+For each $i$, there are $k_i + 1$ choices for $l_i$, making $\paren {k_1 + 1} \paren {k_2 + 1} \cdots \paren {k_r + 1}$ choices in all.
+By the [[Fundamental Theorem of Arithmetic]] and hence the uniqueness of [[Definition:Prime Decomposition|prime decomposition]], each of these choices results in a different number, therefore a distinct [[Definition:Divisor of Integer|divisor]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+From [[Tau of Power of Prime]] we have:
+:$\forall j \in \closedint 1 r: \map \tau {p_j^{k_j} } = k_j + 1$
+The result follows immediately from [[Tau Function is Multiplicative]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{N less than M to the N}
+Tags: Algebra
+
+\begin{theorem}
+:$\forall m, n \in \Z_{>0}: m > 1 \implies n < m^n$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = n
+      | r = \underbrace {1 + 1 + \cdots + 1}_{\text {$n$ times} }
+      | c = 
+}}
+{{eqn | o = <
+      | r = 1 + m + m^2 + \cdots + m^{n - 1}
+      | c = as $m > 1$
+}}
+{{eqn | r = \frac {m^n - 1} {m - 1}
+      | c = [[Sum of Geometric Sequence]]
+}}
+{{eqn | o = \le
+      | r = m^n - 1
+      | c = as $m - 1 \ge 1$
+}}
+{{eqn | o = <
+      | r = m^n
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Basis Representation Theorem}
+Tags: Number Bases, Named Theorems
+
+\begin{theorem}
+Let $b \in \Z: b > 1$.
+For every $n \in \Z_{> 0}$, there exists [[Definition:Exactly One|one and only one]] [[Definition:Sequence|sequence]] $\sequence {r_j}_{0 \mathop \le j \mathop \le t}$ such that:
+: $(1): \quad \displaystyle n = \sum_{k \mathop = 0}^t r_k b^k$
+: $(2): \quad \displaystyle \forall k \in \closedint 0 t: r_k \in \N_b$
+: $(3): \quad r_t \ne 0$
+This unique sequence is called the '''representation of $n$ to the base $b$''', or, informally, we can say '''$n$ is (written) in base $b$'''.
+\end{theorem}
+
+\begin{proof}
+Let $\map {s_b} n$ be the number of ways of representing $n$ to the base $b$.
+We need to show that $\map {s_b} n = 1$ always.
+Now, it is possible that some of the $r_k = 0$ in a particular representation.
+So we may exclude these terms, and it won't affect the representation.
+So, suppose:
+:$n = r_k b^k + r_{k - 1} b^{k - 1} + \cdots + r_t b^t$
+where $r_k \ne 0, r_t \ne 0$.
+Then:
+{{begin-eqn}}
+{{eqn | l = n - 1
+      | r = r_k b^k + r_{k - 1} b^{k - 1} + \cdots + r_t b^t - 1
+      | c = 
+}}
+{{eqn | r = r_k b^k + r_{k - 1} b^{k - 1} + \cdots + \paren {r_t - 1} b^t + b^t - 1
+      | c = 
+}}
+{{eqn | r = r_k b^k + r_{k - 1} b^{k - 1} + \cdots + \paren {r_t - 1} b^t + \sum_{j \mathop = 0}^{t - 1} {\paren {b - 1} b^j}
+      | c = [[Sum of Geometric Sequence]]
+}}
+{{end-eqn}}
+Note that we have already specified that $b > 1$.
+So for each representation of $n$ to the base $b$, we can find a representation of $n - 1$.
+If $n$ has another representation to the base $b$, then the same procedure will generate a new representation of $n - 1$.
+Thus:
+:$(1): \quad \map {s_b} n \le \map {s_b} {n - 1}$
+Note that this holds even if $n$ has no representation at all, because if this is the case, then $\map {s_b} n = 0 \le \map {s_b} {n - 1}$.
+So for $m, n \in \N$ such that $m \ge n$, the inequality $(1)$ implies the following:
+:$\forall m, n: \map {s_b} m \le \map {s_b} {m - 1} \le \ldots \le \map {s_b} {n + 1} \le \map {s_b} n$
+From [[N less than M to the N]] and the fact that $b^n$ has at least one representation (itself), we see:
+:$1 \le \map {s_b} {b^n} \le \map {s_b} n \le \map {s_b} 1 = 1$
+The entries at either end of this inequality are $1$, so all the intermediate entries must also be $1$.
+So $\map {s_b} n = 1$ and the theorem has been proved.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sum of Geometric Sequence}
+Tags: Sums of Sequences, Geometric Sequences, Sum of Geometric Sequence
+
+\begin{theorem}
+Let $x$ be an element of one of the [[Definition:Standard Number Field|standard number fields]]: $\Q, \R, \C$ such that $x \ne 1$.
+Let $n \in \N_{>0}$.
+Then:
+:$\displaystyle \sum_{j \mathop = 0}^{n - 1} x^j = \frac {x^n - 1} {x - 1}$
+\end{theorem}
+
+\begin{proof}
+For all $n \in \N_{> 0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
+:$\displaystyle \sum_{j \mathop = 0}^{n - 1} x^j = \frac {x^n - 1} {x - 1}$
+=== Basis for the Induction ===
+$\map P 1$ is the case:
+{{begin-eqn}}
+{{eqn | l = \dfrac {x^1 - 1} {x - 1}
+      | r = 1
+      | c = 
+}}
+{{eqn | r = 2^0
+      | c = 
+}}
+{{eqn | r = \sum_{j \mathop = 0}^{1 - 1} x^j
+      | c = 
+}}
+{{end-eqn}}
+so $\map P 1$ holds.
+This is our [[Definition:Basis for the Induction|basis for the induction]].
+=== Induction Hypothesis ===
+Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
+So this is our [[Definition:Induction Hypothesis|induction hypothesis]]:
+:$\displaystyle \sum_{j \mathop = 0}^{k - 1} x^j = \frac {x^k - 1} {x - 1}$
+Then we need to show:
+:$\displaystyle \sum_{j \mathop = 0}^k x^j = \frac {x^{k + 1} - 1} {x - 1}$
+=== Induction Step ===
+This is our [[Definition:Induction Step|induction step]]:
+{{begin-eqn}}
+{{eqn | l = \sum_{j \mathop = 0}^k x^j
+      | r = \sum_{j \mathop = 0}^{k - 1} x^j + x^k
+      | c = 
+}}
+{{eqn | r = \frac {x^k - 1} {x - 1} + x^k
+      | c = 
+}}
+{{eqn | r = \frac {x^k - 1 + \paren {x - 1} x^k} {x - 1}
+      | c = 
+}}
+{{eqn | r = \frac {x^k - 1 + x^{k + 1} - x^k} {x - 1}
+      | c = 
+}}
+{{eqn | r = \frac {x^{k + 1} - 1} {x - 1}
+      | c = 
+}}
+{{end-eqn}}
+So $\map P k \implies \map P {k + 1}$ and the result follows by the [[Principle of Mathematical Induction]].
+Therefore:
+:$\displaystyle \forall n \in \N_{> 0}: \sum_{j \mathop = 0}^{n - 1} x^j = \frac {x^n - 1} {x - 1}$
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let $\displaystyle S_n = \sum_{j \mathop = 0}^{n - 1} x^j$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \paren {x - 1} S_n
+      | r = x S_n - S_n
+      | c = 
+}}
+{{eqn | r = x \sum_{j \mathop = 0}^{n - 1} x^j - \sum_{j \mathop = 0}^{n - 1} x^j
+      | c = 
+}}
+{{eqn | r = \sum_{j \mathop = 1}^n x^j - \sum_{j \mathop = 0}^{n - 1} x^j
+      | c = 
+}}
+{{eqn | r = x^n + \sum_{j \mathop = 1}^{n - 1} x^j - \paren {x^0 + \sum_{j \mathop = 1}^{n - 1} x^j}
+      | c = 
+}}
+{{eqn | r = x^n - x^0
+      | c = 
+}}
+{{eqn | r = x^n - 1
+      | c = 
+}}
+{{end-eqn}}
+The result follows.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+From [[Difference of Two Powers]]:
+:$\displaystyle a^n - b^n = \paren {a - b} \paren {a^{n - 1} + a^{n - 2} b + a^{n - 3} b^2 + \ldots + a b^{n - 2} + b^{n - 1} } = \paren {a - b} \sum_{j \mathop = 0}^{n - 1} a^{n - j - 1} b^j$
+Set $a = x$ and $b = 1$:
+:$\displaystyle x^n - 1 = \paren {x - 1} \paren {x^{n - 1} + x^{n - 2} + \cdots + x + 1} = \paren {x - 1} \sum_{j \mathop = 0}^{n - 1} x^j$
+from which the result follows directly.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+=== [[Sum of Geometric Sequence/Proof 4/Lemma|Lemma]] ===
+{{:Sum of Geometric Sequence/Proof 4/Lemma}}{{qed|lemma}}
+Then by the [[Sum of Geometric Sequence/Proof 4/Lemma|lemma]]:
+{{begin-eqn}}
+{{eqn | l = \paren {1 - x} \sum_{i \mathop = 0}^{n - 1} x^i
+      | r = 1 - x^n
+}}
+{{eqn | ll= \leadsto
+      | l = \sum_{i \mathop = 0}^{n - 1} x^i = \frac {1 - x^n} {1 - x}
+      | r = \frac {x^n - 1} {x - 1}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Congruence of Sum of Digits to Base Less 1}
+Tags: Number Theory
+
+\begin{theorem}
+Let $x \in \Z$, and $b \in \N, b > 1$.
+Let $x$ be [[Basis Representation Theorem|written in base $b$]]:
+:$x = \sqbrk {r_m r_{m - 1} \ldots r_2 r_1 r_0}_b$
+Then:
+:$\displaystyle \map {s_b} x = \sum_{j \mathop = 0}^m r_j \equiv x \pmod {b - 1}$
+where $\map {s_b} x$ is the [[Definition:Digit Sum|digit sum]] of $x$ [[Basis Representation Theorem|in base $b$ notation]].
+That is, the [[Definition:Digit Sum|digit sum]] of any [[Definition:Integer|integer]] $x$ [[Basis Representation Theorem|in base $b$ notation]] is [[Definition:Congruence Modulo Integer|congruent to $x$ modulo $b - 1$]].
+\end{theorem}
+
+\begin{proof}
+Let $x \in \Z, x > 0$, and $b \in \N, b > 1$.
+Then from the [[Basis Representation Theorem]], $x$ can be expressed uniquely as:
+:$\displaystyle x = \sum_{j \mathop = 0}^m r_j b^j, r_0, r_1, \ldots, r_m \in \set {0, 1, \ldots, b - 1}$
+Proof by [[Principle of Mathematical Induction|induction]]:
+For all $n \in \N_{>0}$, let $\map P n$ be the [[Definition:Proposition|proposition]] $\displaystyle \sum_{j \mathop = 0}^n r_j \equiv x \pmod {b - 1}$.
+=== Basis for the Induction ===
+$\map P 1$ is trivially true:
+:$\forall x: 0 \le x \le b: x \equiv x \pmod {b - 1}$
+This is our [[Definition:Basis for the Induction|basis for the induction]].
+=== Induction Hypothesis ===
+Now we need to show that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true.
+So this is our [[Definition:Induction Hypothesis|induction hypothesis]]:
+:$\displaystyle \sum_{j \mathop = 0}^k r_j \equiv \sum_{j \mathop = 0}^k r_j b^j \pmod {b - 1}$
+Then we need to show:
+:$\displaystyle \sum_{j \mathop = 0}^{k \mathop + 1} r_j \equiv \sum_{j \mathop = 0}^{k \mathop + 1} r_j b^j \pmod {b - 1}$
+=== Induction Step ===
+This is our [[Definition:Induction Step|induction step]]:
+Let $y$ be expressed as:
+:$\displaystyle y = \sum_{j \mathop = 0}^{k \mathop + 1} {r_j b^j}, r_0, r_1, \ldots, r_{k + 1} \in \set {0, 1, \ldots, b}$
+Then:
+:$\displaystyle y = \sum_{j \mathop = 0}^k r_j b^j + r_{k + 1} b^{k + 1}$
+Now from [[Congruence of Powers]]:
+:$b \equiv 1 \pmod {b - 1} \leadsto b^n \equiv 1^n \pmod {b - 1} \leadsto b^n \equiv 1 \pmod {b - 1}$
+So by [[Definition:Modulo Multiplication|modulo multiplication]]:
+:$r_{k + 1} b^{k + 1} \equiv r_{k + 1} \pmod {b - 1}$
+From the [[Congruence of Sum of Digits to Base Less 1#Induction Hypothesis|induction hypothesis]]:
+:$\displaystyle \sum_{j \mathop = 0}^{k \mathop + 1} r_j \equiv y \pmod {b - 1}$
+Thus by [[Definition:Modulo Addition|modulo addition]]:
+:$\displaystyle \sum_{j \mathop = 0}^{k \mathop + 1} r_j \equiv \sum_{j \mathop = 0}^k r_j + r_{k + 1} \pmod {b - 1}$
+Hence $\map P k \implies \map P {k + 1}$ and the result follows by the [[Principle of Mathematical Induction]]:
+:$\displaystyle \sum_{j \mathop = 0}^n r_j \equiv \sum_{j \mathop = 0}^n r_j b^j \pmod {b - 1}$
+{{Qed}}
+[[Category:Number Theory]]
+9i4cwx5p7k5bks1ofs6jy82lm78qvu2
+\end{proof}<|endoftext|>
+\section{Euler Phi Function of Integer}
+Tags: Euler Phi Function
+
+\begin{theorem}
+Let $n \in \Z_{>0}$, that is, a [[Definition:Strictly Positive Integer|(strictly) positive integer]].
+Let $\phi: \Z_{>0} \to \Z_{>0}$ be the [[Definition:Euler Phi Function|Euler $\phi$-function]].
+Then for any $n \in \Z_{>0}$, we have:
+:$\map \phi n = n \paren {1 - \dfrac 1 {p_1} } \paren {1 - \dfrac 1 {p_2} } \cdots \paren {1 - \dfrac 1 {p_r} }$
+where $p_1, p_2, \ldots, p_r$ are the distinct [[Definition:Prime Number|primes]] dividing $n$.
+Or, more compactly:
+:$\displaystyle \map \phi n = n \prod_{p \mathop \divides n} \paren {1 - \frac 1 p}$
+where $p \divides n$ denotes the [[Definition:Prime Number|primes]] which [[Definition:Divisor of Integer|divide]] $n$.
+\end{theorem}
+
+\begin{proof}
+If $n = 1$ the result holds by inspection.
+Let $n \ge 2$.
+We express $n$ in its [[Definition:Prime Decomposition|prime decomposition]]:
+:$n = p_1^{k_1} p_2^{k_2} \cdots p_r^{k_r}, p_1 < p_2 < \cdots < p_r$
+as it is always possible to do.
+By definition, all primes are [[Definition:Coprime Integers|coprime]] to each other.
+Hence from [[Euler Phi Function is Multiplicative]]:
+:$\map \phi n = \map \phi {p_1^{k_1} } \map \phi {p_2^{k_2} } \cdots \map \phi {p_r^{k_r} }$
+and from [[Euler Phi Function of Prime Power]]:
+:$\map \phi {p^k} = p^k \paren {1 - \dfrac 1 p}$
+So:
+:$\map \phi n = p_1^{k_1} \paren {1 - \dfrac 1 {p_1} } p_2^{k_2} \paren {1 - \dfrac 1 {p_2} } \cdots p_r^{k_r} \paren {1 - \dfrac 1 {p_r} }$
+and the result follows directly from:
+:$n = p_1^{k_1} p_2^{k_2} \cdots p_r^{k_r}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Euler Phi Function is Multiplicative}
+Tags: Multiplicative Functions, Euler Phi Function
+
+\begin{theorem}
+The [[Definition:Euler Phi Function|Euler $\phi$ function]] is a [[Definition:Multiplicative Arithmetic Function|multiplicative function]]:
+:$m \perp n \implies \map \phi {m n} = \map \phi m \, \map \phi n$
+where $m, n \in \Z_{>0}$.
+\end{theorem}
+
+\begin{proof}
+Let $R = \set {r_1, r_2, \ldots, r_{\map \phi m} }$ and $S = \set {s_1, s_2, \ldots, s_{\map \phi n} }$ be the [[Definition:Reduced Residue System|reduced residue systems]] for the respective moduli $m$ and $n$.
+We are to show that the set of $\map \phi m \map \phi n$ [[Definition:Integer|integers]]:
+:$T = \set {n r + m s: r \in R, s \in S}$
+is a [[Definition:Reduced Residue System|reduced residue system]] for modulus $m n$.
+We need to establish the following:
+: Each integer in $T$ is [[Definition:Coprime Integers|prime to]] $m n$
+: No two integers in $T$ is [[Definition:Congruence (Number Theory)|congruent modulo $m n$]]
+: Each integer [[Definition:Coprime Integers|prime to]] $m n$ is [[Definition:Congruence (Number Theory)|congruent modulo $m n$]] to one of these integers in $T$.
+We prove each in turn:
+Let $p$ be a [[Definition:Prime Factor|prime divisor]] of $\gcd \set {n r + m s, m n}$ where $r \in R, s \in S$.
+As $p$ divides $m n$ but $m \perp n$, $p$ either divides $m$ or $n$ but not both, from [[Divisors of Product of Coprime Integers]].
+{{WLOG}}, suppose $p \divides m$.
+Then as $p \divides n r + m s$, we have $p \divides n r$ and hence $p \divides r$.
+But then $p \divides \gcd \set {m, r} = 1$ which is a contradiction.
+Similarly if $p \divides n$.
+So there is no such prime and hence $n r + m s \perp m n$.
+{{qed|lemma}}
+Let $n r + m s = n r' + m s' \pmod {m n}$, where $r, r' \in R, s, s' \in S$.
+Then:
+:$n \paren {r - r'} + m \paren {s - s'} = k \paren {m n}$ for some $k \in \Z$.
+As $m$ divides two of these terms it must divide the third, so:
+:$m \divides n \paren {r - r'}$
+Now $m \perp n$ so by [[Euclid's Lemma]]:
+:$m \divides \paren {r - r'}$
+or:
+:$r \equiv r' \pmod m$
+But $r$ and $r'$ are part of the same [[Definition:Reduced Residue System|reduced residue system]] modulo $m$, so:
+:$r = r'$
+Similarly for $n$: we get:
+:$s = s'$
+Hence distinct elements of $T$ can not be [[Definition:Congruence (Number Theory)|congruent modulo $m n$]].
+{{qed|lemma}}
+Let $k \in \Z: k \perp m n$.
+Since $m \perp n$, from [[Set of Integer Combinations equals Set of Multiples of GCD]] we can write $k = n r' + m s'$ for some $r', s' \in \Z$.
+Suppose there exists some [[Definition:Prime Number|prime number]] $p$ such that $p \divides m$ and $p \divides r'$.
+Such a prime would be a common divisor of both $k$ and $m n$, contradicting $k \perp m n$.
+Hence $r' \perp m$ and so is [[Definition:Congruence (Number Theory)|congruent modulo $m $]] to one of these integers in $R$.
+By the same argument, $s' \perp n$ and so is [[Definition:Congruence (Number Theory)|congruent modulo $n$]] to one of these integers in $S$.
+Writing $r' = r + a m, \, s' = s + b n$ we have:
+:$k = n r' + m s' = n r + m s + m n \paren {a + b} \equiv n r + m s \pmod {m n}$
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Euler Phi Function of Prime Power}
+Tags: Euler Phi Function, Prime Numbers
+
+\begin{theorem}
+Let $p^n$ be a [[Definition:Prime Power|prime power]] for some [[Definition:Prime Number|prime number]] $p > 1$.
+Then:
+:$\map \phi {p^n} = p^n \paren {1 - \dfrac 1 p} = \paren {p - 1} p^{n - 1}$
+where $\phi: \Z_{>0} \to \Z_{>0}$ is the [[Definition:Euler Phi Function|Euler $\phi$ function]].
+\end{theorem}
+
+\begin{proof}
+From [[Euler Phi Function of Prime]]:
+:$\map \phi p = p - 1$
+From [[Prime not Divisor implies Coprime]]:
+:$k \perp p^n \iff p \nmid k$
+There are $p^{n - 1}$ [[Definition:Integer|integers]] $k$ such that $1 \le k \le p^n$ which are [[Definition:Divisor of Integer|divisible]] by $p$:
+:$k \in \set {p, 2 p, 3 p, \ldots, \paren {p^{n - 1} } p}$
+Therefore:
+:$\map \phi {p^n} = p^n - p^{n - 1} = p^n \paren {1 - \dfrac 1 p} = \paren {p - 1} p^{n - 1}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sum of Euler Phi Function over Divisors}
+Tags: Euler Phi Function
+
+\begin{theorem}
+Let $n \in \Z_{>0}$ be a [[Definition:Strictly Positive Integer|strictly positive integer]].
+Then $\displaystyle \sum_{d \mathop \divides n} \map \phi d = n$
+where:
+:$\displaystyle \sum_{d \mathop \divides n}$ denotes the [[Definition:Sum Over Divisors|sum over all of the divisors]] of $n$
+:$\map \phi d$ is the [[Definition:Euler Phi Function|Euler $\phi$ function]], the number of [[Definition:Integer|integers]] less than $d$ that are [[Definition:Coprime Integers|prime to]] $d$.
+That is, the total of all the [[Definition:Euler Phi Function|totients]] of all [[Definition:Divisor of Integer|divisors]] of a number equals that number.
+\end{theorem}
+
+\begin{proof}
+Let us define:
+:$S_d = \set {m \in \Z: 1 \le m \le n, \gcd \set {m, n} = d}$.
+That is, $S_d$ is all the numbers less than or equal to $n$ whose [[Definition:Greatest Common Divisor of Integers|GCD]] with $n$ is $d$.
+Now from [[Integers Divided by GCD are Coprime]] we have:
+: $\gcd \set {m, n} = d \iff \dfrac m d, \dfrac n d \in \Z: \dfrac m d \perp \dfrac n d$
+So the number of integers in $S_d$ equals the number of [[Definition:Positive Integer|positive integers]] no bigger than $\dfrac n d$ which are [[Definition:Coprime Integers|prime to]] $\dfrac n d$.
+That is, by definition of the [[Definition:Euler Phi Function|Euler phi function]]:
+:$\card {S_d} = \map \phi {\dfrac n d}$
+From the definition of the $S_d$, it follows that for all $1 \le m \le n$:
+:$\exists d \divides n: m \in S_d$
+Therefore:
+:$\displaystyle \set {1, \ldots, n} = \bigcup_{d \mathop \divides n} S_d$
+Moreover, it follows from the definition of the $S_d$ that they are [[Definition:Pairwise Disjoint Family|pairwise disjoint]].
+Now from [[Cardinality of Set Union/Corollary|Corollary to Cardinality of Set Union]], it follows that:
+{{begin-eqn}}
+{{eqn | l = n 
+      | r = \sum_{d \mathop \divides n} \card {S_d}
+}}
+{{eqn | r = \sum_{d \mathop \divides n} \map \phi {\dfrac n d}
+}}
+{{end-eqn}}
+But from [[Sum Over Divisors Equals Sum Over Quotients]]:
+:$\displaystyle \sum_{d \mathop \divides n} \map \phi {\dfrac n d} = \sum_{d \mathop \divides n} \map \phi d$
+and hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Möbius Function is Multiplicative}
+Tags: Multiplicative Functions, Möbius Function
+
+\begin{theorem}
+The [[Definition:Möbius Function|Möbius function]] $\mu$  is a [[Definition:Multiplicative Arithmetic Function|multiplicative function]]:
+:$m \perp n \implies \map \mu {m n} = \map \mu m \map \mu n$
+where $m, n \in \Z_{>0}$.
+\end{theorem}
+
+\begin{proof}
+First note that we have $\map \mu 1 = 1$, which agrees with [[Value of Multiplicative Function at One]].
+Let $m, n \in \Z_{>0}$ such that $m \perp n$.
+First, suppose that either $\map \mu m = 0$ or $\map \mu n = 0$.
+Then either $m$ or $n$ has a factor $p^2$ where $p$ is [[Definition:Prime Number|prime]].
+Thus it will follow that $m n$ will also have a factor $p^2$ and hence $\map \mu {m n} = 0$.
+So the result holds when $\map \mu m = 0$ or $\map \mu n = 0$.
+Now suppose that $\map \mu m \ne 0$ and $\map \mu n \ne 0$.
+Let $m = p_1 p_2 \ldots p_r$, $n = q_1 q_2 \ldots q_s$ where all the $p_i, q_j$ are [[Definition:Prime Number|prime]].
+Then:
+:$m n = p_1 p_2 \ldots p_r q_1 q_2 \ldots q_s$
+As $m \perp n$ it follows that:
+:$\forall i, j: p_i \ne q_j$
+Hence there is no prime in $m n$ whose power is higher than $1$, which means that $\map \mu {m n} \ne 0$.
+So:
+:$\map \mu {m n} = \paren {-1}^{r + s} = \paren {-1}^r \paren {-1}^s = \map \mu m \map \mu n$
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sum of Möbius Function over Divisors}
+Tags: Euler Phi Function, Möbius Function
+
+\begin{theorem}
+Let $n \in \Z_{>0}$ be a [[Definition:Strictly Positive Integer|strictly positive integer]].
+Then:
+:$\displaystyle \sum_{d \mathop \divides n} \map \mu d \frac n d = \map \phi n$
+where:
+:$\displaystyle \sum_{d \mathop \divides n}$ denotes the [[Definition:Sum Over Divisors|sum over all of the divisors]] of $n$
+:$\map \phi n$ is the [[Definition:Euler Phi Function|Euler $\phi$ function]], the number of [[Definition:Integer|integers]] less than $n$ that are [[Definition:Coprime Integers|prime to]] $n$
+:$\map \mu d$ is the [[Definition:Möbius Function|Möbius function]].
+Equivalently, this says that:
+:$\phi = \mu * I_{\Z_{>0} }$
+where:
+:$*$ denotes [[Definition:Dirichlet Convolution|Dirichlet convolution]]
+:$I_{\Z_{>0} }$ denotes the [[Definition:Identity Mapping|identity mapping]] on $\Z_{>0}$, that is:
+::$\forall n \in \Z_{>0}: I_{\Z_{>0} }: n \mapsto n$
+{{wtd|Add a link to a page proving this equivalence.}}
+\end{theorem}
+
+\begin{proof}
+=== [[Sum of Möbius Function over Divisors/Lemma|Lemma]] ===
+{{:Sum of Möbius Function over Divisors/Lemma}}
+Let $\map 1 k = 1$ be the [[Definition:Constant Mapping|constant mapping]].
+Then $\phi$ is defined as:
+:$\displaystyle \map \phi n = \sum_{k \mathop \perp n} \map 1 k$
+We have that $\gcd \set {n, k}$ is $1$ if $k \perp n$ and $0$ otherwise.
+{{explain|How can that be?}}
+Thus we can rewrite the above sum as:
+:$\displaystyle \sum_{k \mathop = 1}^n \floor {\frac 1 {\gcd \set {n, k} } }$
+Now we may use the lemma, with $\gcd \set {n, k}$ replacing $n$, to get:
+{{begin-eqn}}
+{{eqn | l = \map \phi n
+      | r = \sum_{k \mathop = 1}^n \paren {\sum_{d \mathop \divides \gcd \set {n, k} } \map \mu d}
+      | c = 
+}}
+{{eqn | r = \sum_{k \mathop = 1}^n \sum_{\binom {d \mathop \divides n} {d \mathop \divides k} } \map \mu d
+      | c = 
+}}
+{{end-eqn}}
+For a fixed [[Definition:Divisor of Integer|divisor]] $d$ of $n$, we must sum over all those $k$ in the range $1 \le k \le n$ which are multiples of $d$.
+If we write $k = q d$, then $1 \le k \le n$ {{iff}} $1 \le q \le \dfrac n d$.  
+Hence the last sum for $\map \phi n$ can be written as:
+{{begin-eqn}}
+{{eqn | l = \phi \left({n}\right)
+      | r = \sum_{d \mathop \divides n} \paren {\sum_{q \mathop = 1}^{\tfrac n d} \map \mu d}
+      | c = 
+}}
+{{eqn | r = \sum_{d \mathop \divides n} \map \mu d \sum_{q \mathop = 1}^{\tfrac n d} \map 1 q
+      | c = 
+}}
+{{eqn | r = \sum_{d \mathop \divides n} \map \mu d \frac n d
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Binomial Coefficient of Prime}
+Tags: Prime Numbers, Binomial Coefficients, Binomial Coefficient of Prime
+
+\begin{theorem}
+Let $p$ be a [[Definition:Prime Number|prime number]].
+Then:
+:$\forall k \in \Z: 0 < k < p: \dbinom p k \equiv 0 \pmod p$
+where $\dbinom p k$ is defined as a [[Definition:Binomial Coefficient|binomial coefficient]].
+\end{theorem}
+
+\begin{proof}
+Because:
+:$\dbinom p k = \dfrac {p \left({p - 1}\right) \left({p - 2}\right) \cdots \left({p - k + 1}\right)} {k!}$
+is an [[Definition:Integer|integer]], we have that:
+:$k! \mathrel \backslash p \left({p - 1}\right) \left({p - 2}\right) \cdots \left({p - k + 1}\right)$
+But because $k < p$ it follows that:
+:$k! \mathop \perp p$
+that is, that:
+:$\gcd \left\{{k!, p}\right\} = 1$
+So by [[Euclid's Lemma]]:
+:$k! \mathrel \backslash \left({p - 1}\right) \left({p - 2}\right) \cdots \left({p - k + 1}\right)$
+Hence:
+:$\dbinom p k = p \dfrac {\left({p - 1}\right) \left({p - 2}\right) \cdots \left({p - k + 1}\right)} {k!}$
+Hence the result.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+[[Lucas' Theorem]] gives:
+:$\dbinom n k \equiv \dbinom {\left \lfloor {n / p} \right \rfloor} {\left \lfloor {k / p} \right \rfloor} \dbinom {n \bmod p} {k \bmod p} \pmod p$
+So, substituting $p$ for $n$:
+:$\dbinom p k \equiv \dbinom {\left \lfloor {p / p} \right \rfloor} {\left \lfloor {k / p} \right \rfloor} \dbinom {p \bmod p} {k \bmod p} \pmod p$
+But $p \bmod p = 0$ by [[Definition:Modulo Operation|definition]].
+Hence, if $0 < k < p$, we have that:
+:$k \bmod p \ne 0$
+and so:
+:$\dbinom {p \bmod p} {k \bmod p} = \dbinom 0 {k \bmod p} = 0$
+by [[Definition:Binomial Coefficient|definition of binomial coefficients]].
+The result follows immediately.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+By the definition of [[Definition:Binomial Coefficient|binomial coefficient]]:
+{{begin-eqn}}
+{{eqn | l = \binom p k
+      | r = \frac {p!} {k! \paren {n - k}!}
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = p!
+      | r = k! \paren {p - k}! \binom p k
+}}
+{{end-eqn}}
+Now, $p$ [[Definition:Divisor of Integer|divides]] the {{LHS}} by [[Divisors of Factorial]].
+So $p$ must also [[Definition:Divisor of Integer|divide]] the {{RHS}}.
+[[Definition:By Hypothesis|By hypothesis]] $k < p$.
+We have that:
+:$k! = k \paren {k - 1} \dotsm \paren 2 \paren 1$
+We also have that $p$ is [[Definition:Prime Number|prime]], and each factor is less than $p$.
+Thus $p$ is not a [[Definition:Factor (Algebra)|factor]] of $k!$.
+Similarly:
+:$\paren {p - k}! = \paren {p - k} \paren {p - k - 1} \dotsm \paren 2 \paren 1$
+It follows, by the same reasoning as above, that $p$ is also not a [[Definition:Factor (Algebra)|factor]] of $\paren {p - k}!$.
+The result then follows from [[Euclid's Lemma]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Prime Power of Sum Modulo Prime}
+Tags: Number Theory, Combinatorics, Prime Numbers, Proofs by Induction
+
+\begin{theorem}
+Let $p$ be a [[Definition:Prime Number|prime number]].
+Then:
+:$\forall n \in \N_{> 0}: \paren {a + b}^{p^n} \equiv a^{p^n} + b^{p^n} \pmod p$
+\end{theorem}
+
+\begin{proof}
+Proof by [[Principle of Mathematical Induction|induction]]:
+For all $n \in \N_{> 0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
+:$\paren {a + b}^{p^n} \equiv a^{p^n} + b^{p^n} \pmod p$
+=== Basis for the Induction ===
+First from [[Power of Sum Modulo Prime]] we have that $\map P 1$ is true:
+:$\paren {a + b}^p \equiv a^p + b^p \pmod p$
+This is our [[Definition:Basis for the Induction|basis for the induction]].
+=== Induction Hypothesis ===
+Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
+So this is our [[Definition:Induction Hypothesis|induction hypothesis]]:
+:$\paren {a + b}^{p^k} \equiv a^{p^k} + b^{p^k} \pmod p$
+Then we need to show:
+:$\paren {a + b}^{p^{k + 1} } \equiv a^{p^{k + 1} } + b^{p^{k + 1} } \pmod p$
+=== Induction Step ===
+This is our [[Definition:Induction Step|induction step]]:
+{{begin-eqn}}
+{{eqn | l = \paren {a + b}^{p^k}
+      | o = \equiv
+      | r = a^{p^k} + b^{p^k}
+      | rr= \pmod p
+      | c = [[Prime Power of Sum Modulo Prime#Induction Hypothesis|Induction Hypothesis]]
+}}
+{{eqn | ll= \leadsto
+      | l = \paren {\paren {a + b}^{p^k} }^p
+      | o = \equiv
+      | r = \paren {a^{p^k} + b^{p^k} }^p
+      | rr= \pmod p
+      | c = [[Congruence of Powers]]
+}}
+{{eqn | ll= \leadsto
+      | l = \paren {\paren {a + b}^{p^k} }^p
+      | o = \equiv
+      | r = \paren {a^{p^k} }^p + \paren {b^{p^k} }^p
+      | rr= \pmod p
+      | c = [[Prime Power of Sum Modulo Prime#Basis for the Induction|Basis for the Induction]]
+}}
+{{eqn | ll= \leadsto
+      | l = \paren {a + b}^{p^{k + 1} }
+      | o = \equiv
+      | r = a^{p^{k + 1} } + b^{p^{k + 1} }
+      | rr= \pmod p
+      | c = 
+}}
+{{end-eqn}}
+So $\map P k \implies \map P {k + 1}$ and the result follows by the [[Principle of Mathematical Induction]].
+Therefore:
+:$\forall n \in \N_{> 0}: \paren {a + b}^{p^n} \equiv a^{p^n} + b^{p^n} \pmod p$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Binomial Coefficient involving Power of Prime}
+Tags: Prime Numbers, Binomial Coefficients, Binomial Coefficient involving Power of Prime
+
+\begin{theorem}
+: $\dbinom {p^n k} {p^n} \equiv k \pmod p$
+where $\dbinom {p^n k} {p^n}$ is a [[Definition:Binomial Coefficient|binomial coefficient]].
+\end{theorem}
+
+\begin{proof}
+From  [[Prime Power of Sum Modulo Prime]] we have:
+:$(1): \quad \paren {a + b}^{p^n} \equiv \paren {a^{p^n} + b^{p^n} } \pmod p$
+We can write this:
+:$\paren {a + b}^{p^n k} = \paren {\paren {a + b}^{p^n} }^k$
+By $(1)$ and [[Congruence of Powers]], we therefore have:
+:$\paren {a + b}^{p^n k} \equiv \paren {a^{p^n} + b^{p^n} }^k \pmod p$
+The coefficient $\dbinom {p^n k} {p^n}$ is the [[Definition:Binomial Coefficient|binomial coefficient]] of $b^{p^n}$ in $\paren {a + b}^{p^n k} = \paren {\paren {a + b}^{p^n} }^k$.
+Expanding $\paren {a^{p^n} + b^{p^n} }^k$ using the [[Binomial Theorem]], we find that the coefficient of $b^{p^n}$, the second term, is $\dbinom k 1 = k$.
+So:
+:$\dbinom {p^n k} {p^n} \equiv k \pmod p$
+{{qed}}
+\end{proof}
+
+\begin{proof}
+[[Lucas' Theorem]] states that for $n, k, p \in \Z$ and $p$ be a [[Definition:Prime Number|prime number]], such that:
+:$n = a_r p^r + \cdots + a_1 p + a_0$
+:$k = b_r p^r + \cdots + b_1 p + b_0$
+then:
+:$\displaystyle \binom n k \equiv \prod_{j \mathop = 0}^r \binom {a_j}{b_j} \pmod p$
+Therefore:
+{{begin-eqn}}
+{{eqn | l = \binom {p^n k} {p^n}
+      | o = \equiv
+      | r = \binom k 1 \prod_{j \mathop = 0}^{n - 1} \binom 0 0 \pmod p
+      | c = [[Lucas' Theorem]]
+}}
+{{eqn | o = \equiv
+      | r = k \prod_{j \mathop = 0}^{n - 1} 1 \pmod p
+      | c = [[Binomial Coefficient with One]], [[Binomial Coefficient with Zero]]
+}}
+{{eqn | o = \equiv
+      | r = k \pmod p
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Cassini's Identity}
+Tags: Fibonacci Numbers, Cassini's Identity
+
+\begin{theorem}
+:$F_{n + 1} F_{n - 1} - F_n^2 = \paren {-1}^n$
+\end{theorem}
+
+\begin{proof}
+We see that:
+: $F_2 F_0 - F_1^2 = 1 \times 0 - 1 = -1 = \left({-1}\right)^1$
+so the proposition holds for $n = 1$.
+We also see that:
+: $F_3 F_1 - F_2^2 = 2 \times 1 - 1 = \left({-1}\right)^2$
+so the proposition holds for $n = 2$.
+Suppose the proposition is true for $n = k$, that is:
+: $F_{k + 1} F_{k - 1} - F_k^2 = \left({-1}\right)^k$
+It remains to be shown that it follows from this that the proposition is true for $n = k + 1$, that is:
+: $F_{k + 2} F_k - F_{k + 1}^2 = \left({-1}\right)^{k + 1}$
+So:
+{{begin-eqn}}
+{{eqn | l = F_{k + 2} F_k - F_{k + 1}^2
+      | r = \left({F_k + F_{k + 1} }\right) F_k - F_{k + 1}^2
+}}
+{{eqn | r = F_k^2 + F_k F_{k + 1} - F_{k + 1}^2
+}}
+{{eqn | r = F_k^2 + F_k F_{k + 1} - F_{k + 1} \left({F_k + F_{k - 1} }\right)
+}}
+{{eqn | r = F_k^2 + F_k F_{k + 1} - F_k F_{k + 1} - F_{k + 1} F_{k - 1}
+}}
+{{eqn | r = F_k^2 - F_{k + 1} F_{k - 1}
+}}
+{{eqn | r = \left({-1}\right) \left({F_{k + 1} F_{k - 1} - F_k^2}\right)
+}}
+{{eqn | r = \left({-1}\right) \left({-1}\right)^k
+}}
+{{eqn | r = \left({-1}\right)^{k + 1}
+}}
+{{end-eqn}}
+By the [[Principle of Mathematical Induction]], the proof is complete.
+{{qed}}
+Note that from the above we have that:
+: $F_{k + 2} F_k - F_{k + 1}^2 = \left({-1}\right)^{k + 1}$
+from which:
+: $F_{n + 1}^2 - F_n F_{n + 2} = \left({-1}\right)^n$
+follows immediately.
+\end{proof}
+
+\begin{proof}
+First we use this [[Cassini's Identity/Lemma|lemma]]:
+{{:Cassini's Identity/Lemma}}
+Then the [[Definition:Determinant of Matrix|determinant]] of both sides is taken.
+The {{LHS}} follows directly from the [[Definition:Determinant of Order 2|order 2 determinant]]:
+:$\begin{bmatrix}
+F_{n + 1}  & F_n     \\
+F_n        & F_{n - 1}
+\end{bmatrix} = F_{n + 1} F_{n - 1} - F_n^2$
+Now for the {{RHS}}:
+==== Basis for the Induction ====
+:$\begin{vmatrix}
+1 & 1 \\
+1 & 0
+\end{vmatrix} = 1 \times 0 - 1 \times 1 = -1 = \left({-1}\right)^1$
+==== Induction Hypothesis ====
+For $k \in \Z_{>0}$, it is assumed that:
+:$\begin{vmatrix}
+1 & 1 \\
+1 & 0
+\end{vmatrix}^k = \left({-1}\right)^k$
+It remains to be shown that:
+:$\begin{vmatrix}
+1 & 1 \\
+1 & 0
+\end{vmatrix}^{k + 1} = \left({-1}\right)^{k + 1}$
+==== Induction Step ====
+The induction step follows from [[Determinant of Matrix Product]]:
+:$\begin{vmatrix}
+1 & 1 \\
+1 & 0
+\end{vmatrix}^{k+1} = \begin{vmatrix}
+1 & 1 \\
+1 & 0
+\end{vmatrix}^k \begin{vmatrix}
+1 & 1 \\
+1 & 0
+\end{vmatrix} = \left({-1}\right)^k \left({-1}\right) = \left({-1}\right)^{k + 1}$
+Hence by [[Principle of Mathematical Induction|induction]]:
+:$\forall n \in \Z_{>0}: \begin{vmatrix}
+1 & 1 \\
+1 & 0
+\end{vmatrix}^n = \left({-1}\right)^n$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Rational Numbers are Countably Infinite}
+Tags: Rational Numbers, Countable Sets, Rational Numbers are Countably Infinite
+
+\begin{theorem}
+The [[Definition:Set|set]] $\Q$ of [[Definition:Rational Number|rational numbers]] is [[Definition:Countably Infinite|countably infinite]].
+\end{theorem}
+
+\begin{proof}
+The [[Definition:Rational Number|rational numbers]] are arranged thus:
+:$\dfrac 0 1, \dfrac 1 1, \dfrac {-1} 1, \dfrac 1 2, \dfrac {-1} 2, \dfrac 2 1, \dfrac {-2} 1, \dfrac 1 3, \dfrac 2 3,  \dfrac {-1} 3, \dfrac {-2} 3, \dfrac 3 1, \dfrac 3 2, \dfrac {-3} 1, \dfrac {-3} 2, \dfrac 1 4, \dfrac 3 4, \dfrac {-1} 4, \dfrac {-3} 4, \dfrac 4 1, \dfrac 4 3, \dfrac {-4} 1, \dfrac {-4} 3 \ldots$
+It is clear that every [[Definition:Rational Number|rational number]] will appear somewhere in this list.
+Thus it is possible to set up a [[Definition:Bijection|bijection]] between each [[Definition:Rational Number|rational number]] and its position in the list, which is an element of $\N$.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let us define the mapping $\phi: \Q \to \Z \times \N$ as follows:
+:$\forall \dfrac p q \in \Q: \phi \left({\dfrac p q}\right) = \left({p, q}\right)$
+where $\dfrac p q$ is [[Definition:Canonical Form of Rational Number|in canonical form]].
+Then $\phi$ is clearly [[Definition:Injection|injective]].
+From [[Cartesian Product of Countable Sets is Countable]], we have that $\Z \times \N$ is [[Definition:Countable|countably infinite]].
+The result follows directly from [[Domain of Injection to Countable Set is Countable]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+For each $n \in \N$, define $S_n$ to be the [[Definition:Set|set]]:
+:$S_n := \left\{{\dfrac m n: m \in \Z}\right\}$
+By [[Integers are Countably Infinite]], each $S_n$ is [[Definition:Countably Infinite|countably infinite]].
+Because each [[Definition:Rational Number|rational number]] can be written down with a [[Definition:Positive Integer|positive]] [[Definition:Denominator|denominator]], it follows that:
+:$\forall q \in \Q: \exists n \in \N: q \in S_n$
+which is to say:
+:$\displaystyle \bigcup_{n \mathop \in \N} S_n = \Q$
+By [[Countable Union of Countable Sets is Countable]], it follows that $\Q$ is [[Definition:Countable|countable]].
+Since $\Q$ is manifestly [[Definition:Infinite Set|infinite]], it is [[Definition:Countably Infinite|countably infinite]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let $Q_\pm = \set {q \in \Q: \pm q > 0}$.
+For every $q \in Q_+$, there exists at least one pair $\tuple {m, n} \in \N \times \N$ such that $q = \dfrac m n$.
+Therefore, we can find an [[Definition:Injection|injection]] $i: Q_+ \to \N \times \N$.
+By [[Cartesian Product of Natural Numbers with Itself is Countable]], $\N \times \N$ is [[Definition:Countable|countable]].
+Hence $Q_+$ is [[Definition:Countable|countable]], by [[Domain of Injection to Countable Set is Countable]].
+The map $-: q \mapsto -q$ provides a [[Definition:Bijection|bijection]] from $Q_-$ to $Q_+$, hence $Q_-$ is also [[Definition:Countable|countable]].
+Hence $\Q$ is [[Definition:Countable Set|countable]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Real Numbers form Ordered Field}
+Tags: Examples of Fields, Ordered Fields, Real Numbers
+
+\begin{theorem}
+The [[Definition:Real Number|set of real numbers]] $\R$ forms an [[Definition:Ordered Field|ordered field]] under [[Definition:Real Addition|addition]] and [[Definition:Real Multiplication|multiplication]]: $\struct {\R, +, \times, \le}$.
+\end{theorem}
+
+\begin{proof}
+From [[Real Numbers form Field]], we have that $\struct {\R, +, \times}$ forms a [[Definition:Field (Abstract Algebra)|field]].
+From [[Ordering Properties of Real Numbers]] we have that $\struct {\R, +, \times, \le}$ is a [[Definition:Ordered Field|ordered field]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Rational Numbers form Subfield of Real Numbers}
+Tags: Rational Numbers, Real Numbers
+
+\begin{theorem}
+The [[Definition:Ordered Field|(ordered) field]] $\struct {\Q, +, \times, \le}$ of [[Definition:Rational Number|rational numbers]] forms a [[Definition:Subfield|subfield]] of the [[Definition:Field of Real Numbers|field of real numbers]] $\struct {\R, +, \times, \le}$.
+That is, the [[Definition:Field of Real Numbers|field of real numbers]] $\struct {\R, +, \times, \le}$ is an [[Definition:Extension of Operation|extension]] of the [[Definition:Rational Number|rational numbers]] $\struct {\Q, +, \times, \le}$.
+\end{theorem}
+
+\begin{proof}
+Recall that [[Rational Numbers form Ordered Field]].
+Then from [[Rational Numbers form Subset of Real Numbers]]:
+:$\Q \subseteq \R$
+Hence the result by definition of [[Definition:Subfield|subfield]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Definition:Rational Number/Canonical Form}
+Tags: Definitions: Rational Numbers, Definitions: Fractions
+
+\begin{theorem}
+Let $r \in \Q$ be a [[Definition:Rational Number|rational number]].
+The '''canonical form of $r$''' is the expression $\dfrac p q$, where:
+:$r = \dfrac p q: p \in \Z, q \in \Z_{>0}, p \perp q$
+where $p \perp q$ denotes that $p$ and $q$ have no [[Definition:Common Divisor of Integers|common divisor]] except $1$.
+\end{theorem}<|endoftext|>
+\section{Ordering Properties of Real Numbers}
+Tags: Real Numbers, Inequalities
+
+\begin{theorem}
+=== [[Trichotomy Law for Real Numbers|Trichotomy Law]] ===
+{{:Trichotomy Law for Real Numbers}}
+=== [[Real Number Ordering is Transitive|Ordering is Transitive]] ===
+{{:Real Number Ordering is Transitive}}
+=== [[Real Number Ordering is Compatible with Addition|Ordering is Compatible with Addition]] ===
+{{:Real Number Ordering is Compatible with Addition}}
+=== [[Real Number Ordering is Compatible with Multiplication|Ordering is Compatible with Multiplication]] ===
+{{:Real Number Ordering is Compatible with Multiplication}}
+\end{theorem}<|endoftext|>
+\section{Order is Preserved on Positive Reals by Squaring}
+Tags: Analysis, Inequalities, Order is Preserved on Positive Reals by Squaring
+
+\begin{theorem}
+:$x < y \iff x^2 < y^2$
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+Assume $x < y$.
+Then:
+{{begin-eqn}}
+{{eqn | l = x < y
+      | o = \implies
+      | r = x \times x < x \times y
+      | c = [[Real Number Ordering is Compatible with Multiplication]]
+}}
+{{eqn | l = x < y
+      | o = \implies
+      | r = x \times y < y \times y
+      | c = [[Real Number Ordering is Compatible with Multiplication]]
+}}
+{{eqn | o = \leadsto
+      | r = x^2 < y^2
+      | c = [[Real Number Ordering is Transitive]]
+}}
+{{end-eqn}}
+So:
+:$x < y \implies x^2 < y^2$
+{{qed|lemma}}
+=== Sufficient Condition ===
+Assume $x^2 < y^2$.
+Then:
+{{begin-eqn}}
+{{eqn | l = x^2
+      | o = <
+      | r = y^2
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = 0
+      | o = <
+      | r = y^2 - x^2
+      | c = [[Real Number Ordering is Compatible with Addition]]
+}}
+{{eqn | ll=\leadsto
+      | l = \paren {y - x} \paren {y + x}
+      | o = >
+      | r = 0
+      | c = [[Difference of Two Squares]]
+}}
+{{eqn | ll= \leadsto
+      | l = \paren {y - x} \paren {y + x} \paren {y + x}^{-1}
+      | o = >
+      | r = 0 \times \paren {y + x}^{-1}
+      | c = as $x + y > 0$
+}}
+{{eqn | ll= \leadsto
+      | l = y - x
+      | o = >
+      | r = 0
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = x
+      | o = <
+      | r = y
+      | c = 
+}}
+{{end-eqn}}
+So:
+:$x^2 < y^2 \implies x < y$
+{{qed}}
+\end{proof}
+
+\begin{proof}
+From [[Real Numbers form Ordered Field]], the [[Definition:Real Number|real numbers]] form an [[Definition:Ordered Field|ordered field]].
+The result follows from [[Order of Squares in Ordered Field]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+From [[Real Numbers form Ordered Field]], the [[Definition:Real Number|real numbers]] form an [[Definition:Ordered Field|ordered field]].
+By definition, an [[Definition:Ordered Field|ordered field]] is a [[Definition:Totally Ordered Ring|totally ordered ring]] without [[Definition:Proper Zero Divisor|proper zero divisors]].
+The result follows from [[Order of Squares in Totally Ordered Ring without Proper Zero Divisors]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+=== Necessary Condition ===
+Let $x < y$.
+Then:
+{{begin-eqn}}
+{{eqn | l = x < y
+      | o = \implies
+      | r = x \times x < x \times y
+      | c = [[Real Number Ordering is Compatible with Multiplication]]
+}}
+{{eqn | l = x < y
+      | o = \implies
+      | r = x \times y < y \times y
+      | c = [[Real Number Ordering is Compatible with Multiplication]]
+}}
+{{eqn | o = \leadsto
+      | r = x^2 < y^2
+      | c = [[Real Number Ordering is Transitive]]
+}}
+{{end-eqn}}
+So:
+:$x < y \implies x^2 < y^2$
+{{qed|lemma}}
+=== Sufficient Condition ===
+Let $x^2 < y^2$.
+{{AimForCont}} $x \ge y$.
+Then:
+{{begin-eqn}}
+{{eqn | l = x \ge y
+      | o = \implies
+      | r = x \times x \ge x \times y
+      | c = [[Real Number Ordering is Compatible with Multiplication]]
+}}
+{{eqn | l = x \ge y
+      | o = \implies
+      | r = x \times y \ge y \times y
+      | c = [[Real Number Ordering is Compatible with Multiplication]]
+}}
+{{eqn | o = \leadsto
+      | r = x^2 \ge y^2
+      | c = [[Real Number Ordering is Transitive]]
+}}
+{{end-eqn}}
+But this [[Definition:Contradiction|contradicts]] our assertion that $x^2 < y^2$.
+Hence by [[Proof by Contradiction]] it follows that:
+:$x < y$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Real Plus Epsilon}
+Tags: Real Analysis, Inequalities
+
+\begin{theorem}
+Let $a, b \in \R$, such that:
+:$\forall \epsilon \in \R_{>0}: a < b + \epsilon$
+where $\R_{>0}$ is the set of [[Definition:Strictly Positive|strictly positive]] [[Definition:Real Number|real numbers]].
+That is:
+:$\epsilon > 0$
+Then:
+:$a \le b$
+\end{theorem}
+
+\begin{proof}
+{{AimForCont}} $a > b$.
+Then:
+:$a - b > 0$
+[[Definition:By Hypothesis|By hypothesis]], we have:
+:$\forall \epsilon > 0: a < b + \epsilon$
+Let $\epsilon = a - b$.
+Then:
+:$a < b + \paren {a - b} \implies a < a$
+The result follows by [[Proof by Contradiction]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Difference of Two Squares}
+Tags: Commutative Rings, Square Function, Polynomial Theory, Difference of Two Squares
+
+\begin{theorem}
+Let $\struct {R, +, \circ}$ be a [[Definition:Commutative Ring|commutative ring]] whose [[Definition:Ring Zero|zero]] is $0_R$.
+Let $x, y \in R$.
+Then:
+: $x \circ x + \paren {- \paren {y \circ y} } = \paren {x + y} \circ \paren {x + \paren {-y} }$
+When $R$ is one of the standard [[Definition:Number|sets of numbers]], that is $\Z, \Q, \R$, and so on, then this translates into:
+:$x^2 - y^2 = \paren {x + y} \paren {x - y}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \paren {x + y} \circ \paren {x + \paren {-y} }
+      | r = x \circ x + y \circ x + x \circ \paren {-y} + y \circ \paren {-y}
+      | c = [[Definition:Distributive Operation|Distributivity]] of $\circ$ over $+$ in a [[Definition:Ring (Abstract Algebra)|ring]]
+}}
+{{eqn | r = x \circ x + x \circ y + x \circ \paren {-y} + y \circ \paren {-y}
+      | c = $R$ is a [[Definition:Commutative Ring|commutative ring]]
+}}
+{{eqn | r = x \circ x + x \circ \paren {y + \paren {-y} } + \paren {-\paren {y \circ y} }
+      | c = various [[Definition:Ring (Abstract Algebra)|ring]] properties
+}}
+{{eqn | r = x \circ x + x \circ 0_R + \paren {-\paren {y \circ y} }
+      | c = 
+}}
+{{eqn | r = x \circ x + \paren {-\paren {y \circ y} }
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+This is a special case of [[Difference of Two Powers]]:
+:$\displaystyle a^n - b^n = \left({a - b}\right) \left({a^{n - 1} + a^{n - 2} b + a^{n - 3} b^2 + \cdots + a b^{n - 2} + b^{n - 1} }\right) = \left({a - b}\right) \sum_{j \mathop = 0}^{n - 1} a^{n - j - 1} b^j$
+The result follows by setting $n = 2$.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+{{:Euclid:Proposition/II/5}}
+:[[File:Euclid-II-5.png|400px]]
+Let $AB$ be cut into equal segments at $C$ and unequal segments at $D$.
+Then the [[Definition:Containment of Rectangle|rectangle contained]] by $AD$ and $DB$ together with the square on $CD$ equals the square on $BC$.
+(That is, let $x = AC, y = CD$. Then $\paren {x + y} \paren {x - y} + y^2 = x^2$.)
+This is proved as follows.
+[[Construction of Square on Given Straight Line|Construct the square $CBFE$]] on $CB$, and join $BE$.
+[[Construction of Parallel Line|Construct $DG$ parallel]] to $CE$ through $G$, and let $DG$ cross $BE$ at $H$.
+[[Construction of Parallel Line|Construct $KM$ parallel]] to $AB$ through $H$.
+[[Construction of Parallel Line|Construct $AK$ parallel]] to $BF$ through $A$.
+From [[Complements of Parallelograms are Equal]]:
+:$\Box CDHL = \Box FGHM$.
+Add the square $DBMH$ to each.
+Then $\Box CBML = \Box DBFG$.
+But as $AC = CB$, from [[Parallelograms with Equal Base and Same Height have Equal Area]] we have that:
+:$\Box ACLK = \Box CBML$
+Add $\Box CDHL$ to each.
+Then $\Box ADHK$ is equal in area to the [[Definition:Gnomon|gnomon]] $CBFGHL$.
+But $\Box ADHK$ is the [[Definition:Containment of Rectangle|rectangle contained]] by $AD$ and $DB$, because $DB = DH$.
+So the [[Definition:Gnomon|gnomon]] $CBFGHL$ is equal in area to the [[Definition:Containment of Rectangle|rectangle contained]] by $AD$ and $DB$.
+Now $\Box LHGE$ is equal to the square on $CD$.
+Add $\Box LHGE$ to each of the [[Definition:Gnomon|gnomon]] $CBFGHL$ and $\Box ADHK$.
+Then the [[Definition:Gnomon|gnomon]] $CBFGHL$ together with $\Box LHGE$ equals the [[Definition:Containment of Rectangle|rectangle contained]] by $AD$ and $DB$ and the square on $CD$.
+But the [[Definition:Gnomon|gnomon]] $CBFGHL$ together with $\Box LHGE$ is the square $CBFE$.
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Mediant is Between}
+Tags: Real Analysis
+
+\begin{theorem}
+Let $a, b, c, d$ be ''any'' [[Definition:Real Number|real numbers]] such that $b > 0, d > 0$.
+Let $r = \dfrac a b < \dfrac c d = s$.
+Then:
+:$r < \dfrac {a + c} {b + d} < s$
+\end{theorem}
+
+\begin{proof}
+Let $r, s \in \R$ be such that:
+:$r < s$
+and:
+:$r = \dfrac a b, s = \dfrac c d$
+where $a, b, c, d$ are [[Definition:Real Number|real numbers]] such that $b > 0, d > 0$.
+Because $b, d > 0$, it follows from [[Real Number Ordering is Compatible with Multiplication]] that:
+:$b d > 0$
+Thus:
+{{begin-eqn}}
+{{eqn | l = \frac a b
+      | o = <
+      | r = \frac c d
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \frac a b \paren {b d}
+      | o = <
+      | r = \frac c d \paren {b d}
+      | c = [[Real Number Ordering is Compatible with Multiplication]]
+}}
+{{eqn | ll= \leadsto
+      | l = a d
+      | o = <
+      | r = b c
+      | c = 
+}}
+{{end-eqn}}
+Then:
+{{begin-eqn}}
+{{eqn | l = a \paren {b + d}
+      | r = a b + a d
+      | c = 
+}}
+{{eqn | o = <
+      | r = a b + b c
+      | c = [[Real Number Ordering is Compatible with Addition]], as $a d < b c$ from above
+}}
+{{eqn | r = \paren {a + c} b
+      | c = 
+}}
+{{end-eqn}}
+From [[Reciprocal of Strictly Positive Real Number is Strictly Positive]]:
+:$\paren {a + c}^{-1} > 0$
+and:
+:$b^{-1} > 0$
+It follows from [[Real Number Ordering is Compatible with Multiplication]] that:
+:$\dfrac a b < \dfrac {a + c} {b + d}$
+The other half is proved similarly.
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Real Numbers Between Epsilons}
+Tags: Real Analysis
+
+\begin{theorem}
+Let $a, b \in \R$ such that $\forall \epsilon \in \R_{>0}: a - \epsilon < b < a + \epsilon$.
+Then $a = b$.
+\end{theorem}
+
+\begin{proof}
+From [[Real Plus Epsilon]]:
+:$b < a + \epsilon \implies b \le a$
+From [[Real Number Ordering is Compatible with Addition]]:
+:$a - \epsilon < b \implies a < b + \epsilon$
+Then from [[Real Plus Epsilon]]:
+:$a < b + \epsilon \implies a \le b$
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Exists Integer Below Any Real Number}
+Tags: Analysis
+
+\begin{theorem}
+Let $x$ be a [[Definition:Real Number|real number]].
+Then there exists an [[Definition:Integer|integer]] less than $x$:
+:$\forall x \in \R: \exists n \in \Z: n < x$
+\end{theorem}
+
+\begin{proof}
+Clearly we may assume without loss of generality that $x < 0$.
+From the [[Archimedean Principle]]:
+:$\exists m \in \N: m > -x$
+By [[Real Numbers form Ordered Field]], we have that $\R$ is an [[Definition:Ordered Field|ordered field]].
+Therefore by property $(3)$ of [[Properties of Ordered Field]], $\Z \owns -m < x$.
+{{qed}}
+[[Category:Analysis]]
+279b5qb10xgfmugmysigl3ksxrndsf3
+\end{proof}<|endoftext|>
+\section{Archimedean Principle}
+Tags: Real Analysis, Number Theory
+
+\begin{theorem}
+Let $x$ be a [[Definition:Real Number|real number]].
+Then there exists a [[Definition:Natural Numbers|natural number]] greater than $x$.
+:$\forall x \in \R: \exists n \in \N: n > x$
+That is, the [[Definition:Natural Numbers|set of natural numbers]] is [[Definition:Bounded Above Set|unbounded above]].
+\end{theorem}
+
+\begin{proof}
+Let $x \in \R$.
+Let $S$ be the set of all [[Definition:Natural Numbers|natural numbers]] less than or equal to $x$:
+:$S = \set {a \in \N: a \le x}$
+It is possible that $S = \O$.
+Suppose $0 \le x$.
+Then by definition, $0 \in S$.
+But $S = \O$, so this is a [[Proof by Contradiction|contradiction]].
+From the [[Trichotomy Law for Real Numbers]] it follows that $0 > x$.
+Thus we have the element $0 \in \N$ such that $0 > x$.
+Now suppose $S \ne \O$.
+Then $S$ is [[Definition:Bounded Above Set|bounded above]] (by $x$, for example).
+Thus by the [[Continuum Property]] of $\R$, $S$ has a [[Definition:Supremum of Set|supremum]] in $\R$.
+Let $s = \map \sup S$.
+Now consider the number $s - 1$.
+Since $s$ is the [[Definition:Supremum of Set|supremum]] of $S$, $s - 1$ cannot be an [[Definition:Upper Bound of Set|upper bound]] of $S$ by definition.
+So $\exists m \in S: m > s - 1 \implies m + 1 > s$.
+But as $m \in \N$, it follows that $m + 1 \in \N$.
+Because $m + 1 > s$, it follows that $m + 1 \notin S$ and so $m + 1 > x$.
+\end{proof}<|endoftext|>
+\section{Real Number is between Floor Functions}
+Tags: Floor Function
+
+\begin{theorem}
+:$\forall x \in \R: \floor x \le x < \floor {x + 1}$
+\end{theorem}
+
+\begin{proof}
+$\floor x$ is defined as:
+:$\floor x = \sup \set {m \in \Z: m \le x}$
+So $\floor x \le x$ by definition.
+From [[Floor plus One]]:
+:$\floor {x + 1} > \floor x$
+Hence by the definition of the [[Definition:Supremum of Set|supremum]]:
+:$\floor {x + 1} > x$
+The result follows.
+{{qed}}
+[[Category:Floor Function]]
+akn94fzyqk42jhnqcxjm9xqmvzfon8w
+\end{proof}<|endoftext|>
+\section{Real Number is between Ceiling Functions}
+Tags: Ceiling Function
+
+\begin{theorem}
+:$\forall x \in \R: \left \lceil {x - 1} \right \rceil \le x < \left \lceil {x} \right \rceil$
+\end{theorem}
+
+\begin{proof}
+$\left \lceil {x} \right \rceil$ is defined as:
+:$\left \lceil {x} \right \rceil = \inf \left({\left\{{m \in \Z: m \ge x}\right\}}\right)$
+So $\left \lceil {x} \right \rceil \ge x$ by definition.
+Now $\left \lceil {x - 1} \right \rceil < \left \lceil {x} \right \rceil$, so by the definition of the [[Definition:Infimum of Set|infimum]]:
+:$\left \lceil {x - 1} \right \rceil > x$
+The result follows.
+{{qed}}
+[[Category:Ceiling Function]]
+d8pg5fzw8p22asnvcvmbkt769zo7dcs
+\end{proof}<|endoftext|>
+\section{Real Number minus Floor}
+Tags: Floor Function
+
+\begin{theorem}
+:$x - \floor x \in \hointr 0 1$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \floor x
+      | o = \le
+      | m = x
+      | mo= <
+      | r = \floor x + 1
+      | c = {{Defof|Floor Function}}
+}}
+{{eqn | ll= \implies
+      | l = \floor x - \floor x
+      | o = \le
+      | m = x - \floor x
+      | mo= <
+      | r = \floor x + 1 - \floor x
+      | c = subtracting $\floor x$ from all parts
+}}
+{{eqn | ll= \implies
+      | l = 0
+      | o = \le
+      | m = x - \floor x
+      | mo= <
+      | r = 1
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = x - \floor x
+      | o = \in
+      | m = \hointr 0 1
+      | c = as required
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Ceiling minus Real Number}
+Tags: Ceiling Function
+
+\begin{theorem}
+:$\forall x \in \R: \left \lceil {x} \right \rceil - x \in \left[{0 \,.\,.\, 1}\right)$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \left \lceil {x} \right \rceil - 1
+      | o = <
+      | r = x \le \left \lceil {x} \right \rceil
+      | c = [[Real Number is between Ceiling Functions]]
+}}
+{{eqn | ll= \implies
+      | l = \left \lceil {x} \right \rceil - 1 - \left \lceil {x} \right \rceil
+      | o = <
+      | r = x - \left \lceil {x} \right \rceil \le \left \lceil {x} \right \rceil - \left \lceil {x} \right \rceil
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = -1
+      | o = <
+      | r = x - \left \lceil {x} \right \rceil \le 0
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = 1
+      | o = >
+      | r = \left \lceil {x} \right \rceil - x \ge 0
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = \left \lceil {x} \right \rceil - x
+      | o = \in
+      | r = \left[{0 \,.\,.\, 1}\right)
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Ceiling Function]]
+f6xkmvsorteylqbus2ipuydihvfmd4a
+\end{proof}<|endoftext|>
+\section{Real Number is Floor plus Difference}
+Tags: Floor Function
+
+\begin{theorem}
+:There exists an [[Definition:Integer|integer]] $n \in \Z$ such that for some $t \in \hointr 0 1$:
+::$x = n + t$
+{{iff}}:
+:$n = \floor x$
+\end{theorem}
+
+\begin{proof}
+=== Sufficient Condition ===
+Let there exist $n \in \Z$ such that $x = n + t$, where $t \in \hointr 0 1$.
+We have that $1 - t > 0$.
+Thus:
+:$0 \le x - n < 1$
+Thus:
+: $n \le x < n + 1$
+That is, $n$ is the [[Definition:Floor Function|floor]] of $x$.
+{{qed|lemma}}
+=== Necessary Condition ===
+Let $n = \floor x$.
+Let $t = x - \floor x$.
+Then $x = n + t$.
+From [[Real Number minus Floor]]:
+:$t = x - \floor x \in \hointr 0 1$
+and so:
+:$x = n + t: t \in \hointr 0 1$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Floor plus One}
+Tags: Floor Function
+
+\begin{theorem}
+Let $x \in \R$.
+Then:
+:$\left \lfloor {x + 1} \right \rfloor = \left \lfloor {x} \right \rfloor + 1$
+where $\left \lfloor {x} \right \rfloor$ is the [[Definition:Floor Function|floor function]] of $x$.
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \left \lfloor {x + 1} \right \rfloor
+      | r = n
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = n
+      | o = \le
+      | r = x + 1 < n + 1
+      | c = Definition of [[Definition:Floor Function|Floor Function]]
+}}
+{{eqn | ll= \implies
+      | l = n - 1
+      | o = \le
+      | r = x < n
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = \left \lfloor {x} \right \rfloor
+      | r = n - 1
+      | c = Definition of [[Definition:Floor Function|Floor Function]]
+}}
+{{eqn | ll= \implies
+      | l = \left \lfloor {x + 1} \right \rfloor
+      | r = \left \lfloor {x} \right \rfloor + 1
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Real Number is Integer iff equals Floor}
+Tags: Floor Function
+
+\begin{theorem}
+:$x = \floor x \iff x \in \Z$
+\end{theorem}
+
+\begin{proof}
+Let $x = \floor x$.
+As $\floor x \in \Z$, then so must $x$ be.
+Now let $x \in \Z$.
+We have:
+:$\floor x = \sup \set {m \in \Z: m \le x}$
+As $x \in \sup \set {m \in \Z: m \le x}$, and there can be no greater $n \in \Z$ such that $n \in \sup \set {m \in \Z: m \le x}$, it follows that:
+:$x = \floor x$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sum of Floor and Floor of Negative}
+Tags: Floor Function
+
+\begin{theorem}
+Let $x \in \R$. Then:
+:$\floor x + \floor {-x} = \begin{cases} 0 & : x \in \Z \\ -1 & : x \notin \Z \end{cases}$
+where $\floor x$ denotes the [[Definition:Floor Function|floor]] of $x$.
+\end{theorem}
+
+\begin{proof}
+Let $x \in \Z$.
+Then from [[Real Number is Integer iff equals Floor]]:
+:$x = \floor x$
+Now $x \in \Z \implies -x \in \Z$, so:
+:$\floor {-x} = -x$
+Thus:
+:$\floor x + \floor {-x} = x + \paren {-x} = x - x = 0$
+Now let $x \notin \Z$.
+From [[Real Number is Floor plus Difference]]:
+:$x = n + t$
+where $n = \floor x$ and $t \in \hointr 0 1$.
+Thus:
+:$-x = -\paren {n + t} = -n - t = -n - 1 + \paren {1 - t}$
+As $t \in \hointr 0 1$, we have:
+:$1 - t \in \hointr 0 1$
+Thus:
+:$\floor {-x} = -n - 1$
+So:
+:$\floor x + \floor {-x} = n + \paren {-n - 1} = n - n - 1 = -1$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Floor defines Equivalence Relation}
+Tags: Floor Function, Equivalence Relations
+
+\begin{theorem}
+Let $x \in \R$ be a [[Definition:Real Number|real number]].
+Let $\floor x$ denote the [[Definition:Floor Function|floor function]] of $x$.
+Let $\RR$ be the [[Definition:Relation|relation]] defined on $\R$ such that:
+:$\forall x, y, \in \R: \tuple {x, y} \in \RR \iff \floor x = \floor y$
+Then $\RR$ is an [[Definition:Equivalence Relation|equivalence]], and $\forall n \in \Z$, the [[Definition:Equivalence Class|$\RR$-class]] of $n$ is the [[Definition:Half-Open Real Interval|half-open interval]] $\hointr n {n + 1}$.
+\end{theorem}
+
+\begin{proof}
+Checking in turn each of the critera for [[Definition:Equivalence Relation|equivalence]]:
+=== Reflexivity ===
+:$\forall x \in \R: \floor x = \floor x$
+Thus the [[Definition:Floor Function|floor function]] is [[Definition:Reflexive Relation|reflexive]].
+{{qed|lemma}}
+=== Symmetry ===
+:$\forall x, y \in \R: \floor x = \floor y \implies \floor y = \floor x$
+Thus the [[Definition:Floor Function|floor function]] is [[Definition:Symmetric Relation|symmetric]].
+{{qed|lemma}}
+=== Transitivity ===
+Let $\floor x = \floor y$ and $\floor y = \floor z$.
+Let $n = \floor x = \floor y = \floor z$, which follows from transitivity of $=$.
+Thus from [[Real Number is Floor plus Difference]]:
+:$x = n + t_x, y = n + t_y, z = n + t_z: t_x, t_y, t_z \in \hointr 0 1$
+Thus:
+:$x = n + t_x, z = n + t_z$
+and:
+:$\floor x = \floor z$
+Thus the [[Definition:Floor Function|floor function]] is [[Definition:Transitive Relation|transitive]].
+{{qed|lemma}}
+Thus we have shown that $\RR$ is an [[Definition:Equivalence Relation|equivalence relation]].
+{{qed|lemma}}
+Now we show that the [[Definition:Equivalence Class|$\RR$-class]] of $n$ is the [[Definition:Half-Open Real Interval|interval]] $\hointr n {n + 1}$.
+Defining $\RR$ as above, with $n \in \Z$:
+{{begin-eqn}}
+{{eqn | l = x
+      | o = \in
+      | r = \eqclass n \RR
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \floor x
+      | r = \floor n
+      | c = 
+}}
+{{eqn | r = n
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \exists t \in \hointr 0 1: x
+      | r = n + t
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = x
+      | o = \in
+      | r = \hointr n {n + 1}
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Ceiling defines Equivalence Relation}
+Tags: Ceiling Function, Equivalence Relations
+
+\begin{theorem}
+Let $\mathcal R$ be the [[Definition:Relation|relation]] defined on $\R$ such that:
+:$\forall x, y, \in \R: \left({x, y}\right) \in \mathcal R \iff \left \lceil {x}\right \rceil = \left \lceil {y}\right \rceil$
+where $\left \lceil {x}\right \rceil$ is the [[Definition:Ceiling Function|ceiling]] of $x$.
+Then $\mathcal R$ is an [[Definition:Equivalence Relation|equivalence]], and $\forall n \in \Z$, the [[Definition:Equivalence Class|$\mathcal R$-class]] of $n$ is the [[Definition:Half-Open Real Interval|half-open interval]] $\left({n - 1 \,.\,.\, n}\right]$.
+\end{theorem}
+
+\begin{proof}
+Checking in turn each of the critera for [[Definition:Equivalence Relation|equivalence]]:
+=== Reflexivity ===
+:$\forall x \in \R: \left \lceil {x}\right \rceil = \left \lceil {x}\right \rceil$
+Thus the [[Definition:Ceiling Function|ceiling function]] is [[Definition:Reflexive Relation|reflexive]].
+{{qed|lemma}}
+=== Symmetry ===
+:$\forall x, y \in \R: \left \lceil {x}\right \rceil = \left \lceil {y}\right \rceil \implies \left \lceil {y}\right \rceil = \left \lceil {x}\right \rceil$
+Thus the [[Definition:Ceiling Function|ceiling function]] is [[Definition:Symmetric Relation|symmetric]].
+{{qed|lemma}}
+=== Transitivity ===
+Let:
+:$\left \lceil {x}\right \rceil = \left \lceil {y}\right \rceil$
+:$\left \lceil {y}\right \rceil = \left \lceil {z}\right \rceil$
+Let:
+:$n = \left \lceil {x}\right \rceil = \left \lceil {y}\right \rceil = \left \lceil {z}\right \rceil$
+which follows from transitivity of $=$.
+Thus:
+:$x = n - t_x, y = n - t_y, z = n - t_z: t_x, t_y, t_z \in \left[{0 \,.\,.\, 1}\right)$
+from [[Real Number is Ceiling minus Difference]].
+So:
+:$x = n - t_x$
+and:
+:$z = n - t_z$
+and so:
+:$\left \lceil {x}\right \rceil = \left \lceil {z}\right \rceil$
+Thus the [[Definition:Ceiling Function|ceiling function]] is [[Definition:Transitive Relation|transitive]].
+{{qed|lemma}}
+Thus we have shown that $\mathcal R$ is an [[Definition:Equivalence Relation|equivalence]].
+Now we show that the [[Definition:Equivalence Class|$\mathcal R$-class]] of $n$ is the [[Definition:Half-Open Real Interval|interval]] $\left({n - 1 \,.\,.\, n}\right]$.
+Defining $\mathcal R$ as above, with $n \in \Z$:
+{{begin-eqn}}
+{{eqn | l = x
+      | o = \in
+      | r = \left[\!\left[{n}\right]\!\right]_\mathcal R
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = \left \lceil {x}\right \rceil
+      | r = \left \lceil {n}\right \rceil = n
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = \exists t \in \left({0 \,.\,.\, 1}\right]: x
+      | r = n - t
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = x
+      | o = \in
+      | r = \left({n - 1 \,.\,.\, n}\right]
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Ceiling Function]]
+[[Category:Equivalence Relations]]
+kktdzbepmt2otz9fgu475n43d93twsn
+\end{proof}<|endoftext|>
+\section{Real Number is Ceiling minus Difference}
+Tags: Ceiling Function
+
+\begin{theorem}
+Let $n$ be a [[Definition:Integer|integer]].
+{{TFAE}}
+:$(1): \quad$ There exists $t \in \hointr 0 1$ such that $x = n - t$
+:$(2): \quad n = \ceiling x$
+\end{theorem}
+
+\begin{proof}
+=== 1 implies 2 ===
+Let $x = n - t$, where $t \in \hointr 0 1$.
+Because $0 \le t < 1$, we have:
+:$0 \leq n - x < 1$
+Thus:
+:$n - 1 < x \le  n$
+That is, $n$ is the [[Definition:Ceiling Function|ceiling]] of $x$.
+{{qed|lemma}}
+=== 2 implies 1 ===
+Now let $n = \ceiling x$.
+Let $t = \ceiling x - x$.
+Then $x = n - t$.
+From [[Ceiling minus Real Number]], $t = \ceiling x - x \in \hointr 0 1$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Cauchy's Inequality}
+Tags: Analysis, Inequalities, Cauchy's Inequality, Cauchy-Bunyakovsky-Schwarz Inequality
+
+\begin{theorem}
+:$\displaystyle \sum {r_i^2} \sum {s_i^2} \ge \left({\sum {r_i s_i}}\right)^2$
+\end{theorem}
+
+\begin{proof}
+For any $\lambda \in \R$, we define $f: \R \to \R$ as the [[Definition:Real Function|function]]:
+:$\displaystyle \map f \lambda = \sum {\paren {r_i + \lambda s_i}^2}$
+Now:
+:$\map f \lambda \ge 0$
+because it is the sum of [[Definition:Square Function|squares]] of [[Definition:Real Number|real numbers]].
+Hence:
+{{begin-eqn}}
+{{eqn | l = \forall \lambda \in \R: \map f \lambda
+      | o = \equiv
+      | r = \sum {\paren {r_i^2 + 2 \lambda r_i s_i + \lambda^2 s_i^2} } \ge 0
+      | c = 
+}}
+{{eqn | o = \equiv
+      | r = \sum {r_i^2} + 2 \lambda \sum {r_i s_i} + \lambda^2 \sum {s_i^2} \ge 0
+      | c = 
+}}
+{{end-eqn}}
+This is a [[Definition:Quadratic Equation|quadratic equation]] in $\lambda$.
+From [[Solution to Quadratic Equation]]:
+:$\displaystyle a \lambda^2 + b \lambda + c = 0: a = \sum {s_i^2}, b = 2 \sum {r_i s_i}, c = \sum {r_i^2}$
+The [[Definition:Discriminant of Quadratic Equation|discriminant]] of this equation (that is $b^2 - 4 a c$) is:
+:$D := \displaystyle 4 \paren {\sum {r_i s_i} }^2 - 4 \sum {r_i^2} \sum {s_i^2}$
+{{AimForCont}} $D$ is [[Definition:Strictly Positive Real Number|(strictly) positive]].
+Then $\map f \lambda = 0$ has two [[Definition:Distinct Elements|distinct]] [[Definition:Real Number|real]] [[Definition:Root of Polynomial|roots]], $\lambda_1 < \lambda_2$, say.
+From [[Sign of Quadratic Function Between Roots]], it follows that $f$ is [[Definition:Strictly Negative Real Number|(strictly) negative]] somewhere between $\lambda_1$ and $\lambda_2$.
+But we have:
+:$\forall \lambda \in \R: \map f \lambda \ge 0$
+From this [[Definition:Contradiction|contradiction]] it follows that:
+:$D \le 0$
+which is the same thing as saying:
+:$\displaystyle \sum {r_i^2} \sum {s_i^2} \ge \paren {\sum {r_i s_i} }^2$
+{{qed}}
+\end{proof}
+
+\begin{proof}
+From the [[Cauchy-Schwarz Inequality/Complex Numbers|Complex Number form of the Cauchy-Schwarz Inequality]], we have:
+:$\displaystyle \sum \left|{w_i}\right|^2 \left|{z_i}\right|^2 \ge \left|{\sum w_i z_i}\right|^2$
+where all of $w_i, z_i \in \C$.
+As elements of $\R$ are also elements of $\C$, it follows that:
+:$\displaystyle \sum \left|{r_i}\right|^2 \left|{s_i}\right|^2 \ge \left|{\sum r_i s_i}\right|^2$
+where all of $r_i, s_i \in \R$.
+But from the [[Definition:Modulus of Complex Number|definition of modulus]], it follows that:
+:$\displaystyle \forall r_i \in \R: \left|{r_i}\right|^2 = r_i^2$
+Thus:
+:$\displaystyle \sum {r_i^2} \sum {s_i^2} \ge \left({\sum {r_i s_i}}\right)^2$
+where all of $r_i, s_i \in \R$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Cancellable Finite Semigroup is Group}
+Tags: Semigroups, Finite Groups, Cancellability
+
+\begin{theorem}
+Let $\struct {S, \circ}$ be a [[Definition:Non-Empty Set|non-empty]] [[Definition:Finite Semigroup|finite semigroup]] in which all [[Definition:Element|elements]] are [[Definition:Cancellable Element|cancellable]].
+Then $\struct {S, \circ}$ is a [[Definition:Group|group]].
+\end{theorem}
+
+\begin{proof}
+As $\struct {S, \circ}$ is a [[Definition:Semigroup|semigroup]], it is already [[Definition:Closed Algebraic Structure|closed]] and [[Definition:Associative|associative]].
+It remains to be shown that it has an [[Definition:Identity Element|identity]] and that every [[Definition:Element|element]] of $S$ has an [[Definition:Inverse Element|inverse]] in $S$.
+First we show that $\struct {S, \circ}$ has an [[Definition:Identity Element|identity]].
+Choose $a \in S$.
+Let the [[Definition:Mapping|mapping]] $\lambda_a: S \to S$ be the [[Definition:Left Regular Representation|left regular representation]] of $\struct {S, \circ}$ with respect to $a$.
+[[Definition:By Hypothesis|By hypothesis]]:
+:all [[Definition:Element|elements]] of $S$ are [[Definition:Cancellable Element|cancellable]]
+:$S$ is [[Definition:Finite Semigroup|finite]].
+By [[Regular Representation wrt Cancellable Element on Finite Semigroup is Bijection]], $\lambda_a$ is a [[Definition:Bijection|bijection]].
+Hence $a \circ e = a$ for some $e \in S$.
+Let $x \in S$.
+Then because of [[Definition:Cancellable Element|cancellability]]:
+{{begin-eqn}}
+{{eqn | l = a \circ e \circ x
+      | r = a \circ x
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = e \circ x
+      | r = x
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = x \circ e \circ x
+      | r = x \circ x
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = x \circ e
+      | r = x
+      | c = 
+}}
+{{end-eqn}}
+Thus $e$ is the [[Definition:Identity Element|identity]].
+The existence of [[Definition:Inverse Element|inverses]] comes from the [[Definition:Surjection|surjectivity]] of $\lambda_a$.
+As $\lambda_a$ is [[Definition:Surjection|surjective]]:
+:$\exists y \in S: \map {\lambda_a} y = e$
+That is:
+:$a \circ y = e$
+So we see that $y$ acts as a [[Definition:Right Inverse Element|right inverse]] for $a$.
+This is the case for any $a \in S$: all of them have some [[Definition:Right Inverse Element|right inverse]].
+So, from [[Right Inverse for All is Left Inverse]], each of these [[Definition:Element|elements]] is also a [[Definition:Left Inverse Element|left inverse]], and therefore an [[Definition:Inverse Element|inverse]].
+Thus $S$ is [[Definition:Closed Algebraic Structure|closed]], [[Definition:Associative|associative]], has an [[Definition:Identity Element|identity]] and every element has an [[Definition:Inverse Element|inverse]].
+So, by definition, $\struct {S, \circ}$ is a [[Definition:Group|group]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Finite Semigroup Equal Elements for Different Powers}
+Tags: Semigroups
+
+\begin{theorem}
+Let $\left({S, \circ}\right)$ be a [[Definition:Finite|finite]] [[Definition:Semigroup|semigroup]].
+Then:
+: $\forall x \in S: \exists m, n \in \N: m \ne n: x^m = x^n$
+\end{theorem}
+
+\begin{proof}
+List the [[Definition:Power of Element of Semigroup|positive powers]] $x, x^2, x^3, \ldots$ of any element $x$ of a finite semigroup $\left({S, \circ}\right)$.
+Since all are elements of $S$, and the [[Definition:Semigroup|semigroup]] has a [[Definition:Finite|finite]] number of elements, it follows from the [[Pigeonhole Principle]] this list must contain repetitions.
+So there must be at least one instance where $x^m = x^n$ for some $m, n \in \N$.
+{{Qed}}
+[[Category:Semigroups]]
+1xgxx01rhazfn21f659gs2m7pt8z620
+\end{proof}<|endoftext|>
+\section{Element has Idempotent Power in Finite Semigroup}
+Tags: Semigroups, Idempotence
+
+\begin{theorem}
+Let $\struct {S, \circ}$ be a [[Definition:Finite Semigroup|finite semigroup]].
+For every [[Definition:Element|element]] in $\struct {S, \circ}$, there is a [[Definition:Power of Element of Semigroup|power]] of that [[Definition:Element|element]] which is [[Definition:Idempotent Element|idempotent]].
+That is:
+:$\forall x \in S: \exists i \in \N: x^i = x^i \circ x^i$
+\end{theorem}
+
+\begin{proof}
+From [[Finite Semigroup Equal Elements for Different Powers]], we have:
+:$\forall x \in S: \exists m, n \in \N: m \ne n: x^m = x^n$
+Let $m > n$.
+Let $n = k, m = k + l$.
+Then:
+: $\forall x \in S: \exists k, l \in \N: x^k = x^{k + l}$
+Now we show that:
+:$x^k = x^{k + l} \implies x^{k l} = x^{k l} \circ x^{k l}$
+That is, that $x^{k l}$ is [[Definition:Idempotent Element|idempotent]].
+First:
+{{begin-eqn}}
+{{eqn | l = x^k
+      | r = x^{k + l}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = x^k \circ x^l
+      | r = x^{k + l} \circ x^l
+      | c = both sides ${} \circ x^l$
+}}
+{{eqn | ll= \leadsto
+      | l = x^{k + l}
+      | r = x^{k + 2 l}
+      | c = [[Index Laws for Semigroup/Sum of Indices|Index Laws for Semigroup: Sum of Indices]]
+}}
+{{eqn | ll= \leadsto
+      | l = x^k
+      | r = x^{k + 2 l}
+      | c = as $x^k = x^{k + l}$
+}}
+{{end-eqn}}
+From here we can easily prove by [[Principle of Mathematical Induction|induction]] that:
+: $\forall n \in \N: x^k = x^{k + n l}$
+In particular:
+: $x^k = x^{k + k l} = x^{k \paren {l + 1} }$
+There are two cases to consider:
+:$(1): \quad$ If $l = 1$, then $x^k = x^{k \paren {l + 1} } = x^{2 k} = x^k \circ x^k$, and $x^{k l} = x^k$ is [[Definition:Idempotent Element|idempotent]].
+:$(2): \quad$ If $l > 1$, then:
+{{begin-eqn}}
+{{eqn | l = x^k
+      | r = x^{k + k l}
+      | c = 
+}}
+{{eqn | r = x^{k \paren {l + 1} }
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = x^k \circ x^{k \paren {l - 1} }
+      | r = x^{k \paren {l + 1} } \circ x^{k \paren {l - 1} }
+      | c = both sides ${} \circ x^{k \paren {l - 1} }$
+}}
+{{eqn | ll= \leadsto
+      | l = x^{k + k l - k}
+      | r = x^{kl + k + k l - k}
+      | c = [[Index Laws for Semigroup/Sum of Indices|Index Laws for Semigroup: Sum of Indices]]
+}}
+{{eqn | ll= \leadsto
+      | l = x^{k l}
+      | r = x^{k l + k l}
+      | c = 
+}}
+{{eqn | r = x^{k l} \circ x^{k l}
+      | c = 
+}}
+{{end-eqn}}
+and again, $x^{k l}$ is [[Definition:Idempotent Element|idempotent]].
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Powers of Ring Elements}
+Tags: Ring Theory, Proofs by Induction
+
+\begin{theorem}
+Let $\struct {R, +, \circ}$ be a [[Definition:Ring (Abstract Algebra)|ring]] whose [[Definition:Ring Zero|zero]] is $0_R$.
+Let $n \cdot x$ be an [[Definition:Integral Multiple|integral multiple]] of $x$:
+:$n \cdot x = \begin {cases}
+0_R & : n = 0 \\
+x & : n = 1 \\
+\paren {n - 1} \cdot x + x & : n > 1
+\end {cases}$
+that is $n \cdot x = x + x + \cdots \paren n \cdots x$.
+For $n < 0$ we use:
+:$-n \cdot x = n \cdot \paren {-x}$
+Then:
+:$\forall n \in \Z: \forall x \in R: \paren {n \cdot x} \circ x = n \cdot \paren {x \circ x} = x \circ \paren {n \cdot x}$
+\end{theorem}
+
+\begin{proof}
+Proof by [[Principle of Mathematical Induction|induction]]:
+For all $n \in \N$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
+:$\paren {n \cdot x} \circ x = n \cdot \paren {x \circ x} = x \circ \paren {n \cdot x}$
+First we verify $\map P 0$.
+When $n = 0$, we have:
+{{begin-eqn}}
+{{eqn | l = \paren {0 \cdot x} \circ x
+      | r = 0_R \circ x
+      | c = 
+}}
+{{eqn | r = 0_R
+      | c = 
+}}
+{{eqn | r = 0 \cdot \paren {x \circ x}
+      | c = 
+}}
+{{eqn | r = x \circ 0_R
+      | c = 
+}}
+{{eqn | r = x \circ \paren {0 \cdot x}
+      | c = 
+}}
+{{end-eqn}}
+So $\map P 0$ holds.
+=== Basis for the Induction ===
+Now we verify $\map P 1$:
+{{begin-eqn}}
+{{eqn | l = \paren {1 \cdot x} \circ x
+      | r = x \circ x
+      | c = 
+}}
+{{eqn | r = 1 \cdot \paren {x \circ x}
+      | c = 
+}}
+{{eqn | r = x \circ \paren {1 \cdot x}
+      | c = 
+}}
+{{end-eqn}}
+So $\map P 1$ holds.
+This is our [[Definition:Basis for the Induction|basis for the induction]].
+=== Induction Hypothesis ===
+Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
+So this is our [[Definition:Induction Hypothesis|induction hypothesis]]:
+:$\paren {k \cdot x} \circ x = k \cdot \paren {x \circ x} = x \circ \paren {k \cdot x}$
+Then we need to show:
+:$\paren {\paren {k + 1} \cdot x} \circ x = \paren {k + 1} \cdot \paren {x \circ x} = x \circ \paren {\paren {k + 1} \cdot x}$
+=== Induction Step ===
+This is our [[Definition:Induction Step|induction step]]:
+{{begin-eqn}}
+{{eqn | l = \paren {\paren {k + 1} \cdot x} \circ x
+      | r = \paren {x + \paren {k \cdot x} } \circ x
+      | c = 
+}}
+{{eqn | r = x \circ x + \paren {k \cdot x} \circ x
+      | c = {{Ring-axiom|D}}
+}}
+{{eqn | r = x \circ x + k \cdot \paren {x \circ x}
+      | c = [[Powers of Ring Elements#Induction Hypothesis|Induction Hypothesis]]
+}}
+{{eqn | r = \paren {k + 1} \cdot \paren {x \circ x}
+      | c = 
+}}
+{{end-eqn}}
+A similar construction shows that $\paren {k + 1} \cdot \paren {x \circ x} = x \circ \paren {\paren {k + 1} \cdot x}$.
+So $\map P k \implies \map P {k + 1}$ and the result follows by the [[Principle of Mathematical Induction]].
+Therefore:
+:$\forall n \in \N: \paren {n \cdot x} \circ x = n \cdot \paren {x \circ x} = x \circ \paren {n \cdot x}$
+{{qed|lemma}}
+The result for $n < 0$ follows directly from [[Powers of Group Elements]].
+{{qed}}
+[[Category:Ring Theory]]
+[[Category:Proofs by Induction]]
+8hv2385mggjxed224pcgpzdu9xdduss
+\end{proof}<|endoftext|>
+\section{Power of Conjugate equals Conjugate of Power}
+Tags: Conjugacy
+
+\begin{theorem}
+Let $\struct {G, \circ}$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
+Let $x, y \in G$ such that $\exists a \in G: x \circ a = a \circ y$.
+That is, let $x$ and $y$ be [[Definition:Conjugate of Group Element|conjugate]].
+Then:
+: $\forall n \in \Z: y^n = \paren {a^{-1} \circ x \circ a}^n = a^{-1} \circ x^n \circ a$
+It follows directly that:
+: $\exists b \in G: \forall n \in \Z: y^n = b \circ x^n \circ b^{-1}$
+In particular:
+: $y^{-1} = \paren {a^{-1} \circ x \circ a}^{-1} = a^{-1} \circ x^{-1} \circ a$
+\end{theorem}
+
+\begin{proof}
+Proof by [[Principle of Mathematical Induction|induction]]:
+For all $n \in \N$, let $\map P n$ be the [[Definition:Proposition|proposition]] $y^n = a^{-1} \circ x^n \circ a$.
+$\map P 0$ is true, as this just says $e =  a^{-1} \circ e \circ a$.
+=== Basis for the Induction ===
+$\map P 1$ is the case $y = a^{-1} \circ x \circ a$, which is how [[Definition:Conjugate of Group Element|conjugacy]] is defined for $x$ and $y$.
+This is our [[Principle of Mathematical Induction#Basis for the Induction|basis for the induction]].
+=== Induction Hypothesis ===
+Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
+So this is our [[Principle of Mathematical Induction#Induction Hypothesis|induction hypothesis]]:
+:$y^k = a^{-1} \circ x^k \circ a$
+Then we need to show:
+:$y^{k + 1} = a^{-1} \circ x^{k + 1} \circ a$
+=== Induction Step ===
+This is our [[Principle of Mathematical Induction#Induction Step|induction step]]:
+{{begin-eqn}}
+{{eqn | l = y^{k + 1}
+      | r = \paren {a^{-1} \circ x \circ a}^{k + 1}
+      | c = 
+}}
+{{eqn | r = \paren {a^{-1} \circ x \circ a}^k \circ \paren {a^{-1} \circ x \circ a}
+      | c = 
+}}
+{{eqn | r = \paren {a^{-1} \circ x^k \circ a} \circ \paren {a^{-1} \circ x \circ a}
+      | c = [[Power of Conjugate equals Conjugate of Power#Induction Hypothesis|Induction Hypothesis]]
+}}
+{{eqn | r = a^{-1} \circ x^k \circ x \circ a
+      | c = 
+}}
+{{eqn | r = a^{-1} \circ x^{k + 1} \circ a
+      | c = 
+}}
+{{end-eqn}}
+So $\map P k \implies \map P {k + 1}$ and the result follows by the [[Principle of Mathematical Induction]].
+Therefore:
+:$\forall n \in \N: y^n = a^{-1} \circ x^n \circ a$
+Now we need to show that if $\map P n$ holds, then $\map P {-n}$ holds.
+That is:
+:$y^{-n} = a^{-1} \circ x^{-n} \circ a$
+Let $n \in \N$.
+Then:
+{{begin-eqn}}
+{{eqn | l = y^{-n}
+      | r = \paren {a^{-1} \circ x \circ a}^{-n}
+      | c = 
+}}
+{{eqn | r = \paren {\paren {a^{-1} \circ x \circ a}^n}^{-1}
+      | c = [[Index Laws for Monoids/Negative Index|Index Laws for Monoids: Negative Index]]
+}}
+{{eqn | r = \paren {a^{-1} \circ x^n \circ a}^{-1}
+      | c = 
+}}
+{{eqn | r = a^{-1} \circ \paren {x^n}^{-1} \circ \paren {a^{-1} }^{-1}
+      | c = [[Inverse of Group Product/General Result|Inverse of Group Product]]
+}}
+{{eqn | r = a^{-1} \circ x^{-n} \circ a
+      | c = 
+}}
+{{end-eqn}}
+Thus $\map P n$ has been shown to hold for all $n \in \Z$.
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Product of Conjugates equals Conjugate of Products}
+Tags: Conjugacy
+
+\begin{theorem}
+Let $\struct {G, \circ}$ be a [[Definition:Group|group]].
+Then:
+:$\forall a, x, y \in G: \paren {a \circ x \circ a^{-1} } \circ \paren {a \circ y \circ a^{-1} } = a \circ \paren {x \circ y} \circ a^{-1}$
+That is, the [[Definition:Product Element|product]] of [[Definition:Conjugate of Group Element|conjugates]] is equal to the [[Definition:Conjugate of Group Element|conjugate]] of the [[Definition:Product Element|product]].
+\end{theorem}
+
+\begin{proof}
+Follows directly from the [[Definition:Group Axioms|group axioms]].
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Power of Product with Inverse}
+Tags: Group Theory
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]] whose [[Definition:Identity (Abstract Algebra)|identity]] is $e$.
+Let $a, b \in G: a b = b a^{-1}$.
+Then:
+: $\forall n \in \Z: a^n b = b a^{-n}$
+\end{theorem}
+
+\begin{proof}
+Proof by [[Principle of Mathematical Induction|induction]]:
+For all $n \in \Z$, let $P \left({n}\right)$ be the [[Definition:Proposition|proposition]] $a^n b = b a^{-n}$.
+$P(0)$ is trivially true, as $a^0 b = e b = b = b e = b a^{-0}$.
+=== Basis for the Induction ===
+$P(1)$ is true, as this is the given relation between $a$ and $b$:
+:$a b = b a^{-1}$
+This is our [[Principle of Mathematical Induction#Basis for the Induction|basis for the induction]].
+=== Induction Hypothesis ===
+Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 1$, then it logically follows that $P \left({k+1}\right)$ is true.
+So this is our [[Principle of Mathematical Induction#Induction Hypothesis|induction hypothesis]]:
+:$a^k b = b a^{-k}$
+Then we need to show:
+:$a^{k+1} b = b a^{-\left({k+1}\right)}$
+=== Induction Step ===
+This is our [[Principle of Mathematical Induction#Induction Step|induction step]]:
+{{begin-eqn}}
+{{eqn | l=a^{k+1} b
+      | r=a \left({a^k b}\right)
+      | c=
+}}
+{{eqn | r=a \left({b a^{-k} }\right)
+      | c=[[Power of Product with Inverse#Induction Hypothesis|Induction Hypothesis]]
+}}
+{{eqn | r=\left({a b}\right) a^{-k}
+      | c=
+}}
+{{eqn | r=\left({b a^{-1} }\right) a^{-k}
+      | c=[[Power of Product with Inverse#Basis for the Induction|Basis for the Induction]]
+}}
+{{eqn | r=b \left({a^{-1} a^{-k} }\right)
+      | c=
+}}
+{{eqn | r=b a^{-\left({k+1}\right)}
+      | c=
+}}
+{{end-eqn}}
+So $P \left({k}\right) \implies P \left({k+1}\right)$ and the result follows by the [[Principle of Mathematical Induction]].
+Therefore $\forall n \in \N: a^n b = b a^{-n}$.
+Now we show that $P \left({-1}\right)$ holds, i.e. that $a^{-1} b = b a$.
+{{begin-eqn}}
+{{eqn | l=a b
+      | r=b a^{-1}
+      | c=
+}}
+{{eqn | ll=\iff
+      | l=a b a
+      | r=b
+      | c=
+}}
+{{eqn | ll=\iff
+      | l=b a
+      | r=a^{-1} b
+      | c=
+}}
+{{end-eqn}}
+... thus showing that $P \left({-1}\right)$ holds.
+The proof that $P \left({n}\right)$ holds for all $n \in \Z: n < 0$ then follows by [[Principle of Mathematical Induction|induction]], similarly to the proof for $n > 0$.
+{{qed}}
+[[Category:Group Theory]]
+fghfsm379raiteyy6tp8lhpqp8i5qbl
+\end{proof}<|endoftext|>
+\section{Powers of Elements in Group Direct Product}
+Tags: Group Theory, Direct Products
+
+\begin{theorem}
+Let $\left({G, \circ_1}\right)$ and $\left({H, \circ_2}\right)$ be [[Definition:Group|group]] whose [[Definition:Identity (Abstract Algebra)|identities]] are $e_G$ and $e_H$.
+Let $\left({G \times H, \circ}\right)$ be the group direct product (either [[Definition:External Direct Product|external]] or [[Definition:Internal Group Direct Product|internal]]) of $G$ and $H$.
+Then:
+: $\forall n \in \Z: \forall g \in G, h \in H: \left({g, h}\right)^n = \left({g^n, h^n}\right)$
+\end{theorem}
+
+\begin{proof}
+Proof by [[Principle of Mathematical Induction|induction]]:
+For all $n \in \N$, let $P \left({n}\right)$ be the [[Definition:Proposition|proposition]] $\forall g \in G, h \in H: \left({g, h}\right)^n = \left({g^n, h^n}\right)$.
+=== Basis for the Induction ===
+$P(0)$ is true, as this says:
+: $\left({g, h}\right)^0 = \left({e_G, e_H}\right)$
+$P(1)$ is true, as this says:
+: $\left({g, h}\right) = \left({g, h}\right)$
+This is our [[Principle of Mathematical Induction#Basis for the Induction|basis for the induction]].
+=== Induction Hypothesis ===
+Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 1$, then it logically follows that $P \left({k+1}\right)$ is true.
+So this is our [[Principle of Mathematical Induction#Induction Hypothesis|induction hypothesis]]:
+:$\left({g, h}\right)^k = \left({g^k, h^k}\right)$
+Then we need to show:
+:$\left({g, h}\right)^{k+1} = \left({g^{k+1}, h^{k+1}}\right)$
+=== Induction Step ===
+This is our [[Principle of Mathematical Induction#Induction Step|induction step]]:
+{{begin-eqn}}
+{{eqn | l=\left({g, h}\right)^{k+1}
+      | r=\left({g, h}\right)^k \circ \left({g, h}\right)
+      | c=
+}}
+{{eqn | r=\left({g^k, h^k}\right) \circ \left({g, h}\right)
+      | c=[[#Induction Hypothesis|Induction Hypothesis]]
+}}
+{{eqn | r=\left({g^k \circ_1 g, h^k \circ_2 h}\right)
+      | c=by definition of [[Definition:External Direct Product|external]] or [[Definition:Internal Group Direct Product|internal]] group product
+}}
+{{eqn | r=\left({g^{k+1}, h^{k+1} }\right)
+      | c=
+}}
+{{end-eqn}}
+So $P \left({k}\right) \implies P \left({k+1}\right)$ and the result follows by the [[Principle of Mathematical Induction]].
+Therefore:
+:$\forall n \in \N: \forall g \in G, h \in H: \left({g, h}\right)^n = \left({g^n, h^n}\right)$
+So we have shown the result holds true for all $n \ge 0$.
+The result for $n < 0$ follows directly from [[Powers of Group Elements/Negative Index|Powers of Group Elements for negative indices]].
+{{qed}}
+[[Category:Group Theory]]
+[[Category:Direct Products]]
+hsrdqb9vr937yff7dzku2el3m4ojkc6
+\end{proof}<|endoftext|>
+\section{Powers of Commutative Elements in Semigroups}
+Tags: Semigroups, Commutativity
+
+\begin{theorem}
+Let $\left ({S, \circ}\right)$ be a [[Definition:Semigroup|semigroup]].
+Let $a, b \in S$ both be [[Definition:Cancellable Element|cancellable elements]] of $S$.
+Then the following results hold:
+\end{theorem}<|endoftext|>
+\section{Powers of Commutative Elements in Monoids}
+Tags: Monoids
+
+\begin{theorem}
+These results are an extension of the results in [[Powers of Commutative Elements in Semigroups]] in which the [[Definition:Domain of Mapping|domain]] of the indices is extended to include all [[Definition:Integer|integers]].
+Let $\left ({S, \circ}\right)$ be a [[Definition:Monoid|monoid]] whose [[Definition:Identity Element|identity]] is $e_S$.
+Let $a, b \in S$ be [[Definition:Invertible Element|invertible elements]] for $\circ$ that also [[Definition:Commute|commute]].
+Then the following results hold.
+\end{theorem}<|endoftext|>
+\section{Powers of Commutative Elements in Groups}
+Tags: Group Theory, Commutativity
+
+\begin{theorem}
+Let $\struct {G, \circ}$ be a [[Definition:Group|group]].
+Let $a, b \in G$ such that $a$ and $b$ [[Definition:Commuting Elements|commute]].
+Then the following results hold:
+\end{theorem}<|endoftext|>
+\section{General Morphism Property for Semigroups}
+Tags: Semigroups, Homomorphisms
+
+\begin{theorem}
+Let $\struct {S, \circ}$ and $\struct {T, *}$ be [[Definition:Semigroup|semigroups]].
+Let $\phi: S \to T$ be a [[Definition:Semigroup Homomorphism|homomorphism]].
+Then:
+:$\forall s_k \in S: \map \phi {s_1 \circ s_2 \circ \cdots \circ s_n} = \map \phi {s_1} * \map \phi {s_2} * \cdots * \map \phi {s_n}$
+Hence it follows that:
+:$\forall n \in \N_{>0}: \forall s \in S: \map \phi {s^n} = \paren {\map \phi s}^n$
+\end{theorem}
+
+\begin{proof}
+$\forall s_k \in S: \map \phi {s_1 \circ s_2 \circ \cdots \circ s_n} = \map \phi {s_1} * \map \phi {s_2} * \cdots * \map \phi {s_n}$ can be proved by [[Principle of Mathematical Induction|induction]].
+For all $n \in \N_{>0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
+:$\map \phi {s_1 \circ s_2 \circ \cdots \circ s_n} = \map \phi {s_1} * \map \phi {s_2} * \cdots * \map \phi {s_n}$
+$P(1)$ is true, as this just says:
+:$\map \phi {s_1} = \map \phi {s_1}$.
+=== Basis for the Induction ===
+$P(2)$ is the case:
+:$\map \phi {s_1 \circ s_2} = \map \phi {s_1} * \map \phi {s_2}$
+This follows from the fact that $\phi$ is a [[Definition:Semigroup Homomorphism|homomorphism]].
+This is our [[Principle of Mathematical Induction#Basis for the Induction|basis for the induction]].
+=== Induction Hypothesis ===
+Now we need to show that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true.
+So this is our [[Principle of Mathematical Induction#Induction Hypothesis|induction hypothesis]]:
+:$\map \phi {s_1 \circ s_2 \circ \cdots \circ s_k} = \map \phi {s_1} * \map \phi {s_2} * \cdots * \map \phi {s_k}$
+Then we need to show:
+:$\map \phi {s_1 \circ s_2 \circ \cdots \circ s_k \circ s_{k + 1} } = \map \phi {s_1} * \map \phi {s_2} * \cdots * \map \phi {s_k} * \map \phi {s_{k + 1} }$
+=== Induction Step ===
+This is our [[Principle of Mathematical Induction#Induction Step|induction step]]:
+{{begin-eqn}}
+{{eqn | l = \map \phi {s_1 \circ s_2 \circ \cdots \circ s_k \circ s_{k + 1} }
+      | r = \map \phi {s_1 \circ s_2 \circ \cdots \circ s_k} * \map \phi {s_{k + 1} }
+      | c = [[General Morphism Property for Semigroups#Basis for the Induction|Basis for the Induction]]
+}}
+{{eqn | r = \map \phi {s_1} * \map \phi {s_2} * \cdots * \map \phi {s_k} * \map \phi {s_{k + 1} }
+      | c = [[General Morphism Property for Semigroups#Induction Hypothesis|Induction Hypothesis]]
+}}
+{{end-eqn}}
+So $\map P k \implies \map P {k + 1}$ and the result follows by the [[Principle of Mathematical Induction]].
+Therefore:
+:$\map \phi {s_1 \circ s_2 \circ \cdots \circ s_n} = \map \phi {s_1} * \map \phi {s_2} * \cdots * \map \phi {s_n}$
+The result for $n \in \N_{>0}$ follows directly from the above, by replacing each occurrence of $s_k$ with $s$.
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Homomorphism of Power of Group Element}
+Tags: Group Homomorphisms
+
+\begin{theorem}
+Let $\struct {G, \circ}$ and $\struct {H, \ast}$ be [[Definition:Group|groups]].
+Let $\phi: S \to T$ be a [[Definition:Group Homomorphism|group homomorphism]].
+Then:
+: $\forall n \in \Z: \forall g \in G: \map \phi {g^n} = \paren {\map \phi g}^n$
+\end{theorem}
+
+\begin{proof}
+The result for $n \in \N_{>0}$ follows directly from [[General Morphism Property for Semigroups]].
+For $n = 0$, we use [[Homomorphism with Cancellable Codomain Preserves Identity]].
+For $n < 0$, we use [[Homomorphism with Identity Preserves Inverses]], along with [[Index Laws for Monoids/Negative Index|Index Laws for Monoids: Negative Index]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Existence of Unique Subsemigroup Generated by Subset}
+Tags: Subsemigroups
+
+\begin{theorem}
+Let $\struct {S, \circ}$ be a [[Definition:Semigroup|semigroup]].
+Let $\O \subset X \subseteq S$.
+Let $\struct {T, \circ}$ be the [[Definition:Generator of Semigroup|subsemigroup generated by $X$]].
+Then $T = \gen X$ exists and is [[Definition:Unique|unique]].
+\end{theorem}
+
+\begin{proof}
+=== Existence ===
+First, we prove that such a [[Definition:Subsemigroup|subsemigroup]] exists.
+Let $\mathbb S$ be the [[Definition:Set|set]] of all [[Definition:Subsemigroup|subsemigroups]] of $S$ which contain $X$.
+$\mathbb S \ne \O$ because $S$ is itself a [[Definition:Subsemigroup|subsemigroup]] of $S$ (trivial), and thus $S \in \mathbb S$.
+Let $T$ be the [[Definition:Set Intersection|intersection]] of all the [[Definition:Element|elements]] of $\mathbb S$.
+By [[Intersection of Subsemigroups]], $T$ is the largest element of $\mathbb S$ contained in each element of $\mathbb S$.
+Thus $T$ is a [[Definition:Subsemigroup|subsemigroup]] of $S$.
+Since $\forall x \in \mathbb S: X \subseteq x$, we see that $X \subseteq T$, so $T \in \mathbb S$.
+=== Smallest ===
+Now to show that $T$ is the smallest such [[Definition:Subsemigroup|subsemigroup]].
+If any $K \le S: X \subseteq K$, then $K \in \mathbb S$ and therefore $T \subseteq K$.
+So $T$ is the smallest [[Definition:Subsemigroup|subsemigroup]] of $S$ containing $X$.
+=== Uniqueness ===
+Now we show that $T$ is unique.
+Suppose $\exists T_1, T_2 \in \mathbb S$ such that $T_1$ and $T_2$ were two such smallest [[Definition:Subsemigroup|subsemigroups]] containing $X$.
+Then, by the definition of '''smallest''', each would be equal in size.
+{{WLOG}}, suppose that $T_1$ is not a [[Definition:Subset|subset]] of $T_2$.
+Then their [[Definition:Set Intersection|intersection]] (by definition containing $X$) would be a [[Definition:Subsemigroup|subsemigroup]] smaller than $T_1$.
+Hence $T_1$ would not be the smallest.
+Therefore each must be the [[Definition:Subset|subset]] of each other.
+By definition of [[Definition:Set Equality/Definition 2|set equality]], they must be the same [[Definition:Set|set]].
+So the smallest subsemigroup, whose existence we have proved above, is [[Definition:Unique|unique]].
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Finite Subgroup Test}
+Tags: Subgroups, Finite Subgroup Test
+
+\begin{theorem}
+Let $\struct {G, \circ}$ be a [[Definition:Group|group]].
+Let $H$ be a [[Definition:Non-Empty Set|non-empty]] [[Definition:Finite Set|finite]] [[Definition:Subset|subset]] of $G$.
+Then:
+:$H$ is a [[Definition:Subgroup|subgroup]] of $G$
+{{iff}}:
+:$\forall a, b \in H: a \circ b \in H$
+That is, a [[Definition:Non-Empty Set|non-empty]] [[Definition:Finite Set|finite]] [[Definition:Subset|subset]] of $G$ is a [[Definition:Subgroup|subgroup]] {{iff}} it is [[Definition:Closed Algebraic Structure|closed]].
+\end{theorem}
+
+\begin{proof}
+Let $H$ be a [[Definition:Finite Set|finite]] [[Definition:Subset|subset]] of $G$ such that $a, b \in H \implies a \circ b \in H$.
+From the [[Two-Step Subgroup Test]], it follows that we only need to show that $a \in H \implies a^{-1} \in H$.
+So, let $a \in H$.
+First it is straightforward to show by [[Principle of Mathematical Induction|induction]] that $\set {x \in G: x = a^n: n \in \N} \subseteq H$.
+That is, $a \in H \implies \forall n \in \N: a^n \in H$.
+Now, since $H$ is [[Definition:Finite Set|finite]], we have that [[Order of Element Divides Order of Finite Group|the order of $a$ is finite]].
+Let the [[Definition:Order of Group Element|order of $a$]] be $m$.
+From [[Inverse Element is Power of Order Less 1]] we have that $a^{m-1} = a^{-1}$.
+As $a^{m-1} \in H$ (from above) the result follows.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+=== Sufficient Condition ===
+Let $H$ be a [[Definition:Subgroup|subgroup]] of $G$.
+Then:
+:$\forall a, b \in H: a \circ b \in H$
+by definition of [[Definition:Subgroup|subgroup]].
+{{qed|lemma}}
+=== Necessary Condition ===
+Let $H$ be a [[Definition:Non-Empty Set|non-empty]] [[Definition:Finite Set|finite]] [[Definition:Subset|subset]] of $G$ such that:
+:$\forall a, b \in H: a \circ b \in H$
+Let $x \in H$.
+We have [[Definition:By Hypothesis|by hypothesis]] that $H$ is [[Definition:Closed Algebraic Structure|closed]] under $\circ$.
+Thus all [[Definition:Element|elements]] of $\set {x, x^2, x^3, \ldots}$ are in $H$.
+But $H$ is [[Definition:Finite Set|finite]].
+Therefore it must be the case that:
+:$\exists r, s \in \N: x^r = x^s$
+for $r < s$.
+So we can write:
+{{begin-eqn}}
+{{eqn | l = x^r
+      | r = x^s
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = x^r \circ e
+      | r = x^r \circ x^{s - r}
+      | c = {{Defof|Identity Element}}, [[Powers of Group Elements/Sum of Indices|Powers of Group Elements: Sum of Indices]]
+}}
+{{eqn | n = 1
+      | ll= \leadsto
+      | l = e
+      | r = x^{s - r}
+      | c = [[Cancellation Laws]]
+}}
+{{eqn | ll= \leadsto
+      | l = e
+      | o = \in
+      | r = H
+      | c = as $H$ is [[Definition:Closed Algebraic Structure|closed]] under $\circ$
+}}
+{{end-eqn}}
+Then we have:
+{{begin-eqn}}
+{{eqn | l = e
+      | r = x^{s - r}
+      | c = which is $(1)$
+}}
+{{eqn | ll= \leadsto
+      | l = e
+      | r = x \circ x^{s - r - 1}
+      | c = as $H$ is [[Definition:Closed Algebraic Structure|closed]] under $\circ$
+}}
+{{eqn | ll= \leadsto
+      | l = x^{-1} \circ e
+      | r = x^{-1} \circ x \circ x^{s - r - 1}
+      | c = 
+}}
+{{eqn | n = 2
+      | ll= \leadsto
+      | l = x^{-1}
+      | r = x^{s - r - 1}
+      | c = {{Defof|Inverse Element}}, {{Defof|Identity Element}}
+}}
+{{end-eqn}}
+But we have that:
+{{begin-eqn}}
+{{eqn | l = r
+      | o = <
+      | r = s
+      | c = [[Definition:By Hypothesis|by hypothesis]]
+}}
+{{eqn | ll= \leadsto
+      | l = s - r
+      | o = <
+      | r = 0
+}}
+{{eqn | ll= \leadsto
+      | l = s - r - 1
+      | o = \le
+      | r = 0
+}}
+{{eqn | ll= \leadsto
+      | l = x^{s - r - 1}
+      | o = \in
+      | r = \set {e, x, x^2, x^3, \ldots}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = x^{s - r - 1}
+      | o = \in
+      | r = H
+      | c = as all [[Definition:Element|elements]] of $\set {e, x, x^2, x^3, \ldots}$ are in $H$
+}}
+{{end-eqn}}
+So from $(2)$:
+:$x^{-1} = x^{s - r - 1}$
+it follows that:
+: $x^{-1} \in H$
+and from the [[Two-Step Subgroup Test]] it follows that $H$ is a [[Definition:Subgroup|subgroup]] of $G$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Powers of Element form Subgroup}
+Tags: Subgroups
+
+\begin{theorem}
+Let $\struct {G, \circ}$ be a [[Definition:Group|group]].
+Then:
+:$\forall a \in G: H = \set {a^n: n \in \Z} \le G$
+That is, the [[Definition:Subset|subset]] of $G$ comprising all [[Definition:Element|elements]] possible as powers of $a \in G$ is a [[Definition:Subgroup|subgroup]] of $G$.
+\end{theorem}
+
+\begin{proof}
+Clearly $a \in H$, so $H \ne \O$.
+Let $x, y \in H$.
+{{begin-eqn}}
+{{eqn | o = 
+      | r = x, y \in H
+      | c = 
+}}
+{{eqn | o = \leadsto
+      | r = \exists m, n \in \Z: x = a^m, y = a^n
+      | c = 
+}}
+{{eqn | o = \leadsto
+      | r = x^{-1} y = \paren {a^m}^{-1} a^n
+      | c = 
+}}
+{{eqn | o = \leadsto
+      | r = x^{-1} y = a^{-m} a^n = a^{n-m}
+      | c = 
+}}
+{{eqn | o = \leadsto
+      | r = x^{-1} y \in H
+      | c = 
+}}
+{{end-eqn}}
+Thus by the [[One-Step Subgroup Test]]:
+: $H \le G$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Existence of Unique Subgroup Generated by Subset}
+Tags: Group Theory
+
+\begin{theorem}
+Let $\struct {G, \circ}$ be a [[Definition:Group|group]].
+Let $\O \subset S \subseteq G$.
+Let $\struct {H, \circ}$ be the [[Definition:Generator of Subgroup|subgroup generated by $S$]].
+Then $H = \gen S$ exists and is [[Definition:Unique|unique]].
+Also, $\struct {H, \circ}$ is the [[Definition:Set Intersection|intersection]] of all of the [[Definition:Subgroup|subgroups]] of $G$ which contain the set $S$:
+:$\displaystyle \gen S = \bigcap_i {H_i}: S \subseteq H_i \le G$
+\end{theorem}
+
+\begin{proof}
+=== Existence ===
+First, we prove that such a [[Definition:Subgroup|subgroup]] exists.
+Let $\mathbb S$ be the [[Definition:Set|set]] of all [[Definition:Subgroup|subgroups]] of $G$ which contain $S$.
+$\mathbb S \ne \O$ because [[Group is Subgroup of Itself|$G$ is itself a subgroup]] of $G$, and thus $G \in \mathbb S$.
+Let $H$ be the [[Definition:Set Intersection|intersection]] of all the [[Definition:Element|elements]] of $\mathbb S$.
+By [[Intersection of Subgroups is Subgroup]], $H$ is the largest element of $\mathbb S$ contained in each element of $\mathbb S$.
+Thus $H$ is a [[Definition:Subgroup|subgroup]] of $G$.
+Since $\forall x \in \mathbb S: S \subseteq x$, we see that $S \subseteq H$, so $H \in \mathbb S$.
+=== Smallest ===
+Now to show that $H$ is the smallest such [[Definition:Subgroup|subgroup]].
+If any $K \le G: S \subseteq K$, then $K \in \mathbb S$ and therefore $H \subseteq K$.
+So $H$ is the smallest [[Definition:Subgroup|subgroup]] of $G$ containing $S$.
+=== Uniqueness ===
+Now we show that $H$ is unique.
+Suppose $\exists H_1, H_2 \in \mathbb S$ such that $H_1$ and $H_2$ were two such smallest [[Definition:Subgroup|subgroups]] containing $S$.
+Then, by the definition of "smallest", each would be equal in size.
+If one is not a subset of the other, then their intersection (by definition containing $S$) would be a smaller subgroup and hence neither $H_1$ nor $H_2$ would be the smallest.
+Hence one must be a [[Definition:Subset|subset]] of the other.
+By definition of [[Definition:Set Equality/Definition 2|set equality]], that means they must be the same set.
+So the smallest subgroup, whose existence we have proved above, is unique.
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Homomorphism of Generated Group}
+Tags: Group Homomorphisms
+
+\begin{theorem}
+Let $\left({G, \circ}\right)$ and $\left({H, \circ}\right)$ be [[Definition:Group|groups]].
+Let $\phi: G \to H$ and $\psi: G \to H$ be [[Definition:Group Homomorphism|homomorphisms]]. 
+Let $\left \langle {S} \right \rangle = G$ be the [[Definition:Generator of Group|group generated by $S$]].
+Let $\forall x \in S: \phi \left({x}\right) = \psi \left({x}\right)$
+Then $\phi = \psi$.
+\end{theorem}
+
+\begin{proof}
+Let $H = \left\{{x \in G: \phi \left({x}\right) = \psi \left({x}\right)}\right\}$.
+From [[Elements of Group with Equal Images under Homomorphisms form Subgroup]], $H$ is a [[Definition:Subgroup|subgroup]] of $G$.
+But from the definition of the [[Definition:Generator of Group|group generated by $S$]], the smallest subgroup that contains $S$ is $G$ itself.
+Thus $G = \left\{{x \in G: \phi \left({x}\right) = \psi \left({x}\right)}\right\}$ and so $\phi = \psi$.
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Set of Words Generates Group}
+Tags: Group Theory
+
+\begin{theorem}
+Let $S \subseteq G$ where $G$ is a [[Definition:Group|group]].
+Let $\hat S$ be defined as $S \cup S^{-1}$, where $S^{-1}$ is the [[Definition:Inverse of Subset of Group|set of all the inverses]] of all the elements of $S$.
+Then $\gen S = \map W {\hat S}$, where $\map W {\hat S}$ is the [[Definition:Word (Abstract Algebra)|set of words]] of $\hat S$.
+\end{theorem}
+
+\begin{proof}
+Let $H = \gen S$ where $S \subseteq G$.
+$H$ must certainly include $\hat S$, because any [[Definition:Group|group]] containing $s \in S$ must also contain $s^{-1}$.
+Thus $\hat S \subseteq H$.
+By the [[Definition:Group Axioms|closure axiom]], $H$ must also contain all [[Definition:Product Element|products]] of a [[Definition:Finite Set|finite number]] of [[Definition:Element|elements]] of $\hat S$.
+Thus $\map W {\hat S} \subseteq H$.
+Now we prove that $\map W {\hat S} \le G$.
+By the [[Two-Step Subgroup Test]]:
+Let $x, y \in \map W {\hat S}$.
+	
+As $x$ and $y$ are both [[Definition:Product Element|products]] of a [[Definition:Finite Set|finite number]] of [[Definition:Element|elements]] of $\hat S$, it follows that so is their [[Definition:Product Element|product]] $x y$.
+Thus $x y \in \map W {\hat S}$ and {{GroupAxiom|0}} is satisfied.
+Let $x = s_1 s_2 \ldots s_n \in \map W {\hat S}$.
+Then $x^{-1} = s_n^{-1} \ldots s_2^{-1} s_1^{-1} \in \map W {\hat S}$.
+Thus the conditions of the [[Two-Step Subgroup Test]] are fulfilled, and $\map W {\hat S} \le G$.
+Thus $\map W {\hat S}$ is the [[Definition:Subgroup|subgroup]] of $G$ [[Definition:Generator of Group|generated by $S$]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Subset Product is Subset of Generator}
+Tags: Group Theory, Subset Products
+
+\begin{theorem}
+Let $\struct {G, \circ}$ be a [[Definition:Group|group]].
+Let $X, Y \subseteq \struct {G, \circ}$.
+Then $X \circ Y \subseteq \gen {X, Y}$ where:
+:$X \circ Y$ is the [[Definition:Subset Product|Subset Product]] of $X$ and $Y$ in $G$.
+:$\gen {X, Y}$ is the [[Definition:Generator of Subgroup|subgroup of $G$ generated by $X$ and $Y$]].
+\end{theorem}
+
+\begin{proof}
+It is clear from [[Set of Words Generates Group]] that $\map W {\hat X \cup \hat Y} = \gen {X, Y}$.
+It is equally clear that $X \circ Y \subseteq \map W {\hat X \cup \hat Y}$.
+{{qed}}
+[[Category:Group Theory]]
+[[Category:Subset Products]]
+3cekksmrfte9xtmhqnc5h4td9k4wcc6
+\end{proof}<|endoftext|>
+\section{Order of Subset Product with Singleton}
+Tags: Subset Products, Singletons, Order of Subset Product with Singleton
+
+\begin{theorem}
+Let $\struct {G, \circ}$ be a [[Definition:Group|group]].
+Let $X, Y \subseteq \struct {G, \circ}$ such that $X$ is a [[Definition:Singleton|singleton]]:
+:$X = \set x$
+Then:
+:$\order {X \circ Y} = \order Y = \order {Y \circ X}$
+where $\order S$ is defined as the [[Definition:Order of Structure|order of $S$]].
+\end{theorem}
+
+\begin{proof}
+From [[Regular Representations of Subset Product]], we have that the [[Definition:Left Regular Representation|left regular representation]] of $\struct {S, \circ}$ with respect to $a$ is:
+:$\lambda_x \sqbrk S = \set x \circ S = x \circ S$
+The result then follows directly from [[Regular Representation of Invertible Element is Permutation]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Product of Subset with Intersection}
+Tags: Subset Products, Set Intersection, Product of Subset with Intersection
+
+\begin{theorem}
+Let $\struct {G, \circ}$ be an [[Definition:Algebraic Structure|algebraic structure]].
+Let $X, Y, Z \subseteq G$.
+Then:
+:$X \circ \paren {Y \cap Z} \subseteq \paren {X \circ Y} \cap \paren {X \circ Z}$
+:$\paren {Y \cap Z} \circ X \subseteq \paren {Y \circ X} \cap \paren {Z \circ X}$
+where $X \circ Y$ denotes the [[Definition:Subset Product|subset product]] of $X$ and $Y$.
+\end{theorem}
+
+\begin{proof}
+Consider the [[Definition:Relation|relation]] $\RR \subseteq G \times G$ defined as:
+:$\forall g, h \in G: \tuple {g, h} \in \RR \iff \exists g \in X$
+Then:
+:$\forall S \subseteq G: X \circ S = \RR \sqbrk S$
+Then:
+{{begin-eqn}}
+{{eqn | l = X \circ \paren {Y \cap Z}
+      | r = \RR \sqbrk {Y \cap Z}
+      | c = 
+}}
+{{eqn | o = \subseteq
+      | r = \RR \sqbrk Y \cap \RR \sqbrk Z
+      | c = [[Image of Intersection under Relation]]
+}}
+{{eqn | r = \paren {X \circ Y} \cap \paren {X \circ Z}
+      | c = 
+}}
+{{end-eqn}}
+Next, consider the [[Definition:Relation|relation]] $\RR \subseteq G \times G$ defined as:
+:$\forall g, h \in G: \tuple {g, h} \in \RR \iff \exists h \in X$
+Then:
+:$\forall S \subseteq G: S \circ X = \RR \sqbrk S$
+Then:
+{{begin-eqn}}
+{{eqn | l = \paren {Y \cap Z} \circ X
+      | r = \RR \sqbrk {Y \cap Z}
+      | c = 
+}}
+{{eqn | o = \subseteq
+      | r = \RR \sqbrk Y \cap \RR \sqbrk Z
+      | c = [[Image of Intersection under Relation]]
+}}
+{{eqn | r = \paren {Y \circ X} \cap \paren {Z \circ X}
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let $x \in X, t \in Y \cap Z$.
+By the definition of [[Definition:Set Intersection|intersection]], $t \in Y$ and $t \in Z$.
+Consider $X \circ \paren {Y \cap Z}$.
+We have $x \circ t \in X \circ \paren {Y \cap Z}$ by definition of [[Definition:Subset Product|subset product]].
+As $t \in Y$ and $t \in Z$, we also have $x \circ t \in X \circ Y$ and $x \circ t \in X \circ Z$.
+The result follows.
+Similarly, consider $\paren {Y \cap Z} \circ X$.
+Then we have $t \circ x \in \paren {Y \cap Z} \circ X$ by definition of [[Definition:Subset Product|subset product]].
+As $t \in Y$ and $t \in Z$, we also have $t \circ x \in Y \circ X$ and $t \circ x \in Z \circ X$.
+Again, the result follows.
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Order of Subgroup Product}
+Tags: Subgroups, Subset Products, Order of Subgroup Product
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]].
+Let $H$ and $K$ be [[Definition:Subgroup|subgroups]] of $G$.
+Then:
+:$\order {H K} = \dfrac {\order H \order K} {\order {H \cap K} }$
+where:
+:$H K$ denotes [[Definition:Subset Product|subset product]]
+:$\order H$ denotes the [[Definition:Order of Structure|order of $H$]].
+\end{theorem}
+
+\begin{proof}
+From [[Intersection of Subgroups is Subgroup]], we have that $H \cap K \le H$.
+Let the number of [[Definition:Left Coset|left cosets]] of $H \cap K$ in $H$ be $r$.
+Then the [[Definition:Left Coset Space|left coset space]] of $H \cap K$ in $H$ is:
+: $\set {x_1 \paren {H \cap K}, x_2 \paren {H \cap K}, \ldots, x_r \paren {H \cap K} }$
+So each element of $H$ is in $x_i \paren {H \cap K}$ for some $1 \le i \le r$.
+Also, if $i \ne j$, we have:
+:$x_j^{-1} x_i \notin H \cap K$
+Let $h k \in H K$.
+We can write $h = x_i g$ for some $1 \le i \le r$ and some $g \in K$.
+Thus:
+:$h k = x_i \paren {g k}$
+Since $g, k \in K$, this shows $h k \in x_i K$.
+{{AimForCont}} the [[Definition:Left Coset|left cosets]] $x_i K$ are not all [[Definition:Disjoint Sets|disjoint]].
+Then by [[Left Coset Space forms Partition]]:
+:$x_i K = x_j K$ for some $i, j$.
+So by [[Left Congruence Class Modulo Subgroup is Left Coset]]:
+:$x_j^{-1} x_i \in K$
+Since $x_i, x_j \in H$, we have:
+:$x_j^{-1} x_i \in H \cap K$
+which [[Definition:Contradiction|contradicts]] the definition.
+Therefore the [[Definition:Left Coset|left cosets]] $x_i K$ are [[Definition:Disjoint Sets|disjoint]] for $1 \le i \le r$.
+This leads us to:
+:$\dfrac {\order H} {\order {H \cap K} } = \dfrac {\order {H K} } {\order K} = r$
+whence the result.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+=== [[Order of Subgroup Product/Lemma|Lemma]] ===
+{{:Order of Subgroup Product/Lemma}}{{qed|lemma}}
+We have that $H K$ is the [[Definition:Set Union|union]] of all [[Definition:Left Coset|left cosets]] $h K$ with $h \in H$:
+:$\displaystyle H K = \bigcup_{h \mathop \in H} h K$
+From [[Left Coset Space forms Partition]], unequal $h K$ are [[Definition:Disjoint Sets|disjoint]].
+From [[Cosets are Equivalent]], each $h K$ contains $\order K$ [[Definition:Element|elements]].
+From the [[Order of Subgroup Product/Lemma|Lemma]], the number of different such [[Definition:Left Coset|left cosets]] is:
+:$\index H {H \cap K}$
+where $\index H {H \cap K}$ denotes the [[Definition:Index of Subgroup|index]] of $H \cap K$ in $H$.
+From [[Lagrange's Theorem (Group Theory)|Lagrange's Theorem]]:
+$\index H {H \cap K} = \dfrac {\order H} {\order {H \cap K} }$
+Hence:
+:$\order {H K} = \dfrac {\order H \order K} {\order {H \cap K} }$
+{{qed}}
+\end{proof}
+
+\begin{proof}
+The number of [[Definition:Product Element|product elements]] $h k$ that can be formed where $h \in H$ and $k \in K$ is $\order H \order K$, although this may include duplication:
+:$h_1 k_1 = h_2 k_2$ may be possible for $h_1, h_2 \in H, k_1, k_2 \in K$.
+So, consider the [[Definition:Cartesian Product|Cartesian product]] $H \times K$.
+From [[Cardinality of Cartesian Product]]:
+:$\size {H \times K} = \order H \times \order K$
+Let us define a [[Definition:Relation|relation]] $\sim$ on $H \times K$ as:
+:$\tuple {h_1, k_1} \sim \tuple {h_2, k_2} \iff h_1 k_1 = h_2 k_2$
+As $\sim$ is based on the [[Definition:Equals|equality relation]] it is seen that $\sim$ is an [[Definition:Equivalence Relation|equivalence relation]]:
+:[[Definition:Reflexive Relation|Reflexivity]]: $h_1 k_1 = h_1 k_1$
+:[[Definition:Symmetric Relation|Symmetry]]: $h_1 k_1 = h_2 k_2 \implies h_2 k_2 = h_1 k_1$
+:[[Definition:Transitive Relation|Transitivity]]: $h_1 k_1 = h_2 k_2,  h_2 k_2 = h_3 k_3 \implies h_1 k_1 = h_3 k_3$
+Each [[Definition:Equivalence Class|equivalence class]] of $\sim$ corresponds to a particular [[Definition:Element|element]] of $H K$.
+Hence $\size {H K}$ is the number of [[Definition:Equivalence Class|equivalence classes]] of $\sim$.
+It remains to be shown that each of these [[Definition:Equivalence Class|equivalence classes]] contains exactly $\order {H \cap K}$ [[Definition:Element|elements]].
+Let $E$ be the [[Definition:Equivalence Class|equivalence class]] of $\tuple {h k}$.
+We aim to prove that:
+:$(1): \quad E = \set {\tuple {h x^{-1}, x k}: x \in H \cap K}$
+Let $x \in H \cap K$.
+Then:
+:$a x^{-1} \in H$
+and:
+:$x k \in K$
+so:
+:$\tuple {h x^{-1}, x k} \in H \times K$
+[[Definition:Converse|Conversely]]:
+:$\tuple {h, k} \sim \tuple {h_1, k_1} \implies h k = h_1 k_1$
+and so:
+:$x = h_1^{-1} h = k_1 k^{-1} \in A \cap B$
+Thus:
+:$h_1 = h x^{-1}$
+and:
+:$k_1 = x k$
+and $(1)$ is seen to hold.
+Finally:
+:$\tuple {h x^{-1}, x k} = \tuple {h y^{-1}, y k} \implies x = y$
+Thus $E$ has exactly $\order {H \cap K}$ [[Definition:Element|elements]] and the proof is complete.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Index of Intersection of Subgroups}
+Tags: Subgroups, Index of Subgroups, Set Intersection, Index of Intersection of Subgroups
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]].
+Let $H, K$ be [[Definition:Subgroup|subgroups]] of [[Definition:Finite Index|finite index]] of $G$.
+Then:
+:$\index G {H \cap K} \le \index G H \index G K$
+where $\index G H$ denotes the [[Definition:Index of Subgroup|index of $H$ in $G$]].
+Note that here the symbol $\le$ is being used with its meaning '''less than or equal to'''.
+Equality holds {{iff}} $H K = \set {h k: h \in H, k \in K} = G$.
+\end{theorem}
+
+\begin{proof}
+Note that $H \cap K$ is a [[Definition:Subgroup|subgroup]] of $H$.
+From [[Tower Law for Subgroups]], we have:
+:$\index G {H \cap K} = \index G H \index H {H \cap K}$
+From [[Index in Subgroup]], also:
+:$\index H {H \cap K} \le \index G K$
+Combining these results yields the desired inequality.
+Again from [[Index in Subgroup]], it follows that:
+:$\index H {H \cap K} = \index G K$
+{{iff}} $H K = G$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Intersection of Subgroups of Prime Order}
+Tags: Prime Groups, Subgroups
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
+Let $H$ and $K$ be [[Definition:Subset|subsets]] of $G$ such that:
+:$\order H = \order K = p$
+:$H \ne K$
+:$p$ is [[Definition:Prime Number|prime]].
+Then:
+: $H \cap K = \set e$
+That is, the [[Definition:Set Intersection|intersection]] of two unequal [[Definition:Subgroup|subgroups]] of a [[Definition:Group|group]], both of whose [[Definition:Order of Structure|order]] is the same [[Definition:Prime Number|prime]], consists solely of the [[Definition:Identity Element|identity]].
+\end{theorem}
+
+\begin{proof}
+From [[Intersection of Subgroups is Subgroup]]:
+:$H \cap K \le G$
+and:
+:$H \cap K \le H$
+where $\le$ denotes [[Definition:Subgroup|subgrouphood]].
+So:
+{{begin-eqn}}
+{{eqn | l = H \cap K
+      | o = \le
+      | r = H
+      | c = [[Intersection of Subgroups is Subgroup]]
+}}
+{{eqn | ll= \leadsto
+      | l = \order {H \cap K}
+      | o = \divides
+      | r = \order H
+      | c = [[Lagrange's Theorem (Group Theory)|Lagrange's Theorem]]
+}}
+{{eqn | ll= \leadsto
+      | l = \order {H \cap K}
+      | o = \divides
+      | r = p
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \order {H \cap K}
+      | r = 1 \text{ or } p
+      | c = $p$ is [[Definition:Prime Number|prime]]
+}}
+{{end-eqn}}
+Because $H \ne K$ and $\order H = \order k$, it follows that $H \nsubseteq K$.
+So:
+{{begin-eqn}}
+{{eqn | l = H \cap K
+      | o = \ne
+      | r = H
+      | c = [[Intersection with Subset is Subset]]
+}}
+{{eqn | ll= \leadsto
+      | l = H \cap K
+      | o = \subset
+      | r = H
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \order {H \cap K}
+      | o = <
+      | r = \order H = p
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \order {H \cap K}
+      | r = 1
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = H \cap K
+      | r = \set e
+      | c = {{Defof|Trivial Group}}
+}}
+{{end-eqn}}
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Tower Law for Subgroups}
+Tags: Subgroups, Tower Law for Subgroups, Named Theorems, Index of Subgroups
+
+\begin{theorem}
+Let $\struct {G, \circ}$ be a [[Definition:Group|group]].
+Let $H$ be a [[Definition:Subgroup|subgroup]] of $G$ with [[Definition:Finite Index|finite index]].
+Let $K$ be a [[Definition:Subgroup|subgroup]] of $H$.
+Then:
+:$\index G K = \index G H \index H K$
+where $\index G H$ denotes the [[Definition:Index of Subgroup|index of $H$ in $G$]].
+\end{theorem}
+
+\begin{proof}
+Let $p = \index G H$, $q = \index H K$.
+By hypothesis these numbers are [[Definition:Finite|finite]].
+Therefore, there exist $g_1, \ldots, g_p \in G$ such that $G$ is a [[Definition:Disjoint Union (Set Theory)|disjoint union]]:
+$\displaystyle G = \bigsqcup_{i \mathop = 1}^p g_i H$
+Similarly, there exist $h_1,\ldots,h_q \in H$ such that $H$ is a [[Definition:Disjoint Union (Set Theory)|disjoint union]]:
+$\displaystyle H = \bigsqcup_{j \mathop = 1}^q h_j K$
+Thus:
+{{begin-eqn}}
+{{eqn | l = G
+      | r = \bigsqcup_{i \mathop = 1}^p g_i H
+}}
+{{eqn | r = \bigsqcup_{i \mathop = 1}^p g_i \bigsqcup_{j \mathop = 1}^q h_j K
+}}
+{{eqn | r = \bigsqcup_{i \mathop = 1}^p \bigsqcup_{j \mathop = 1}^q g_i \paren {h_j K}
+      | c = [[Product of Subset with Union]]
+}}
+{{eqn | r = \bigsqcup_{i \mathop = 1}^p \bigsqcup_{j \mathop = 1}^q \paren {g_i h_j} K
+      | c = [[Subset Product within Semigroup is Associative/Corollary|Subset Product within Semigroup is Associative: Corollary]]
+}}
+{{end-eqn}}
+This expression for $G$ is the [[Definition:Disjoint Union (Set Theory)|disjoint union]] of $p q$ [[Definition:Coset|cosets]].
+Therefore the number of elements of the [[Definition:Coset Space|coset space]] is:
+:$\index G K = p q = \index G H \index H K$
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Assume $G$ is [[Definition:Finite Group|finite]].
+Then:
+{{begin-eqn}}
+{{eqn | l = \index G H
+      | r = \frac {\order G} {\order H}
+      | c = [[Lagrange's Theorem (Group Theory)|Lagrange's Theorem]]
+}}
+{{eqn | l = \index G K
+      | r = \frac {\order G} {\order K}
+      | c = [[Lagrange's Theorem (Group Theory)|Lagrange's Theorem]]
+}}
+{{eqn | ll= \leadsto
+      | l = \index G K
+      | r = \frac {\order H} {\order K} \times \index G H
+      | c = 
+}}
+{{end-eqn}}
+Since $K \le H$, from [[Lagrange's Theorem (Group Theory)|Lagrange's Theorem]] we have that $\dfrac {\order H} {\order K} = \index H K$.
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Morphism from Integers to Group}
+Tags: Group Isomorphisms, Group Epimorphisms, Integers, Ideal Theory
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
+Let $g \in G$.
+Let $\phi: \Z \to G$ be the [[Definition:Mapping|mapping]] defined as:
+:$\forall n \in \Z: \map \phi n = g^n$.
+Then:
+: If $g$ has [[Definition:Infinite Order Element|infinite order]], then $\phi$ is a [[Definition:Group Isomorphism|group isomorphism]] from $\struct {\Z, +}$ to $\gen g$.
+: If $g$ has [[Definition:Finite Order Element|finite order]] such that $\order g = m$, then $\phi$ is a [[Definition:Group Epimorphism|group epimorphism]] from $\struct {\Z, +}$ to $\gen g$ whose [[Definition:Kernel of Group Homomorphism|kernel]] is the [[Definition:Principal Ideal of Ring|principal ideal]] $\paren m$.
+:Thus $\gen g$ is [[Definition:Group Isomorphism|isomorphic]] to $\struct {\Z, +}$, and $m$ is the smallest [[Definition:Strictly Positive Integer|(strictly) positive integer]] such that $g^m = e$.
+\end{theorem}
+
+\begin{proof}
+By [[Epimorphism from Integers to Cyclic Group]], $\phi$ is an [[Definition:Group Epimorphism|epimorphism]] from $\struct {\Z, +}$ onto $\gen g$.
+By [[Kernel of Group Homomorphism is Subgroup]], the [[Definition:Kernel of Group Homomorphism|kernel]] $K$ of $G$ is a [[Definition: Subgroup|subgroup]] of $\struct {\Z, +}$.
+Therefore by [[Subgroup of Integers is Ideal]] and [[Principal Ideals of Integers]], $\exists m \in \N_{>0}: K = \paren m$.
+Thus $\gen g \rangle \cong \struct {\Z, +}$.
+By [[Canonical Epimorphism from Integers by Principal Ideal]]:
+:$\forall m \in \N_{>0}: \order {\Z_m} = m$
+So, if $\gen g$ is [[Definition:Finite|finite]], and if $\gen g \cong \struct {\Z, +}$, then $m = \order g$.
+Furthermore, $m$ is the smallest [[Definition:Strictly Positive Integer|(strictly) positive integer]] such that $g^m = e$, since $m$ is the smallest (strictly) positive integer in $\paren m$ from [[Principal Ideals of Integers]].
+If $\gen g$ is [[Definition:Infinite|infinite]], then $m = 0$ and so $\phi$ is a [[Definition:Group Isomorphism|(group) isomorphism]] from $\struct {\Z, +}$ onto $\gen g$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Identity is Only Group Element of Order 1}
+Tags: Identity Elements, Order of Group Elements
+
+\begin{theorem}
+In every [[Definition:Group|group]], the [[Definition:Identity Element|identity]], and only the [[Definition:Identity Element|identity]], has [[Definition:Order of Group Element|order]] $1$.
+\end{theorem}
+
+\begin{proof}
+Let $G$ be a [[Definition:Group|group]] with [[Definition:Identity Element|identity]] $e$.
+Then:
+:$e^1 = e$
+and:
+:$\forall a \in G: a \ne e: a^1 = a \ne e$.
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Group Element is Self-Inverse iff Order 2}
+Tags: Order of Group Elements
+
+\begin{theorem}
+Let $\struct {S, \circ}$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
+An [[Definition:Element|element]] $x \in \struct {S, \circ}$ is [[Definition:Self-Inverse Element|self-inverse]] {{iff}}:
+:$\order x = 2$
+\end{theorem}
+
+\begin{proof}
+Let $x \in G: x \ne e$.
+{{begin-eqn}}
+{{eqn | l = \order x
+      | r = 2
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = x \circ x
+      | r = e
+      | c = {{Defof|Order of Group Element}}
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = x
+      | r = x^{-1}
+      | c = [[Equivalence of Definitions of Self-Inverse]]
+}}
+{{end-eqn}}
+{{qed}}
+{{expand|Free generalisation to monoids}}
+\end{proof}<|endoftext|>
+\section{Powers of Infinite Order Element}
+Tags: Order of Group Elements
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
+Let $a \in G$ have [[Definition:Infinite Order Element|infinite order]] in $G$.
+Then:
+:$\forall m, n \in \Z: m \ne n \implies a^m \ne a^n$
+\end{theorem}
+
+\begin{proof}
+Let $m, n \in \Z$.
+Then:
+{{begin-eqn}}
+{{eqn | l = a^m
+      | r = a^n
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = a^m \paren {a^n}^{-1}
+      | r = e
+      | c = 
+}}
+{{eqn | lo= \land
+      | l = a^n \paren {a^m}^{-1}
+      | r = e
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = a^{m - n}
+      | r = a^{n - m} = e
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = a^{\size {m - n} }
+      | r = e
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \size {m - n}
+      | r = 0
+      | c = {{Defof|Infinite Order Element}}: there exists no $x \in \Z_{>0}$ such that $a^x = e$
+}}
+{{eqn | ll= \leadsto
+      | l = m
+      | r = n
+      | c = 
+}}
+{{end-eqn}}
+The result follows from [[Rule of Transposition|transposition]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Element of Finite Group is of Finite Order}
+Tags: Finite Groups, Order of Group Elements, Element of Finite Group is of Finite Order
+
+\begin{theorem}
+In any [[Definition:Finite Group|finite group]], each [[Definition:Element|element]] has [[Definition:Finite Order Element|finite order]].
+\end{theorem}
+
+\begin{proof}
+Follows as a direct [[Definition:Corollary|corollary]] to the result [[Powers of Infinite Order Element]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let $G$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
+From [[Element has Idempotent Power in Finite Semigroup]], for every [[Definition:Element|element]] in a ''finite'' [[Definition:Semigroup|semigroup]], there is a [[Definition:Power of Group Element|power]] of that [[Definition:Element|element]] which is [[Definition:Idempotent Element|idempotent]].
+As $G$, being a [[Definition:Group|group]], is also a [[Definition:Semigroup|semigroup]], the same applies to $G$.
+That is:
+:$\forall x \in G: \exists n \in \N_{>0}: x^n \circ x^n = x^n$
+From [[Identity is only Idempotent Element in Group]], it follows that:
+:$x^n \circ x^n = x^n \implies x^n = e$
+So $x$ has [[Definition:Finite Order Element|finite order]].
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Inverse Element is Power of Order Less 1}
+Tags: Inverse Elements, Order of Group Elements
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
+Let $g \in G$ be of [[Definition:Finite Order Element|finite order]].
+Then:
+: $\order g = n \implies g^{n - 1} = g^{-1}$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \order g
+      | r = n
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = g^n
+      | r = e
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = g^n g^{-1}
+      | r = e g^{-1}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = g^{n - 1}
+      | r = g^{-1}
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Equal Powers of Finite Order Element}
+Tags: Order of Group Elements
+
+\begin{theorem}
+:$g^r = g^s \iff k \divides \paren {r - s}$
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+Suppose that $k \divides \paren {r - s}$.
+From the definition of [[Definition:Divisor of Integer|divisor]]:
+:$k \divides \left({r - s}\right) \implies \exists t \in \Z: r - s = k t$
+So:
+:$g^{r - s} = g^{k t}$
+Thus:
+{{begin-eqn}}
+{{eqn | l = g^r
+      | r = g^{s + k t}
+      | c = 
+}}
+{{eqn | r = g^s g^{k t}
+      | c = 
+}}
+{{eqn | r = g^s \paren {g^k}^t
+      | c = 
+}}
+{{eqn | r = g^s \paren e^t
+      | c = 
+}}
+{{eqn | r = g^s
+      | c = 
+}}
+{{end-eqn}}
+{{qed|lemma}}
+=== Sufficient Condition ===
+Let $g^r = g^s$.
+Then:
+: $g^{r - s} = g^r g^{-s} = g^s g^{-s} = e$
+ 
+By the [[Division Theorem]]:
+: $r - s = q k + t$
+for some $q \in \Z, 0 \le t < k$.
+Thus:
+:$e = g^{r - s} = g^{k q + t} = \paren {g^k}^q g^t = e^q g^t = g^t$
+So by the definition of $k$:
+:$\paren {t < k} \land \paren {e = g^t} \implies t = 0$
+So:
+:$r - s = q k + 0 = q k \implies k \divides \paren {r - s}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Order of Element Divides Order of Finite Group}
+Tags: Finite Groups, Order of Group Elements
+
+\begin{theorem}
+In a [[Definition:Finite Group|finite group]], the [[Definition:Order of Group Element|order of a group element]] [[Definition:Divisor of Integer|divides]] the [[Definition:Order of Group|order of its group]]:
+:$\forall x \in G: \order x \divides \order G$
+\end{theorem}
+
+\begin{proof}
+Let $G$ be a [[Definition:Group|group]].
+Let $x \in G$.
+By definition, the [[Definition:Order of Group Element/Definition 2|order of $x$]] is the [[Definition:Order of Group|order]] of the [[Definition:Generated Subgroup|subgroup generated by $x$]].
+Therefore, by [[Lagrange's Theorem (Group Theory)|Lagrange's Theorem]], $\order x$ is a [[Definition:Divisor of Integer|divisor]] of $\order G$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Recurrence Relation for Number of Derangements on Finite Set}
+Tags: Counting Arguments, Number of Derangements on Finite Set
+
+\begin{theorem}
+The number of [[Definition:Derangement|derangements]] $D_n$ on a [[Definition:Finite Set|finite set]] $S$ of [[Definition:Cardinality|cardinality]] $n$ is:
+:$D_n = \paren {n - 1} \paren {D_{n - 1} + D_{n - 2} }$
+where $D_1 = 0$, $D_2 = 1$.
+\end{theorem}
+
+\begin{proof}
+Let $\card S = 1$ such that $S = \set s$, say.
+Then $\map f s = s$ is the only [[Definition:Permutation|permutation]] on $S$.
+This by definition is not a [[Definition:Derangement|derangement]].
+Thus:
+:$D_1 = 0$
+Let $\card S = 2$.
+Then $S = \set {s, t}$, say.
+There are two [[Definition:Permutation|permutations]] on $S$:
+:$f = \set {\tuple {s, s}, \tuple {t, t} }$
+and:
+:$g = \set {\tuple {s, t}, \tuple {t, s} }$
+and only the latter is a [[Definition:Derangement|derangement]].
+So:
+:$D_2 = 1$
+Let $\card S > 2$.
+Let $f: S \to S$ be a [[Definition:Derangement|derangement]].
+We aim to count the total number of such $f$. 
+{{WLOG}}, we can take:
+:$S = \set {1, 2, \ldots, n + 1}$
+Now, consider an arbitrary $s \in S$ such that $s \ne 1$.
+Let $\map f s = 1$.
+By the [[Sum Rule for Counting]], the total number of $f$ will be: 
+:$\paren {\text {the number of $f$ where } \map f 1 \ne s} + \paren {\text {the number of $f$ where} \map f 1 = s}$
+=== Case 1  ===
+:$\map f 1 \ne s$
+Take:
+:$T_1 = \set {1, 2, \dotsc, s - 1, s + 1, \dotsc, n + 1} = S \setminus \set s$
+Define the [[Definition:Derangement|derangement]] $g_1: T_1 \to T_1$ by:
+:$\forall t \in T_1: \map {g_1} t = \map f t$
+Then $g_1$ is a [[Definition:Derangement|derangement]] on a set of [[Definition:Cardinality|cardinality]] $n$.
+Thus there are $D_n$ such $g_1$. 
+Note that:
+:$f = g_1 \cup \map f s$
+We have that $s$ can be chosen in $n$ ways.
+By the [[Product Rule for Counting]] there are in total $n D_n$ such $f$ where $\map f s \ne 1$.
+=== Case 2 ===
+:$\map f 1 = s$
+Take:
+:$T_2 = \set {2, 3, \dotsc, s - 1, s + 1, \dotsc, n + 1} = S \setminus \set {1, s}$
+Define the [[Definition:Derangement|derangement]] $g_2: T_2 \to T_2$ by:
+:$\forall t \in T_2: \map {g_2} t = \map f t$
+Then $g_2$ is a [[Definition:Derangement|derangement]] on a [[Definition:Set|set]] of [[Definition:Cardinality|cardinality]] $n - 1$.
+So there are $D_{n - 1}$ such $g_2$. 
+Note that:
+:$f = g_2 \cup \set s \cup \map f s$
+We have that $s$ can be chosen in $n$ ways
+By the [[Product Rule for Counting]] there are in total $n D_{n - 1}$ such $f$ where $\map f s = 1$.
+Summing the results from both cases, we get the total number of [[Definition:Derangement|derangements]] $f$ on a [[Definition:Set|set]] of [[Definition:Cardinality|cardinality]] $n + 1$ is:
+:$D_{n + 1} = n D_n + n D_{n - 1} = n \paren {D_n + D_{n - 1} }$
+as was to be proved.
+{{qed}}
+[[Category:Counting Arguments]]
+[[Category:Number of Derangements on Finite Set]]
+r2mrcuj4x7ja1j53ganx2y98kpw4oic
+\end{proof}<|endoftext|>
+\section{Element to Power of Group Order is Identity}
+Tags: Group Theory
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$ and whose [[Definition:Order of Structure|order]] is $n$.
+Then:
+:$\forall g \in G: g^n = e$
+\end{theorem}
+
+\begin{proof}
+Let $G$ be a [[Definition:Group|group]] such that $\order G = n$.
+Let $g \in G$ and let $\order G = k$.
+From [[Order of Element Divides Order of Finite Group]]:
+:$k \divides n$
+So:
+:$\exists m \in \Z_{>0}: k m = n$
+Thus:
+{{begin-eqn}}
+{{eqn | l = g^n
+      | r = \paren {g^k}^m
+      | c = [[Powers of Group Elements/Product of Indices|Powers of Group Elements: Product of Indices]]
+}}
+{{eqn | r = e^m
+      | c = {{Defof|Order of Group Element}}: $g^k = e$
+}}
+{{eqn | r = e
+      | c = [[Power of Identity is Identity]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Boolean Group is Abelian}
+Tags: Abelian Groups, Boolean Groups, Boolean Group is Abelian
+
+\begin{theorem}
+Let $G$ be a [[Definition:Boolean Group|Boolean group]].
+Then $G$ is [[Definition:Abelian Group|abelian]].
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Boolean Group|Boolean group]], all [[Definition:Element|elements]] of $G$, other than the [[Definition:Identity Element|identity]], have [[Definition:Order of Group Element|order]] $2$.
+By [[Group Element is Self-Inverse iff Order 2]] and [[Identity is Self-Inverse]], all elements of $G$ are [[Definition:Self-Inverse Element|self-inverse]].
+The result follows directly from [[All Elements Self-Inverse then Abelian]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let $ a, b \in G$.
+By definition of [[Definition:Boolean Group|Boolean group]]:
+:$\forall x \in G: x^2 = e$
+where $e$ is the [[Definition:Identity Element|identity]] of $G$.
+Then:
+{{begin-eqn}}
+{{eqn | l = a b
+      | r = a e b 
+      | c = [[Definition:Group Axioms|Group Axiom $G2$]]: Properties of [[Definition:Identity Element|Identity]]
+}}
+{{eqn | r = a \left({a b}\right)^2 b
+      | c = as $\forall x \in G: x^2 = e$
+}}
+{{eqn | r = a \left({a b}\right) \left({a b}\right) b
+      | c = 
+}}
+{{eqn | r = \left({a a}\right) \left({b a}\right) \left({b b}\right)
+      | c = [[Definition:Group Axioms|Group Axiom $G1$]]: [[Definition:Associative|Associativity]]
+}}
+{{eqn | r = a^2 \left({b a}\right) b^2
+      | c = 
+}}
+{{eqn | r = e \left({b a}\right) e
+      | c = as $\forall x \in G: x^2 = e$
+}}
+{{eqn | r = b a
+      | c = 
+}}
+{{end-eqn}}
+Thus $a b = b a$ and therefore $G$ is [[Definition:Abelian Group|abelian]]. 
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Order of Group Element equals Order of Inverse}
+Tags: Order of Group Elements, Inverse Elements
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
+Then:
+: $\forall x \in G: \order x = \order {x^{-1} }$
+where $\order x$ denotes the [[Definition:Order of Group Element|order]] of $x$.
+\end{theorem}
+
+\begin{proof}
+By [[Powers of Group Elements/Negative Index|Powers of Group Elements: Negative Index]]:
+: $\paren {x^k}^{-1} = x^{-k} = \paren {x^{-1} }^k$
+Hence:
+{{begin-eqn}}
+{{eqn | l = x^k
+      | r = e
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \paren {x^{-1} }^k
+      | r = e^{-1}
+      | c = 
+}}
+{{eqn | r = e
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \order {x^{-1} }
+      | o = \le
+      | r = \order x
+      | c = {{Defof|Order of Group Element}}
+}}
+{{end-eqn}}
+Similarly:
+{{begin-eqn}}
+{{eqn | l = \paren {x^{-1} }^k
+      | r = e
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \paren {\paren {x^{-1} }^{-1} }^k
+      | r = e^{-1}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = x^k
+      | r = e^{-1}
+      | c = 
+}}
+{{eqn | r = e
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \order x
+      | o = \le
+      | r = \order {x^{-1} }
+      | c = {{Defof|Order of Group Element}}
+}}
+{{end-eqn}}
+A similar argument shows that if $x$ is of [[Definition:Infinite Order Element|infinite order]], then so must $x^{-1}$ be.
+Hence the result.
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Even Order Group has Order 2 Element}
+Tags: Finite Groups, Order of Group Elements, Even Order Group has Order 2 Element
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
+Let $G$ be of even [[Definition:Order of Structure|order]].
+Then:
+:$\exists x \in G: \order x = 2$
+That is:
+:$\exists x \in G: x \ne e: x^2 = e$
+\end{theorem}
+
+\begin{proof}
+In any [[Definition:Group|group]] $G$, the [[Identity is Self-Inverse|identity element $e$ is self-inverse]] with [[Identity is Only Group Element of Order 1]], and is the only such.
+That leaves an [[Definition:Odd Integer|odd number]] of [[Definition:Element|elements]].
+Each element in $x \in G: \order x > 2$ can be paired off with its [[Definition:Inverse Element|inverse]], as $\order {x^{-1} } = \order x > 2$ from [[Order of Group Element equals Order of Inverse]].
+Hence there must be at least one [[Definition:Element|element]] which has not been paired off with any of the others which is therefore [[Definition:Self-Inverse Element|self-inverse]].
+The result follows from [[Group Element is Self-Inverse iff Order 2]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Order of Conjugate Element equals Order of Element}
+Tags: Order of Group Elements, Conjugacy
+
+\begin{theorem}
+Let $\struct {G, \circ}$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
+Then
+:$\forall a, x \in \struct {G, \circ}: \order {x \circ a \circ x^{-1} } = \order a$
+where $\order a$ denotes the [[Definition:Order of Group Element|order of $a$ in $G$]].
+\end{theorem}
+
+\begin{proof}
+Let $\order a = k$.
+Then $a^k = e$, and:
+: $\forall n \in \N_{>0}: n < k \implies a^n \ne e$
+by definition of the [[Definition:Order of Group Element|order of $a$ in $G$]]
+We have:
+{{begin-eqn}}
+{{eqn | l = \paren {x \circ a \circ x^{-1} }^k
+      | r = x \circ a^k \circ x^{-1}
+      | c = [[Power of Conjugate equals Conjugate of Power]]
+}}
+{{eqn | r = x \circ e \circ x^{-1}
+      | c = 
+}}
+{{eqn | r = x \circ x^{-1}
+      | c = 
+}}
+{{eqn | r = e
+      | c = 
+}}
+{{end-eqn}}
+Thus $\order {x \circ a \circ x^{-1} } \le \order a$.
+Now suppose $a^n = y, y \ne e$.
+Then:
+: $x \circ a^n \circ x^{-1} = x \circ y \circ x^{-1}$
+If $x \circ y = e$, then:
+: $x \circ a^n \circ x^{-1} = x^{-1}$
+If $y \circ x^{-1} = e$, then:
+: $x \circ a^n \circ x^{-1} = x$
+So:
+: $a^n \ne e \implies x \circ a^n \circ x^{-1} = \paren {x \circ a \circ x^{-1} }^n \ne e$
+Thus:
+: $\order {x \circ a \circ x^{-1} } \ge \order a$
+and the result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Order of Homomorphic Image of Group Element}
+Tags: Group Homomorphisms
+
+\begin{theorem}
+Let $G$ and $H$ be [[Definition:Group|groups]] whose [[Definition:Identity Element|identities]] are $e_G$ and $e_H$ respectively.
+Let $\phi: G \to H$ be a [[Definition:Group Homomorphism|homomorphism]].
+Let $g \in G$ be of [[Definition:Order of Group Element|finite order]].
+Then:
+:$\forall g \in G: \order {\map \phi g} \divides \order g$
+where $\divides$ denotes [[Definition:Divisor of Integer|divisibility]].
+\end{theorem}
+
+\begin{proof}
+Let $\phi: G \to H$ be a [[Definition:Group Homomorphism|homomorphism]].
+Let $\order g = n, \order {\map \phi g} = m$.
+{{begin-eqn}}
+{{eqn | l = \paren {\map \phi g}^n
+      | r = \map \phi {g^n}
+      | c = [[Homomorphism of Power of Group Element]]
+}}
+{{eqn | r = \map \phi {e_G}
+      | c = 
+}}
+{{eqn | r = e_H
+      | c = [[Homomorphism to Group Preserves Identity]]
+}}
+{{end-eqn}}
+It follows from [[Element to Power of Multiple of Order is Identity]] that $m \divides n$.
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{No Group has Two Order 2 Elements}
+Tags: Order of Group Elements
+
+\begin{theorem}
+A [[Definition:Group|group]] can not contain exactly two [[Definition:Element|elements]] of [[Definition:Order of Group Element|order]] $2$.
+\end{theorem}
+
+\begin{proof}
+Let $\struct {G, \circ}$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
+Suppose:
+: $s, t \in \struct {G, \circ}\: s \ne t, \order s = \order t = 2$
+That is. they are [[Definition:Self-Inverse Element|self-inverse]]:
+:$s^2 = e = t^2$
+The [[Definition:Identity Element|identity]] is of [[Definition:Order of Group Element|order]] $1$.
+Hence $s$ nor $t$ is the [[Definition:Identity Element|identity]]
+Hence, as $s \ne t$, then $s \circ t \in G$ is [[Definition:Distinct Elements|distinct]] from both $s$ and $t$.
+Also $s \circ t \ne e$ because $s \ne t^{-1}$.
+Suppose $s$ and $t$ [[Definition:Commute|commute]].
+Then $\paren {s \circ t}^2 = e$ from [[Self-Inverse Elements Commute iff Product is Self-Inverse]].
+Thus there is a third element (at least) in $G$ which is of [[Definition:Order of Group Element|order]] $2$.
+Now suppose $s$ and $t$ do ''not'' [[Definition:Commute|commute]].
+Then from [[Commutation Property in Group]], $s \circ t \circ s^{-1}$ is another [[Definition:Element|element]] of $G$ [[Definition:Distinct Elements|distinct]] from both $s$ and $t$.
+But from [[Order of Conjugate Element equals Order of Element]]:
+:$\order {s \circ t \circ s^{-1} } = \order t$
+and thus $s \circ t \circ s^{-1}$ is another [[Definition:Element|element]] of [[Definition:Order of Group Element|order]] $2$.
+Thus there is a third [[Definition:Element|element]] (at least) in $G$ which is of [[Definition:Order of Group Element|order]] $2$.
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Odd Order Group Element is Square}
+Tags: Order of Group Elements
+
+\begin{theorem}
+Let $\struct {G, \circ}$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
+Let $x \in G$.
+Then:
+:$\exists y \in G: y^2 = x$
+{{iff}}:
+:the [[Definition:Order of Group Element|order]]  $\order x$ is [[Definition:Odd Integer|odd]]
+\end{theorem}
+
+\begin{proof}
+Let $\order x$ be [[Definition:Odd Integer|odd]].
+Then:
+:$\exists n \in \Z_{> 0}: x^{2 n - 1} = e$
+from the definition of the [[Definition:Order of Group Element|order of an element]].
+Conversely, suppose that 
+:$\exists n \in \Z_{> 0}: x^{2 n - 1} = e$
+Then $\order x$ is a [[Definition:Divisor of Integer|divisor]] of $2 n - 1$ from [[Element to Power of Multiple of Order is Identity]].
+Hence $\order x$ is [[Definition:Odd Integer|odd]].
+So $\order x$ is [[Definition:Odd Integer|odd]] {{iff}} $\exists n \in \Z_{> 0}: x^{2 n - 1} = e$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \order x
+      | r = 2 n - 1
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = x^{2 n - 1}
+      | r = e
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = x^{2 n}
+      | r = x
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \paren {x^n}^2
+      | r = x
+      | c = 
+}}
+{{end-eqn}}
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Order of Isomorphic Image of Group Element}
+Tags: Group Isomorphisms
+
+\begin{theorem}
+Let $G$ and $H$ be [[Definition:Group|groups]] whose [[Definition:Identity Element|identities]] are $e_G$ and $e_H$.
+Let $\phi: G \to H$ be a [[Definition:Group Isomorphism|group isomorphism]].
+Then:
+:$a \in G \implies \order {\map \phi a} = \order a$
+\end{theorem}
+
+\begin{proof}
+First, suppose $a$ is of [[Definition:Finite Order Element|finite order]].
+By definition, $\phi$ is [[Definition:Bijection|bijective]], therefore [[Definition:Injection|injective]].
+The result then follows from [[Order of Homomorphic Image of Group Element]].
+{{qed|lemma}}
+Now suppose $a$ is of [[Definition:Infinite Order Element|infinite order]].
+Suppose $\map \phi a$ is of [[Definition:Finite Order Element|finite order]].
+Consider the mapping $\phi^{-1}: H \to G$.
+Let $b = \map \phi a$.
+Let $\order b = m$.
+Then:
+:$\order a = \order {\map {\phi^{-1} } b} = m$
+and that would mean $a$ was of [[Definition:Finite Order Element|finite order]].
+The result follows by [[Rule of Transposition|transposition]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Non-Trivial Group has Non-Trivial Cyclic Subgroup}
+Tags: Cyclic Groups
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity element]] is $e$.
+Let $g \in G$.
+If $g$ has [[Definition:Infinite Order Element|infinite order]], then $\gen g$ is an [[Definition:Infinite Cyclic Group|infinite cyclic group]].
+If $\order g = n$, then $\gen g$ is a [[Definition:Cyclic Group|cyclic group]] with $n$ [[Definition:Element|elements]].
+Thus, every [[Definition:Group|group]] which is [[Definition:Non-Trivial Group|non-trivial]] has at least one [[Definition:Cyclic Group|cyclic]] [[Definition:Subgroup|subgroup]] which is also [[Definition:Non-Trivial Group|non-trivial]].
+In the case that $G$ is itself [[Definition:Cyclic Group|cyclic]], that [[Definition:Cyclic Group|cyclic]] [[Definition:Subgroup|subgroup]] may of course be itself.
+\end{theorem}
+
+\begin{proof}
+=== Infinite Order ===
+Suppose that $g$ has [[Definition:Infinite Order Element|infinite order]].
+We have that $\gen g$ consists of all possible [[Powers of Group Elements|powers of $g$]].
+So $\gen g$ can contain a [[Definition:Finite Set|finite number of elements]] only if some of these were equal.
+Then we would have:
+:$\exists i, j \in \Z, i < j: g^i = g^j$
+and so:
+:$g^{j - i} = g^{i - i} = g^0 = e$
+which would mean that $g$ was of [[Definition:Finite Order Element|finite order]].
+This contradiction leads to the conclusion that $\gen g$ must be an [[Definition:Infinite Cyclic Group|infinite cyclic group]].
+{{qed|lemma}}
+=== Finite Order ===
+If $\order g = n$, we have from [[Equal Powers of Finite Order Element]] that there are exactly $n$ different [[Definition:Element|elements]] of $G$ of the form $g^i$.
+Hence $\gen g$ is a [[Definition:Cyclic Group|cyclic group]] with $n$ elements.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Epimorphism from Integers to Cyclic Group}
+Tags: Integers, Group Epimorphisms, Cyclic Groups
+
+\begin{theorem}
+Let $\gen a = \struct {G, \circ}$ be a [[Definition:Cyclic Group|cyclic group]].
+Let $f: \Z \to G$ be a [[Definition:Mapping|mapping]] defined as:
+$\forall n \in \Z: \map f n = a^n$.
+Then $f$ is a [[Definition:Group Epimorphism|(group) epimorphism]] from $\struct {\Z, +}$ onto $\gen a$.
+\end{theorem}
+
+\begin{proof}
+By [[Powers of Element form Subgroup]]:
+:$\forall n \in \N: a^n \in \gen a$
+Hence by the [[Index Laws for Monoids/Negative Index|Index Law for Monoids: Negative Index]]:
+:$\forall n \in \Z: a^n \in \gen a$
+Also, by [[Index Laws for Monoids/Sum of Indices|Index Law for Monoids: Sum of Indices]], $f$ is a [[Definition:Group Homomorphism|homomorphism]] from $\struct {\Z, +}$ into $\struct {G, \circ}$.
+By [[Homomorphism Preserves Subsemigroups]], its [[Definition:Codomain of Mapping|codomain]] $f \sqbrk \Z$ is therefore a [[Definition:Subgroup|subgroup]] of $\gen a$ containing $a$.
+By [[Existence of Unique Subgroup Generated by Subset/Singleton Generator|Existence of Unique Subgroup Generated by Subset: Singleton Generator]], $\gen a$ is the smallest subgroup of $G$ containing $a$.
+So:
+:$f \sqbrk \Z = \gen a$
+Therefore $f$ is an [[Definition:Group Epimorphism|epimorphism]] from $\struct {\Z, +}$ onto $\struct {G, \circ}$.
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Cyclic Group is Abelian}
+Tags: Cyclic Groups, Abelian Groups, Cyclic Group is Abelian
+
+\begin{theorem}
+Let $G$ be a [[Definition:Cyclic Group|cyclic group]].
+Then $G$ is [[Definition:Abelian Group|abelian]].
+\end{theorem}
+
+\begin{proof}
+We have that [[Integers under Addition form Abelian Group]].
+The result then follows from combining:
+: [[Epimorphism from Integers to Cyclic Group]]
+: [[Epimorphism Preserves Commutativity]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let $G$ be a [[Definition:Cyclic Group|cyclic group]].
+All elements of $G$ are of the form $a^n$, where $n \in \Z$.
+Let $x, y \in G: x = a^p, y = a^q$.
+From [[Powers of Group Elements/Sum of Indices|Powers of Group Elements: Sum of Indices]]:
+: $x y = a^p a^q = a^{p + q} = a^{q + p} = a^q a^p = y x$
+Thus:
+: $\forall x, y \in G: x y = y x$
+and $G$ is by definition [[Definition:Abelian Group|abelian]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Cyclic Groups of Same Order are Isomorphic}
+Tags: Cyclic Groups, Group Isomorphisms
+
+\begin{theorem}
+Two [[Definition:Cyclic Group|cyclic groups]] of the same [[Definition:Order of Structure|order]] are [[Definition:Group Isomorphism|isomorphic]] to each other.
+\end{theorem}
+
+\begin{proof}
+Let $G_1$ and $G_2$ be [[Definition:Cyclic Group|cyclic groups]], both of [[Definition:Finite Group|finite order]] $k$.
+Let $G_1 = \gen a, G_2 = \gen b$.
+Then, by the definition of a [[Definition:Cyclic Group|cyclic group]]:
+:$\order a = \order b = k$
+Also, by definition:
+:$G_1 = \set {a^0, a^1, \ldots, a^{k - 1} }$
+and:
+:$G_2 = \set {b^0, b^1, \ldots, b^{k - 1} }$
+Let us set up the obvious [[Definition:Bijection|bijection]]:
+:$\phi: G_1 \to G_2: \map \phi {a^n} = b^n$
+The next task is to show that $\phi$ is an [[Definition:Group Isomorphism|isomorphism]].
+Note that $\map \phi {a^n} = b^n$ holds for all $n \in \Z$, not just where $0 \le n < k$, as follows:
+Let $n \in \Z: n = q k + r, 0 \le r < k$, by the [[Division Theorem]].
+Then, by [[Element to Power of Remainder]]:
+:$a^n = a^r, b^n = b^r$
+Thus:
+:$\map \phi {a^n} = \map \phi {a^r} = b^r = b^n$
+Now let $x, y \in G_1$.
+Since $G_1 = \gen a$, it follows that:
+:$\exists s, t \in \Z: x = a^s, y = a^t$
+Thus:
+{{begin-eqn}}
+{{eqn | l = \map \phi {x y}
+      | r = \map \phi {a^s a^t}
+      | c = 
+}}
+{{eqn | r = \map \phi {a^{s + t} }
+      | c = 
+}}
+{{eqn | r = b^{s + t}
+      | c = 
+}}
+{{eqn | r = b^s b^t
+      | c = 
+}}
+{{eqn | r = \map \phi {a^s} \map \phi {a^t}
+      | c = 
+}}
+{{eqn | r = \map \phi x \map \phi y
+      | c = 
+}}
+{{end-eqn}}
+So $\phi$ is a [[Definition:Group Homomorphism|homomorphism]].
+As $\phi$ is [[Definition:Bijection|bijective]], $\phi$ is an [[Definition:Group Isomorphism|isomorphism]] from $G_1$ to $G_2$.
+Thus $G_1 \cong G_2$ and the result is proved.
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Order of Subgroup of Cyclic Group}
+Tags: Cyclic Groups
+
+\begin{theorem}
+Let $C_n = \gen g$ be the [[Definition:Cyclic Group|cyclic group]] of [[Definition:Order of Structure|order]] $n$ which is [[Definition:Generator of Cyclic Group|generated]] by $g$ whose [[Definition:Identity Element|identity]] is $e$.
+Let $a \in C_n: a = g^i$.
+Let $H = \gen a$.
+Then:
+:$\order H = \dfrac n {\gcd \set {n, i} }$
+where:
+:$\order H$ denotes the [[Definition:Order of Structure|order]] of $H$
+:$\gcd \set {n, i}$ denotes the [[Definition:Greatest Common Divisor|greatest common divisor]] of $n$ and $i$.
+\end{theorem}
+
+\begin{proof}
+The fact that $H$ is [[Definition:Cyclic Group|cyclic]] follows from [[Subgroup of Cyclic Group is Cyclic]].
+We need to show that $H$ has $\dfrac n d$ elements.
+Let $\order H = k$.
+By [[Non-Trivial Group has Non-Trivial Cyclic Subgroup]]:
+:$k = \order a$
+where $\order a$ denotes the [[Definition:Order of Group Element|order]] of $a$.
+That is:
+:$a^k = e$
+We have that $a = g^i$.
+So:
+{{begin-eqn}}
+{{eqn | l = a^k
+      | r = e
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \paren {g^i}^k
+      | r = e = g^n
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = n
+      | o = \divides
+      | r = i k
+      | c = {{Defof|Order of Group Element}}
+}}
+{{end-eqn}}
+We now need to calculate the smallest $k$ such that:
+: $n \divides i k$
+where $\divides$ denotes [[Definition:Divisor of Integer|divisibility]].
+That is, the smallest $t \in \N$ such that $n t = i k$.
+Let $d = \gcd \set {n, i}$.
+Thus:
+:$t = \dfrac {k \paren {i / d} } {n / d}$
+From [[Integers Divided by GCD are Coprime]], $\dfrac n d$ and $\dfrac i d$ are [[Definition:Coprime Integers|coprime]].
+Thus from [[Euclid's Lemma]]:
+:$\dfrac n d \divides k$
+As $a \divides b \implies a \le b$, the smallest value of $k$ such that $\dfrac k {\paren {n / d} } \in \Z$ is $\dfrac n d$.
+Hence the result.
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Number of Powers of Cyclic Group Element}
+Tags: Cyclic Groups
+
+\begin{theorem}
+Let $G$ be a [[Definition:Cyclic Group|cyclic group]] of [[Definition:Order of Structure|order]] $n$, [[Definition:Generator of Cyclic Group|generated]] by $g$.
+Let $d \divides n$.
+Then the element $g^{n/d}$ has $d$ distinct [[Definition:Power of Group Element|powers]].
+\end{theorem}
+
+\begin{proof}
+Follows directly from [[Order of Subgroup of Cyclic Group]]:
+:$\order {\gen {g^{n/d} } } = \dfrac n {\gcd \set {n, n/d} } = d$
+Thus from [[List of Elements in Finite Cyclic Group]]:
+:$\gen {g^{n/d} } = \set {e, g^{n/d}, \paren {g^{n/d} }^2, \ldots, \paren {g^{n/d} }^{d - 1} }$
+and the result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Subgroup of Finite Cyclic Group is Determined by Order}
+Tags: Cyclic Groups
+
+\begin{theorem}
+Let $G = \gen g$ be a [[Definition:Cyclic Group|cyclic group]] whose [[Definition:Order of Structure|order]] is $n$ and whose [[Definition:Identity Element|identity]] is $e$.
+Let $d \divides n$, where $\divides$ denotes [[Definition:Divisor of Integer|divisibility]].
+Then there exists exactly one [[Definition:Subgroup|subgroup]] $G_d = \gen {g^{n / d} }$ of $G$ with $d$ [[Definition:Element|elements]].
+\end{theorem}
+
+\begin{proof}
+Let $G$ be [[Definition:Generated Subgroup|generated]] by $g$, such that $\order g = n$.
+From [[Number of Powers of Cyclic Group Element]], $g^{n/d}$ has $d$ distinct [[Definition:Power of Group Element|powers]].
+Thus $\gen {g^{n / d} }$ has $d$ [[Definition:Element|elements]].
+Now suppose $H$ is another [[Definition:Subgroup|subgroup]] of $G$ of [[Definition:Order of Structure|order]] $d$.
+Then by [[Subgroup of Cyclic Group is Cyclic]], $H$ is [[Definition:Cyclic Group|cyclic]].
+Let $H = \gen y$ where $y \in G$.
+Thus $\order y = d$.
+Thus $\exists r \in \Z: y = g^r$.
+Since $\order y = d$, it follows that $y^d = g^{r d} = e$.
+From [[Equal Powers of Finite Order Element]]:
+: $n \divides r d$
+Thus:
+{{begin-eqn}}
+{{eqn | lo= \exists k \in \N:
+      | l = k n
+      | r = r d
+      | c = 
+}}
+{{eqn | r = k \paren {\dfrac n d} d
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = r
+      | r = k \paren {\dfrac n d}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \dfrac n d
+      | o = \divides
+      | r = r
+      | c = 
+}}
+{{end-eqn}}
+Thus $y$ is a power of $g^{n / d}$.
+Hence $H$ is a [[Definition:Subgroup|subgroup]] of $\gen {g^{n / d} }$.
+Since both $H$ and $\gen {g^{n / d} }$ have [[Definition:Order of Structure|order]] $d$, they must be equal.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Cyclic Group Elements whose Powers equal Identity}
+Tags: Cyclic Groups
+
+\begin{theorem}
+Let $G$ be a [[Definition:Cyclic Group|cyclic group]] whose [[Definition:Identity Element|identity]] is $e$ and whose [[Definition:Order of Structure|order]] is $n$.
+Let $d \divides n$.
+Then there exist exactly $d$ elements $x \in G$ satisfying the equation $x^d = e$.
+These are the elements of the group $G_d$ [[Definition:Generator of Cyclic Group|generated by]] $g^{n / d}$: 
+:$G_d = \gen {g^{n / d} }$
+\end{theorem}
+
+\begin{proof}
+From the argument in [[Subgroup of Finite Cyclic Group is Determined by Order]], it follows that $x$ satisfies the equation $x^d = e$ {{iff}} $x$ is a [[Definition:Power of Group Element|power]] of $g^{n/d}$.
+Thus there are $d$ solutions to this equation.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Prime Group is Cyclic}
+Tags: Cyclic Groups, Prime Groups
+
+\begin{theorem}
+Let $p$ be a [[Definition:Prime Number|prime number]].
+Let $G$ be a [[Definition:Group|group]] whose [[Definition:Order of Group|order]] is $p$.
+Then $G$ is [[Definition:Cyclic Group|cyclic]].
+\end{theorem}
+
+\begin{proof}
+Let $a \in G: a \ne e$ where $e$ is the [[Definition:Identity Element|identity]] of $G$.
+From [[Group of Prime Order p has p-1 Elements of Order p]], $a$ has [[Definition:Order of Group|order]] $p$.
+Hence by definition, $a$ [[Definition:Generator of Cyclic Group|generates]] $G$.
+Hence also by definition, $G$ is [[Definition:Cyclic Group|cyclic]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Group of Order less than 6 is Abelian}
+Tags: Abelian Groups
+
+\begin{theorem}
+All [[Definition:Group|groups]] with less than $6$ [[Definition:Element|elements]] are [[Definition:Abelian Group|abelian]].
+\end{theorem}
+
+\begin{proof}
+Let $G$ be a non-[[Definition:Abelian Group|abelian group]].
+From [[Non-Abelian Group has Order Greater than 4]], the [[Definition:Order of Structure|order]] of $G$ must be at least $5$.
+But $5$ is a [[Definition:Prime Number|prime number]].
+By [[Prime Group is Cyclic]] it follows that a [[Definition:Group|group]] of [[Definition:Order of Group|order]] $5$ is [[Definition:Cyclic Group|cyclic]].
+By [[Cyclic Group is Abelian]] this group is [[Definition:Abelian Group|abelian]].
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Power of Generator of Cyclic Group is Generator iff Power is Coprime with Order}
+Tags: Cyclic Groups, Coprime Integers
+
+\begin{theorem}
+Let $C_n$ be the [[Definition:Cyclic Group|cyclic group]] of [[Definition:Order of Structure|order]] $n$.
+Let $C_n = \gen a$, that is, that $C_n$ is [[Definition:Generator of Cyclic Group|generated]] by $a$.
+Then:
+:$C_n = \gen {a^k} \iff k \perp n$
+That is, $C_n$ is also [[Definition:Generator of Cyclic Group|generated]] by $a^k$ {{iff}} $k$ is [[Definition:Coprime Integers|coprime]] to $n$.
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+Let $k \perp n$.
+Then by [[Integer Combination of Coprime Integers]]:
+:$\exists u, v \in \Z: 1 = u k + v n$
+So $\forall m \in \Z$, we have:
+{{begin-eqn}}
+{{eqn | l =a^m
+      | r = a^{m u k + m v n}
+      | c = 
+}}
+{{eqn | r = a^{m u k}
+      | c = as $a^{m v n} = e$
+}}
+{{eqn | r = \paren {a^k}^{m u}
+      | c = 
+}}
+{{end-eqn}}
+Thus $a^k$ [[Definition:Generator of Cyclic Group|generate]] $C_n$.
+{{qed|lemma}}
+=== Sufficient Condition ===
+Let $C_n = \gen {a^k}$.
+That is, let $a^k$ [[Definition:Generator of Cyclic Group|generate]] $C_n$.
+{{begin-eqn}}
+{{eqn | lo= \exists u \in \Z:
+      | l = a
+      | r = \paren {a^k}^u
+      | c = as $a$ is an [[Definition:Element|element]] of the [[Definition:Generator of Cyclic Group|group generated]] by $a^k$
+}}
+{{eqn | ll= \leadsto
+      | l = u k
+      | o = \equiv
+      | r = 1
+      | rr= \pmod n
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | lo= \exists u, v \in \Z:
+      | l = 1
+      | r = u k + v n
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = k
+      | o = \perp
+      | r = n
+      | c = [[Integer Combination of Coprime Integers]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Order of Conjugate of Subgroup}
+Tags: Conjugacy
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]].
+Let $H$ be a [[Definition:Subgroup|subgroup]] of $G$ such that $H$ is of [[Definition:Order of Structure|finite order]].
+Then $\order {H^a} = \order H$.
+\end{theorem}
+
+\begin{proof}
+From the definition of [[Definition:Conjugate of Group Subset|Conjugate of Group Subet]] we have $H^a = a H a^{-1}$.
+From [[Set Equivalence of Regular Representations]]:
+:$\order {a H a^{-1} } = \order {a H} = \order H$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Subgroup of Index 2 is Normal}
+Tags: Normal Subgroups
+
+\begin{theorem}
+A [[Definition:Subgroup|subgroup]] of [[Definition:Index of Subgroup|index $2$]] is always [[Definition:Normal Subgroup|normal]].
+\end{theorem}
+
+\begin{proof}
+Suppose $H \le G$ such that $\index G H = 2$.
+Thus $H$ has two [[Definition:Left Coset|left cosets]] (and two [[Definition:Right Coset|right cosets]]) in $G$.
+If $g \in H$, then $g H = H = H g$.
+If $g \notin H$, then $g H = G \setminus H$ as there are only two [[Definition:Coset|cosets]] and the [[Congruence Class Modulo Subgroup is Coset|cosets partition $G$]].
+For the same reason, $g \notin H \implies H g = G \setminus H$.
+That is, $g H = H g$.
+The result follows from the definition of [[Definition:Normal Subgroup|normal subgroup]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Intersection of Normal Subgroups is Normal}
+Tags: Normal Subgroups
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]].
+Let $I$ be an [[Definition:Indexing Set|indexing set]].
+Let $\family {N_i}_{i \mathop \in I}$ be a [[Definition:Non-Empty Set|non-empty]] [[Definition:Indexed Family|indexed family]] of [[Definition:Normal Subgroup|normal subgroups]] of $G$.
+Then $\displaystyle \bigcap_{i \mathop \in I} N_i$ is a [[Definition:Normal Subgroup|normal subgroup]] of $G$.
+\end{theorem}
+
+\begin{proof}
+Let $\displaystyle N = \bigcap_{i \mathop \in I} N_i$.
+From [[Intersection of Subgroups is Subgroup]], $N \le G$.
+Suppose $H \in \set {N_i: i \in I}$.
+We have that $N \subseteq H$.
+Thus from [[Subgroup is Superset of Conjugate iff Normal]]:
+:$a N a^{-1} \subseteq a H a^{-1} \subseteq H$
+Thus $a N a^{-1}$ is a [[Definition:Subset|subset]] of each one of the [[Definition:Subgroup|subgroups]] in $\set {N_i: i \in I}$, and hence in their [[Definition:Intersection of Family|intersection]] $N$.
+That is, $a N a^{-1} \subseteq N$.
+The result follows by [[Subgroup is Superset of Conjugate iff Normal]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Union of Conjugacy Classes is Normal}
+Tags: Normal Subgroups
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]].
+Let $H \le G$.
+Then $H$ is [[Definition:Normal Subgroup|normal]] in $G$ {{iff}} $H$ is a [[Definition:Set Union|union]] of [[Definition:Conjugacy Class|conjugacy classes]] of $G$.
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = H
+      | o = \lhd
+      | r = G
+      | c = where $\lhd$ denotes that $H$ is [[Definition:Normal Subgroup|normal]] in $G$
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \forall g \in G: g H g^{-1}
+      | o = \subseteq
+      | r = H
+      | c = {{Defof|Normal Subgroup}}
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \forall x \in H: \forall g \in G: g x g^{-1}
+      | o = \in
+      | r = H
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \forall x \in H: \conjclass x
+      | o = \subseteq
+      | r = H
+      | c = where $\conjclass x$ is the [[Definition:Conjugacy Class|conjugacy class]] of $x \in G$
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = H
+      | r = \bigcup_{x \mathop \in H} \conjclass x
+      | c = 
+}}
+{{end-eqn}}
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Unique Subgroup of a Given Order is Normal}
+Tags: Normal Subgroups
+
+\begin{theorem}
+Let a [[Definition:Group|group]] $G$ have only one [[Definition:Subgroup|subgroup]] of a given [[Definition:Order of Structure|order]].
+Then that [[Definition:Subgroup|subgroup]] is [[Definition:Normal Subgroup|normal]].
+\end{theorem}
+
+\begin{proof}
+Let $H \le G$, where $\le$ denotes that $H$ is a [[Definition:Subgroup|subgroup]] of $G$.
+Let $H$ be the only [[Definition:Subgroup|subgroup]] of $G$ whose [[Definition:Order of Structure|order]] is $\order H$.
+Let $g \in G$.
+From [[Conjugate of Subgroup is Subgroup]]:
+:$g H g^{-1} \le G$
+From [[Order of Conjugate of Subgroup]]:
+: $\order {g H g^{-1} } = \order H$
+But $H$ is the only [[Definition:Subgroup|subgroup]] of $G$ of [[Definition:Order of Structure|order]] $\order H$.
+Hence any [[Definition:Subgroup|subgroup]] whose [[Definition:Order of Structure|order]] is $\order H$ must in fact ''be'' $H$.
+That is, $g H g^{-1} = H$.
+The result follows from [[Subgroup equals Conjugate iff Normal]].
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Group Generated by Normal Intersection is Normal}
+Tags: Normal Subgroups
+
+\begin{theorem}
+Let $I$ be an [[Definition:Indexing Set|indexing set]], and $\left\{{N_i: i \in I}\right\}$ be a set of [[Definition:Normal Subgroup|normal subgroups]] of the [[Definition:Group|group]] $G$.
+Then $\left \langle {N_i: i \in I} \right \rangle$ is a [[Definition:Normal Subgroup|normal subgroup]] of $G$.
+\end{theorem}
+
+\begin{proof}
+By definition, $\left \langle {N_i: i \in I} \right \rangle$ is the [[Definition:Set Intersection|intersection]] of all the [[Definition:Subgroup|subgroups]] of $G$ which contain every $N_i$.
+For each $H \le G$, the [[Definition:Conjugate of Group Subset|conjugate]] $g H g^{-1}$ contains each $g N_i g^{-1}$.
+Since each $N_i \lhd G$, it follows that:
+: $N_i \subseteq g H g^{-1}$
+Thus it follows that:
+: $\left \langle {N_i: i \in I} \right \rangle \lhd G$
+{{qed}}
+[[Category:Normal Subgroups]]
+bae4k1vl8whwigmiqi2pq8hjl1rthii
+\end{proof}<|endoftext|>
+\section{Smallest Normal Subgroup containing Set}
+Tags: Normal Subgroups
+
+\begin{theorem}
+Let $S \subseteq G$ where $G$ is a [[Definition:Group|group]].
+Then there exists a unique smallest [[Definition:Normal Subgroup|normal subgroup]] of $G$ which contains $S$.
+\end{theorem}
+
+\begin{proof}
+Let $\Bbb S$ be the [[Definition:Set|set]] of all [[Definition:Normal Subgroup|normal subgroups]] of $G$ that contain $S$.
+$\Bbb S \ne \varnothing$, since $S \subseteq G \lhd G$.
+Let $N = \bigcap H: H \in \Bbb S$, that is, the [[Definition:Set Intersection|intersection]] of all [[Definition:Element|elements]] of $\Bbb S$.
+By [[Intersection of Normal Subgroups is Normal]], $N \lhd G$, and by the definition of [[Definition:Set Intersection|intersection]], $S \subseteq N$.
+By the method of construction, $N$ is the smallest such subgroup.
+By the definition of "smallest", $N$ is unique.
+{{qed}}
+[[Category:Normal Subgroups]]
+syea0j00dnpyhtp2qtjk5ukvyhh4kzb
+\end{proof}<|endoftext|>
+\section{Conjugate of Set with Inverse Closed for Inverses}
+Tags: Conjugacy
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]].
+Let $S \subseteq G$.
+Let $\hat S = S \cup S^{-1}$, where $S^{-1}$ is the [[Definition:Inverse of Subset of Group|set of all the inverses]] of all the elements of $S$.
+Let $\tilde S = \set {a s a^{-1}: s \in \hat S, a \in G}$.
+That is, $\tilde S$ is the [[Definition:Set|set]] containing all the [[Definition:Conjugate of Group Element|conjugates]] of the [[Definition:Element|elements]] of $S$ and all their [[Definition:Inverse Element|inverses]].
+Then:
+: $\forall x \in \tilde S: x^{-1} \in \tilde S$
+\end{theorem}
+
+\begin{proof}
+Let $x \in \tilde S$.
+That is:
+:$\exists s \in \hat S: x = a s a^{-1}$
+Then:
+{{begin-eqn}}
+{{eqn | l = x^{-1}
+      | r = \paren {a s a^{-1} }^{-1}
+      | c = 
+}}
+{{eqn | r = a s^{-1} a^{-1}
+      | c = [[Power of Conjugate equals Conjugate of Power]]
+}}
+{{end-eqn}}
+Since $s^{-1} \in \hat S$, it follows that $x^{-1} \in \tilde S$.
+{{Qed}}
+[[Category:Conjugacy]]
+o56izk8q21jzbzwty9uvh72whozxfbh
+\end{proof}<|endoftext|>
+\section{Conjugate of Set with Inverse is Closed}
+Tags: Conjugacy
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]].
+Let $S \subseteq G$.
+Let $\hat S = S \cup S$.
+Let $\tilde S = \left\{{a s a^{-1}: s \in \hat S, a \in G}\right\}$.
+Let $W \left({\tilde S}\right)$ be the [[Definition:Word (Abstract Algebra)|set of words]] of $\tilde S$.
+Then $\forall w \in W \left({\tilde S}\right): \forall a \in G: a w a^{-1} \in W \left({\tilde S}\right)$.
+\end{theorem}
+
+\begin{proof}
+Let $w \in W \left({\tilde S}\right)$.
+From the [[Definition:Word (Abstract Algebra)|definition]] of $W \left({\tilde S}\right)$, we have:
+$w = \left({a_1 s_1 a_1^{-1}}\right) \left({a_2 s_2 a_2^{-1}}\right) \cdots \left({a_n s_n a_n^{-1}}\right), n \in \N^*, a_i \in G, s_1 \in \hat S, 1 \le i \le n$.
+Thus:
+{{begin-eqn}}     
+{{eqn      | l=a w a^{-1}
+           | r=a \left({a_1 s_1 a_1^{-1} }\right) \left({a_2 s_2 a_2^{-1} }\right) \cdots \left({a_n s_n a_n^{-1} }\right) a^{-1}
+           | c=
+}}
+{{eqn      | r=a \left({a_1 s_1 a_1^{-1} }\right) a^{-1} a \left({a_2 s_2 a_2^{-1} }\right) a^{-1} \cdots  a \left({a_n s_n a_n^{-1} }\right) a^{-1}
+           | c=
+}}
+{{eqn      | r=\left({\left({a a_1}\right) s_1 \left({a a_1}\right)^{-1} }\right) \left({\left({a a_2}\right) s_2 \left({a a_2}\right)^{-1} }\right) \cdots  \left({\left({a a_n}\right) s_n \left({a a_n}\right)^{-1} }\right)
+           | c=
+}}
+{{end-eqn}}
+As $G$ is a group, all of the $a a_i \in G$.
+The result follows.
+{{qed}}
+[[Category:Conjugacy]]
+jompv56uogazp88n2mpvhk944cn1tqz
+\end{proof}<|endoftext|>
+\section{Generator of Normal Subgroup}
+Tags: Normal Subgroups
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]].
+Let $S \subseteq G$.
+Let $\hat S = S \cup S^{-1}$, where $S^{-1}$ is the [[Definition:Inverse of Subset of Group|set of all the inverses]] of all the elements of $S$.
+Let $\tilde S = \left\{{a s a^{-1}: s \in \hat S, a \in G}\right\}$.
+Let $W \left({\tilde S}\right)$ be the [[Definition:Word (Abstract Algebra)|set of words]] of $\tilde S$.
+Let $N$ be the smallest [[Definition:Normal Subgroup|normal subgroup]] of $G$ that contains $S$.
+Then $N = \left \langle {S} \right \rangle = W \left({\tilde S}\right)$.
+\end{theorem}
+
+\begin{proof}
+Let $N$ be the smallest [[Definition:Normal Subgroup|normal subgroup]] of $G$ that contains $S$, where $S \subseteq G$.
+$N$ must certainly include $\hat S$, because any group containing $s \in S$ must also contain $s^{-1}$. 
+Therefore, $N$ must be the smallest normal subgroup containing $\hat S$.
+Since $N \lhd G$, it follows that $\forall a \in G: \forall s \in \hat S: a s a^{-1} \in N$.
+Thus, $\forall x \in \tilde S: x \in N$.
+Thus $\tilde S \subseteq N$.
+By the [[Definition:Group Axioms|closure axiom]], $N$ must also contain all products of any finite number of elements of $\tilde S$.
+Thus $W \left({\tilde S}\right) \subseteq N$.
+Now we prove that $W \left({\tilde S}\right) \lhd G$.
+$\tilde S \ne \varnothing$, as $e \left({s s^{-1}}\right) e^{-1} = e \in \tilde S$.
+By [[Conjugate of Set with Inverse Closed for Inverses]], $\tilde S$ is closed under taking inverses.
+So from [[Set of Words Generates Group/Corollary|Set of Words Generates Group: Corollary]]:
+: $W \left({\tilde S}\right) \le G$
+From [[Conjugate of Set with Inverse is Closed]]:
+: $\forall w \in W \left({\tilde S}\right): \forall a \in G: a w a^{-1} \in W \left({\tilde S}\right)$
+From [[Subgroup is Normal iff Contains Conjugate Elements]]:
+: $W \left({\tilde S}\right) \lhd G$.
+The result follows by the definition of the minimality of $N$.
+{{qed}}
+[[Category:Normal Subgroups]]
+tmf3it242yg20gtqnpz2aoa5sswhrjt
+\end{proof}<|endoftext|>
+\section{Subset Product with Normal Subgroup as Generator}
+Tags: Normal Subgroups, Subset Products
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
+Let:
+:$H$ be a [[Definition:Subgroup|subgroup]] of $G$
+:$N$ be a [[Definition:Normal Subgroup|normal subgroup]] of $G$.
+Then:
+:$N \lhd \gen {N, H} = N H = H N \le G$
+where:
+:$\le$ denotes [[Definition:Subgroup|subgroup]]
+:$\lhd$ denotes [[Definition:Normal Subgroup|normal subgroup]]
+:$\gen {N, H}$ denotes a [[Definition:Generator of Subgroup|subgroup generator]]
+:$N H$ denotes [[Definition:Subset Product|subset product]].
+\end{theorem}
+
+\begin{proof}
+From [[Subset Product is Subset of Generator]]:
+:$N H \subseteq \gen {N, H}$
+From [[Subset Product with Normal Subgroup is Subgroup]]:
+:$N H = H N \le G$
+Then by the definition of a [[Definition:Generator of Subgroup|subgroup generator]], $\gen {N, H}$ is the smallest [[Definition:Subgroup|subgroup]] containing $N H$ and so:
+:$\gen {N, H} = N H = H N \le G$
+From [[Normal Subgroup of Subset Product of Subgroups]] we have that:
+:$N \lhd N H$
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Subset Product of Normal Subgroups is Normal}
+Tags: Normal Subgroups, Subset Products
+
+\begin{theorem}
+Let $\struct {G, \circ}$ be a [[Definition:Group|group]].
+Let $N$ and $N'$ be [[Definition:Normal Subgroup|normal subgroups]] of $G$.
+Then $N N'$ is also a [[Definition:Normal Subgroup|normal subgroup]] of $G$.
+\end{theorem}
+
+\begin{proof}
+From [[Subset Product with Normal Subgroup is Subgroup]], we already have that $N N'$ is a [[Definition:Subgroup|subgroup]] of $G$.
+Let $n n' \in N N'$, so that $n \in N, n' \in N'$.
+Let $g \in G$.
+From [[Subgroup is Normal iff Contains Conjugate Elements]]:
+:$g n g^{-1}\in N, g n' g^{-1}\in N'$
+So:
+:$\paren {g n g^{-1} } \paren {g n' g^{-1} } = g n n' g^{-1} \in N N'$
+So $N N'$ is [[Definition:Normal Subgroup|normal]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Prime Group is Simple}
+Tags: Simple Groups, Prime Groups
+
+\begin{theorem}
+[[Definition:Group|Groups]] of [[Definition:Prime Number|prime]] [[Definition:Order of Structure|order]] are [[Definition:Simple Group|simple]].
+\end{theorem}
+
+\begin{proof}
+Follows directly from [[Prime Group has no Proper Subgroups]]: a group of prime order has only itself and the [[Definition:Trivial Group|trivial group]] as [[Definition:Subgroup|subgroups]].
+From [[Trivial Subgroup and Group Itself are Normal]], these subgroups are normal.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Prime Group has no Proper Subgroups}
+Tags: Subgroups, Prime Groups
+
+\begin{theorem}
+A [[Definition:Non-Trivial Group|nontrivial group]] $G$ has no [[Definition:Proper Subgroup|proper subgroups]] except the [[Definition:Trivial Group|trivial group]] {{iff}} $G$ is [[Definition:Finite Group|finite]] and its [[Definition:Order of Structure|order]] is [[Definition:Prime Number|prime]].
+\end{theorem}
+
+\begin{proof}
+=== Sufficient Condition ===
+Suppose $G$ is [[Definition:Finite Group|finite]] and of [[Definition:Prime Number|prime]] [[Definition:Order of Structure|order]] $p$.
+From [[Lagrange's Theorem (Group Theory)|Lagrange's Theorem]], the [[Definition:Order of Structure|order]] of any [[Definition:Subgroup|subgroup]] of $G$ must [[Definition:Divisor of Integer|divide]] the [[Definition:Order of Structure|order]] $p$ of $G$.
+From the definition of [[Definition:Prime Number|prime]], any [[Definition:Subgroup|subgroups]] of $p$ can therefore only have [[Definition:Order of Structure|order]] $1$ or $p$.
+Hence $G$ can have only itself and the [[Definition:Trivial Group|trivial group]] as [[Definition:Subgroup|subgroups]].
+{{qed|lemma}}
+=== Necessary Condition ===
+Suppose $G$ is ''not'' [[Definition:Finite Group|finite]] and [[Definition:Prime Group|prime]].
+Let the [[Definition:Identity Element|identity]] of $G$ be $e$.
+Let $h \in G$ be an [[Definition:Element|element]] of $G$ such that $h \ne e$.
+Then $H = \gen h$ is a [[Definition:Cyclic Group|cyclic subgroup]] of $G$.
+If $H \ne G$ then $H$ is a [[Definition:Non-Trivial Group|non-trivial]] [[Definition:Proper Subgroup|proper subgroup]] of $G$, and the proof is complete.
+Otherwise, $H = G$ is a [[Definition:Cyclic Group|cyclic group]] and there are two possibilities:
+:$(1): \quad G$ is [[Definition:Infinite Cyclic Group|infinite]]
+:$(2): \quad G$ is [[Definition:Finite Cyclic Group|finite]] (and of non-[[Definition:Prime Group|prime]] [[Definition:Order of Structure|order]]).
+First, suppose $G$ is [[Definition:Infinite Cyclic Group|infinite]].
+From [[Infinite Cyclic Group is Isomorphic to Integers]], $G$ is [[Definition:Group Isomorphism|isomorphic]] to $\struct {\Z, +}$.
+From [[Subgroups of Additive Group of Integers]], $\struct {\Z, +}$ has [[Definition:Proper Subgroup|proper subgroups]], for example: $\gen 2$.
+Because $G \cong \struct {\Z, +}$, then so does $G$ have [[Definition:Proper Subgroup|proper subgroups]], and the proof is complete.
+Suppose $G$ is [[Definition:Finite Group|finite]], and of [[Definition:Order of Structure|order]] $n$ where $n$ is not [[Definition:Prime Group|prime]] .
+Then:
+: $\exists d \in \N: d \divides n, 1 < d < n$
+From [[Subgroup of Finite Cyclic Group is Determined by Order]], $G$ has a [[Definition:Proper Subgroup|proper subgroup]] of [[Definition:Order of Structure|order]] $d$ and again, the proof is complete.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Quotient Group of Cyclic Group}
+Tags: Cyclic Groups, Quotient Groups, Quotient Group of Cyclic Group
+
+\begin{theorem}
+Let $G$ be a [[Definition:Cyclic Group|cyclic group]] which is [[Definition:Generator of Cyclic Group|generated by $g$]].
+Let $H$ be a [[Definition:Subgroup|subgroup]] of $G$.
+Then $g H$ [[Definition:Generator of Cyclic Group|generates]] $G / H$.
+\end{theorem}
+
+\begin{proof}
+Let $G$ be a [[Definition:Cyclic Group|cyclic group]] [[Definition:Generator of Cyclic Group|generated by $g$]].
+Let $H \le G$.
+We need to show that every element of $G / H$ is of the form $\left({g H}\right)^k$ for some $k \in \Z$.
+Suppose $x H \in G / H$.
+Then, since $G$ is generated by $g$, $x = g^k$ for some $k \in \Z$.
+But $\left({g H}\right)^k = \left({g^k}\right) H = x H$.
+So $g H$ [[Definition:Generator of Cyclic Group|generates]] $G / H$.
+{{Qed}}
+\end{proof}
+
+\begin{proof}
+Let $H$ be a [[Definition:Subgroup|subgroup]] of the [[Definition:Cyclic Group|cyclic group]] $G = \left \langle {g} \right \rangle$.
+Then by [[Homomorphism of Powers/Integers|Homomorphism of Powers for Integers]]:
+:$\forall n \in \Z: q_H \left({g^n}\right) = \left({q_H \left({g}\right)}\right)^n = \left({g H}\right)^n$
+As $G = \left\{{g^n: n \in \Z}\right\}$, we conclude that:
+:$G / H = q_H \left({G}\right) = \left\{{\left({g H}\right)^n: n \in \Z}\right\}$
+Thus, by [[Epimorphism from Integers to Cyclic Group]], $g H$ generates $G / H$.
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Order of Element in Quotient Group}
+Tags: Quotient Groups
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]], and let $H$ be a [[Definition:Normal Subgroup|normal subgroup]] of $G$.
+Let $G / H$ be the [[Definition:Quotient Group|quotient group]] of $G$ by $H$.
+The [[Definition:Order of Group Element|order]] of $a H \in G / H$ [[Definition:Divisor of Integer|divides]] the [[Definition:Order of Group Element|order]] of $a \in G$.
+\end{theorem}
+
+\begin{proof}
+Let $G$ be a [[Definition:Group|group]] with [[Definition:Normal Subgroup|normal subgroup]] $H$.
+Let $G / H$ be the [[Definition:Quotient Group|quotient of $G$ by $H$]].
+From [[Quotient Group Epimorphism is Epimorphism]], $G / H$ is a [[Definition:Homomorphic Image|homomorphic image]] of $G$.
+Let $q_H: G \to G / H$ given by $\map f a = a H$ be that [[Definition:Quotient Epimorphism|quotient epimorphism]].
+Let $a \in G$ such that $a^n = e$ for some [[Definition:Integer|integer]] $n$.
+Then, by the [[Definition:Morphism Property|morphism property]] of $q_H$:
+{{begin-eqn}}
+{{eqn | l = \map {q_H} {a^n}
+      | r = \paren {\map {q_H} a}^n
+      | c = 
+}}
+{{eqn | r = \paren {a H}^n
+      | c = 
+}}
+{{eqn | r = a^n H
+      | c = {{Defof|Coset Product}}
+}}
+{{eqn | r = e H
+      | c = [[Definition:By Hypothesis|by hypothesis]]
+}}
+{{eqn | r = H
+      | c = 
+}}
+{{end-eqn}}
+Hence $\order H$ [[Definition:Divisor of Integer|divides]] $n$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Abelian Quotient Group}
+Tags: Abelian Groups, Quotient Groups
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]].
+Let $H$ be a [[Definition:Normal Subgroup|normal subgroup]] of $G$.
+Let $G / H$ denote the [[Definition:Quotient Group|quotient group]] of $G$ by $H$.
+Then $G / H$ is [[Definition:Abelian Group|abelian]] {{iff}} $H$ contains every element of $G$ of the form $a b a^{-1} b^{-1}$ where $a, b \in G$.
+\end{theorem}
+
+\begin{proof}
+Let $G / H$ be [[Definition:Abelian Group|abelian]].
+Then:
+{{begin-eqn}}
+{{eqn | lo= \forall a H, b H \in G / H:
+      | l = a H b H
+      | r = b H a H
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = a b H
+      | r = b a H
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = a b \paren {b a}^{-1}
+      | o = \in
+      | r = H
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = a b a^{-1} b^{-1}
+      | o = \in
+      | r = H
+      | c = 
+}}
+{{end-eqn}}
+The argument reverses.
+Suppose that $\forall a, b \in G: a b a^{-1} b^{-1} \in H$.
+{{begin-eqn}}
+{{eqn | lo= \forall a, b \in G:
+      | l = a b a^{-1} b^{-1}
+      | o = \in
+      | r = H
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = a b \paren {b a}^{-1}
+      | o = \in
+      | r = H
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = a b H
+      | r = b a H
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = a H b H
+      | r = b H a H
+      | c = 
+}}
+{{end-eqn}}
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Intersection of Quotient Groups}
+Tags: Quotient Groups
+
+\begin{theorem}
+Let $N \lhd G$ be a [[Definition:Normal Subgroup|normal subgroup]] of $G$.
+Let:
+:$N \le A \le G$
+:$N \le B \le G$
+For a [[Definition:Subgroup|subgroup]] $H$ of $G$, let $\alpha$ be the [[Definition:Bijection|bijection]] defined as:
+:$\map \alpha H = \set {h N: h \in H}$
+Then:
+:$\map \alpha {A \cap B} = \map \alpha A \cap \map \alpha B$
+\end{theorem}
+
+\begin{proof}
+From the proof of the [[Correspondence Theorem (Group Theory)|Correspondence Theorem]]:
+:$\map \alpha H \subseteq G / N$
+Then:
+{{begin-eqn}}
+{{eqn | l = \map \alpha A \cap \map \alpha B
+      | r = \set {g N: g \in A \cap B}
+      | c = 
+}}
+{{eqn | r = \paren {A \cap B} / N
+      | c = 
+}}
+{{eqn | r = \map \alpha {A \cap B}
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Generator of Quotient Groups}
+Tags: Quotient Groups, Generated Subgroups
+
+\begin{theorem}
+Let $N \lhd G$ be a [[Definition:Normal Subgroup|normal subgroup]] of $G$.
+Let:
+:$N \le A \le G$
+:$N \le B \le G$
+For a [[Definition:Subgroup|subgroup]] $H$ of $G$, let $\alpha$ be the [[Definition:Bijection|bijection]] defined as:
+:$\map \alpha H = \set {h N: h \in H}$
+Then:
+:$\map \alpha {\gen {A, B} } = \gen {\map \alpha A, \map \alpha B}$
+where $\gen {A, B}$ denotes the [[Definition:Generated Subgroup|subgroup of $G$ generated]] by $\set {A, B}$.
+\end{theorem}
+
+\begin{proof}
+From the proof of the [[Correspondence Theorem (Group Theory)|Correspondence Theorem]]:
+:$\map \alpha H \subseteq G / N$
+Then:
+{{begin-eqn}}
+{{eqn | l = \map \alpha {\gen {A, B} }
+      | r = \set {h N: h \in \gen {A, B} }
+      | c = 
+}}
+{{eqn | r = \set {h N \in \gen {A / N, B / N} }
+      | c = 
+}}
+{{eqn | r = \gen {\map \alpha A, \map \alpha B}
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Quotient of Group by Center Cyclic implies Abelian}
+Tags: Quotient Groups, Cyclic Groups, Abelian Groups, Centers of Groups
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]].
+Let $\map Z G$ be the [[Definition:Center of Group|center]] of $G$.
+Let $G / \map Z G$ be the [[Definition:Quotient Group|quotient group]] of $G$ by $\map Z G$.
+Let $G / \map Z G$ be [[Definition:Cyclic Group|cyclic]].
+Then $G$ is [[Definition:Abelian Group|abelian]], so $G = \map Z G$.
+That is, the group $G / \map Z G$ cannot be a [[Definition:Cyclic Group|cyclic group]] which is [[Definition:Non-Trivial Group|non-trivial]].
+\end{theorem}
+
+\begin{proof}
+Suppose $G / \map Z G$ is [[Definition:Cyclic Group|cyclic]].
+Then by definition:
+:$\exists \tau \in G / \map Z G: G / \map Z G = \gen \tau$
+Since $\tau$ is a [[Definition:Coset|coset]] by $\map Z G$:
+:$\exists t \in G: \tau = t \map Z G$
+Thus each [[Definition:Coset|coset]] of $\map Z G$ in $G$ is equal to $\paren {t \map Z G}^i = t^i \map Z G$ for some $i \in \Z$.
+Now let $x, y \in G$.
+Suppose $x \in t^m \map Z G, y \in t^n \map Z G$.
+Then $x = t^m z_1, y = t^n z_2$ for some $z_1, z_2 \in \map Z G$.
+Thus:
+{{begin-eqn}}
+{{eqn | l = x y
+      | r = t^m z_1 t^n z_2
+      | c = 
+}}
+{{eqn | r = t^m t^n z_1 z_2
+      | c = $z_1$ [[Definition:Commute|commutes]] with all $t \in G$ since it is in the [[Definition:Center of Group|center]]
+}}
+{{eqn | r = t^{m + n} z_1 z_2
+      | c = 
+}}
+{{eqn | r = t^{n + m} z_2 z_1
+      | c = 
+}}
+{{eqn | r = t^n t^m z_2 z_1
+      | c = 
+}}
+{{eqn | r = t^n z_2 t^m z_1
+      | c = $z_2$ [[Definition:Commute|commutes]] with all $t \in G$ since it is in the [[Definition:Center of Group|center]]
+}}
+{{eqn | r = y x
+      | c = 
+}}
+{{end-eqn}}
+This holds for all $x, y \in G$, and thus $G$ is [[Definition:Abelian Group|abelian]].
+Thus by [[Group equals Center iff Abelian]] $\map Z G = G$.
+Therefore [[Quotient of Group by Itself]] it follows that $G / \map Z G$ is the [[Definition:Trivial Group|trivial group]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Centralizer is Normal Subgroup of Normalizer}
+Tags: Normal Subgroups, Normalizers, Quotient Groups
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]].
+Let $H \le G$ be a [[Definition:Subgroup|subgroup]] of $G$. 
+Let $\map {C_G} H$ be the [[Definition:Centralizer of Subgroup|centralizer]] of $H$ in $G$.
+Let $\map {N_G} H$ be the [[Definition:Normalizer|normalizer]] of $H$ in $G$.
+Let $\Aut H$ be the [[Definition:Automorphism Group of Group|automorphism group]] of $H$.
+Then:
+:$(1): \quad \map {C_G} H \lhd \map {N_G} H$
+:$(2): \quad \map {N_G} H / \map {C_G} H \cong K$
+where:
+:$\map {N_G} H / \map {C_G} H$ is the [[Definition:Quotient Group|quotient group]] of $\map {N_G} H$ by $\map {C_G} H$
+:$K$ is a [[Definition:Subgroup|subgroup]] of $\Aut H$.
+\end{theorem}
+
+\begin{proof}
+In order to invoke the [[First Isomorphism Theorem for Groups]], we must construct a [[Definition:Group Homomorphism|group homomorphism]] $\phi: \map {N_G} H \to \Aut H$.
+Consider the mapping $\phi: x \mapsto \paren {g \mapsto x g x^{-1}}$.
+From [[Inner Automorphism is Automorphism]], $g \mapsto x g x^{-1}$ is an [[Definition:Group Automorphism|automorphism]] of $G$, so $\phi$ is well-defined.
+To see that $\phi$ is a [[Definition:Group Homomorphism|homomorphism]], notice that for any $x, y \in \map {N_G} H$:
+{{begin-eqn}}
+{{eqn | l = \map \phi x \map \phi y
+      | r = \paren {g \mapsto x g x^{-1} } \circ \paren {g \mapsto y g y^{-1} }
+      | c = where $\circ$ denote [[Definition:Composition of Mappings|composition of maps]]
+}}
+{{eqn | r = g \mapsto x \paren {y g y^{-1} } x^{-1}
+}}
+{{eqn | r = g \mapsto \paren {x y} g \paren {x y}^{-1}
+      | c = [[Inverse of Group Product]]
+}}
+{{eqn | r = \map \phi {x y}
+}}
+{{end-eqn}}
+Hence $\phi$ is a [[Definition:Group Homomorphism|homomorphism]].
+Now we prove that $\ker \phi = \map {C_G} H$.
+Note that for $x \in \map {N_G} H$:
+{{begin-eqn}}
+{{eqn | l = x
+      | o = \in
+      | r = \ker \phi
+}}
+{{eqn | ll = \leadstoandfrom
+      | l = g
+      | r = x g x^{-1}
+      | rr = \forall g \in H
+      | c = $g \mapsto g$ is the [[Definition:Identity Element of Group|identity]] of $\Aut H$
+}}
+{{eqn | ll = \leadstoandfrom
+      | l = g x
+      | r = x g
+      | rr = \forall g \in H
+}}
+{{eqn | ll = \leadstoandfrom
+      | l = x
+      | o = \in
+      | r = \map {C_G} H
+      | c = {{Defof|Centralizer of Subgroup}}
+}}
+{{end-eqn}}
+Hence $\ker \phi = \map {C_G} H$.
+By [[Kernel is Normal Subgroup of Domain]]:
+:$\map {C_G} H \lhd \map {N_G} H$
+By [[First Isomorphism Theorem for Groups]]:
+:$\map {N_G} H / \map {C_G} H \cong \Img \phi$
+By [[Image of Group Homomorphism is Subgroup]]:
+:$\Img \phi \le \Aut H$
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Number of Distinct Conjugate Subsets is Index of Normalizer}
+Tags: Normalizers, Conjugacy, Number of Distinct Conjugate Subsets is Index of Normalizer
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]].
+Let $S$ be a [[Definition:Subset|subset]] of $G$.
+Let $\map {N_G} S$ be the [[Definition:Normalizer|normalizer]] of $S$ in $G$.
+Let $\index G {\map {N_G} S}$ be the [[Definition:Index of Subgroup|index of $\map {N_G} S$ in $G$]].
+The number of distinct [[Definition:Subset|subsets]] of $G$ which are [[Definition:Conjugate of Group Subset|conjugates]] of $S \subseteq G$ is $\index G {\map {N_G} S}$.
+\end{theorem}
+
+\begin{proof}
+Let $G$ [[Definition:Group Action|act on]] its [[Definition:Power Set|power set]] $\powerset G$ by the rule:
+:$\forall g \in G, S \in \powerset G: g * S := S^{g^{-1} } = \set {x \in G: g^{-1} x g \in S}$
+That is, the [[Definition:Conjugacy Action on Subsets|conjugacy action on subsets]].
+From [[Conjugacy Action on Subsets is Group Action]], $*$ is a  [[Definition:Group Action|group action]].
+The [[Definition:Orbit (Group Theory)|orbit]] of $S \in \powerset G$ is the [[Definition:Conjugacy Class|conjugacy class]] of $S$.
+That is, $\Orb S$ is the [[Definition:Set|set]] of [[Definition:Distinct|distinct]] [[Definition:Subset|subsets]] of $G$ which are [[Definition:Conjugate of Group Subset|conjugate]] to $S$
+The [[Definition:Stabilizer|stabilizer]] of $S$ is its [[Definition:Normalizer|normalizer]] $\map {N_G} S$.
+From the [[Orbit-Stabilizer Theorem]]:
+:$\order {\Orb x} = \index G {\Stab x} = \dfrac {\order G} {\order {\Stab x} }$
+from which the result follows directly.
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Subset has 2 Conjugates then Normal Subgroup}
+Tags: Conjugacy, Normal Subgroups
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]].
+Let $S$ be a [[Definition:Subset|subset]] of $G$.
+Let $S$ have exactly two [[Definition:Conjugate of Group Subset|conjugates]] in $G$.
+Then $G$ has a [[Definition:Proper Subgroup|proper]] [[Definition:Non-Trivial Group|non-trivial]] [[Definition:Normal Subgroup|normal subgroup]].
+\end{theorem}
+
+\begin{proof}
+{{MissingLinks}}
+Consider the [[Definition:Centralizer|centralizer]] $\map {C_G} S$ of $S$ in $G$.
+From [[Centralizer of Group Subset is Subgroup]], $\map {C_G} S$ is a [[Definition:Subgroup|subgroup]] of $G$.
+If $\map {C_G} S = G$, then $S$ has no [[Definition:Conjugate of Group Subset|conjugate]] but itself.
+{{explain|Link to that result.}}
+So, in order for $S$ to have exactly two [[Definition:Conjugate of Group Subset|conjugates]] in $G$, it is necessary for $\map {C_G} S$ to be a [[Definition:Proper Subgroup|proper subgroup]].
+Let $e$ be the [[Definition:Identity Element|identity]] of $G$.
+If $\map {C_G} S = \set e$, then for there to be exactly two [[Definition:Conjugate of Group Subset|conjugates]] of $S$: 
+:$\forall a \ne b \in G \setminus \set e: b x b^{-1} = a x a^{-1}$
+But:
+{{begin-eqn}}
+{{eqn | l = b x b^{-1}
+      | r = a x a^{-1}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \paren {a^{-1} b} x b^{-1}
+      | r = x a^{-1}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \paren {a^{-1} b} x \paren {a^{-1} b}^{-1}
+      | r = x
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = a^{-1} b
+      | o = \in
+      | r = \map {C_G} S
+      | c = 
+}}
+{{end-eqn}}
+This implies either that $\map {C_G} S$ is actually [[Definition:Non-Trivial Group|nontrivial]], or that $a^{-1}b = e \iff a = b$, a [[Definition:Contradiction|contradiction]].
+Thus $\map {C_G} S$ is a [[Definition:Non-Trivial Group|nontrivial]] [[Definition:Proper Subgroup|proper subgroup]] of $G$.
+We have that there are exactly $2$ [[Definition:Conjugacy Class|conjugacy classes]] of $S$.
+These are in [[Definition:One-to-One Correspondence|one-to-one correspondence]] with [[Definition:Coset|cosets]] of $S$.
+Thus the [[Definition:Index of Subgroup|index]] $\index G {\map {C_G} S}$ of the [[Definition:Centralizer of Group Subset|centralizer]] is:
+:$\index G {\map {C_G} S} = 2$
+From [[Subgroup of Index 2 is Normal]]:
+:$\map {C_G} S$ is a [[Definition:Proper Subgroup|proper]] [[Definition:Non-Trivial Group|nontrivial]] [[Definition:Normal Subgroup|normal subgroup]] of $G$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Element of Group Not Conjugate to Proper Subgroup}
+Tags: Conjugacy
+
+\begin{theorem}
+Let $G$ be a [[Definition:Finite Group|finite group]].
+Let $H$ be a [[Definition:Proper Subgroup|proper subgroup]] of $G$.
+Then there is at least one [[Definition:Element|element]] of $G$ not contained in $H$ or in any of its [[Definition:Conjugate of Group Subset|conjugates]].
+\end{theorem}
+
+\begin{proof}
+Let $S \subseteq G$ be defined by:
+:$S := \set {g \in G: \exists h \in H, a \in G: g = a h a^{-1} }$
+It is required to show that $S \ne G$.
+Let $\map {N_G} H$ be the [[Definition:Normalizer|normalizer]] of $H$ in $G$.
+By [[Subgroup is Subgroup of Normalizer]], $H \le \map {N_G} H$.
+Therefore, by definition of [[Definition:Index of Subgroup|index]]:
+:$\index G {\map {N_G} H} \le \index G H$
+Each of the [[Definition:Conjugate of Group Subset|conjugates]] of $H$ has $\order H$ elements.
+By [[Number of Distinct Conjugate Subsets is Index of Normalizer]], there are $\index G {\map {N_G} H}$ [[Definition:Conjugate of Group Subset|conjugates]].
+Therefore:
+:$\card S \le \order H \index G {\map {N_G} H}$
+with equality {{iff}} all [[Definition:Conjugate of Group Subset|conjugates]] of $H$ are [[Definition:Disjoint Sets|disjoint]].
+Combining this with the earlier inequality, we find that:
+:$\card S \le \order H \index G H = \order G$
+Now if $H$ is [[Definition:Normal Subgroup|normal]] in $G$, then by [[Normalizer of Subgroup is Largest Subgroup containing that Subgroup as Normal Subgroup]]:
+:$\index G {\map {N_G} H} < \index G H$
+for $\map {N_G} H = G$ in that case, and $H$ is a [[Definition:Proper Subgroup|proper subgroup]].
+If $H$ is not [[Definition:Normal Subgroup|normal]] in $G$, then it has multiple [[Definition:Conjugate of Group Subset|conjugates]].
+By [[Conjugate of Subgroup is Subgroup]], they all are [[Definition:Subgroup|subgroups]] of $G$.
+In particular, then, the [[Definition:Identity Element|identity]] of $G$ is common to all of them.
+Hence they are not [[Definition:Disjoint Sets|disjoint]], and:
+:$\card S < \order H \index G {\map {N_G} H}$
+In either case, it follows that:
+:$\card S < \order G$
+as desired.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Second Isomorphism Theorem}
+Tags: Named Theorems, Isomorphisms, Isomorphism Theorems
+
+\begin{theorem}
+=== [[Second Isomorphism Theorem/Groups|Groups]] ===
+{{:Second Isomorphism Theorem/Groups}}
+=== [[Second Isomorphism Theorem/Rings|Rings]] ===
+{{:Second Isomorphism Theorem/Rings}}
+This result is also referred to by some sources as the '''first isomorphism theorem'''.
+\end{theorem}<|endoftext|>
+\section{Third Isomorphism Theorem}
+Tags: Isomorphisms, Named Theorems, Isomorphism Theorems
+
+\begin{theorem}
+=== [[Third Isomorphism Theorem/Groups|Groups]] ===
+{{:Third Isomorphism Theorem/Groups}}
+=== [[Third Isomorphism Theorem/Rings|Rings]] ===
+{{:Third Isomorphism Theorem/Rings}}
+\end{theorem}<|endoftext|>
+\section{Parity Group is Group}
+Tags: Parity Group, Groups of Order 2
+
+\begin{theorem}
+The [[Definition:Parity Group|parity group]] is in fact a [[Definition:Group|group]].
+\end{theorem}
+
+\begin{proof}
+We can completely describe the parity group by showing its [[Definition:Cayley Table|Cayley table]]:
+:$\begin{array}{r|rr}
+\struct {\set {1, -1}, \times} & 1 & -1\\
+\hline
+1 & 1 & -1 \\
+-1 & -1 & 1 \\
+\end{array} \qquad
+\begin{array}{r|rr}
+\struct {\Z_2, +_2} & \eqclass 0 2 & \eqclass 1 2 \\
+\hline
+\eqclass 0 2 & \eqclass 0 2 & \eqclass 1 2 \\
+\eqclass 1 2 & \eqclass 1 2 & \eqclass 0 2 \\
+\end{array}$
+From [[Prime Group is Cyclic]], there is only one [[Definition:Group|group]] of [[Definition:Order of Group|order $2$]], up to [[Definition:Group Isomorphism|isomorphism]].
+Thus both instantiations of the [[Definition:Parity Group|parity group]] are [[Definition:Group Isomorphism|isomorphic]] to $C_2$, the [[Definition:Cyclic Group|cyclic group]] of [[Definition:Order of Group|order $2$]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Inverse of Inner Automorphism}
+Tags: Inner Automorphisms
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]].
+Let $x \in G$.
+Let $\kappa_x$ be the [[Definition:Inner Automorphism|inner automorphism of $G$ given by $x$]].
+Then:
+: $\paren {\kappa_x}^{-1} = \kappa_{x^{-1} }$
+\end{theorem}
+
+\begin{proof}
+Let $G$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
+Let $x \in G$.
+Let $\kappa_x \in \Inn G$.
+Then from the definition of [[Definition:Inner Automorphism|inner automorphism]]:
+:$\forall g \in G: \map {\kappa_x} g = x g x^{-1}$
+As $G$ is a [[Definition:Group|group]]:
+:$x \in G \implies x^{-1} \in G$
+So:
+:$\kappa_{x^{-1} } \in \Inn G$
+and is defined as:
+:$\forall g \in G: \map {\kappa_{x^{-1} } } g = x^{-1} g \paren {x^{-1} }^{-1} = x^{-1} g x$
+Now we need to show that:
+:$\kappa_x \circ \kappa_{x^{-1} } = I_G = \kappa_{x^{-1} } \circ \kappa_x$
+where $I_G: G \to G$ is the [[Definition:Identity Mapping|identity mapping]].
+So:
+{{begin-eqn}}
+{{eqn | ll= \forall g \in G:
+      | l = \map {\kappa_x \circ \kappa_{x^{-1} } } g
+      | r = \map {\kappa_x} {\map {\kappa_{x^{-1} } } g}
+      | c = {{Defof|Composition of Mappings}}
+}}
+{{eqn | r = \map {\kappa_x } {x^{-1} g x}
+      | c = Definition of $\kappa_{x^{-1} }$
+}}
+{{eqn | r = x \paren {x^{-1} g x} x^{-1}
+      | c = Definition of $\kappa_x$
+}}
+{{eqn | r = g
+      | c = [[Definition:Group|Group Properties]]
+}}
+{{eqn | r = x^{-1} \paren {x g x^{-1} } x
+      | c = [[Definition:Group|Group Properties]]
+}}
+{{eqn | r = \map {\kappa_{x^{-1} } } {x g x^{-1} }
+      | c = Definition of $\kappa_{x^{-1} }$
+}}
+{{eqn | r = \map {\kappa_{x^{-1} } } {\map {\kappa_x} g}
+      | c = Definition of $\kappa_x$
+}}
+{{eqn | r = \map {\kappa_{x^{-1} } \circ \kappa_x} g
+      | c = {{Defof|Composition of Mappings}}
+}}
+{{end-eqn}}
+Thus:
+:$\forall g \in G: \map {\kappa_x \circ \kappa_{x^{-1} } } g = \map {I_G} g = \map {\kappa_{x^{-1} } \circ \kappa_x} g$
+Hence the result.
+{{qed}}
+[[Category:Inner Automorphisms]]
+se1c1eh10u9lnbhu4qlivuoylagn5kf
+\end{proof}<|endoftext|>
+\section{Inner Automorphism Group is Isomorphic to Quotient Group with Center}
+Tags: Quotient Groups, Inner Automorphisms, Group Isomorphisms
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]].
+Let $\Inn G$ be the [[Definition:Inner Automorphism Group|inner automorphism group]] of $G$.
+Let $\map Z G$ be the [[Definition:Center of Group|center]] of $G$.
+Let $G / \map Z G$ be the [[Definition:Quotient Group|quotient group]] of $G$ by $\map Z G$.
+Then $G / \map Z G \cong \Inn G$.
+\end{theorem}
+
+\begin{proof}
+Let $G$ be a [[Definition:Group|group]].
+Let the [[Definition:Mapping|mapping]] $\kappa: G \to \Inn G$ be defined as:
+:$\map \kappa a = \kappa_a$
+where $\kappa_a$ is the [[Definition:Inner Automorphism|inner automorphism]] of $G$ given by $a$.
+From [[Kernel of Inner Automorphism Group is Center]]:
+:$\map \ker \kappa = \map Z G$
+and also that:
+:$\Img \kappa = \Inn G$
+From the [[First Isomorphism Theorem for Groups|First Isomorphism Theorem]]:
+:$\Img \kappa \cong G / \map \ker \kappa$
+Thus, as $\map \ker \kappa = \map Z G$ and $\Img \kappa = \Inn G$:
+:$G / \map Z G \cong \Inn G$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Conjugates of Elements in Centralizer}
+Tags: Conjugacy, Centralizers, Cosets
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]].
+Let $\map {C_G} a$ be the [[Definition:Centralizer of Group Element|centralizer]] of $a$ in $G$.
+Then $\forall g, h \in G: g a g^{-1} = h a h^{-1}$ {{iff}} $g$ and $h$ belong to the same [[Definition:Left Coset|left coset]] of $\map {C_G} a$.
+\end{theorem}
+
+\begin{proof}
+The [[Definition:Centralizer of Group Element|centralizer]] of $a$ in $G$ is defined as:
+:$\map {C_G} a = \set {x \in G: x \circ a = a \circ x}$
+Let $g, h \in G$.
+Then:
+{{begin-eqn}}
+{{eqn | l = g a g^{-1}
+      | r = h a h^{-1}
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = g^{-1} \paren {g a g^{-1} } h
+      | r = g^{-1} \paren {h a h^{-1} } h
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = a g^{-1} h
+      | r = g^{-1} h a
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = g^{-1} h
+      | o = \in
+      | r = \map {C_G} a
+      | c = {{Defof|Centralizer of Group Element}}
+}}
+{{end-eqn}}
+By [[Elements in Same Left Coset iff Product with Inverse in Subgroup]]:
+:$g$ and $h$ belong to the same [[Definition:Left Coset|left coset]] of $\map {C_G} a$ {{iff}} $g^{-1} h \in \map {C_G} a$.
+The result follows.
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Number of Conjugates is Number of Cosets of Centralizer}
+Tags: Conjugacy, Centralizers, Cosets
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]].
+Let $\map {C_G} a$ be the [[Definition:Centralizer of Group Element|centralizer]] of $a$ in $G$.
+Then the number of different [[Definition:Conjugate of Group Element|conjugates]] of $a$ in $G$ equals the number of different [[Definition:Left Coset|(left) cosets]] of $\map {C_G} a$:
+:$\card {\conjclass a} = \index G {\map {C_G} a}$
+where:
+:$\conjclass a$ is the [[Definition:Conjugacy Class|conjugacy class]] of $a$ in $G$
+:$\index G {\map {C_G} a}$ is the [[Definition:Index of Subgroup|index of $\map {C_G} a$ in $G$]].
+Consequently:
+:$\card {\conjclass a} \divides \order G$
+\end{theorem}
+
+\begin{proof}
+Let $x, y \in \conjclass a$.
+By definition of $\conjclass a$:
+:$x a x^{-1} = y a y^{-1}$
+By [[Conjugates of Elements in Centralizer]], this is the case {{iff}} $x$ and $y$ belong to the same [[Definition:Left Coset|left coset]] of $\map {C_G} a$.
+Hence:
+:$\card {\conjclass a} = \index G {\map {C_G} a}$
+It follows from [[Lagrange's Theorem (Group Theory)|Lagrange's Theorem]] that:
+:$\card {\conjclass a} \divides \order G$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Size of Conjugacy Class is Index of Normalizer}
+Tags: Conjugacy Classes, Normalizers
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]].
+Let $x \in G$.
+Let $\conjclass x$ be the [[Definition:Conjugacy Class|conjugacy class]] of $x$ in $G$.
+Let $\map {N_G} x$ be the [[Definition:Normalizer|normalizer of $x$ in $G$]].
+Let $\index G {\map {N_G} x}$ be the [[Definition:Index of Subgroup|index of $\map {N_G} x$ in $G$]].
+The number of elements in $\conjclass x$ is $\index G {\map {N_G} x}$.
+\end{theorem}
+
+\begin{proof}
+The number of elements in $\conjclass x$ is the number of [[Definition:Conjugate of Group Subset|conjugates]] of the set $\set x$.
+From [[Number of Distinct Conjugate Subsets is Index of Normalizer]], the number of distinct [[Definition:Subset|subsets]] of a $G$ which are [[Definition:Conjugate of Group Subset|conjugates]] of $S \subseteq G$ is $\index G {\map {N_G} S}$.
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Conjugacy Class of Element of Center is Singleton}
+Tags: Conjugacy Classes, Centers of Groups, Singletons
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]].
+Let $\map Z G$ denote the [[Definition:Center of Group|center]] of $G$.
+The [[Definition:Element|elements]] of $\map Z G$ form [[Definition:Singleton|singleton]] [[Definition:Conjugacy Class|conjugacy classes]], and the elements of $G \setminus \map Z G$ belong to multi-element [[Definition:Conjugacy Class|conjugacy classes]].
+\end{theorem}
+
+\begin{proof}
+Let $\conjclass a$ be the [[Definition:Conjugacy Class|conjugacy class]] of $a$ in $G$.
+{{begin-eqn}}
+{{eqn | l = a
+      | o = \in
+      | r = \map Z G
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \forall x \in G: x a
+      | r = a x
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \forall x \in G: x a x^{-1}
+      | r = a
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \conjclass a
+      | r = \set a
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Conjugacy Class Equation}
+Tags: Conjugacy, Normal Subgroups, Named Theorems, Normalizers, Conjugacy Class Equation, Conjugacy Classes
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]].
+Let $\order G$ denote the [[Definition:Order of Structure|order]] of $G$.
+Let $\map Z G$ denote the [[Definition:Center of Group|center]] of $G$.
+Let $x \in G$.
+Let $\map {N_G} x$ denote the [[Definition:Normalizer|normalizer of $x$ in $G$]].
+Let $\index G {\map {N_G} x}$ denote the [[Definition:Index of Subgroup|index of $\map {N_G} x$ in $G$]].
+Let $m$ be the number of [[Definition:Singleton|non-singleton]] [[Definition:Conjugacy Class|conjugacy classes]] of $G$.
+Let $x_j: j \in \set {1, 2, \ldots, m}$ be arbitrary [[Definition:Element|elements]] of those [[Definition:Conjugacy Class|conjugacy classes]].
+Then:
+:$\displaystyle \order G = \order {\map Z G} + \sum_{j \mathop = 1}^m \index G {\map {N_G} {x_j} }$
+\end{theorem}
+
+\begin{proof}
+From [[Conjugacy Class of Element of Center is Singleton]], all [[Definition:Element|elements]] of $\map Z G$ form their own [[Definition:Singleton|singleton]] [[Definition:Conjugacy Class|conjugacy classes]].
+=== Abelian Group ===
+Suppose $G$ is [[Definition:Abelian Group|abelian]].
+Then from [[Group equals Center iff Abelian]] we have:
+:$\map Z G = G$
+So there are as many [[Definition:Conjugacy Class|conjugacy classes]] as there are [[Definition:Element|elements]] in $\map Z G$ and hence in $G$.
+So in this case the result certainly holds.
+{{qed|lemma}}
+=== Non-Abelian Group ===
+Now suppose $G$ is non-abelian.
+Thus:
+:$\map Z G \ne G$
+and therefore:
+:$G \setminus \map Z G \ne \O$
+From [[Conjugacy Class of Element of Center is Singleton]], all the non-[[Definition:Singleton|singleton]] [[Definition:Conjugacy Class|conjugacy classes]] of $G$ are in $G \setminus \map Z G$.
+From the way the theorem has been worded, there are $m$ of them.
+Let us choose one element from each of the non-[[Definition:Singleton|singleton]] [[Definition:Conjugacy Class|conjugacy classes]] and call them $x_1, x_2, \ldots, x_m$.
+Thus, these [[Definition:Conjugacy Class|conjugacy classes]] can be written:
+:$\conjclass {x_1}, \conjclass {x_2}, \ldots, \conjclass {x_m}$
+So:
+:$\displaystyle \order {G \setminus \map Z G} = \sum_{j \mathop = 1}^m \order {\conjclass {x_j} }$
+or:
+:$\displaystyle \order G - \order {\map Z G} = \sum_{j \mathop = 1}^m \order {\conjclass {x_j} }$
+From [[Size of Conjugacy Class is Index of Normalizer]]:
+:$\order {\conjclass {x_j} } = \index G {\map {N_G} {x_j} }$
+and the result follows.
+{{Qed}}
+\end{proof}
+
+\begin{proof}
+Let the distinct [[Definition:Orbit (Group Theory)|orbits]] of $G$ under the [[Definition:Conjugacy Action|conjugacy action]] be:
+:$\Orb {x_1}, \Orb {x_2}, \ldots, \Orb {x_s}$
+Then from the [[Partition Equation]]:
+:$\order G = \order {\Orb {x_1} } + \order {\Orb {x_2} } + \cdots + \order {\Orb {x_s} }$
+From the [[Orbit-Stabilizer Theorem]]:
+:$\order {\Orb {x_i} } \divides \order G, i = 1, \ldots, s$
+The result follows from the definition of the [[Definition:Conjugacy Action|conjugacy action]].
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Group of Order Prime Squared is Abelian}
+Tags: Order of Groups, Abelian Groups, P-Groups, Groups of Order p^2
+
+\begin{theorem}
+A [[Definition:Group|group]] whose [[Definition:Order of Structure|order]] is the [[Definition:Square Number|square]] of a [[Definition:Prime Number|prime]] is [[Definition:Abelian Group|abelian]].
+\end{theorem}
+
+\begin{proof}
+Let $G$ be a [[Definition:Group|group]] of order $p^2$, where $p$ is [[Definition:Prime Number|prime]].
+Let $\map Z G$ be the [[Definition:Center of Group|center]] of $G$.
+By [[Center of Group is Subgroup]], $\map Z G$ is a [[Definition:Subgroup|subgroup]] of $G$.
+By [[Lagrange's Theorem (Group Theory)|Lagrange's Theorem]]:
+:$\order {\map Z G} \divides \order G$
+It follows that $\order {\map Z G} = 1, p$ or $p^2$.
+By [[Center of Group of Prime Power Order is Non-Trivial]]:
+:$\order {\map Z G} \ne 1$
+Suppose $\order {\map Z G} = p$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \order {\map Z G}
+      | r = \index G {\map Z G}
+      | c = {{Defof|Index of Subgroup}}
+}}
+{{eqn | r = \order G / \order {\map Z G}
+      | c = [[Lagrange's Theorem (Group Theory)|Lagrange's Theorem]]
+}}
+{{eqn | r = p^2 / p
+      | c = 
+}}
+{{eqn | r = p
+      | c = 
+}}
+{{end-eqn}}
+So $G / \map Z G$ is [[Definition:Non-Trivial Group|non-trivial]], and of [[Definition:Prime Number|prime]] [[Definition:Order of Structure|order]].
+From [[Prime Group is Cyclic]], $G / \map Z G$ is a [[Definition:Cyclic Group|cyclic group]].
+But by [[Quotient of Group by Center Cyclic implies Abelian]], that cannot be the case.
+Therefore $\order {\map Z G} = p^2$ and therefore $\map Z G = G$.
+Therefore $G$ is [[Definition:Abelian Group|abelian]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Center of Group of Prime Power Order is Non-Trivial}
+Tags: Group Theory, P-Groups, Centers of Groups
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]] whose [[Definition:Order of Structure|order]] is the [[Definition:Prime Power|power of a prime]].
+Then the [[Definition:Center of Group|center]] of $G$ is [[Definition:Trivial Group|non-trivial]]:
+:$\forall G: \order G = p^r: p \in \mathbb P, r \in \N_{>0}: \map Z G \ne \set e$
+\end{theorem}
+
+\begin{proof}
+Suppose $G$ is [[Definition:Abelian Group|abelian]].
+From [[Group equals Center iff Abelian]]:
+: $\map Z G = G$
+and the result is seen to be true as $G$ is itself [[Definition:Non-Trivial Group|non-trivial]].
+From [[Prime Group is Cyclic]] and [[Cyclic Group is Abelian]], this will always be the case for $r = 1$.
+So, suppose $G$ is non-[[Definition:Abelian Group|abelian]].
+Thus $\map Z G \ne G$ and therefore $G \setminus \map Z G \ne \O$.
+Let $\conjclass {x_1}, \conjclass {x_2}, \ldots, \conjclass {x_m}$ be the [[Definition:Conjugacy Class|conjugacy classes]] into which $G \setminus \map Z G$ is [[Definition:Partition (Set Theory)|partitioned]].
+From [[Conjugacy Class of Element of Center is Singleton]], all of these will have more than one element.
+From the [[Conjugacy Class Equation]]:
+:$\displaystyle \order {\map Z G} = \order G - \sum_{j \mathop = 1}^m \order {\conjclass {x_j} }$
+From [[Number of Conjugates is Number of Cosets of Centralizer]]:
+:$\order {\conjclass {x_j} } \divides \order G$
+Let $\map {N_G} x$ be the [[Definition:Normalizer|normalizer of $x$ in $G$]].
+Then:
+:$\forall j: 1 \le j \le m: \index G {\map {N_G} {x_j} } > 1 \implies p \divides \index G {\map {N_G} {x_j} }$
+Since $p \divides \order G$, it follows that:
+: $p \divides \order {\map Z G}$
+and the result follows.
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Center of Group of Order Prime Cubed}
+Tags: Normal Subgroups, Centers of Groups, P-Groups
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]] of [[Definition:Order of Structure|order]] $p^3$, where $p$ is a [[Definition:Prime Number|prime]].
+Let $\map Z G$ be the [[Definition:Center of Group|center]] of $G$.
+Then $\order {\map Z G} \ne p^2$.
+\end{theorem}
+
+\begin{proof}
+{{AimForCont}} $\order {\map Z G} = p^2$.
+Then $\order {G / \map Z G} = p$.
+From [[Prime Group is Cyclic]] it follows that $G / \map Z G$ is [[Definition:Cyclic Group|cyclic]].
+From [[Quotient of Group by Center Cyclic implies Abelian]] it follows that $G$ is [[Definition:Abelian Group|abelian]].
+From [[Group equals Center iff Abelian]] it follows that $\order {\map Z G} = p^3$.
+The result follows by [[Proof by Contradiction]].
+{{qed}}
+{{expand|This results generalizes naturally to a group of order $p^n$: its center is not of order $p^{n-1}$.}}
+[[Category:Normal Subgroups]]
+[[Category:Centers of Groups]]
+[[Category:P-Groups]]
+37p385xpkzyb8mfwohstvgjm9u679z5
+\end{proof}<|endoftext|>
+\section{Prime Power Group has Non-Trivial Proper Normal Subgroup}
+Tags: Normal Subgroups, P-Groups
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]], whose [[Definition:Identity Element|identity]] is $e$, such that $\order G = p^n: n > 1, p \in \mathbb P$.
+Then $G$ has a [[Definition:Proper Subgroup|proper]] [[Definition:Normal Subgroup|normal subgroup]] which is [[Definition:Non-Trivial Group|non-trivial]].
+\end{theorem}
+
+\begin{proof}
+From [[Center of Group is Normal Subgroup]], $\map Z G \lhd G$.
+By [[Center of Group of Prime Power Order is Non-Trivial]], $\map Z G$ is [[Definition:Non-Trivial Group|non-trivial]].
+If $\map Z G$ is a [[Definition:Proper Subgroup|proper subgroup]], the proof is finished.
+Otherwise, $\map Z G = G$.
+Then $G$ is [[Definition:Abelian Group|abelian]] by [[Group equals Center iff Abelian]].
+However, then any $a \in G: a \ne e$ generates a [[Definition:Non-Trivial Group|non-trivial]] [[Definition:Normal Subgroup|normal subgroup]] $\gen a$, as [[Subgroup of Abelian Group is Normal]].
+If $\order a = \order {\gen a} < \order G$, the proof is complete.
+Otherwise:
+:$\order a = \order G = p^n$
+Then $\order {a^p} = p^{n - 1}$ from [[Subgroup of Finite Cyclic Group is Determined by Order]], so $a^p$ generates that [[Definition:Non-Trivial Group|non-trivial]] [[Definition:Normal Subgroup|normal subgroup]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Composition Series of Group of Prime Power Order}
+Tags: Cyclic Groups, Normal Series, P-Groups
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$, and whose [[Definition:Order of Structure|order]] is a [[Definition:Prime Power|prime power]]:
+:$\order G = p^n, p \in \mathbb P, n \ge 1$
+Then $G$ has a [[Definition:Composition Series|composition series]]:
+:$\set e = G_0 \subset G_1 \subset \ldots \subset G_n = G$
+such that $\order {G_k} = p^k$, $G_k \lhd G_{k + 1}$ and $G_{k + 1} / G_k$ is [[Definition:Cyclic Group|cyclic]] and of [[Definition:Order of Structure|order]] $p$.
+\end{theorem}
+
+\begin{proof}
+To be proved by [[Second Principle of Mathematical Induction|induction]] on $n$.
+Let $P_n$ be the [[Definition:Proposition|proposition]] for $\order G = p^n$.
+=== Basis for the Induction ===
+$P_1$ is trivially true because:
+:$\set e = G_0 \subset G_1 = G$
+From [[Prime Group is Cyclic]], a [[Definition:Group|group]] whose [[Definition:Order of Structure|order]] is [[Definition:Prime Number|prime]] is [[Definition:Cyclic Group|cyclic]].
+=== Induction Hypothesis ===
+Suppose $P_k$ is true for all [[Definition:Group|groups]] of order $p^k$ for all $k < n$.
+We need to show that this implies $P_{k + 1}$ is true.
+=== Induction Step ===
+Let $G$ be a [[Definition:Group|group]] of [[Definition:Order of Structure|order]] $p^n$.
+By [[Prime Power Group has Non-Trivial Proper Normal Subgroup]], $G$ has a [[Definition:Proper Subgroup|proper]] [[Definition:Trivial Subgroup|non-trivial]] [[Definition:Normal Subgroup|normal subgroup]].
+There will be a [[Definition:Finite Set|finite number]] of these, so we are free to pick one of maximal [[Definition:Order of Structure|order]].
+We call this $H$, such that $\order H = p^t, t < n$.
+We need to show that $t = n - 1$.
+Suppose $t < n - 1$.
+Then $G / H$ is a [[Definition:Group|group]] of [[Definition:Order of Structure|order]] $p^{n - t} \ge p^2$.
+Again by [[Prime Power Group has Non-Trivial Proper Normal Subgroup]], $G / H$ has a [[Definition:Proper Subgroup|proper]] [[Definition:Trivial Subgroup|non-trivial]] [[Definition:Normal Subgroup|normal subgroup]], which we will call $N$.
+Let $H' = \set {g \in G: g H \in N}$.
+We now show that $H \lhd G$.
+Let $g, g' \in H'$.
+Then $g H, g' H \in N$.
+Since $N < G / H$:
+: $\paren {g H} \paren {g' H} = g g' H \in N$
+and so $g g' \in N$.
+If $g \in H$, then $g H \in N$.
+Since $N < G / H$:
+:$\paren {g H}^{-1} = g^{-1} H \in N$
+and so $g^{-1} \in H'$.
+Next:
+{{begin-eqn}}
+{{eqn | l = \paren {H'}^a
+      | r = \set {g \in G: a g a^{-1} \in H'}
+      | c = 
+}}
+{{eqn | r = \set {g \in G: a g a^{-1} H \in N}
+      | c = 
+}}
+{{eqn | r = \set {g \in G: \paren {a H} \paren {g H} \paren {a H}^{-1} \in N}
+      | c = 
+}}
+{{eqn | r = \set {g \in G: \paren {g H} \in N^{a H} }
+      | c = {{Defof|Conjugate of Group Subset}}
+}}
+{{eqn | r = \set {g \in G: \paren {g H} \in N}
+      | c = as $N$ is [[Definition:Normal Subgroup|normal]]
+}}
+{{eqn | r = H'
+      | c = 
+}}
+{{end-eqn}}
+So clearly $H' / H = N$, therefore:
+:$\dfrac {\order {H'} } {\order H} = \index {H'} N = \order N \ge p$
+So:
+:$\order {H'} \ge p \order H$
+contradicting the maximality of $\order H$.
+It follows that $t = n - 1$.
+Finally, we set $G_{n - 1} = H$ and use induction to show that $P_{n - 1}$ holds.
+Since $G / H = G / G_{n - 1}$ is a [[Definition:Group|group]] of [[Definition:Order of Structure|order]] $p$, then it is automatically [[Definition:Cyclic Group|cyclic]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sum Rule for Counting}
+Tags: Combinatorics, Counting Arguments
+
+\begin{theorem}
+Let there be:
+: $r_1$ different [[Definition:Object|objects]] in the [[Definition:Set|set]] $S_1$
+: $r_2$ different [[Definition:Object|objects]] in the set $S_2$
+: $\ldots$
+: $r_m$ different [[Definition:Object|objects]] in the set $S_m$.
+Let $\displaystyle \bigcap_{i \mathop = 1}^m S_i = \varnothing$.
+Then the number of ways to select an object from one of the $m$ sets is $\displaystyle \sum_{i \mathop = 1}^m r_i$.
+\end{theorem}
+
+\begin{proof}
+A direct application of [[Cardinality of Set Union]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Product Rule for Counting}
+Tags: Combinatorics, Counting Arguments, Product Rule for Counting
+
+\begin{theorem}
+Let it be possible to choose an [[Definition:Element|element]] $\alpha$ from a given [[Definition:Set|set]] $S$ in $m$ different ways.
+Let it be possible to choose an [[Definition:Element|element]] $\beta$ from a given [[Definition:Set|set]] $T$ in $n$ different ways.
+Then the [[Definition:Ordered Pair|ordered pair]] $\tuple {\alpha, \beta}$ can be chosen from the [[Definition:Cartesian Product|cartesian product]] $S \times T$ in $m n$ different ways.
+\end{theorem}
+
+\begin{proof}
+{{handwaving}}
+The validity of this rule follows directly from the definition of [[Definition:Integer Multiplication|multiplication of integers]].
+The product $a b$ (for $a, b \in \N_{>0}$) is the number of [[Definition:Sequence|sequences]] $\sequence {A, B}$, where $A$ can be any one of $a$ items and $B$ can be any one of $b$ items.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Commutativity of Group Direct Product}
+Tags: Group Direct Products
+
+\begin{theorem}
+Let $\struct {G, \circ_g}$ and $\struct {H, \circ_h}$ be [[Definition:Group|groups]].
+Let $\struct {G \times H, \circ}$ be the [[Definition:Group Direct Product|group direct product]] of $\struct {G, \circ_g}$ and $\struct {H, \circ_h}$, where the operation $\circ$ is defined as:
+:$\tuple {g_1, h_1} \circ \tuple {g_2, h_2} = \tuple {g_1 \circ_g g_2, h_1 \circ_h h_2}$
+Let $\struct {H \times G, \star}$ be the [[Definition:Group Direct Product|group direct product]] of $\struct {H, \circ_h}$ and $\struct {G, \circ_g}$, where the operation $\star$ is defined as:
+:$\tuple {h_1, g_1} \star \tuple {h_2, g_2} = \tuple {h_1 \circ_h h_2, g_1 \circ_g g_2}$
+The [[Definition:Group Direct Product|group direct product]] $\struct {G \times H, \circ}$ is [[Definition:Group Isomorphism|isomorphic]] to the $\struct {H \times G, \star}$.
+\end{theorem}
+
+\begin{proof}
+The [[Definition:Mapping|mapping]] $\theta: G \times H \to H \times G$ defined as:
+:$\forall g \in G, h \in H: \map \theta {g, h} = \tuple {h, g}$
+is to be shown to be a [[Definition:Group Homomorphism|group homomorphism]], and that $\theta$ is [[Definition:Bijection|bijective]], as follows:
+=== Injective ===
+{{begin-eqn}}
+{{eqn | l = \map \theta {g_1, h_1}
+      | r = \map \theta {g_2, h_2}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \tuple {h_1, g_1}
+      | r = \tuple {h_2, g_2}
+      | c = Definition of $\theta$
+}}
+{{eqn | ll= \leadsto
+      | l = \tuple {g_1, h_1}
+      | r = \tuple {g_2, h_2}
+      | c = [[Equality of Ordered Pairs]]
+}}
+{{end-eqn}}
+Thus $\theta$ is [[Definition:Injection|injective]] by definition.
+{{qed|lemma}}
+=== Surjective ===
+{{begin-eqn}}
+{{eqn | l = \tuple {h, g}
+      | o = \in
+      | r = H \times G
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \tuple {g, h}
+      | o = \in
+      | r = G \times H
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | lo= \exists \tuple {g, h} \in G \times H:
+      | l = \tuple {h, g}
+      | r = \map \theta {g, h}
+      | c = 
+}}
+{{end-eqn}}
+Thus $\theta$ is [[Definition:Surjection|surjective]] by definition.
+{{qed|lemma}}
+=== Group Homomorphism ===
+Let $\tuple {g_1, h_1}, \tuple {g_2, h_2} \in G \times H$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \map \theta {\tuple {g_1, h_1} \circ \tuple {g_2, h_2} }
+      | r = \map \theta {g_1 \circ_g g_2, h_1 \circ_h h_2}
+      | c = Definition of $\circ$
+}}
+{{eqn | r = \tuple {h_1 \circ_h h_2, g_1 \circ_g g_2}
+      | c = Definition of $\theta$
+}}
+{{eqn | r = \tuple {h_1, g_1} \star \tuple {h_2, g_2}
+      | c = Definition of $\star$
+}}
+{{eqn | r = \map \theta {g_1, h_1} \star \map \theta {g_2, h_2}
+      | c = Definition of $\theta$
+}}
+{{end-eqn}}
+thus proving that $\theta$ is a [[Definition:Group Homomorphism|homomorphism]].
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{External Direct Product of Abelian Groups is Abelian Group}
+Tags: Group Direct Products, Abelian Groups
+
+\begin{theorem}
+Let $\struct {G, \circ_1}$ and $\struct {H, \circ_2}$ be [[Definition:Group|groups]].
+Then the [[Definition:Group Direct Product|group direct product]] $\struct {G \times H, \circ}$ is [[Definition:Abelian Group|abelian]] {{iff}} both $\struct {G, \circ_1}$ and $\struct {H, \circ_2}$ are [[Definition:Abelian Group|abelian]].
+\end{theorem}
+
+\begin{proof}
+Let $\struct {G, \circ_1}$ and $\struct {H, \circ_2}$ be [[Definition:Group|groups]] whose [[Definition:Identity Element|identities]] are $e_G$ and $e_H$ respectively.
+Suppose $\struct {G, \circ_1}$ and $\struct {H, \circ_2}$ are both [[Definition:Abelian Group|abelian]].
+Then from [[External Direct Product Commutativity]], $\struct {G \times H, \circ}$ is also [[Definition:Abelian Group|abelian]].
+Now suppose that $\struct {G \times H, \circ}$ is [[Definition:Abelian Group|abelian]].
+Then:
+{{begin-eqn}}
+{{eqn | l = \tuple {g_1 \circ_1 g_2, e_H}
+      | r = \tuple {g_1 \circ_1 g_2, e_H \circ_2 e_H}
+      | c = Definition of $e_H$
+}}
+{{eqn | r = \tuple {g_1, e_H} \circ \tuple {g_2, e_H}
+      | c = {{Defof|Group Direct Product}}
+}}
+{{eqn | r = \tuple {g_2, e_H} \circ \tuple {g_1, e_H}
+      | c = as $\struct {G \times H, \circ}$ is [[Definition:Abelian Group|abelian]]
+}}
+{{eqn | r = \tuple {g_2 \circ_1 g_1, e_H \circ_2 e_H}
+      | c = {{Defof|Group Direct Product}}
+}}
+{{eqn | r = \tuple {g_2 \circ_1 g_1, e_H}
+      | c = Definition of $e_H$
+}}
+{{end-eqn}}
+Thus:
+: $g_1 \circ_1 g_2 = g_2 \circ_1 g_1$
+and $\struct {G, \circ_1}$ is seen to be [[Definition:Abelian Group|abelian]].
+The same argument holds for $\struct {H, \circ_2}$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Associativity of Group Direct Product}
+Tags: Group Direct Products, Associativity
+
+\begin{theorem}
+The [[Definition:Group Direct Product|group direct product]] $G \times \paren {H \times K}$ is [[Definition:Group Isomorphism|(group) isomorphic]] to $\paren {G \times H} \times K$.
+\end{theorem}
+
+\begin{proof}
+Let $G, H, K$ be [[Definition:Group|groups]].
+The [[Definition:Mapping|mapping]] $\theta: G \times \paren {H \times K} \to \paren {G \times H} \times K$ defined as:
+:$\forall g \in G, h \in H, k \in K: \map \theta {\tuple {g, \tuple {h, k} } } = \tuple {\tuple {g, h}, k}$
+is to be shown to be a [[Definition:Group Homomorphism|group homomorphism]], and that $\theta$ is [[Definition:Bijection|bijective]], as follows:
+=== Injective ===
+Let $\map \theta {\tuple {g_1, \tuple {h_1, k_1} } } = \map \theta {\tuple {g_2, \tuple {h_2, k_2} } }$.
+By the definition of $\theta$:
+:$\tuple {\tuple {g_1, h_1}, k_1} = \tuple {\tuple {g_2, h_2}, k_2}$
+By [[Equality of Ordered Pairs]]:
+:$\tuple {g_1, h_1} = \tuple {g_2, h_2}$
+:$k_1 = k_2$
+and consequently:
+:$g_1 = g_2, h_1 = h_2$
+Thus:
+:$\tuple {g_1, \tuple {h_1, k_1} } = \tuple {g_2, \tuple {h_2, k_2} }$
+and so $\theta$ is seen to be [[Definition:Injection|injective]].
+=== Surjective ===
+If $\tuple {\tuple {g, h}, k} \in \paren {G \times H} \times K$, then $g \in G, h \in H, k \in K$.
+Thus:
+:$\tuple {g, \tuple {h, k} } = \in G \times \paren {H \times K}$
+and:
+:$\map \theta {\tuple {g, \tuple {h, k} } } = \tuple {\tuple {g, h}, k}$
+and so $\theta$ is seen to be [[Definition:Surjection|surjective]].
+=== Group Homomorphism ===
+Now let $\tuple {g_1, \tuple {h_1, k_1} }, \tuple {g_2, \tuple {h_2, k_2} } \in G \times \tuple {H \times K}$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \map \theta {\tuple {g_1, \tuple {h_1, k_1} } \tuple {g_2, \tuple {h_2, k_2} } }
+      | r = \map \theta {\tuple {g_1 g_2, \tuple {h_1, k_1} \tuple {h_2, k_2} } }
+      | c = 
+}}
+{{eqn | r = \map \theta {\tuple {g_1 g_2, \tuple {h_1 h_2, k_1 k_2} } }
+      | c = 
+}}
+{{eqn | r = \tuple {\tuple {g_1 g_2, h_1 h_2}, k_1 k_2}
+      | c = 
+}}
+{{eqn | r = \tuple {\tuple {g_1, h_1} \tuple {g_2, h_2}, k_1 k_2}
+      | c = 
+}}
+{{eqn | r = \tuple {\tuple {g_1, h_1}, k_1} \tuple {\tuple {g_2, h_2}, k_2}
+      | c = 
+}}
+{{eqn | r = \map \theta {\tuple {g_1, \tuple {h_1, k_1} } } \map \theta {\tuple {g_2, \tuple {h_2, k_2} } }
+      | c = 
+}}
+{{end-eqn}}
+showing that $\theta$ is a [[Definition:Group Homomorphism|(group) homomorphism]].
+The result follows.
+{{qed}}
+[[Category:Group Direct Products]]
+[[Category:Associativity]]
+0e07ovfdbn3d1ksqoxy8c7rugsa4xz5
+\end{proof}<|endoftext|>
+\section{Group Direct Product of Cyclic Groups}
+Tags: Group Direct Products, Cyclic Groups, Group Direct Product of Cyclic Groups
+
+\begin{theorem}
+Let $G$ and $H$ both be [[Definition:Finite Group|finite]] [[Definition:Cyclic Group|cyclic groups]] with [[Definition:Order of Group|orders]] $n = \order G$ and $m = \order H$ respectively.
+Then their [[Definition:Group Direct Product|group direct product]] $G \times H$ is [[Definition:Cyclic Group|cyclic]] {{iff}} $g$ and $h$ are [[Definition:Coprime Integers|coprime]], that is, $g \perp h$.
+\end{theorem}
+
+\begin{proof}
+Let $G$ and $H$ be [[Definition:Group|groups]] whose [[Definition:Identity Element|identities]] are $e_G$ and $e_H$ respectively.
+=== Necessary condition ===
+Suppose:
+: $(1): \quad \order G = n, G = \gen x$
+: $(2): \quad \order H = m, H = \gen y$
+: $(3): \quad m \perp n$
+Then:
+{{begin-eqn}}
+{{eqn | l = \order {\tuple {x, y} }
+      | r = k
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \tuple {x, y}^k
+      | r = e_{G \times H} = \tuple {x^k, y^k}
+      | c = [[Powers of Elements in Group Direct Product]]
+}}
+{{eqn | ll= \leadsto
+      | l = x^k
+      | r = e_G, y^k = e_H
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = n
+      | o = \divides
+      | r = k, m \divides k
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \lcm \set {n, m}
+      | o = \divides
+      | r = k
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = n m
+      | o = \divides
+      | r = k
+      | c = as $n \perp m$
+}}
+{{end-eqn}}
+But then:
+:$\tuple {x, y}^{n m} = e_{G \times H} = \tuple {x^{n m}, y^{n m} }$
+Thus:
+:$k \divides n m$
+So:
+:$\order {\tuple {x, y} } = n m \implies \gen {\tuple {x, y} } = G \times H$
+{{qed|lemma}}
+=== Sufficient condition ===
+Suppose that $G \times H$ is [[Definition:Cyclic Group|cyclic]].
+Let $\tuple {x, y}$ be a [[Definition:Generator of Group|generator]] of $G \times H$.
+By [[Cardinality of Cartesian Product]] the [[Definition:Order of Group|order]] of $G \times H$ is:
+:$\order G \cdot \order H = g h$
+Therefore by [[Order of Cyclic Group equals Order of Generator]]:
+:$\order {\tuple {x, y} } = g h$
+On the other hand, by [[Order of Group Element in Group Direct Product]] we have:
+:$\order {\tuple {x, y} } = \lcm \set {\tuple {\order x, \order y} }$
+Next we claim that $x$ generates $G$.
+Let $x' \in G$.
+Then:
+:$\tuple {x', e_H} \in G \times H$
+so there exists $k \in \N$ such that:
+:$\tuple {x, y}^k = \tuple {x^k, y^k} = \tuple {x', e_H}$
+and therefore $x^k = x'$.
+Thus the powers of $x$ generate the whole group $G$.
+In the same way, it is seen that $y$ generates $H$.
+Therefore by [[Order of Cyclic Group equals Order of Generator]]:
+:$\order x = g$
+:$\order y = h$
+Thus we have that:
+:$g h = \order {\tuple {x, y} } = \lcm \set {g, h}$
+Moreover by [[Product of GCD and LCM]] we have that:
+:$\lcm \set {g, h} = \dfrac {g h} {\gcd \set {g, h} }$
+These two equalities imply that:
+:$\gcd \set {g, h} = 1$
+That is, $g$ and $h$ are [[Definition:Coprime Integers|coprime]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Group Direct Product of Infinite Cyclic Groups}
+Tags: Group Direct Products, Cyclic Groups
+
+\begin{theorem}
+The [[Definition:Group Direct Product|group direct product]] of two [[Definition:Infinite Cyclic Group|infinite cyclic groups]] is not [[Definition:Cyclic Group|cyclic]].
+\end{theorem}
+
+\begin{proof}
+Let $G_1 = \struct {G_1, \circ_1}$ and $G_2 = \struct {G_2, \circ_2}$ be [[Definition:Infinite Cyclic Group|infinite cyclic groups]].
+Let $G = \struct {G, \circ} = G_1 \times G_2$.
+Let $G_1 = \gen {g_1}, G_2 = \gen {g_2}$.
+From [[Generators of Infinite Cyclic Group]]:
+:$g_1$ and $g_1^{-1}$ are the only [[Definition:Generator of Cyclic Group|generators]] of $G_1$
+:$g_2$ and $g_2^{-1}$ are the only [[Definition:Generator of Cyclic Group|generators]] of $G_2$.
+So a [[Definition:Generator of Group|generator]] of $G$ must be of the form $\tuple { g_1^{\pm 1}, g_2^{\pm 1} }$ to have any hope to generate all of $G$.
+{{WLOG}}, suppose that $G = \gen {\tuple {g_1, g_2} }$.
+Let $e_1$ be the [[Definition:Identity Element|identity element]] of $G_1$.
+Let $x_2 \in G_2$.
+From the definition of an [[Definition:Infinite Cyclic Group|infinite cyclic group]], both $g_1$ and $g_2$ are of [[Definition:Order of Group Element|infinite order]].
+Suppose now $\tuple {e_1, x_2} \in \gen {\tuple {g_1, g_2} }$.
+Then there is an $i \in \Z$ such that we have:
+:$\tuple {g_1, g_2}^i = \tuple {g_1^i, g_2^i} = \tuple {e_1, x_2}$
+However, as $g_1$ is of [[Definition:Order of Group Element|infinite order]], it must be that $i = 0$. 
+Hence $x_2 = g_2^0 = e_2$, where $e_2$ is the [[Definition:Identity Element|identity element]] of $G_2$.
+It follows that $\tuple {e_1, g_2} \notin \gen {\tuple {g_1, g_2} }$.
+Therefore:
+:$G \ne \gen {\tuple {g_1, g_2} }$
+It follows that $G$ cannot be generated by one [[Definition:Element|element]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Order of Group Element in Group Direct Product}
+Tags: Group Direct Products
+
+\begin{theorem}
+Let $G$ and $H$ be [[Definition:Finite Group|finite groups]].
+Let $g \in G: \order g = m, h \in H: \order h = n$.
+Then $\order {\tuple {g, h} }$ in $G \times H$ is $\lcm \set {m, n}$.
+\end{theorem}
+
+\begin{proof}
+Let $G$ and $H$ be a [[Definition:Group|groups]] whose [[Definition:Identity Element|identities]] are $e_G$ and $e_H$.
+Let $l = \lcm \set {m, n}$.
+From the definition of [[Definition:Lowest Common Multiple of Integers|lowest common multiple]]:
+:$\exists x, y \in \Z: l = m x = n y$
+From the definition of [[Definition:Order of Group Element|order of an element]]:
+:$g^m = e_G, h^n = e_H$
+Thus:
+{{begin-eqn}}
+{{eqn | l = \tuple {g, h}^l
+      | r = \tuple {g^l, h^l}
+      | c = 
+}}
+{{eqn | r = \tuple {\paren {g^m}^x, \paren {h^n}^y}
+      | c = 
+}}
+{{eqn | r = \tuple {e_G^x, e_H^y}
+      | c = 
+}}
+{{eqn | r = \tuple {e_G, e_H}
+      | c = 
+}}
+{{end-eqn}}
+Now suppose $\exists k \in \Z: \tuple {g, h}^k = \tuple {e_G, e_H}$.
+It follows that:
+:$\tuple {g, h}^k = \tuple {g^k, h^k} = \tuple {e_G, e_H}$
+Thus:
+:$g^k = e_G, h^k = e_H$
+It follows from $g^k = e_G$ and $\order g = m$, by [[Absolute Value of Integer is not less than Divisors]], that:
+:$m \divides k$
+Similarly it follows that:
+:$n \divides k$
+So $k$ is a [[Definition:Positive Integer|positive]] [[Definition:Common Multiple|common multiple]] of both $m$ and $n$.
+Since $l$ is the [[Definition:Lowest Common Multiple of Integers|lowest common multiple]]:
+:$l \le k$
+Therefore $l$ is the smallest such that $\tuple {g, h}^l = \tuple {e_G, e_H}$.
+The result follows by definition of the [[Definition:Order of Group Element|order of an element]].
+{{qed}}
+[[Category:Group Direct Products]]
+qn3x5oeiex39ck218cbpsj3tjzn6b74
+\end{proof}<|endoftext|>
+\section{Subgroup Product is Internal Group Direct Product iff Surjective}
+Tags: Internal Group Direct Products
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]].
+Let $\sequence {H_n}$ be a [[Definition:Sequence|sequence]] of [[Definition:Subgroup|subgroups]] of $G$.
+Let $\displaystyle \phi: \prod_{k \mathop = 1}^n H_k \to G$ be a [[Definition:Mapping|mapping]] defined by:
+:$\displaystyle \map \phi {h_1, h_2, \ldots, h_n} = \prod_{k \mathop = 1}^n h_k$
+Then $\phi$ is [[Definition:Surjection|surjective]] {{iff}}:
+: $\displaystyle G = \prod_{k \mathop = 1}^n H_k$
+That is, {{iff}} $G$ is the [[Definition:Internal Group Direct Product|internal group direct product]] of $H_1, H_2, \ldots, H_n$.
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+Let $\phi$ be a [[Definition:Surjection|surjection]].
+Then $\Img \phi$ consists of all the products $\displaystyle \prod_{k \mathop = 1}^n h_k$ such that:
+:$\forall k \in \closedint 1 n: h_k \in H_k$
+Thus, as $\phi$ is [[Definition:Surjection|surjective]], every [[Definition:Element|element]] of $G$ must be representable in this form.
+Using the notation:
+:$\displaystyle \prod_{k \mathop = 1}^n H_k = \set {\prod_{k \mathop = 1}^n h_k: \forall k \in \closedint 1 n: h_k \in H_k}$
+the result follows immediately.
+Thus every [[Definition:Element|element]] of $G$ is the product of an [[Definition:Element|element]] of each of $H_k$.
+{{qed|lemma}}
+=== Sufficient Condition ===
+Let $\displaystyle G = \prod_{k \mathop = 1}^n H_k$.
+If this is the case, then $g$ can be written as:
+:$\displaystyle g = \prod_{k \mathop = 1}^n h_k: \forall k \in \closedint 1 n: h_k \in H_k$
+Thus:
+:$g = \map \phi {h_1, h_2, \ldots, h_n}$
+and $\phi$ is [[Definition:Surjection|surjective]].
+{{qed}}
+[[Category:Internal Group Direct Products]]
+19kysqrglzqel7vkr389ikpxd1jxabb
+\end{proof}<|endoftext|>
+\section{Internal Group Direct Product is Injective}
+Tags: Internal Group Direct Products, Injections
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
+Let $H_1, H_2$ be [[Definition:Subgroup|subgroups]] of $G$.
+Let $\phi: H_1 \times H_2 \to G$ be a [[Definition:Mapping|mapping]] defined by:
+:$\map \phi {h_1, h_2} = h_1 h_2$
+Then $\phi$ is [[Definition:Injection|injective]] {{iff}}:
+:$H_1 \cap H_2 = \set e$
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+Let $\phi$ be an [[Definition:Injection|injection]].
+Let $\map \phi {h_1, h_2} = \map \phi {k_1, k_2}$.
+As $\phi$ is [[Definition:Injection|injective]], this means that:
+:$\tuple {h_1, h_2} = \tuple {k_1, k_2}$
+and thus:
+:$h_1 = k_1, h_2 = k_2$
+From the definition of $\phi$, this means:
+:$h_1 h_2 = k_1 k_2$
+Thus, each element of $G$ that can be expressed as a product of the form $h_1 h_2$ can be thus expressed ''uniquely''.
+Now, suppose $h \in H_1 \cap H_2$.
+We have:
+{{begin-eqn}}
+{{eqn | l = h = h e
+      | o = :
+      | r = h \in H_1, e \in H_2
+      | c = 
+}}
+{{eqn | l = h = e h
+      | o = :
+      | r = e \in H_1, h \in H_2
+      | c = 
+}}
+{{end-eqn}}
+Thus we see that:
+{{begin-eqn}}
+{{eqn | l = \map \phi {h, e}
+      | r = \map \phi {e, h}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \tuple {h, e}
+      | r = \tuple {e, h}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = h
+      | r = e
+      | c = 
+}}
+{{end-eqn}}
+Thus:
+:$H_1 \cap H_2 = \set e$
+{{qed|lemma}}
+=== Sufficient Condition ===
+Let $H_1 \cap H_2 = \set e$.
+Let:
+:$\map \phi {h_1, h_2} = \map \phi {k_1, k_2}$
+Then:
+:$h_1 h_2 = k_1 k_2: h_1, k_1 \in H_1, h_2, k_2 \in H_2$
+Thus:
+:$k_1^{-1} h_1 = k_2 h_2^{-1}$
+But:
+:$k_1^{-1} h_1 \in H_1$ and $k_2 h_2^{-1} \in H_2$
+As they are equal, we have:
+:$k_1^{-1} h_1 = k_2 h_2^{-1} \in H_1 \cap H_2 = \set e$
+It follows that:
+:$h_1 = k_1, h_2 = k_2$
+and thus:
+:$\tuple {h_1, h_2} = \tuple {k_1, k_2}$
+Thus $\phi$ is [[Definition:Injection|injective]] and the result follows.
+{{qed}}
+[[Category:Internal Group Direct Products]]
+[[Category:Injections]]
+8dkzl8zwmhesfwrzsqdn1xfvrontyp8
+\end{proof}<|endoftext|>
+\section{Internal Group Direct Product Isomorphism}
+Tags: Internal Group Direct Products, Group Isomorphisms, Normal Subgroups
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]].
+Let $H_1, H_2$ be [[Definition:Subgroup|subgroups]] of $G$.
+Let $\phi: H_1 \times H_2 \to G$ be the [[Definition:Mapping|mapping]] defined by $\map \phi {h_1, h_2} := h_1 h_2$.
+If $\phi$ is a [[Definition:Group Isomorphism|(group) isomorphism]], then both $H_1$ and $H_2$ are [[Definition:Normal Subgroup|normal subgroups]] of $G$.
+\end{theorem}
+
+\begin{proof}
+$\phi$ is an [[Definition:Group Isomorphism|isomorphism]], so in particular a [[Definition:Group Homomorphism|(group) homomorphism]].
+Thus by [[Induced Group Product is Homomorphism iff Commutative]], every [[Definition:Element|element]] of $H_1$ [[Definition:Commuting Elements|commutes]] with every [[Definition:Element|element]] of $H_2$.
+Now suppose $a \in G$.
+As $\phi$ is an [[Definition:Group Isomorphism|isomorphism]], it follows that $\phi$ is [[Definition:Surjection|surjective]].
+Thus by [[Subgroup Product is Internal Group Direct Product iff Surjective]]:
+:$\exists h_1 \in H_1, h_2 \in H_2: a = h_1 h_2$
+Now any [[Definition:Element|element]] of $a H_1 a^{-1}$ is in the form $a h a^{-1}$ for some $h \in H_1$.
+Thus:
+{{begin-eqn}}
+{{eqn | l = a h a^{-1}
+      | r = \paren {h_1 h_2} h \paren {h_1 h_2}^{-1}
+      | c = 
+}}
+{{eqn | r = h_1 h_2 h h_2^{-1} h_1^{-1}
+      | c = 
+}}
+{{eqn | r = h_1 h h_2 h_2^{-1} h_1^{-1}
+      | c = $h \in H_1$ and $h_2 \in H_2$ [[Definition:Commuting Elements|commute]]
+}}
+{{eqn | r = h_1 h h_1^{-1} \in H_1
+      | c = 
+}}
+{{end-eqn}}
+Thus $a H_1 a^{-1} \subseteq H_1$, and $H_1$ is therefore a [[Definition:Normal Subgroup|normal subgroup]] of $G$.
+Similarly, any [[Definition:Element|element]] of $a H_2 a^{-1}$ is in the form $a h a^{-1}$ for some $h \in H_2$.
+Thus:
+{{begin-eqn}}
+{{eqn | l = a h a^{-1}
+      | r = \paren {h_1 h_2} h \paren {h_1 h_2}^{-1}
+      | c = 
+}}
+{{eqn | r = h_1 h_2 h h_2^{-1} h_1^{-1}
+      | c = 
+}}
+{{eqn | r = h_1 h' h_1^{-1}
+      | c = as $h' = h_2 h h_2^{-1} \in H_2$
+}}
+{{eqn | r = h' h_1 h_1^{-1}
+      | c = $h_1 \in H_1$ and $h' \in H_2$ [[Definition:Commuting Elements|commute]]
+}}
+{{eqn | r = h' \in H_2
+      | c = 
+}}
+{{end-eqn}}
+Thus $a H_2 a^{-1} \subseteq H_2$ and $H_2$ is [[Definition:Normal Subgroup|normal]] as well.
+{{qed}}
+[[Category:Internal Group Direct Products]]
+[[Category:Group Isomorphisms]]
+[[Category:Normal Subgroups]]
+tal7o18xi8em9ebt61g1ghj7qd3spdp
+\end{proof}<|endoftext|>
+\section{Internal Group Direct Product of Normal Subgroups}
+Tags: Internal Group Direct Products, Normal Subgroups, Group Homomorphisms
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
+Let $H_1, H_2$ be [[Definition:Subgroup|subgroups]] of $G$.
+Let $\phi: H_1 \times H_2 \to G$ be a [[Definition:Mapping|mapping]] defined by $\map \phi {h_1, h_2} = h_1 h_2$.
+Let $H_1$ and $H_2$ be [[Definition:Normal Subgroup|normal subgroups]] of $G$, and let $H_1 \cap H_2 = \set e$.
+Then $\phi$ is a [[Definition:Group Homomorphism|(group) homomorphism]].
+\end{theorem}
+
+\begin{proof}
+Let $H_1$ and $H_2$ be [[Definition:Normal Subgroup|normal subgroups]] of $G$.
+Let $h_1 \in H_1, h_2 \in H_2$.
+Consider $x \in G: x = h_1 h_2 h_1^{-1} h_2^{-1}$.
+{{begin-eqn}}
+{{eqn | l = x
+      | r = h_1 h_2 h_1^{-1} h_2^{-1}
+      | c = 
+}}
+{{eqn | r = \paren {h_1 h_2 h_1^{-1} } h_2^{-1}
+      | c = 
+}}
+{{end-eqn}}
+As $H_2$ is [[Definition:Normal Subgroup|normal]], we have $h_1 h_2 h_1^{-1} \in H_2$ and thus $x \in H_2$.
+Similarly, we can show that $x \in H_1$ and so $x \in H_1 \cap H_2$ and thus $x = e$.
+From [[Product of Commuting Elements with Inverses]], $h_1 h_2 h_1^{-1} h_2^{-1} = e$ {{iff}} $h_1$ and $h_2$ [[Definition:Commuting Elements|commute]].
+Thus $h_1 h_2 = h_2 h_1$. 
+As $h_1$ and $h_2$ are arbitrary [[Definition:Element|elements]] of $H_1$ and $H_2$, it follows that every [[Definition:Element|element]] of $H_1$ [[Definition:Commuting Elements|commutes]] with every [[Definition:Element|element]] of $H_2$.
+Thus from [[Induced Group Product is Homomorphism iff Commutative]], $\phi$ is a homomorphism.
+{{qed}}
+[[Category:Internal Group Direct Products]]
+[[Category:Normal Subgroups]]
+[[Category:Group Homomorphisms]]
+dr1zduk92t4fsu9xcfdmjshzwap7ogp
+\end{proof}<|endoftext|>
+\section{Internal Direct Product Theorem}
+Tags: Internal Group Direct Products, Normal Subgroups, Named Theorems, Internal Direct Product Theorem
+
+\begin{theorem}
+Let $\struct {G, \circ}$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
+Let $H_1, H_2 \le G$.
+Then $G$ is the [[Definition:Internal Group Direct Product|internal group direct product]] of $H_1$ and $H_2$ {{iff}}:
+:$(1): \quad G = H_1 \circ H_2$
+:$(2): \quad H_1 \cap H_2 = \set e$
+:$(3): \quad H_1, H_2 \lhd G$
+where $H_1 \lhd G$ denotes that $H_1$ is a [[Definition:Normal Subgroup|normal subgroup]] of $G$.
+\end{theorem}
+
+\begin{proof}
+=== Sufficient Condition ===
+Let $G$ be the [[Definition:Internal Group Direct Product|internal group direct product]] of $H_1$ and $H_2$.
+:$(1): \quad$ From [[Subgroup Product is Internal Group Direct Product iff Surjective]], $G = H_1 \circ H_2$.
+:$(2): \quad$ From [[Internal Group Direct Product is Injective]], $H_1$ and $H_2$ are [[Definition:Independent Subgroups|independent subgroups]] of $G$.
+:$(3): \quad$ From [[Internal Group Direct Product Isomorphism]], $H_1 \lhd G$ and $H_2 \lhd G$.
+{{qed|lemma}}
+=== Necessary Condition ===
+Let $C: H_1 \times H_2 \to G$ be the [[Definition:Mapping|mapping]] defined as:
+:$\forall \tuple {h_1, h_2} \in H_1 \times H_2: \map C {h_1, h_2} = h_1 \circ h_2$
+Suppose the three conditions hold.
+:$(1): \quad$ From [[Subgroup Product is Internal Group Direct Product iff Surjective]], $C$ is [[Definition:Surjection|surjective]].
+:$(2): \quad$ From [[Internal Group Direct Product is Injective]], $C$ is [[Definition:Injection|injective]].
+:$(3): \quad$ From [[Internal Group Direct Product of Normal Subgroups]], $C$ is a [[Definition:Group Homomorphism|group homomorphism]].
+Putting these together, we see that $C$ is a [[Definition:Bijection|bijective]] [[Definition:Group Homomorphism|homomorphism]], and therefore an [[Definition:Group Isomorphism|isomorphism]].
+So by definition, $G$ is the [[Definition:Internal Group Direct Product|internal group direct product]] of $H_1$ and $H_2$.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+A specific instance of the [[Internal Direct Product Theorem/General Result|general result]], with $n = 2$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Inclusion Mapping is Surjection iff Identity}
+Tags: Surjections, Inclusion Mappings, Identity Mappings
+
+\begin{theorem}
+Let $T$ be a [[Definition:Set|set]].
+Let $S\subseteq T$ be a [[Definition:Subset|subset]].
+Let $i_S: S \to T$ be the [[Definition:Inclusion Mapping|inclusion mapping]].
+Then:
+:$i_S: S \to T$ is [[Definition:Surjection|surjective]] {{iff}} $i_S: S \to T = I_S: S \to S$
+where $I_S: S \to S$ denotes the [[Definition:Identity Mapping|identity mapping]] on $S$.
+Alternatively, this theorem can be worded as:
+:$i_S: S \to S = I_S: S \to S$
+It follows directly that from [[Surjection by Restriction of Codomain]], the [[Definition:Surjective Restriction|surjective restriction]] of $i_S: S \to T$ to $i_S: S \to \Img {i_S}$ is itself the [[Definition:Identity Mapping|identity mapping]].
+\end{theorem}
+
+\begin{proof}
+It is apparent from the definitions of both the [[Definition:Inclusion Mapping|inclusion mapping]] and the [[Definition:Identity Mapping|identity mapping]] that:
+:$(1): \quad \Dom {i_S} = S = \Dom {I_S}$
+:$(2): \quad \forall s \in S: \map {i_S} s = s = \map {I_S} s$
+=== Necessary Condition ===
+Let $i_S: S \to T = I_S: S \to S$.
+From [[Equality of Mappings]], we have that the [[Definition:Codomain of Mapping|codomain]] of $i_S$ equals the [[Definition:Codomain of Mapping|codomain]] of $I_S$.
+Thus the [[Definition:Codomain of Mapping|codomain]] of $i_S$ equals the [[Definition:Codomain of Mapping|codomain]] of $I_S$ equals $S$ and thus $T = S$.
+So $\forall s \in S: s = \map {i_S} s$ and so $i_S$ is [[Definition:Surjection|surjective]].
+{{qed|lemma}}
+=== Sufficient Condition ===
+Now let $i_S: S \to T$ be a [[Definition:Surjection|surjection]].
+Then:
+:$\forall s \in T: s = \map {i_S} s \implies s \in S$
+and therefore:
+:$T \subseteq S$
+Thus:
+:$T = S$
+and so the [[Definition:Codomain of Mapping|codomain]] of $i_S$ equals the [[Definition:Codomain of Mapping|codomain]] of $I_S$ which equals $S$.
+Thus $i_S: S \to T = I_S: S \to S$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Cyclic Group of Order 6}
+Tags: Examples of Cyclic Groups, Groups of Order 6
+
+\begin{theorem}
+Let $C_n$ be the [[Definition:Cyclic Group|cyclic group]] of [[Definition:Order of Structure|order]] $n$.
+Then:
+: $C_2 \times C_3 \cong C_6$
+: $C_6$ is the [[Definition:Internal Group Direct Product|internal group direct product]] of $C_2$ and $C_3$.
+\end{theorem}
+
+\begin{proof}
+From [[Group Direct Product of Cyclic Groups]] noting that $2 \perp 3$:
+:$C_2 \times C_3 \cong C_6$
+* $C_6$ is the [[Definition:Internal Group Direct Product|internal group direct product]] of $C_2$ and $C_3$:
+Let $\left({C_6, \circ}\right) = \left \langle {x} \right \rangle$, let $\left({C_2, \circ}\right) = \left \langle {x^3} \right \rangle$, and let $\left({C_3, \circ}\right) = \left \langle {x^2} \right \rangle$.
+From [[Subgroup of Abelian Group is Normal]], $C_2 \lhd C_6$ and $C_3 \lhd C_6$.
+We can factorise the elements of $C_6$ thus:
+:$e = e \circ e, x = x^3 \circ \left({x^2}\right)^2, x^2 = e \circ x^2, x^3 = x^3 \circ e, x^4 = e \circ \left({x^2}\right)^2, x^5 = x^2 \circ x^3$ 
+Hence the result.
+{{qed}}
+[[Category:Examples of Cyclic Groups]]
+[[Category:Groups of Order 6]]
+krip2kmx54xcemcyjmzixdpxw4ne1rd
+\end{proof}<|endoftext|>
+\section{Internal Direct Product Generated by Subgroups}
+Tags: Internal Group Direct Products
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
+Let $\sequence {H_n}$ be a [[Definition:Sequence|sequence]] of [[Definition:Subgroup|subgroups]] of $G$.
+Then:
+:the [[Definition:Generator of Subgroup|subgroup generated by $\displaystyle \bigcup_{k \mathop = 1}^n H_k$]] is the [[Definition:Internal Group Direct Product|internal group direct product]] of $\sequence {H_n}$
+{{iff}}:
+:$\sequence {H_n}$ is an [[Definition:Independent Subgroups|independent sequence of subgroups]] such that every element of $H_i$ [[Definition:Commute|commutes]] with every element of $H_j$ whenever $1 \le i < j \le n$.
+\end{theorem}
+
+\begin{proof}
+In the following, the notation $\closedint m n$ is to be understood to mean a [[Definition:Integer Interval|(closed) integer interval]]:
+:$\closedint m n := \set {x \in \Z: m \le x \le n}$
+for $m, n \in \Z$.
+For each $k \in \closedint 1 n$, let $\displaystyle L_k = \prod_{j \mathop = 1}^k H_j$ be the [[Definition:Finite Cartesian Product|cartesian product]] of the subgroups $H_1, H_2, \ldots, H_k$ of $G$.
+Let $\displaystyle C_k: L_k \to G: \map {C_k} {x_1, x_2, \ldots, x_k} = \prod_{j \mathop = 1}^k x_j$.
+=== Necessary Condition ===
+Suppose that the [[Definition:Generator of Subgroup|subgroup generated by $\displaystyle \bigcup_{k \mathop = 1}^n H_k$]] is the [[Definition:Internal Group Direct Product/General Definition|internal group direct product]] of $\sequence {H_n}$.
+We have that [[Internal Group Direct Product is Injective]].
+Hence by definition $C_k$ is a [[Definition:Group Monomorphism|monomorphism]].
+It follows from [[Kernel is Trivial iff Monomorphism]] that the [[Definition:Kernel of Group Homomorphism|kernel]] of $C_n$ is $\set {\tuple {e, e, \ldots, e} }$.
+Therefore $\sequence {H_n}$ is an [[Definition:Independent Subgroups|independent sequence]].
+Let $x \in H_i$ and $y \in H_j$ where $1 \le i < j \le n$.
+For each $k \in \closedint 1 n$, let $x_k$ and $y_k$ be defined as:
+:$x_k = \begin{cases}
+e & : k \ne i \\
+x & : k = i
+\end{cases}
+\qquad
+y_k = \begin{cases}
+e & : k \ne j \\
+y & : k = j
+\end{cases}$
+Then:
+{{begin-eqn}}
+{{eqn | l = x y
+      | r = \paren {y_i x_i} \paren {y_j x_j}
+      | c = 
+}}
+{{eqn | r = \prod_{k \mathop = 1}^n y_k x_k
+      | c = 
+}}
+{{eqn | r = \map {C_n} {y_1 x_1, \ldots, y_n x_n}
+      | c = 
+}}
+{{eqn | r = \map {C_n} {\tuple {y_1, \ldots, y_n} \tuple {x_1, \ldots, x_n} }
+      | c = 
+}}
+{{eqn | r = \map {C_n} {y_1, \ldots, y_n} \map {C_n} {x_1, \ldots, x_n}
+      | c = 
+}}
+{{eqn | r = y x
+      | c = 
+}}
+{{end-eqn}}
+{{qed|lemma}}
+=== Sufficient Condition ===
+Suppose that $\sequence {H_n}$ is an [[Definition:Independent Subgroups|independent sequence of subgroups]] such that every element of $H_i$ [[Definition:Commute|commutes]] with every element of $H_j$ whenever $1 \le i < j \le n$.
+Let $S$ be the set of all $k \in \closedint 1 n$ such that $C_k: L_k \to G$ is a [[Definition:Group Homomorphism|(group) homomorphism]].
+Clearly $1 \in S$.
+Now let $k \in S$ such that $k < n$.
+Let $\tuple {x_1, \ldots, x_k, x_{k + 1} }, \tuple {y_1, \ldots, y_k, y_{k + 1} } \in L_k$.
+By the [[General Associativity Theorem]] and [[Element Commutes with Product of Commuting Elements]]:
+{{begin-eqn}}
+{{eqn | l = \map {C_{k + 1} } {\tuple {x_1, \ldots, x_k, x_{k + 1} } \tuple {y_1, \ldots, y_k, y_{k + 1} } }
+      | r = \map {C_{k + 1} } {x_1 y_1, \ldots, x_k y_k, x_{k + 1} y_{k + 1} }
+      | c = 
+}}
+{{eqn | r = \paren {x_1 y_1} \cdots \paren {x_k y_k} \paren {x_{k + 1} y_{k + 1} }
+      | c = 
+}}
+{{eqn | r = \map {C_k} {\paren {x_1 y_1} \cdots \paren {x_k y_k} } \paren {x_{k + 1} y_{k + 1} }
+      | c = 
+}}
+{{eqn | r = \map {C_k} {\tuple {x_1, \ldots, x_k} \tuple {y_1, \ldots, y_k} } \paren {x_{k + 1} y_{k + 1} }
+      | c = 
+}}
+{{eqn | r = \map {C_k} {x_1, \ldots, x_k} \map {C_k} {y_1, \ldots, y_k} \paren {x_{k + 1} y_{k + 1} }
+      | c = 
+}}
+{{eqn | r = \paren {\prod_{k \mathop = 1}^n{x_j} } \paren {\prod_{k \mathop = 1}^n{y_j} } x_{k + 1} y_{k + 1}
+      | c = 
+}}
+{{eqn | r = \paren {\paren {\prod_{k \mathop = 1}^n x_j} x_{k + 1} } \paren {\paren {\prod_{k \mathop = 1}^n y_j} y_{k + 1} }
+      | c = 
+}}
+{{eqn | r = \map {C_{k + 1} } {x_1, \ldots, x_k, x_{k + 1} } \map {C_{k + 1} } {y_1, \ldots, y_k, y_{k + 1} }
+      | c = 
+}}
+{{end-eqn}}
+Thus $C_{k + 1}: L_{k + 1} \to G$ is a [[Definition:Group Homomorphism|homomorphism]], so $k + 1 \in S$.
+So by [[Principle of Mathematical Induction|induction]]:
+:$S = \closedint 1 n$
+In particular, $C_n: L_n \to G$ is a [[Definition:Group Homomorphism|homomorphism]].
+By [[Morphism Property Preserves Closure]] and [[Homomorphism Preserves Subsemigroups]], the [[Definition:Codomain of Mapping|codomain]] $\displaystyle \prod_{k \mathop = 1}^n H_k$ of $C_n$ is therefore a [[Definition:Subgroup|subgroup]] of $G$ containing $\displaystyle \bigcup_{k \mathop = 1}^n H_k$.
+However, any [[Definition:Subgroup|subgroup]] containing $\displaystyle \bigcup_{k \mathop = 1}^n H_k$ must clearly contain $\displaystyle \prod_{k \mathop = 1}^n H_k$.
+Therefore, $\displaystyle \prod_{k \mathop = 1}^n H_k$ is the [[Definition:Subgroup|subgroup]] of $G$ generated by $\displaystyle \bigcup_{k \mathop = 1}^n H_k$.
+As $\sequence {H_n}$ is an [[Definition:Independent Subgroups|independent sequence of subgroups]], the [[Definition:Kernel of Group Homomorphism|kernel]] of $C_n$ is $\set {\tuple {e, \ldots, e} }$.
+Hence by the [[Quotient Theorem for Group Epimorphisms]], $C_n$ from $L_n$ to the subgroup of $G$ generated by $\displaystyle \bigcup_{k \mathop = 1}^n H_k$ is an [[Definition:Group Isomorphism|isomorphism]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Internal Group Direct Product Commutativity}
+Tags: Internal Group Direct Products, Internal Direct Products, Commutativity, Internal Group Direct Product Commutativity
+
+\begin{theorem}
+Let $\struct {G, \circ}$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
+Let $H_1, H_2 \le G$.
+Let $\struct {G, \circ}$ be the [[Definition:Internal Group Direct Product|internal group direct product]] of $H_1$ and $H_2$.
+Then:
+:$\forall h_1 \in H_1, h_2 \in H_2: h_1 \circ h_2 = h_2 \circ h_1$
+\end{theorem}
+
+\begin{proof}
+Let $\sqbrk {x, y}$ denote the [[Definition:Commutator of Group Elements|commutator]] of $x, y \in G$:
+:$\sqbrk {x, y} := x^{-1} y^{-1} x y$
+We have that:
+{{begin-eqn}}
+{{eqn | n = 1
+      | l = y x \sqbrk {x, y}
+      | r = y x x^{-1} y^{-1} x y
+      | c = {{Defof|Commutator of Group Elements}}
+}}
+{{eqn | r = y y^{-1} x y
+      | c = {{GroupAxiom|3}}
+}}
+{{eqn | r = x y
+      | c = {{GroupAxiom|3}}
+}}
+{{end-eqn}}
+Let $h_1 \in H_1$, $h_2 \in H_2$.
+We have:
+{{begin-eqn}}
+{{eqn | l = \sqbrk {h_1, h_2}
+      | r = {h_1}^{-1} {h_2}^{-1} h_1 h_2
+      | c = 
+}}
+{{eqn | r = {h_1}^{-1} \paren { {h_2}^{-1} h_1 h_2}
+      | c = {{GroupAxiom|1}}
+}}
+{{eqn | o = \in
+      | r = {h_1}^{-1} H_1
+      | c = {{Defof|Normal Subgroup}}
+}}
+{{eqn | ll= \leadsto
+      | l = \sqbrk {h_1, h_2}
+      | o = \in
+      | r = H_1
+      | c = as ${h_1}^{-1} H_1 = H_1$
+}}
+{{end-eqn}}
+and:
+{{begin-eqn}}
+{{eqn | l = \sqbrk {h_1, h_2}
+      | r = {h_1}^{-1} {h_2}^{-1} h_1 h_2
+      | c = 
+}}
+{{eqn | r = \paren { {h_1}^{-1} {h_2}^{-1} h_1} h_2
+      | c = {{GroupAxiom|1}}
+}}
+{{eqn | o = \in
+      | r = H_2 h_2
+      | c = {{Defof|Normal Subgroup}}
+}}
+{{eqn | ll= \leadsto
+      | l = \sqbrk {h_1, h_2}
+      | o = \in
+      | r = H_2
+      | c = as $H_2 h_2 = H_2$
+}}
+{{end-eqn}}
+Thus:
+:$\sqbrk {h_1, h_2} \in H_1 \cap H_2$
+But as $H_1 \cap H_2 = \set e$, it follows that:
+:$\sqbrk {h_1, h_2} = e$
+It follows from [[Commutator is Identity iff Elements Commute]] that:
+:$h_1 h_2 = h_2 h_1$
+and the result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Internal and External Group Direct Products are Isomorphic}
+Tags: Group Direct Products, Internal Group Direct Products, Group Isomorphisms
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
+Then $G$ is the [[Definition:Group Direct Product/Finite Product|(external) group direct product]] of $G_1, G_2, \ldots, G_n$ {{iff}} $G$ is the [[Definition:Internal Group Direct Product/General Definition|internal group direct product]] of $N_1, N_2, \ldots, N_n$ such that:
+:$\forall i \in \N_n: N_i \cong G_i$
+where:
+:$\cong$ denotes [[Definition:Group Isomorphism|(group) isomorphism]]
+:$\N_n$ denotes $\set {1, 2, \ldots, n}$
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+Let $G$ be the [[Definition:Group Direct Product/Finite Product|external direct product]] of [[Definition:Group|groups]] $G_1, G_2, \ldots, G_n$.
+For all $i \in \N_n$, let $N_i$ be defined as the set:
+:$N_i = \set e \times \cdots \times \set e \times G_i \times \set e \times \cdots \times \set e$
+of elements which have entry $e$ everywhere except possibly in the $i$th co-ordinate.
+It remains to be checked that:
+:$(1): \quad N_i$ is [[Definition:Group Isomorphism|isomorphic]] to $G_i$
+:$(2): \quad N_i$ is a [[Definition:Normal Subgroup|normal subgroup]] of $G$
+:$(3): \quad$ Every element of $G$ has a unique expression:
+:$\tuple {g_1, \ldots, g_n} = \tuple {g_1, e, \ldots, e} \tuple {e, g_2, e \ldots e} \ldots \tuple {e, \ldots, g_n}$
+as a product of elements of $N_1, \ldots, N_n$.
+==== $N_i$ is [[Definition:Group Isomorphism|isomorphic]] to $G_i$ ====
+{{finish}}
+==== $N_i$ is a [[Definition:Normal Subgroup|normal subgroup]] of $G$ ====
+{{finish}}
+==== Every element of $G$ has a unique expression ====
+{{finish}}
+=== Sufficient Condition ===
+Now let $G$ be the [[Definition:Internal Group Direct Product|internal group direct product]] of $N_1, N_2, \ldots, N_n$.
+We define a [[Definition:Mapping|mapping]] $\theta: G \to N_1 \times N_2 \times \cdots \times N_n$ by:
+:$\map \theta {g_1 g_2 \ldots g_n} = \tuple {g_1, g_2, \ldots, g_n}$
+The fact that $\theta$ is a [[Definition:Bijection|bijection]] follows from the definitions.
+It remains to be shown that $\theta$ is a [[Definition:Group Homomorphism|homomorphism]].
+The proof proceeds by [[Principle of Mathematical Induction|induction]].
+For all $n \in \Z_{\ge 2}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
+:$\map \theta {g_1 g_2 \ldots g_n} \map \theta {h_1 h_2 \ldots h_n} = \map \theta {g_1 g_2 \ldots g_n h_1 h_2 \ldots h_n}$
+==== Basis for the Induction ====
+$\map P 2$ is the case:
+{{begin-eqn}}
+{{eqn | l = \map \theta {g_1 g_2} \map \theta {h_1 h_2}
+      | r = \tuple {g_1, g_2} \tuple {h_1, h_2}
+      | c = Definition of $\theta$
+}}
+{{eqn | r = \tuple {g_1 h_1, g_2 h_2}
+      | c = 
+}}
+{{eqn | r = \map \theta {g_1 h_1 g_2 h_2}
+      | c = Definition of $\theta$
+}}
+{{eqn | r = \map \theta {g_1 g_2 h_1 h_2}
+      | c = [[Internal Group Direct Product Commutativity]]: $h_1 g_2 = g_2 h_1$
+}}
+{{end-eqn}}
+Thus $\map P 2$ is seen to hold.
+This is the [[Principle of Mathematical Induction#Basis for the Induction|basis for the induction]].
+==== Induction Hypothesis ====
+Now it needs to be shown that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
+So this is the [[Principle of Mathematical Induction#Induction Hypothesis|induction hypothesis]]:
+:$\map \theta {g_1 g_2 \ldots g_k} \map \theta {h_1 h_2 \ldots h_k} = \map \theta {g_1 g_2 \ldots g_k h_1 h_2 \ldots h_k}$
+from which it is to be shown that:
+:$\map \theta {g_1 g_2 \ldots g_{k + 1} } \map \theta {h_1 h_2 \ldots h_{k + 1} } = \map \theta {g_1 g_2 \ldots g_{k + 1} h_1 h_2 \ldots h_{k + 1} }$
+==== Induction Step ====
+This is the [[Principle of Mathematical Induction#Induction Step|induction step]]:
+{{begin-eqn}}
+{{eqn | o = 
+      | r = \map \theta {g_1 g_2 \ldots g_{k + 1} } \map \theta {h_1 h_2 \ldots h_{k + 1} }
+      | c = 
+}}
+{{eqn | r = \map \theta {\paren {g_1 g_2 \ldots g_k} g_{k + 1} } \map \theta {\paren {h_1 h_2 \ldots h_k} h_{k + 1} }
+      | c = {{GroupAxiom|1}}
+}}
+{{eqn | r = \map \theta {\paren {g_1 g_2 \ldots g_k} g_{k + 1} \paren {h_1 h_2 \ldots h_k} h_{k + 1} }
+      | c = [[Internal and External Group Direct Products are Isomorphic#Basis for the Induction|Basis for the Induction]]
+}}
+{{eqn | r = \map \theta {g_1 g_2 \ldots g_k g_{k + 1} h_1 h_2 \ldots h_k h_{k + 1} }
+      | c = {{GroupAxiom|1}}
+}}
+{{end-eqn}}
+So $\map P k \implies \map P {k + 1}$ and the result follows by the [[Principle of Mathematical Induction]].
+Therefore:
+:$\forall n \in \Z_{>1}: \map \theta {g_1 g_2 \ldots g_n} \map \theta {h_1 h_2 \ldots h_n} = \map \theta {g_1 g_2 \ldots g_n h_1 h_2 \ldots h_n}$
+So $\theta$ is a [[Definition:Group Homomorphism|homomorphism]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Direct Product of Central Subgroup with Inverse Isomorphism is Central Subgroup}
+Tags: Quotient Groups, Isomorphisms, Central Subgroups
+
+\begin{theorem}
+Let $G$ and $H$ be [[Definition:Group|groups]].
+Let $\map Z G$ denote the [[Definition:Center of Group|center]] of $G$.
+Let $Z$ and $W$ be [[Definition:Central Subgroup|central subgroups]] of $G$ and $H$ respectively.
+Let:
+:$Z \cong W$
+where $\cong$ denotes [[Definition:Group Isomorphism|isomorphism]].
+Let such a [[Definition:Group Isomorphism|group isomorphism]] be $\theta: Z \to W$.
+Let $X$ be the [[Definition:Set|set]] defined as:
+:$X = \set {\tuple {x, \map \theta x^{-1} }: x \in Z}$
+Then $X$ is a [[Definition:Central Subgroup|central subgroup]] of $G \times H$.
+\end{theorem}
+
+\begin{proof}
+First note that by [[Group Homomorphism Preserves Inverses]]:
+:$\forall x \in G: \map \phi {x^{-1} } = \paren {\map \phi x}^{-1} = \map \phi x^{-1}$
+and so there is no [[Definition:Amphiboly|amphiboly]] in the notation used.
+It is established that $\tuple {x, \map \theta x^{-1} } \in G \times H$:
+:$x \in G$
+:$\map \theta x \in H$ and so $\map \theta x^{-1} \in H$.
+Let:
+:$e_G$ be the [[Definition:Identity Element|identity element]] of $G$
+:$e_H$ be the [[Definition:Identity Element|identity element]] of $H$
+We have that $e_G \in Z$, and so:
+{{begin-eqn}}
+{{eqn | l = \tuple {e_G, \map \theta {e_G}^{-1} }
+      | r = \tuple {e_G, \map \theta {e_G} }
+      | c = as $e_G^{-1} = e_G$
+}}
+{{eqn | r = \tuple {e_G, e_H}
+      | c = [[Group Homomorphism Preserves Identity]]
+}}
+{{end-eqn}}
+Thus $\tuple {e_G, e_H} \in X$ and so $X \ne \O$.
+Then we note that:
+{{begin-eqn}}
+{{eqn | l = \tuple {x, \map \theta x^{-1} } \tuple {x^{-1}, \map \theta x}
+      | r = \tuple {x x^{-1}, \map \theta x^{-1} \map \theta x}
+      | c = {{Defof|Group Direct Product}}
+}}
+{{eqn | r = \tuple {e_G, e_H}
+      | c = as $\theta$ is a [[Definition:Group Isomorphism|group isomorphism]]
+}}
+{{eqn | r = \tuple {x^{-1} x, \map \theta x \map \theta x^{-1} }
+      | c = as $\theta$ is a [[Definition:Group Isomorphism|group isomorphism]]
+}}
+{{eqn | r = \tuple {x^{-1}, \map \theta x} \tuple {x, \map \theta x^{-1} }
+      | c = {{Defof|Group Direct Product}}
+}}
+{{end-eqn}}
+and so $\tuple {x^{-1}, \map \theta x}$ is the [[Definition:Inverse Element|inverse]] of $\tuple {x, \map \theta x^{-1} }$ in $H$.
+Let $x, y \in G$.
+Then $x^{-1} \in G$ from {{GroupAxiom|0}}.
+Hence if:
+:$\tuple {x, \map \theta x^{-1} } \in X$
+it follows that:
+:$\tuple {x^{-1}, \map \theta x} \in X$
+Let $x, y^{-1} \in G$.
+Then:
+:$\tuple {x, \map \theta x^{-1} } \in X$
+and:
+:$\tuple {y^{-1}, \map \theta y} \in X$
+Thus:
+{{begin-eqn}}
+{{eqn | l = \tuple {x, \map \theta x^{-1} } \tuple {y^{-1}, \map \theta y}
+      | r = \tuple {x y^{-1}, \map \theta x^{-1} \map \theta y}
+      | c = {{Defof|Group Direct Product}}
+}}
+{{eqn | r = \tuple {x y^{-1}, \map \theta {x^{-1} y} }
+      | c = as $\theta$ is a [[Definition:Group Isomorphism|group isomorphism]]
+}}
+{{eqn | r = \tuple {x y^{-1}, \map \theta {\paren {y^{-1} x}^{-1} } }
+      | c = [[Inverse of Group Product]]
+}}
+{{eqn | r = \tuple {x y^{-1}, \map \theta {\paren {x y^{-1} }^{-1} } }
+      | c = $x$ and $y$ are in the [[Definition:Center of Group|center]] of $G$, so [[Definition:Commuting Elements|commute]]
+}}
+{{eqn | r = \tuple {x y^{-1}, \map \theta {x y^{-1} }^{-1} }
+      | c = 
+}}
+{{end-eqn}}
+So:
+:$\tuple {x, \map \theta x^{-1} } \tuple {y^{-1}, \map \theta y} \in X$
+and it follows from the [[One-Step Subgroup Test]] that $X$ is a [[Definition:Subgroup|subgroup]] of $G \times H$.
+{{qed|lemma}}
+We have that $Z$ is a [[Definition:Central Subgroup|central subgroup]] of $G$.
+We also have that the [[Definition:Image of Mapping|image]] of $\theta$ is $W$.
+From [[Direct Product of Central Subgroups]], $Z \times W$ is a [[Definition:Central Subgroup|central subgroup]] of $G \times H$.
+It follows that $X$ is a [[Definition:Central Subgroup|central subgroup]] of $G \times H$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Pullback of Quotient Group Isomorphism is Subgroup}
+Tags: Pullbacks of Quotient Group Isomorphisms, Quotient Groups, Group Isomorphisms, Subgroups
+
+\begin{theorem}
+Let $\struct {G, \circ}$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity element]] is $e_G$.
+Let $\struct {H, *}$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity element]] is $e_H$.
+Let $N \lhd G, K \lhd H$ be [[Definition:Normal Subgroup|normal subgroups]] of $G$ and $H$ respectively.
+Let:
+:$G / N \cong H / K$
+where:
+:$G / N$ denotes the [[Definition:Quotient Group|quotient]] of $G$ by $N$
+:$\cong$ denotes [[Definition:Group Isomorphism|group isomorphism]].
+Let $\theta: G / N \to H / K$ be such a [[Definition:Group Isomorphism|group isomorphism]].
+Let $G \times^\theta H$ be the [[Definition:Pullback of Quotient Group Isomorphism|pullback of $G$ and $H$ via $\theta$]].
+Then $G \times^\theta H$ is a [[Definition:Subgroup|subgroup]] of $G \times H$.
+\end{theorem}
+
+\begin{proof}
+This result is proved by an application of the [[Two-Step Subgroup Test]]:
+=== Condition $(1)$ ===
+From the definition of [[Definition:Pullback of Quotient Group Isomorphism|pullback]]:
+:$\tuple {e_G, e_H} \in G \times^\theta H$
+{{iff}}:
+:$\map \theta {e_G \circ N} = e_H * K$
+By [[Coset by Identity]], $e_G \circ N, e_H * K$ are the [[Definition:Identity Element|identities]] of $G / N$ and $H / K$
+From [[Group Homomorphism Preserves Identity]]:
+:$\map \theta {e_G \circ N} = e_H * K$
+So $\tuple {e_G, e_H} \in G \times^\theta H$
+Thus $G \times^\theta H$ is [[Definition:Non-Empty Set|non-empty]].
+{{qed|lemma}}
+=== Condition $(2)$ ===
+Let $\tuple {g, h}$ and  $\tuple {g', h'}$ be [[Definition:Element|elements]] of $G \times^\theta H$.
+It follows by definition of $\theta$ that:
+:$\map \theta {g \circ N} = h * K$
+and:
+:$\map \theta {g' \circ N} = h' * K$
+By the [[Definition:Morphism Property|morphism property]]:
+:$\map \theta {g \circ N} * \map \theta {g' \circ N} = \map \theta {g \circ N \circ g' \circ N} = \map \theta {\paren {g \circ g'} \circ N}$
+Hence:
+{{begin-eqn}}
+{{eqn | l = \paren {h * h'} * K
+      | r = h * K * h' * K
+}}
+{{eqn | l = 
+      | r = \map \theta {g \circ N} * \map \theta {g' \circ N}
+}}
+{{eqn | l = 
+      | r = \map \theta {\paren {g \circ g'} \circ N}
+}}
+{{end-eqn}}
+Thus:
+:$\tuple {g \circ g', h * h'} \in G \times^\theta H$
+Hence $G \times^\theta H$ is closed under the operation.
+{{qed|lemma}}
+=== Condition $(3)$ ===
+Let $\tuple {g, h} \in G \times^\theta H$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \map \theta {g \circ N}
+      | r = h * K
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \map \theta {g \circ N}^{-1}
+      | r = h^{-1} * K
+      | c = 
+}}
+{{end-eqn}}
+Then:
+{{begin-eqn}}
+{{eqn | l = \map \theta {g \circ N}^{-1}
+      | r = \map \theta {g^{-1} \circ N}
+      | c = [[Group Homomorphism Preserves Inverses]]
+}}
+{{eqn | ll= \leadsto
+      | l = \map \theta {g^{-1} \circ N}
+      | r = h^{-1} * K
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \tuple {g^{-1}, h^{-1} }
+      | o = \in
+      | r = G \times^\theta H
+      | c = 
+}}
+{{end-eqn}}
+Thus $G \times^\theta H$ is closed under inverses.
+{{qed|lemma}}
+Therefore by the [[Two-Step Subgroup Test]]:
+:$G \times^\theta H \le G \times H$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Group/Examples/x+y over 1+xy}
+Tags: Examples of Groups, Examples of Groups: x+y over 1+xy
+
+\begin{theorem}
+Let $G = \set {x \in \R: -1 < x < 1}$ be the [[Definition:Set|set]] of all [[Definition:Real Number|real numbers]] whose [[Definition:Absolute Value|absolute value]] is less than $1$.
+Let $\circ: G \times G \to \R$ be the [[Definition:Binary Operation|binary operation]] defined as:
+:$\forall x, y \in G: x \circ y = \dfrac {x + y} {1 + x y}$
+The [[Definition:Algebraic Structure|algebraic structure]] $\struct {G, \circ}$ is a [[Definition:Group|group]].
+\end{theorem}
+
+\begin{proof}
+Let $-1 < x, y, z < 1$.
+We check the [[Definition:Group Axioms|group axioms]] in turn:
+=== $\text G 1$: Associativity ===
+{{begin-eqn}}
+{{eqn | l = \paren {x \circ y} \circ z
+      | r = \frac {\frac {x + y} {1 + x y} + z} {1 + \frac {x + y} {1 + xy} z}
+      | c = 
+}}
+{{eqn | r = \frac {x + y + z + x y z} {1 + x y + x z + y z}
+      | c = 
+}}
+{{eqn | l = x \circ \paren {y \circ z}
+      | r = \frac {x + \frac {y + z} {1 + y z} } {1 + x \frac {y + z} {1 + y z} }
+      | c = 
+}}
+{{eqn | r = \frac {x + y + z + x y z} {1 + x y + x z + y z}
+      | c = 
+}}
+{{eqn | r = \paren {x \circ y} \circ z
+      | c = 
+}}
+{{end-eqn}}
+Thus $\circ$ has been shown to be [[Definition:Associative|associative]].
+{{qed|lemma}}
+=== $\text G 2$: Identity ===
+{{begin-eqn}}
+{{eqn | l = x
+      | r = 0
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {x + y} {1 + x y}
+      | r = \frac {0 + y} {1 + 0 y}
+      | c = 
+}}
+{{eqn | r = \frac y 1
+      | c = 
+}}
+{{eqn | r = y
+      | c = 
+}}
+{{end-eqn}}
+Similarly, putting $y = 0$ we find $x \circ y = x$.
+So $0$ is the [[Definition:Identity Element|identity]].
+{{qed|lemma}}
+=== $\text G 3$: Inverses ===
+{{begin-eqn}}
+{{eqn | l = x \circ -x
+      | r = \frac {x + \paren {-x} } {1 + x \paren {-x} }
+      | c = 
+}}
+{{eqn | r = 0
+      | c = 
+}}
+{{end-eqn}}
+Similarly, putting $x = -y$ gives us $\paren {-y} \circ y = 0$.
+So each $x$ has an [[Definition:Inverse Element|inverse]] $-x$.
+{{qed|lemma}}
+=== $\text G 0$: Closure ===
+First note that:
+:$-1 < x, y < 1 \implies x y > -1 \implies 1 + x y > 0$
+Next:
+{{begin-eqn}}
+{{eqn | l = -1
+      | o = <
+      | r = x, y < 1
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = 0
+      | o = <
+      | r = \paren {1 - x} \paren {1 - y}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = 0
+      | o = <
+      | r = 1 + x y - \paren {x + y}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = 0
+      | o = <
+      | r = \frac {1 + x y - \paren {x + y} } {1 + x y}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = 0
+      | o = <
+      | r = \frac {1 + x y} {1 + x y} - \frac {x + y} {1 + x y}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = 0
+      | o = <
+      | r = 1 - \frac {x + y} {1 + x y}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {x + y} {1 + x y}
+      | o = <
+      | r = 1
+      | c = 
+}}
+{{end-eqn}}
+Finally:
+{{begin-eqn}}
+{{eqn | l = -1
+      | o = <
+      | r = x, y < 1
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = 0
+      | o = <
+      | r = \paren {1 + x} \paren {1 + y}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = 0
+      | o = <
+      | r = 1 + x y + \paren {x + y}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = 0
+      | o = <
+      | r = \frac {1 + x y + \paren {x + y} } {1 + x y}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = 0
+      | o = <
+      | r = \frac {1 + x y} {1 + x y} + \frac {x + y} {1 + x y}
+      | c = 
+}}
+{{eqn | ll=\leadsto
+      | l = 0
+      | o = <
+      | r = 1 + \frac {x + y} {1 + x y}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = -1
+      | o = <
+      | r = \frac {x + y} {1 + x y}
+      | c = 
+}}
+{{end-eqn}}
+Thus:
+:$-1 < x, y < 1 \implies -1 < x \circ y < 1$
+and we see that in this range, $\circ$ is [[Definition:Closed Algebraic Structure|closed]].
+{{qed|lemma}}
+Thus the given [[Definition:Set|set]] and [[Definition:Binary Operation|operation]] form a [[Definition:Group|group]].
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Group/Examples/inv x = 1 - x}
+Tags: Examples of Groups, Group: Examples: inv x = 1 - x
+
+\begin{theorem}
+Let $S = \set {x \in \R: 0 < x < 1}$.
+Then an operation $\circ$ can be found such that $\struct {S, \circ}$ is a [[Definition:Group|group]] such that the [[Definition:Inverse Element|inverse]] of $x \in S$ is $1 - x$.
+\end{theorem}
+
+\begin{proof}
+Define $f: \openint 0 1 \to \R$ by:
+:$\map f x := \map \ln {\dfrac {1 - x} x}$
+Let us show that $f$ is a [[Definition:Bijection|bijection]] by constructing an [[Definition:Inverse Mapping|inverse mapping]] $g: \R \to \openint 0 1$:
+:$\map g z := \dfrac 1 {1 + \exp z}$
+=== [[Group/Examples/inv x = 1 - x/Lemma 1|Lemma 1]] ===
+{{:Group/Examples/inv x = 1 - x/Lemma 1}}
+=== [[Group/Examples/inv x = 1 - x/Lemma 2|Lemma 2]] ===
+{{:Group/Examples/inv x = 1 - x/Lemma 2}}
+{{finish|Might also be necessary to demonstrate the domain of $f$ is the image set of $g$ and vice versa.}}
+Thus $f$ is a [[Definition:Bijection|bijection]].
+Let $\struct {\R, +}$ be the [[Definition:Additive Group of Real Numbers|additive group on $\R$]].
+Now define $\circ := +_f$ to be the [[Definition:Operation Induced by Injection|operation induced on $\openint 0 1$ by $f$ and $+$]]:
+:$x \circ y := \map {f^{-1} } {\map f x + \map f y}$
+Let us determine the behaviour of $\circ$ more explicitly:
+{{begin-eqn}}
+{{eqn | l = x \circ y
+      | r = \map g {\map f x + \map f y}
+}}
+{{eqn | r = \frac 1 {1 + \map \exp {\map \log {\frac {1 - x} x} + \map \log {\frac {1 - y} y} } }
+      | c = 
+}}
+{{eqn | r = \frac 1 {1 + \map \exp {\map \log {\frac {1 - x} x} } \map \exp {\map \log {\frac {1 - y} y} } }
+      | c = [[Exponential of Sum]]
+}}
+{{eqn | r = \frac 1 {1 + \paren {\frac {1 - x} x} \paren {\frac {1 - y} y} }
+      | c = [[Exponential of Natural Logarithm]]
+}}
+{{end-eqn}}
+{{MissingLinks}}
+We see that $\circ$ is [[Definition:Commutative Operation|commutative]].
+Let $x = \dfrac 1 2$, $\dfrac {1 - x} x = 1$ so that:
+{{begin-eqn}}
+{{eqn | l = \dfrac 1 2 \circ y
+      | r = \dfrac 1 {1 + \paren {\frac {1 - y} y} }
+}}
+{{eqn | r = \frac 1 {\frac y y + \frac {1 - y} y}
+}}
+{{eqn | r = \frac 1 {\frac 1 y}
+}}
+{{eqn | r = y
+}}
+{{end-eqn}}
+so that $\dfrac 1 2$ is the [[Definition:Identity Element|identity element]] for $\circ$.
+Furthermore, putting $y = 1 - x$, the following obtains:
+{{begin-eqn}}
+{{eqn | l = \frac 1 {1 + \paren {\frac {1 - x} x} \paren {\frac x {1 - x} } }
+      | r = \frac 1 {1 + 1}
+}}
+{{eqn | r = \frac 1 2
+}}
+{{end-eqn}}
+establishing $1 - x$ to be the [[Definition:Inverse Element|inverse]] of $x$, as desired.
+That $\circ$ in fact determines a [[Definition:Group|group]] on $\openint 0 1$ follows from [[Pullback of Group is Group]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Group/Examples/Self-Inverse and Cancellable Elements}
+Tags: Examples of Groups
+
+\begin{theorem}
+Let $S$ be a [[Definition:Set|set]] with an [[Definition:Binary Operation|operation]] which assigns to each $\tuple {a, b} \in S \times S$ an element $a \ast b \in S$ such that:
+: $(1): \quad \exists e \in S: a \ast b = e \iff a = b$
+: $(2): \quad \forall a, b, c \in S: \paren {a \ast c} \ast \paren {b \ast c} = a \ast b$
+Then $\struct {S, \circ}$ is a [[Definition:Group|group]], where $\circ$ is defined as $a \circ b = a \ast \paren {e \ast b}$.
+\end{theorem}
+
+\begin{proof}
+We verify the [[Definition:Group Axioms|group axioms]], in the following order (for convenience):
+=== G0: Closure ===
+Let $a, b \in S$. Then from the definition of $\ast$, we have $a \ast b \in S$, and hence also $a \circ b = a \ast \paren {e \ast b} \in S$.
+This proves [[Definition:Closed Algebraic Structure|closure]] of $\circ$.
+{{qed|lemma}}
+=== G2: Identity ===
+We assert that $e$ is the identity with respect to $\circ$.
+We verify this as follows (let $a \in S$):
+{{begin-eqn}}
+{{eqn | l = a \circ e
+      | r = a \ast \paren {e \ast e}
+      | c = Definition of $\circ$
+}}
+{{eqn | r = a \ast e
+      | c = Property $(1)$ of $\ast$
+}}
+{{eqn | l = e \circ a
+      | r = e \ast \paren {e \ast a}
+      | c = Definition of $\circ$
+}}
+{{eqn | r = \paren {a \ast a} \ast \paren {e \ast a}
+      | c = Property $(1)$ of $\ast$
+}}
+{{eqn | r = a \ast e
+      | c = Property $(2)$ of $\ast$
+}}
+{{end-eqn}}
+It follows that $a \circ e = e \circ a = a \ast e$.
+It remains to prove the following identity:
+: $(3): \quad a \ast e = a$
+By property $(1)$, it suffices to prove $a \ast \paren {a \ast e} = e$:
+{{begin-eqn}}
+{{eqn | l = a \ast \paren {a \ast e}
+      | r = \paren {a \ast e} \ast \paren {\paren {a \ast e} \ast e}
+      | c = Property $(2)$ of $\ast$
+}}
+{{eqn | r = \paren {a \ast e} \ast \paren {\paren {a \ast e} \ast \paren {e \ast e} }
+      | c = Property $(1)$ of $\ast$
+}}
+{{eqn | r = \paren {a \ast e} \ast \paren {a \ast e}
+      | c = Property $(2)$ of $\ast$
+}}
+{{eqn | r = e
+      | c = Property $(1)$ of $\ast$
+}}
+{{end-eqn}}
+So indeed $e$ is the [[Definition:Identity Element|identity]] for $\circ$.
+{{qed|lemma}}
+=== G3: Inverses ===
+For any element $a \in S$, we claim that $e \ast a$ is the [[Definition:Inverse Element|inverse]] for $a$.
+This is verified as follows:
+{{begin-eqn}}
+{{eqn | l = \paren {e \ast a} \circ a
+      | r = \paren {e \ast a} \ast \paren {e \ast a}
+      | c = Definition of $\circ$
+}}
+{{eqn | r = e
+      | c = Property $(1)$ of $\ast$
+}}
+{{eqn | l = a \circ \paren {e \ast a}
+      | r = a \ast \paren {e \ast \paren {e \ast a} }
+      | c = Definition of $\circ$
+}}
+{{eqn | r = a \ast \paren {e \circ a}
+      | c = Definition of $\circ$
+}}
+{{eqn | r = a \ast a
+      | c = $e$ is the identity element for $\circ$
+}}
+{{eqn | r = e
+      | c = Property $(1)$ of $\ast$
+}}
+{{end-eqn}}
+We conclude that $e \ast a$ is indeed the [[Definition:Inverse Element|inverse]] for $a$.
+{{qed|lemma}}
+=== G1: Associativity ===
+Lastly, we prove that $\circ$ is [[Definition:Associative|associative]], i.e.:
+:$a \circ \paren {b \circ c} = \paren {a \circ b} \circ c$
+To this end, we observe the following property of $\ast$:
+{{begin-eqn}}
+{{eqn | n = 4
+      | l = e \ast \paren {a \ast b}
+      | r = \paren {b \ast b} \ast \paren {a \ast b}
+      | c = Property $(1)$ of $\ast$
+}}
+{{eqn | r = b \ast a
+      | c = Property $(2)$ of $\ast$
+}}
+{{end-eqn}}
+Subsequently, we compute:
+{{begin-eqn}}
+{{eqn | l = \paren {a \circ b} \circ c
+      | r = \paren {a \ast \paren {e \ast b} } \ast \paren {e \ast c}
+      | c = Definition of $\circ$
+}}
+{{eqn | r = \paren {a \ast \paren {e \ast b} } \ast \paren {\paren {e \ast c} \ast e}
+      | c = Property $(3)$ of $\ast$
+}}
+{{eqn | r = \paren {a \ast \paren {e \ast b} } \ast \paren {\paren {\paren {e \ast c} \ast b} \ast \paren {e \ast b} }
+      | c = Property $(2)$ of $\ast$
+}}
+{{eqn | r = a \ast \paren {\paren {e \ast c} \ast b}
+      | c = Property $(2)$ of $\ast$
+}}
+{{eqn | r = a \ast \paren {e \ast \paren {b \ast \paren {e \ast c} } }
+      | c = Property $(4)$ of $\ast$
+}}
+{{eqn | r = a \circ \paren {b \circ c}
+      | c = Definition of $\circ$
+}}
+{{end-eqn}}
+This somewhat cumbersome calculation shows that we indeed have [[Definition:Associative|associativity]].
+{{qed|lemma}}
+It follows that $\struct {S, \circ}$ is indeed a [[Definition:Group|group]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Complex Numbers under Addition form Abelian Group}
+Tags: Additive Group of Complex Numbers, Complex Addition, Examples of Abelian Groups, Examples of Infinite Groups
+
+\begin{theorem}
+Let $\C$ be the set of [[Definition:Complex Number|complex numbers]].
+The [[Definition:Algebraic Structure|structure]] $\struct {\C, +}$ is an [[Definition:Infinite Group|infinite]] [[Definition:Abelian Group|abelian group]].
+\end{theorem}
+
+\begin{proof}
+Taking the [[Definition:Group Axioms|group axioms]] in turn:
+=== $\text G 0$: Closure ===
+[[Complex Addition is Closed]].
+{{qed|lemma}}
+=== $\text G 1$: Associativity ===
+[[Complex Addition is Associative]].
+{{qed|lemma}}
+=== $\text G 2$: Identity ===
+From [[Complex Addition Identity is Zero]], we have that the [[Definition:Identity Element|identity element]] of $\struct {\C, +}$ is the [[Definition:Complex Number|complex number]] $0 + 0 i$:
+:$\paren {x + i y} + \paren {0 + 0 i} = \paren {x + 0} + i \paren {y + 0} = x + i y$
+and similarly for $\paren {0 + 0 i} + \paren {x + i y}$.
+{{qed|lemma}}
+=== $\text G 3$: Inverses ===
+From [[Inverse for Complex Addition]], the [[Definition:Inverse Element|inverse]] of $x + i y \in \struct {\C, +}$ is $-x - i y$.
+{{qed|lemma}}
+=== $\text C$: Commutativity ===
+	
+[[Complex Addition is Commutative]].
+{{qed|lemma}}
+=== Infinite ===
+[[Complex Numbers are Uncountable]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Non-Zero Complex Numbers under Multiplication form Abelian Group}
+Tags: Complex Multiplication, Examples of Abelian Groups, Examples of Infinite Groups
+
+\begin{theorem}
+Let $\C_{\ne 0}$ be the set of [[Definition:Complex Number|complex numbers]] without [[Definition:Zero (Number)|zero]], that is:
+:$\C_{\ne 0} = \C \setminus \set 0$
+The [[Definition:Algebraic Structure|structure]] $\struct {\C_{\ne 0}, \times}$ is an [[Definition:Infinite Group|infinite]] [[Definition:Abelian Group|abelian group]].
+\end{theorem}
+
+\begin{proof}
+Taking the [[Definition:Group Axioms|group axioms]] in turn:
+=== $\text G 0$: Closure ===
+[[Non-Zero Complex Numbers Closed under Multiplication]].
+{{qed|lemma}}
+=== $\text G 1$: Associativity ===
+[[Complex Multiplication is Associative]].
+{{qed|lemma}}
+=== $\text G 2$: Identity ===
+From [[Complex Multiplication Identity is One]], the identity element of $\struct {\C_{\ne 0}, \times}$ is the [[Definition:Complex Number|complex number]] $1 + 0 i$.
+{{qed|lemma}}
+=== $\text G 3$: Inverses ===
+From [[Inverse for Complex Multiplication]], the inverse of $x + i y \in \struct {\C_{\ne 0}, \times}$ is:
+:$\dfrac 1 z = \dfrac {x - i y} {x^2 + y^2} = \dfrac {\overline z} {z \overline z}$
+where $\overline z$ is the [[Definition:Complex Conjugate|complex conjugate]] of $z$.
+{{qed|lemma}}
+=== $\text C$: Commutativity ===
+	
+[[Complex Multiplication is Commutative]].
+{{qed|lemma}}
+=== Infinite ===
+[[Complex Numbers are Uncountable]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Real Numbers under Addition form Abelian Group}
+Tags: Additive Group of Real Numbers
+
+\begin{theorem}
+Let $\R$ be the set of [[Definition:Real Number|real numbers]].
+The [[Definition:Algebraic Structure|structure]] $\struct {\R, +}$ is an [[Definition:Infinite Group|infinite]] [[Definition:Abelian Group|abelian group]].
+\end{theorem}
+
+\begin{proof}
+Taking the [[Definition:Group Axioms|group axioms]] in turn:
+=== $\text G 0$: Closure ===
+[[Real Addition is Closed]].
+{{qed|lemma}}
+=== $\text G 1$: Associativity ===
+[[Real Addition is Associative]].
+{{qed|lemma}}
+=== $\text G 2$: Identity ===
+From [[Real Addition Identity is Zero]], we have that the [[Definition:Identity Element|identity element]] of $\struct {\R, +}$ is the real number $0$.
+{{qed|lemma}}
+=== $\text G 3$: Inverses ===
+From [[Inverses for Real Addition]], we have that the [[Definition:Inverse Element|inverse]] of $x \in \struct {\R, +}$ is $-x$.
+{{qed|lemma}}
+=== $\text C$: Commutativity ===
+	
+[[Real Addition is Commutative]].
+{{qed|lemma}}
+=== Infinite ===
+[[Real Numbers are Uncountable|Real Numbers are Uncountably Infinite]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Non-Zero Real Numbers under Multiplication form Abelian Group}
+Tags: Real Multiplication, Examples of Abelian Groups, Examples of Infinite Groups, Non-Zero Real Numbers under Multiplication form Abelian Group
+
+\begin{theorem}
+Let $\R_{\ne 0}$ be the set of [[Definition:Real Number|real numbers]] without [[Definition:Zero (Number)|zero]]:
+:$\R_{\ne 0} = \R \setminus \set 0$
+The [[Definition:Algebraic Structure|structure]] $\struct {\R_{\ne 0}, \times}$ is an [[Definition:Uncountable Group|uncountable]] [[Definition:Abelian Group|abelian group]].
+\end{theorem}
+
+\begin{proof}
+Taking the [[Definition:Group Axioms|group axioms]] in turn:
+=== $\text G 0$: Closure ===
+From [[Non-Zero Real Numbers Closed under Multiplication/Proof 2|Non-Zero Real Numbers Closed under Multiplication: Proof 2]], $\R_{\ne 0}$ is [[Definition:Closed Algebraic Structure|closed]] under [[Definition:Real Multiplication|multiplication]].
+Note that [[Non-Zero Real Numbers Closed under Multiplication/Proof 2|proof 2]] needs to be used specifically here, as [[Non-Zero Real Numbers Closed under Multiplication/Proof 1|proof 1]] rests on this result.
+{{qed|lemma}}
+=== $\text G 1$: Associativity ===
+[[Real Multiplication is Associative]].
+{{qed|lemma}}
+=== $\text G 2$: [[Real Multiplication Identity is One|Identity]] ===
+{{:Real Multiplication Identity is One}}
+{{qed|lemma}}
+=== $\text G 3$: [[Inverses for Real Multiplication|Inverses]] ===
+{{:Inverses for Real Multiplication}}
+{{qed|lemma}}
+=== $\text C$: Commutativity ===
+	
+[[Real Multiplication is Commutative]].
+{{qed|lemma}}
+=== Infinite ===
+[[Real Numbers are Uncountable|Real Numbers are Uncountably Infinite]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+We have [[Real Numbers under Multiplication form Monoid]].
+From [[Inverses for Real Multiplication]], the non-[[Definition:Zero (Number)|zero]] [[Definition:Number|numbers]] are exactly the [[Definition:Invertible Element|invertible elements]] of [[Definition:Real Multiplication|real multiplication]].
+Thus from [[Invertible Elements of Monoid form Subgroup of Cancellable Elements]], the non-[[Definition:Zero (Number)|zero]] [[Definition:Real Number|real numbers]] under [[Definition:Real Multiplication|multiplication]] form a [[Definition:Group|group]].
+From:
+:[[Real Multiplication is Commutative]]
+:[[Subset Product within Commutative Structure is Commutative]]
+it follows that this [[Definition:Group|group]] is also [[Definition:Abelian Group|Abelian]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+From [[Non-Zero Real Numbers under Multiplication form Group]], $\struct {\R_{\ne 0}, \times}$ forms a [[Definition:Group|group]].
+{{qed|lemma}}
+From [[Real Multiplication is Commutative]] it follows that $\struct {\R_{\ne 0}, \times}$ is [[Definition:Abelian Group|abelian]].
+{{qed|lemma}}
+From [[Real Numbers are Uncountable|Real Numbers are Uncountably Infinite]] it follows that $\struct {\R_{\ne 0}, \times}$ is an [[Definition:Uncountable Group|uncountable]] [[Definition:Abelian Group|abelian group]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Rational Numbers under Addition form Abelian Group}
+Tags: Rational Addition, Examples of Abelian Groups, Examples of Infinite Groups
+
+\begin{theorem}
+Let $\Q$ be the set of [[Definition:Rational Number|rational numbers]].
+The [[Definition:Algebraic Structure|structure]] $\struct {\Q, +}$ is a [[Definition:Countably Infinite|countably infinite]] [[Definition:Abelian Group|abelian group]].
+\end{theorem}
+
+\begin{proof}
+The [[Definition:Rational Number|rational numbers]] are, by definition, the [[Definition:Field of Quotients|field of quotients]] of the [[Definition:Integral Domain|integral domain]] $\struct {\Z, +, \times}$ of [[Definition:Integer|integers]].
+Hence by definition, $\struct {\Q, +, \times}$ is a [[Definition:Field (Abstract Algebra)|field]].
+The fact that $\struct {\Q, +}$ forms an [[Definition:Abelian Group|abelian group]] follows directly from the definition of a [[Definition:Field (Abstract Algebra)|field]].
+From [[Rational Numbers are Countably Infinite]], we have that $\struct {\Q, +}$ is [[Definition:Countably Infinite|countably infinite]].
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Non-Zero Rational Numbers under Multiplication form Abelian Group}
+Tags: Rational Multiplication, Examples of Abelian Groups, Examples of Infinite Groups
+
+\begin{theorem}
+Let $\Q_{\ne 0}$ be the [[Definition:Set|set]] of non-[[Definition:Zero (Number)|zero]] [[Definition:Rational Number|rational numbers]]:
+:$\Q_{\ne 0} = \Q \setminus \set 0$
+The [[Definition:Algebraic Structure|structure]] $\struct {\Q_{\ne 0}, \times}$ is a [[Definition:Countably Infinite Group|countably infinite]] [[Definition:Abelian Group|abelian group]].
+\end{theorem}
+
+\begin{proof}
+From the [[Definition:Rational Number|definition of rational numbers]], the [[Definition:Algebraic Structure|structure]] $\struct {\Q, + \times}$ is constructed as the [[Definition:Field of Quotients|field of quotients]] of the [[Definition:Integral Domain|integral domain]] $\struct {\Z, +, \times}$ of [[Definition:Integer|integers]].
+Hence from [[Multiplicative Group of Field is Abelian Group]], $\struct {\Q_{\ne 0}, \times}$ is an [[Definition:Abelian Group|abelian group]].
+From [[Rational Numbers are Countably Infinite]], we have that $\struct {\Q_{\ne 0}, \times}$ is a [[Definition:Countably Infinite Group|countably infinite group]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Integers under Multiplication form Countably Infinite Commutative Monoid}
+Tags: Integer Multiplication
+
+\begin{theorem}
+The [[Definition:Set|set]] of [[Definition:Integer|integers]] under [[Definition:Integer Multiplication|multiplication]] $\struct {\Z, \times}$ is a [[Definition:Countably Infinite Set|countably infinite]] [[Definition:Commutative Monoid|commutative monoid]].
+\end{theorem}
+
+\begin{proof}
+First we note that [[Integers under Multiplication form Monoid]].
+{{qed|lemma}}
+Then we have: 
+=== Commutativity ===
+[[Integer Multiplication is Commutative]].
+{{qed|lemma}}
+=== Infinite ===
+[[Integers are Countably Infinite]].
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Additive Group of Rationals is Subgroup of Reals}
+Tags: Rational Addition, Additive Group of Real Numbers, Normal Subgroups, Additive Group of Rational Numbers
+
+\begin{theorem}
+Let $\struct {\Q, +}$ be the [[Definition:Additive Group of Rational Numbers|additive group of rational numbers]].
+Let $\struct {\R, +}$ be the [[Definition:Additive Group of Real Numbers|additive group of real numbers]].
+Then $\struct {\Q, +}$ is a [[Definition:Normal Subgroup|normal subgroup]] of $\struct {\R, +}$.
+\end{theorem}
+
+\begin{proof}
+From the definition of [[Definition:Real Number|real numbers]], $\Q$ is a [[Definition:Subset|subset]] of $\R$.
+As $\struct {\R, +}$ is a [[Definition:Group|group]], and $\struct {\Q, +}$ is a [[Definition:Group|group]], it follows from the definition of [[Definition:Subgroup|subgroup]] that $\struct {\Q, +}$ is a subgroup of $\struct {\R, +}$.
+As $\struct {\R, +}$ is [[Definition:Abelian Group|abelian]], it follows from [[Subgroup of Abelian Group is Normal]] that $\struct {\Q, +}$ is [[Definition:Normal Subgroup|normal]] in $\struct {\R, +}$.
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Additive Group of Integers is Subgroup of Rationals}
+Tags: Additive Group of Integers, Additive Group of Rational Numbers, Normal Subgroups
+
+\begin{theorem}
+Let $\struct {\Z, +}$ be the [[Definition:Additive Group of Integers|additive group of integers]].
+Let $\struct {\Q, +}$ be the [[Definition:Additive Group of Rational Numbers|additive group of rational numbers]].
+Then $\struct {\Z, +}$ is a [[Definition:Subgroup|subgroup]] of $\struct {\Q, +}$.
+\end{theorem}
+
+\begin{proof}
+Recall that [[Integers form Integral Domain]].
+The set $\Q$ of [[Definition:Rational Number|rational numbers]] is defined as the [[Definition:Rational Number|field of quotients of the integers]].
+The fact that the [[Definition:Integer|integers]] are a [[Definition:Subgroup|subgroup]] of the [[Definition:Rational Number|rationals]] follows from the work done in proving the [[Existence of Field of Quotients]] from an [[Definition:Integral Domain|integral domain]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Additive Group of Reals is Normal Subgroup of Complex}
+Tags: Additive Group of Real Numbers, Additive Group of Complex Numbers, Examples of Normal Subgroups
+
+\begin{theorem}
+Let $\struct {\R, +}$ be the [[Definition:Additive Group of Real Numbers|additive group of real numbers]].
+Let $\struct {\C, +}$ be the [[Definition:Additive Group of Complex Numbers|additive group of complex numbers]].
+Then $\struct {\R, +}$ is a [[Definition:Normal Subgroup|normal subgroup]] of $\struct {\C, +}$.
+\end{theorem}
+
+\begin{proof}
+From [[Additive Group of Reals is Subgroup of Complex]], $\struct {\R, +}$ is a [[Definition:Subgroup|subgroup]] of $\struct {\C, +}$.
+Then from [[Complex Numbers under Addition form Abelian Group]], $\struct {\C, +}$ is [[Definition:Abelian Group|abelian]].
+The result follows from [[Subgroup of Abelian Group is Normal]].
+{{Qed}}
+[[Category:Additive Group of Real Numbers]]
+[[Category:Additive Group of Complex Numbers]]
+[[Category:Examples of Normal Subgroups]]
+1u96lp0tx8g2ml9xdr96nzf90lau0jp
+\end{proof}<|endoftext|>
+\section{Multiplicative Group of Reals is Subgroup of Complex}
+Tags: Multiplicative Group of Real Numbers, Multiplicative Group of Complex Numbers, Normal Subgroups
+
+\begin{theorem}
+Let $\struct {\R_{\ne 0}, \times}$ be the [[Definition:Multiplicative Group of Real Numbers|multiplicative group of real numbers]].
+Let $\struct {\C_{\ne 0}, \times}$ be the [[Definition:Multiplicative Group of Complex Numbers|multiplicative group of complex numbers]].
+Then $\struct {\R_{\ne 0}, \times}$ is a [[Definition:Normal Subgroup|normal subgroup]] of $\struct {\C_{\ne 0}, \times}$.
+\end{theorem}
+
+\begin{proof}
+Let $x, y \in \C_{\ne 0}$ such that $x = x_1 + 0 i, y = y_1 + 0 i$.
+As $x$ and $y$ are [[Definition:Wholly Real|wholly real]], we have that $x, y \in \R_{\ne 0}$.
+Then $x + y = x_1 y_1 - 0 \cdot 0 + \left({0 \cdot y_1 + x_1 \cdot 0}\right) i = x_1 y_1 + 0i$ which is also [[Definition:Wholly Real|wholly real]].
+Also, the inverse of $x$ is $\dfrac 1 x = \dfrac 1 {x_1} + 0 i$ which is also [[Definition:Wholly Real|wholly real]].
+Thus by the [[Two-Step Subgroup Test]], $\struct {\R_{\ne 0}, \times}$ is a [[Definition:Subgroup|subgroup]] of $\struct {\C_{\ne 0}, \times}$.
+From [[Non-Zero Complex Numbers under Multiplication form Abelian Group]], $\struct {\C_{\ne 0}, \times}$ is [[Definition:Abelian Group|abelian]].
+The result follows from [[Subgroup of Abelian Group is Normal]].
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Multiplicative Group of Rationals is Subgroup of Reals}
+Tags: Normal Subgroups, Multiplicative Group of Rational Numbers, Multiplicative Group of Real Numbers
+
+\begin{theorem}
+Let $\struct {\Q_{\ne 0}, \times}$ be the [[Definition:Multiplicative Group of Rational Numbers|multiplicative group of rational numbers]].
+Let $\struct {\R_{\ne 0}, \times}$ be the [[Definition:Multiplicative Group of Real Numbers|multiplicative group of real numbers]].
+Then $\struct {\Q_{\ne 0}, \times}$ is a [[Definition:Normal Subgroup|normal subgroup]] of $\left({\R_{\ne 0}, \times}\right)$.
+\end{theorem}
+
+\begin{proof}
+From the definition of [[Definition:Real Number|real numbers]], it follows that $\Q_{\ne 0}$ is a [[Definition:Subset|subset]] of $\R_{\ne 0}$.
+As $\struct {\R_{\ne 0}, \times}$ is a [[Definition:Group|group]], and $\struct {\Q_{\ne 0}, \times}$ is a [[Definition:Group|group]], it follows from the definition of [[Definition:Subgroup|subgroup]] that $\struct {\Q_{\ne 0}, \times}$ is a subgroup of $\struct {\R_{\ne 0}, \times}$.
+As $\struct {\R_{\ne 0}, \times}$ is [[Definition:Abelian Group|abelian]], it follows from [[Subgroup of Abelian Group is Normal]] that $\struct {\Q_{\ne 0}, \times}$ is [[Definition:Normal Subgroup|normal]] in $\struct {\R_{\ne 0}, \times}$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Circle Group is Infinite Abelian Group}
+Tags: Circle Group, Complex Numbers, Circle Group is Infinite Abelian Group
+
+\begin{theorem}
+The [[Definition:Circle Group|circle group]] $\struct {K, \times}$ is an [[Definition:Uncountable Set|uncountably infinite]] [[Definition:Abelian Group|abelian group]] under the operation of [[Definition:Complex Multiplication|complex multiplication]].
+\end{theorem}
+
+\begin{proof}
+We note that $K \ne \varnothing$ as the [[Complex Multiplication Identity is One|identity element $1 + 0 i \in K$]].
+Since all $z \in K$ have [[Definition:Complex Modulus|modulus]] $1$, they have, for some $\theta \in \hointr 0 {2 \pi}$, the [[Definition:Polar Form of Complex Number|polar form]]:
+:$z = \map \exp {i \theta} = \cos \theta + i \sin \theta$
+Conversely, if a [[Definition:Complex Number|complex number]] has such a [[Definition:Polar Form of Complex Number|polar form]], it has [[Definition:Complex Modulus|modulus]] $1$.
+Observe the following property of the [[Definition:Complex Exponential Function|complex exponential function]]:
+:$\forall a, b \in \C: \map \exp {a + b} = \map \exp a \map \exp b$
+We must show that if $x, y \in K$ then $x \cdot y^{-1} \in K$.
+Let $x, y \in K$ be arbitrary.
+Choose suitable $s, t \in \hointr 0 {2 \pi}$ such that:
+:$x = \map \exp {i s}$
+:$y = \map \exp {i t}$
+We compute:
+{{begin-eqn}}
+{{eqn | l = \map \exp {i t} \map \exp {-i t}
+      | r = \map \exp {i \paren {t - t} }
+}}
+{{eqn | r = \map \exp 0
+}}
+{{eqn | r = 1
+}}
+{{end-eqn}}
+So $y^{-1} = \map \exp {-i t}$.
+We note that this lies in $K$.
+Furthermore, we have:
+{{begin-eqn}}
+{{eqn | l = x y
+      | r = \map \exp {i s} \map \exp {-2 \pi i t}
+}}
+{{eqn | r = \map \exp {i \paren {s - t} }
+}}
+{{end-eqn}}
+We conclude that $x y \in K$.
+By the [[Two-Step Subgroup Test]], $K$ is a subgroup of $\C$ under [[Definition:Complex Multiplication|complex multiplication]].
+{{handwaving|actually, instead of $\hointr 0 {2 \pi}$ we should use $\R / 2 \pi \Z$ for formal correctness, but I haven't thought of a nice way to do so. Also the property of $\exp$ is only established for the reals so far.}}
+That the operation $\times$ on $K$ is [[Definition:Commutative Operation|commutative]] follows from [[Complex Multiplication is Commutative]] and [[Restriction of Commutative Operation is Commutative]].
+That is, $\times$ is [[Definition:Commutative Operation|commutative]] on $K$ because it is already [[Definition:Commutative Operation|commutative]] on $\C$.
+Finally we have that the [[Circle Group is Uncountably Infinite]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+First we note that $K \subseteq \C$.
+So to show that $K$ is a [[Definition:Group|group]] it is sufficient to show that $K$ is a [[Definition:Subgroup|subgroup]] of the [[Definition:Multiplicative Group of Complex Numbers|multiplicative group of complex numbers]] $\struct {\C_{\ne 0}, \times}$.
+From [[Complex Multiplication Identity is One]], the [[Definition:Identity Element|identity element]] $1 + 0 i$ is in $K$.
+Thus $K \ne \O$.
+We now show that $z, w \in K \implies z w \in K$:
+{{begin-eqn}}
+{{eqn | l = z, w
+      | o = \in
+      | r = K
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \cmod z
+      | r = 1 = \cmod w
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \cmod {z w}
+      | r = \cmod z \cmod w
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = z w
+      | o = \in
+      | r = K
+      | c = 
+}}
+{{end-eqn}}
+Next we see that $z \in K \implies z^{-1} \in K$:
+:$\cmod z = 1 \implies \cmod {\dfrac 1 z} = 1$
+Thus by the [[Two-Step Subgroup Test]]:
+:$K \le \C_{\ne 0}$
+Thus $K$ is a [[Definition:Group|group]].
+From [[Complex Multiplication is Commutative]] it also follows from [[Subgroup of Abelian Group is Abelian]] that $K$ is an [[Definition:Abelian Group|abelian group]].
+Finally we have that the [[Circle Group is Uncountably Infinite]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Taking the [[Definition:Group Axioms|group axioms]] in turn:
+=== $\text G 0$: Closure ===
+{{begin-eqn}}
+{{eqn | l = z, w
+      | o = \in
+      | r = K
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \cmod z
+      | r = 1 = \cmod w
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \cmod {z w}
+      | r = \cmod z \cmod w
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = z w
+      | o = \in
+      | r = K
+      | c = 
+}}
+{{end-eqn}}
+So $\struct {\mathbb S, \cdot}$ is [[Definition:Closed Algebraic Structure|closed]].
+{{qed|lemma}}
+=== $\text G 1$: Associativity ===
+[[Complex Multiplication is Associative]].
+{{qed|lemma}}
+=== $\text G 2$: Identity ===
+From [[Complex Multiplication Identity is One]] we have that the [[Definition:Identity Element|identity element]] of $K$ is $1 + 0 i$.
+{{qed|lemma}}
+=== $\text G 3$: Inverses ===
+We have that:
+:$\cmod z = 1 \implies \dfrac 1 {\cmod z} = \cmod {\dfrac 1 z} = 1$
+But:
+:$z \times \dfrac 1 z = 1 + 0 i$
+So the [[Definition:Inverse Element|inverse]] of $z$ is $\dfrac 1 z$.
+{{qed|lemma}}
+=== $\text C$: Commutative ===
+We have that [[Complex Multiplication is Commutative]].
+We also have from [[Restriction of Commutative Operation is Commutative]] that $\times$ is likewise [[Definition:Commutative Operation|commutative]] on $K$.
+{{qed|lemma}}
+=== Uncountably Infinite ===
+[[Circle Group is Uncountably Infinite]].
+{{Qed|lemma}}
+All the criteria are satisfied, and the result follows.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Consider the [[Definition:Complex Modulus|complex modulus]] function $\left\vert{\cdot}\right\vert: \C \to \R, z \mapsto \left\vert{z}\right\vert$.
+By [[Complex Modulus is Norm]], we have that $\left\vert{z}\right\vert \ge 0$ for all $z \in \C$, and:
+:$\left\vert{z}\right\vert = 0 \iff z = 0$
+From [[Group of Units of Field]] and [[Complex Numbers form Field]], we have $\C^\times = \C \setminus \left\{{0}\right\}$.
+By above observation, the [[Definition:Complex Modulus|modulus]] has a [[Definition:Restriction of Mapping|restriction]] to $\C^\times$:
+:$\left\vert{\cdot}\right\vert: \C^\times \to \R_{>0}, z \mapsto \left\vert{z}\right\vert$
+From $\left\vert{1}\right\vert = 1$ and [[Modulus of Product]], it follows that it is in fact a [[Definition:Group Homomorphism|group homomorphism]]:
+:$\phi: \left({C^\times, \cdot}\right) \to \left({\R_{>0}, \cdot}\right), z \mapsto \left\vert{z}\right\vert$
+Now $K$ is by definition the [[Definition:Kernel of Group Homomorphism|kernel]] of $\phi$.
+Hence, by [[Kernel of Group Homomorphism is Subgroup]], $\left({K, \times}\right)$ is a [[Definition:Subgroup|subgroup]] of $\left({\C^\times, \cdot}\right)$.
+By [[Subgroup of Abelian Group is Abelian]] and [[Complex Multiplication is Commutative]], it is also [[Definition:Abelian Group|abelian]].
+Finally, note that $\left({K, \times}\right)$ is [[Definition:Uncountable Set|uncountable]] from [[Circle Group is Uncountably Infinite]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Homomorphism from Reals to Circle Group}
+Tags: Circle Group, Examples of Group Homomorphisms
+
+\begin{theorem}
+Let $\struct {\R, +}$ be the [[Definition:Additive Group of Real Numbers|additive group of real numbers]].
+Let $\struct {K, \times}$ be the [[Definition:Circle Group|circle group]].
+Let $\phi: \struct {\R, +} \to \struct {K, \times}$ be the [[Definition:Mapping|mapping]] defined as:
+:$\forall x \in \R: \map \phi x = e^{i x}$
+Then $\phi$ is a [[Definition:Group Homomorphism|(group) homomorphism]].
+\end{theorem}
+
+\begin{proof}
+Let $x, y \in \R$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \map \phi x \times \map \phi y
+      | r = e^{i x} e^{i y}
+      | c = 
+}}
+{{eqn | r = e^{i \paren {x + y} }
+      | c = 
+}}
+{{eqn | r = \map \phi {x + y}
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Integers under Addition form Infinite Cyclic Group}
+Tags: Examples of Cyclic Groups, Integer Addition, Additive Group of Integers
+
+\begin{theorem}
+The [[Definition:Additive Group of Integers|additive group of integers]] $\struct {\Z, +}$ is an [[Definition:Infinite Cyclic Group|infinite cyclic group]] which is [[Definition:Generator of Cyclic Group|generated by]] the element $1 \in \Z$.
+\end{theorem}
+
+\begin{proof}
+By [[Epimorphism from Integers to Cyclic Group]] and [[Definition:Integer Multiplication|integer multiplication]]:
+:$\forall n \in \Z: n = \map {+^n} 1 \in \gen 1$
+Thus:
+:$\struct {\Z, +} = \gen 1$
+and thus, by the definition of a [[Definition:Cyclic Group|cyclic group]], is [[Definition:Cyclic Group|cyclic]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Generators of Additive Group of Integers}
+Tags: Integer Addition, Cyclic Groups, Additive Group of Integers
+
+\begin{theorem}
+The only [[Definition:Generator of Cyclic Group|generators]] of the [[Definition:Additive Group of Integers|additive group of integers]] $\struct {\Z, +}$ are $1$ and $-1$.
+\end{theorem}
+
+\begin{proof}
+From [[Integers under Addition form Infinite Cyclic Group]], $\struct {\Z, +}$ is an [[Definition:Infinite Cyclic Group|infinite cyclic group]] generated by $1$.
+From [[Generators of Infinite Cyclic Group]], there is only one other [[Definition:Generator of Cyclic Group|generator]] of such a group, and that is the inverse of that generator.
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Inverse of Generator of Cyclic Group is Generator}
+Tags: Cyclic Groups, Inverse of Generator of Cyclic Group is Generator
+
+\begin{theorem}
+Let $\gen g = G$ be a [[Definition:Cyclic Group|cyclic group]].
+Then:
+:$G = \gen {g^{-1} }$
+where $g^{-1}$ denotes the [[Definition:Inverse Element|inverse]] of $g$.
+Thus, in general, a [[Definition:Generator of Cyclic Group|generator]] of a [[Definition:Cyclic Group|cyclic group]] is not [[Definition:Unique|unique]].
+\end{theorem}
+
+\begin{proof}
+Let $\gen g = G$.
+Then from [[Set of Words Generates Group]]:
+:$\map W {\set {g, g^{-1} } } = G$
+But:
+:$\gen {g^{-1} } = \map W {\set {g, g^{-1} } }$
+and the result follows.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let $C_n = \gen g$ be the [[Definition:Cyclic Group|cyclic group]] of [[Definition:Order of Structure|order $n$]].
+By definition, $g^n = e$.
+We have that $n - 1$ is [[Definition:Coprime Integers|coprime]] to $n$.
+So it follows from that [[Power of Generator of Cyclic Group is Generator iff Power is Coprime with Order]] that:
+: $C_n = \gen {g^{n - 1} }$
+But from [[Inverse Element is Power of Order Less 1]]:
+:$g^{n - 1} = g^{-1}$
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Generators of Infinite Cyclic Group}
+Tags: Cyclic Groups, Infinite Cyclic Group
+
+\begin{theorem}
+Let $\gen g = G$ be an [[Definition:Infinite Cyclic Group|infinite cyclic group]].
+Then the only other [[Definition:Generator of Cyclic Group|generator]] of $G$ is $g^{-1}$.
+Thus an [[Definition:Infinite Cyclic Group|infinite cyclic group]] has exactly $2$ [[Definition:Generator of Cyclic Group|generators]].
+\end{theorem}
+
+\begin{proof}
+By definition, the [[Definition:Infinite Cyclic Group|infinite cyclic group]] with [[Definition:Generator of Cyclic Group|generator]] $g$ is:
+:$\gen g = \set {\ldots, g^{-2}, g^{-1}, e, g, g^2, \ldots}$
+where $e$ denotes the [[Definition:Identity Element|identity]] $e = g^0$.
+The fact that $g^{-1}$ [[Definition:Generator of Cyclic Group|generates]] $G$ is shown by [[Inverse of Generator of Cyclic Group is Generator]].
+Futhermore:
+:$\gen e = \set e \ne G$
+By definition of [[Definition:Infinite Cyclic Group|infinite cyclic group]]:
+:$g^i \ne g^j$ for all $i \ne j$
+Let $n \in Z \setminus \set {-1, 0, 1}$.
+Then:
+:$\gen {g^n} = \set {\ldots, g^{-2 n}, g^{-n}, e, g^n, g^{2 n}, \ldots}$
+But since $\order n > 1$, none of these elements is equal to $g$, because $1 \notin n \Z$.
+So:
+:$g \notin \gen {g^n} \implies \gen {g^n} \ne \gen g$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Subgroup of Integers is Ideal}
+Tags: Subgroups, Integers, Ideal Theory
+
+\begin{theorem}
+Let $\struct {\Z, +}$ be the [[Definition:Additive Group of Integers|additive group of integers]].
+Every [[Definition:Subgroup|subgroup]] of $\struct {\Z, +}$ is an [[Definition:Ideal of Ring|ideal]] of the [[Definition:Ring (Abstract Algebra)|ring]] $\struct {\Z, +, \times}$.
+\end{theorem}
+
+\begin{proof}
+Let $H$ be a subgroup of $\struct {\Z, +}$.
+Let $n \in \Z, h \in H$.
+Then from the definition of [[Definition:Cyclic Group|cyclic group]] and [[Index Laws for Monoids/Negative Index|Negative Index Law for Monoids]]:
+:$n h = n \cdot h \in \gen h \subseteq H$
+The result follows.
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Additive Group of Integers is Subgroup of Reals}
+Tags: Additive Group of Integers, Additive Group of Real Numbers
+
+\begin{theorem}
+Let $\struct {\Z, +}$ be the [[Definition:Additive Group of Integers|additive group of integers]].
+Let $\struct {\R, +}$ be the [[Definition:Additive Group of Real Numbers|additive group of real numbers]].
+Then $\struct {\Z, +}$ is a [[Definition:Subgroup|subgroup]] of $\struct {\R, +}$.
+\end{theorem}
+
+\begin{proof}
+From [[Additive Group of Integers is Subgroup of Rationals]], $\struct {\Z, +}$ is a [[Definition:Subgroup|subgroup]] of $\struct {\Q, +}$.
+From [[Additive Group of Rationals is Subgroup of Reals]], $\struct {\Q, +}$ is a [[Definition:Subgroup|subgroup]] of $\struct {\R, +}$.
+Thus $\struct {\Z, +}$ is a [[Definition:Subgroup|subgroup]] of $\struct {\R, +}$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Quotient Group of Reals by Integers is Circle Group}
+Tags: Integers, Real Numbers, Complex Numbers, Quotient Groups, Circle Group
+
+\begin{theorem}
+Let $\struct {\Z, +}$ be the [[Definition:Additive Group of Integers|additive group of integers]].
+Let $\struct {\R, +}$ be the [[Definition:Additive Group of Real Numbers|additive group of real numbers]].
+Let $K$ be the [[Definition:Circle Group|circle group]].
+Then the [[Definition:Quotient Group|quotient group]] of $\struct {\R, +}$ by $\struct {\Z, +}$ is [[Definition:Isomorphism (Abstract Algebra)|isomorphic]] to $K$.
+\end{theorem}
+
+\begin{proof}
+Define $\phi: \R / \Z \to K$ by:
+:$\map \phi {x + \Z} = \map \exp {2 \pi i x}$
+Then $\phi$ is [[Definition:Well-Defined Mapping|well-defined]].
+For, if $x + \Z = y + \Z$, then $y = x + n$ for some $n \in \Z$, and:
+:$\map \exp {2 \pi i \paren {x + n} } = \map \exp {2 \pi i x}$
+by [[Complex Exponential Function has Imaginary Period]].
+Moreover, by [[Exponential of Sum/Complex Numbers|Exponential of Sum]]:
+:$\map \exp {x + y + \Z} = \map \exp {2 \pi i \paren {x + y} } = \map \exp {2 \pi i x} \, \map \exp {2 \pi i y}$
+meaning $\phi$ is a [[Definition:Group Homomorphism|group homomorphism]].
+By [[Euler's Formula]]:
+:$\map \exp {2 \pi i x} = \map \cos {2 \pi i x} + i \, \map \sin {2 \pi i x}$
+so that, by [[Sine and Cosine are Periodic on Reals]]:
+:$\map \phi x = 1$ {{iff}} $x + \Z = 0 + \Z$
+Hence, by [[Kernel is Trivial iff Monomorphism/Group|Kernel is Trivial iff Monomorphism]], $\phi$ is a [[Definition:Group Monomorphism|monomorphism]].
+From [[Definition:Polar Form of Complex Number|polar form for complex numbers]], it follows that all $z \in \C$ with $\cmod z = 1$ are of the form:
+:$z = \map \exp {2 \pi i x}$
+for some $x \in \R$.
+Hence $\phi$ is also an [[Definition:Group Epimorphism|epimorphism]].
+Thus $\phi: \R / \Z \to K$ is a [[Definition:Group Isomorphism|group isomorphism]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Integers Modulo m under Addition form Cyclic Group}
+Tags: Additive Groups of Integers Modulo m, Modulo Addition
+
+\begin{theorem}
+Let $\Z_m$ be the set of [[Definition:Integers Modulo m|integers modulo $m$]].
+Let $+_m$ be the [[Definition:Binary Operation|operation]] of [[Definition:Modulo Addition|addition modulo $m$]].
+Let $\struct {\Z_m, +_m}$ denote the [[Definition:Additive Group of Integers Modulo m|additive group of integers modulo $m$]].
+Then $\struct {\Z_m, +_m}$ is a [[Definition:Cyclic Group|cyclic group]] of [[Definition:Order of Structure|order]] $m$, [[Definition:Generator of Cyclic Group|generated]] by the element $\eqclass 1 m \in \Z_m$.
+\end{theorem}
+
+\begin{proof}
+From the definition of [[Definition:Integers Modulo m|integers modulo $m$]], we have:
+:$\Z_m = \dfrac \Z {\RR_m} = \set {\eqclass 0 m, \eqclass 1 m, \ldots, \eqclass {m - 1} m}$
+It is established that [[Modulo Addition is Well-Defined]]:
+:$\eqclass a m +_m \eqclass b m = \eqclass {a + b} m$
+The [[Definition:Group Axioms|group axioms]] are fulfilled:
+: '''{{GroupAxiom|0}}''': [[Modulo Addition is Closed|Addition modulo $m$ is closed]].
+: '''{{GroupAxiom|1}}''': [[Modulo Addition is Associative|Addition modulo $m$ is associative]].
+: '''{{GroupAxiom|2}}''': The [[Modulo Addition has Identity|identity element of $\struct {\Z_m, +_m}$]] is $\eqclass 0 m$.
+: '''{{GroupAxiom|3}}''': The [[Modulo Addition has Inverses|inverse of $\eqclass k m \in \Z_m$]] is $-\eqclass k m = \eqclass {-k} m = \eqclass {n - k} m$.
+: Commutativity: [[Modulo Addition is Commutative|Addition modulo $m$ is commutative]].
+From [[Integers under Addition form Infinite Cyclic Group]] and [[Quotient Group of Cyclic Group]], $\struct {\dfrac \Z {\RR_m}, +_m}$ is [[Definition:Cyclic Group|cyclic]] [[Definition:Order of Group|order]] $m$.
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Integers Modulo m under Multiplication form Commutative Monoid}
+Tags: Modulo Multiplication, Examples of Monoids
+
+\begin{theorem}
+The [[Definition:Algebraic Structure|structure]]:
+:$\struct {\Z_m, \times}$
+(where $\Z_m$ is the set of [[Definition:Integers Modulo m|integers modulo $m$]]) is a [[Definition:Commutative Monoid|commutative monoid]].
+\end{theorem}
+
+\begin{proof}
+[[Modulo Multiplication is Closed|Multiplication modulo $m$ is closed]].
+[[Modulo Multiplication is Associative|Multiplication modulo $m$ is associative]].
+[[Modulo Multiplication has Identity|Multiplication modulo $m$ has an identity]]:
+:$\forall k \in \Z: \eqclass k m \eqclass 1 m = \eqclass k m = \eqclass 1 m \eqclass k m$
+This [[Identity of Monoid is Unique|identity is unique]].
+[[Modulo Multiplication is Commutative|Multiplication modulo $m$ is commutative]].
+Thus all the conditions are fulfilled for this to be a [[Definition:Commutative Monoid|commutative monoid]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Multiplicative Inverse in Ring of Integers Modulo m}
+Tags: Ring of Integers Modulo m, Multiplicative Inverse in Ring of Integers Modulo m
+
+\begin{theorem}
+Let $\struct {\Z_m, +_m, \times_m}$ be the [[Definition:Ring of Integers Modulo m|ring of integers modulo $m$]].
+Then $\eqclass k m \in \Z_m$ has an [[Definition:Multiplicative Inverse|inverse]] in $\struct {\Z_m, \times_m}$ {{iff}} $k \perp m$.
+\end{theorem}
+
+\begin{proof}
+First, suppose $k \perp m$.
+That is:
+:$\gcd \set {k, m} = 1$
+Then, by [[Bézout's Identity]]:
+:$\exists u, v \in \Z: u k + v m = 1$
+Thus:
+:$\eqclass {u k + v m} m = \eqclass {u k} m = \eqclass u m \eqclass k m = \eqclass 1 m$
+Thus $\eqclass u m$ is an [[Definition:Multiplicative Inverse|inverse]] of $\eqclass k m$.
+Suppose that:
+:$\exists u \in \Z: \eqclass u m \eqclass k m = \eqclass {u k} m = 1$.
+Then:
+:$u k \equiv 1 \pmod m$
+and:
+:$\exists v \in \Z: u k + v m = 1$
+Thus from [[Bézout's Identity]]:
+:$k \perp m$
+{{qed}}
+\end{proof}
+
+\begin{proof}
+From [[Ring of Integers Modulo m is Ring]], $\left({\Z_m, +_m, \times_m}\right)$ is a [[Definition:Commutative and Unitary Ring|commutative ring with unity $\left[\!\left[{1}\right]\!\right]_m$]].
+Thus by definition $\left({\Z_m, \times_m}\right)$ is a [[Definition:Commutative Monoid|commutative monoid]].
+The result follows from [[Multiplicative Inverse in Monoid of Integers Modulo m]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Reduced Residue System is Subset of Set of All Residue Classes}
+Tags: Residue Classes, Reduced Residue Systems
+
+\begin{theorem}
+Let $\Z_m$ be the set of [[Definition:Set of Residue Classes|set of residue classes modulo $m$]].
+Let $\Z'_m$ be the [[Definition:Reduced Residue System|reduced residue system modulo $m$]].
+Then:
+:$\forall m \in \Z_{> 1}: \O \subset \Z'_m \subset \Z_m$
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Reduced Residue System|reduced residue system modulo $m$]]:
+:$\Z'_m = \set {x \in \Z_m: x \perp m}$
+From [[Subset of Set with Propositional Function]]:
+:$\Z'_m \subseteq \Z_m$
+We have that:
+:$\gcd \set {m, 0} = m$
+Thus it follows that:
+:$m > 1 \implies \gcd \set {m, 0} \ne 1$
+So:
+:$\eqclass 0 m \notin \Z'_m$
+However:
+:$\eqclass 0 m \in \Z_m$
+so:
+:$\Z'_m \ne \Z_m$
+Thus:
+:$\Z'_m \subset \Z_m$
+Then:
+:$\eqclass 1 m = 1 \implies 1 \perp m$
+So:
+:$\forall m \in \Z: \eqclass 1 m \in \Z'_m$
+Thus:
+:$\Z'_m \ne \O$
+and therefore:
+:$\O \subset \Z'_m$
+{{qed}}
+[[Category:Residue Classes]]
+[[Category:Reduced Residue Systems]]
+6472l2hgwcmn9kek2nctjmqg87ke7g6
+\end{proof}<|endoftext|>
+\section{Reduced Residue System under Multiplication forms Abelian Group}
+Tags: Reduced Residue Systems, Examples of Abelian Groups, Multiplicative Groups of Reduced Residues, Reduced Residue System under Multiplication forms Abelian Group
+
+\begin{theorem}
+Let $\Z_m$ be the set of [[Definition:Set of Residue Classes|set of residue classes modulo $m$]].
+Let $\struct {\Z'_m, \times}$ denote the [[Definition:Multiplicative Group of Reduced Residues|multiplicative group of reduced residues modulo $m$]].
+Then $\struct {\Z'_m, \times}$ is an [[Definition:Abelian Group|abelian group]], precisely equal to the [[Definition:Group of Units|group of units]] of $\Z_m$.
+\end{theorem}
+
+\begin{proof}
+From [[Ring of Integers Modulo m is Ring]], $\struct {\Z_m, +, \times}$ forms a [[Definition:Commutative and Unitary Ring|(commutative) ring with unity]].
+Then we have that the [[Group of Units is Group|units of a ring with unity form a group]].
+By [[Multiplicative Inverse in Ring of Integers Modulo m]] we have that the elements of $\struct {\Z'_m, \times}$ are precisely those that have [[Definition:Inverse Element|inverses]], and are therefore the [[Definition:Unit of Ring|units]] of $\struct {\Z_m, +, \times}$.
+The fact that $\struct {\Z'_m, \times}$ is [[Definition:Abelian Group|abelian]] follows from [[Restriction of Commutative Operation is Commutative]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Taking the [[Definition:Group Axioms|group axioms]] in turn:
+=== $\text G 0$: Closure ===
+From [[Modulo Multiplication on Reduced Residue System is Closed]]:
+:$\struct {\Z'_m, \times}$ is [[Definition:Closed Algebraic Structure|closed]].
+{{qed|lemma}}
+=== $\text G 1$: Associativity ===
+We have that [[Modulo Multiplication is Associative]].
+{{qed|lemma}}
+=== $\text G 2$: Identity ===
+From [[Modulo Multiplication has Identity]], $\eqclass 1 m$ is the [[Definition:Identity Element|identity element]] of $\struct {\Z'_m, \times_m})$.
+{{qed|lemma}}
+=== $\text G 3$: Inverses ===
+From [[Multiplicative Inverse in Monoid of Integers Modulo m]], $\eqclass k m \in \Z_m$ has an [[Definition:Inverse Element|inverse]] in $\struct {\Z_m, \times_m}$ {{iff}} $k$ is [[Definition:Coprime Integers|coprime]] to $m$.
+Thus every element of $\struct {\Z'_m, \times_m}$ has an [[Definition:Inverse Element|inverse]].
+{{qed|lemma}}
+All the [[Definition:Group Axioms|group axioms]] are thus seen to be fulfilled, and so $\struct {\Z'_m, \times_m}$ is a [[Definition:Group|group]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Taking the [[Definition:Finite Group Axioms|finite group axioms]] in turn:
+=== $\text {FG} 0$: Closure ===
+From [[Modulo Multiplication on Reduced Residue System is Closed]]:
+:$\struct {\Z'_m, \times}$ is [[Definition:Closed Algebraic Structure|closed]].
+{{qed|lemma}}
+=== $\text {FG} 1$: Associativity ===
+We have that [[Modulo Multiplication is Associative]].
+{{qed|lemma}}
+=== $\text {FG} 2$: Finiteness ===
+The order of $\struct {\Z'_m, \times}$ is $\map \phi n$ by definition, where $\map \phi n$ denotes the [[Definition:Euler Phi Function|Euler $\phi$ function]].
+As $\map \phi n < n$ it follows that $\struct {\Z'_m, \times}$ is of [[Definition:Finite Order (Structure)|finite order]].
+{{qed|lemma}}
+=== $\text {FG} 3$: Cancellability ===
+We have that [[Modulo Multiplication on Reduced Residue System is Cancellable]].
+{{qed|lemma}}
+Thus all the [[Definition:Finite Group Axioms|finite group axioms]] are fulfilled, and $\struct {\Z'_m, \times}$ is a [[Definition:Group|group]].
+It remains to note that [[Modulo Multiplication is Commutative]] to confirm that $\struct {\Z'_m, \times}$ is [[Definition:Abelian Group|abelian]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Ring of Integers Modulo Prime is Field}
+Tags: Ring of Integers Modulo m, Galois Fields, Ring of Integers Modulo Prime is Field
+
+\begin{theorem}
+Let $m \in \Z: m \ge 2$.
+Let $\struct {\Z_m, +, \times}$ be the [[Definition:Ring of Integers Modulo m|ring of integers modulo $m$]].
+Then:
+:$m$ is [[Definition:Prime Number|prime]]
+{{iff}}:
+:$\struct {\Z_m, +, \times}$ is a [[Definition:Field (Abstract Algebra)|field]].
+\end{theorem}
+
+\begin{proof}
+=== Prime Modulus ===
+$\struct {\Z_m, +, \times}$ is a [[Definition:Commutative Ring with Unity|commutative ring with unity]] by definition.
+From [[Reduced Residue System under Multiplication forms Abelian Group]], $\struct {\Z'_m, \times}$ is an [[Definition:Abelian Group|abelian group]].
+$\Z'_m$ consists of all the elements of $\Z_m$ [[Definition:Coprime Integers|coprime]] to $m$.
+Now when $m$ is [[Definition:Prime Number|prime]], we have, from [[Reduced Residue System Modulo Prime]]:
+:$\Z'_m = \set {\eqclass 1 m, \eqclass 2 m, \ldots, \eqclass {m - 1} m}$
+That is:
+:$\Z'_m = \Z_m \setminus \set {\eqclass 0 m}$
+where $\setminus$ denotes [[Definition:Set Difference|set difference]].
+Hence the result.
+{{qed|lemma}}
+=== Composite Modulus ===
+Now suppose $m \in \Z: m \ge 2$ is [[Definition:Composite Number|composite]].
+From [[Ring of Integers Modulo Composite is not Integral Domain]], $\struct {\Z_m, +, \times}$ is not an [[Definition:Integral Domain|integral domain]].
+From [[Field is Integral Domain]] $\struct {\Z_m, +, \times}$ is not a [[Definition:Field (Abstract Algebra)|field]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let $p$ be [[Definition:Prime Number|prime]].
+From [[Irreducible Elements of Ring of Integers]], we have that $p$ is [[Definition:Irreducible Element of Ring|irreducible]] in the [[Definition:Ring of Integers|ring of integers]] $\struct {\Z, +, \times}$.
+From [[Ring of Integers is Principal Ideal Domain]], $\struct {\Z, +, \times}$ is a [[Definition:Principal Ideal Domain|principal ideal domain]].
+Thus by [[Principal Ideal of Principal Ideal Domain is of Irreducible Element iff Maximal]], $\ideal p$ is a [[Definition:Maximal Ideal of Ring|maximal ideal]] of $\struct {\Z, +, \times}$.
+Hence by [[Maximal Ideal iff Quotient Ring is Field]], $\Z / \ideal p$ is a [[Definition:Field (Abstract Algebra)|field]].
+But $\Z / \ideal p$ is exactly $\struct {\Z_p, +, \times}$.
+{{qed|lemma}}
+Let $p$ be [[Definition:Composite Number|composite]].
+Then $p$ is not [[Definition:Irreducible Element of Ring|irreducible]] in $\struct {\Z, +, \times}$.
+Thus by [[Principal Ideal of Principal Ideal Domain is of Irreducible Element iff Maximal]], $\ideal p$ is not a [[Definition:Maximal Ideal of Ring|maximal ideal]] of $\struct {\Z, +, \times}$.
+Hence by [[Maximal Ideal iff Quotient Ring is Field]], $\Z / \ideal p$ is not a [[Definition:Field (Abstract Algebra)|field]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let $m$ be [[Definition:Prime Number|prime]].
+From [[Ring of Integers Modulo Prime is Integral Domain]], $\struct {\Z_m, +, \times}$ is an [[Definition:Integral Domain|integral domain]].
+From [[Finite Integral Domain is Galois Field]], $\struct {\Z_m, +, \times}$ is a [[Definition:Field (Abstract Algebra)|field]].
+{{qed|lemma}}
+Now suppose $m \in \Z: m \ge 2$ is [[Definition:Composite Number|composite]].
+From [[Ring of Integers Modulo Composite is not Integral Domain]], $\struct {\Z_m, +, \times}$ is not an [[Definition:Integral Domain|integral domain]].
+From [[Field is Integral Domain]] $\struct {\Z_m, +, \times}$ is not a [[Definition:Field (Abstract Algebra)|field]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Subgroups of Additive Group of Integers}
+Tags: Subgroups, Additive Group of Integers, Additive Groups of Integer Multiples, Subgroups of Additive Group of Integers
+
+\begin{theorem}
+Let $\struct {\Z, +}$ be the [[Definition:Additive Group of Integers|additive group of integers]].
+Let $n \Z$ be the [[Definition:Additive Group of Integer Multiples|additive group of integer multiples]] of $n$.
+Every [[Definition:Non-Trivial Group|non-trivial]] [[Definition:Subgroup|subgroup]] of $\struct {\Z, +}$ has the form $n \Z$.
+\end{theorem}
+
+\begin{proof}
+First we note that, from [[Integer Multiples under Addition form Infinite Cyclic Group]], $\struct {n \Z, +}$ is an [[Definition:Infinite Group|infinite]] [[Definition:Cyclic Group|cyclic group]].
+From [[Cyclic Group is Abelian]], it follows that $\struct {n \Z, +}$ is an [[Definition:Infinite Group|infinite]] [[Definition:Abelian Group|abelian group]].
+Let $H$ be a [[Definition:Non-Trivial Group|non-trivial]] [[Definition:Subgroup|subgroup]] of $\struct {\Z, +}$.
+Because $H$ is [[Definition:Non-Trivial Group|non-trivial]]:
+:$\exists m \in \Z: m \in H: m \ne 0$
+Because $H$ is itself a [[Definition:Group|group]]:
+:$-m \in H$
+So either $m$ or $-m$ is [[Definition:Positive Integer|positive]] and therefore in $\Z_{>0}$.
+Thus:
+:$H \cap \Z_{>0} \ne \O$
+From the [[Well-Ordering Principle]], $H \cap \Z_{>0}$ has a [[Definition:Smallest Element|smallest element]], which we can call $n$.
+It follows from [[Subgroup of Infinite Cyclic Group is Infinite Cyclic Group]] that:
+: $\forall a \in \Z: a n \in H$
+Thus:
+: $n \Z \subseteq H$
+{{AimForCont}}:
+:$\exists m \in \Z: m \in H \setminus n \Z$
+Then $m \ne 0$, and also $-m \in H \setminus n \Z$.
+Assume $m > 0$, otherwise we consider $-m$.
+By the [[Division Theorem]]:
+:$m = q n + r$
+If $r = 0$, then $m = q n \in n \Z$, so $0 \le r < n$.
+Now this means $r = m - q n \in H$ and $0 \le r < n$.
+This would mean $n$ was not the [[Definition:Smallest Element|smallest element]] of $H \cap \Z$.
+Hence, by [[Proof by Contradiction]], there can be no such $m \in H \setminus n \Z$.
+Thus:
+:$H \setminus n \Z = \O$
+Thus from [[Set Difference with Superset is Empty Set]]:
+:$H \subseteq n \Z$
+Thus we have $n \Z \subseteq H$ and $H \subseteq n \Z$.
+Hence:
+:$H = n \Z$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Integer Multiples under Addition form Infinite Cyclic Group}
+Tags: Sets of Integer Multiples, Additive Groups of Integer Multiples, Examples of Cyclic Groups
+
+\begin{theorem}
+Let $n \Z$ be the [[Definition:Set of Integer Multiples|set of integer multiples]] of $n$.
+Then $\struct {n \Z, +}$ is a [[Definition:Countably Infinite|countably infinite]] [[Definition:Cyclic Group|cyclic group]].
+It is generated by $n$ and $-n$:
+:$n \Z = \gen n$
+:$n \Z = \gen {-n}$
+Hence $\struct {n \Z, +}$ can be justifiably referred to as the [[Definition:Additive Group of Integer Multiples|additive group of integer multiples]].
+\end{theorem}
+
+\begin{proof}
+From [[Integer Multiples under Addition form Subgroup of Integers]], $\struct {n \Z, +}$ is a [[Definition:Subgroup|subgroup]] of the [[Definition:Additive Group of Integers|additive group of integers]] $\struct {\Z, +}$.
+From [[Integers under Addition form Infinite Cyclic Group]], $\struct {\Z, +}$ is a [[Definition:Cyclic Group|cyclic group]].
+So by [[Subgroup of Cyclic Group is Cyclic]], $\struct {n \Z, +}$ is a [[Definition:Cyclic Group|cyclic group]].
+The final assertions follow from [[Subgroup of Infinite Cyclic Group is Infinite Cyclic Group]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Quotient Group of Integers by Multiples}
+Tags: Quotient Groups, Modulo Arithmetic, Additive Group of Integers, Additive Groups of Integer Multiples, Additive Groups of Integers Modulo m
+
+\begin{theorem}
+Let $\struct {\Z, +}$ be the [[Definition:Additive Group of Integers|additive group of integers]].
+Let $\struct {m \Z, +}$ be the [[Definition:Additive Group of Integer Multiples|additive group of integer multiples]] of $m$.
+Let $\struct {\Z_m, +_m}$ be the [[Definition:Additive Group of Integers Modulo m|additive group of integers modulo $m$]].
+Then the [[Definition:Quotient Group|quotient group]] of $\struct {\Z, +}$ by $\struct {m \Z, +}$ is $\struct {\Z_m, +_m}$.
+Thus:
+:$\index \Z {m \Z} = m$
+\end{theorem}
+
+\begin{proof}
+From [[Subgroups of Additive Group of Integers]], $\struct {m \Z, +}$ is a [[Definition:Subgroup|subgroup]] of $\struct {\Z, +}$.
+From [[Subgroup of Abelian Group is Normal]], $\struct {m \Z, +}$ is [[Definition:Normal Subgroup|normal]] in $\struct {\Z, +}$.
+Therefore the [[Definition:Quotient Group|quotient group]] $\dfrac {\struct {\Z, +} } {\struct {m \Z, +} }$ is defined.
+Now $\Z$ modulo $m \Z$ is [[Definition:Congruence Modulo Subgroup|Congruence Modulo a Subgroup]].
+This is merely [[Definition:Congruence Modulo Integer|congruence modulo an integer]].
+Thus the [[Definition:Quotient Set|quotient set]] $\Z / m \Z$ is $\Z_m$.
+The [[Definition:Left Coset|left coset]] of $k \in \Z$ is denoted $k + m \Z$, which is the same thing as $\eqclass k m$ from the definition of [[Definition:Residue Class|residue class]].
+So $\index \Z {m \Z} = m$ follows from the definition of [[Definition:Index of Subgroup|Subgroup Index]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Euler's Theorem}
+Tags: Number Theory
+
+\begin{theorem}
+Let $a, m \in \Z$ be [[Definition:Coprime Integers|coprime integers]]: $a \perp m$.
+Let $\map \phi m$ be the [[Definition:Euler Phi Function|Euler $\phi$ function]] of $m$.
+Then:
+:$a^{\map \phi m} \equiv 1 \pmod m$
+\end{theorem}
+
+\begin{proof}
+Let $\eqclass a m$ denote the [[Definition:Residue Class|residue class modulo $m$ of $a$]].
+Since $a \perp m$, it follows by [[Reduced Residue System under Multiplication forms Abelian Group]] that $\eqclass a m$ belongs to the [[Definition:Abelian Group|abelian group]] $\struct {\Z'_m, \times}$.
+Let $k = \order {\eqclass a m}$ where $\order {\, \cdot \,}$ denotes the [[Definition:Order of Group Element|order of a group element]].
+By [[Order of Element Divides Order of Finite Group]]:
+:$k \divides \order {\Z'_m}$
+By the definition of the [[Definition:Euler Phi Function|Euler $\phi$ function]]:
+:$\order {\Z'_m} = \map \phi m$
+Thus:
+{{begin-eqn}}
+{{eqn | l = \eqclass a m^k
+      | r = \eqclass a m
+      | c = {{Defof|Order of Group Element}}
+}}
+{{eqn | ll= \leadsto
+      | l = \eqclass a m^{\map \phi m}
+      | r = \eqclass {a^{\map \phi m} } m
+      | c = [[Congruence of Powers]]
+}}
+{{eqn | r = \eqclass 1 m
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = a^{\map \phi m}
+      | o = \equiv
+      | r = 1 \pmod m
+      | c = {{Defof|Residue Class}}
+}}
+{{end-eqn}}
+{{qed}}
+{{Namedfor|Leonhard Paul Euler|cat = Euler}}
+\end{proof}<|endoftext|>
+\section{Symmetry Group is Group}
+Tags: Symmetry Groups
+
+\begin{theorem}
+Let $P$ be a [[Definition:Geometric Figure|geometric figure]].
+Let $S_P$ be the [[Definition:Set|set]] of all [[Definition:Symmetry Mapping|symmetries]] of $P$.
+Let $\circ$ denote [[Definition:Composition of Mappings|composition of mappings]].
+The [[Definition:Symmetry Group|symmetry group]] $\struct {S_P, \circ}$ is indeed a [[Definition:Group|group]].
+\end{theorem}
+
+\begin{proof}
+By definition, a [[Definition:Symmetry Mapping|symmetry mapping]] is a [[Definition:Bijection|bijection]], and hence a [[Definition:Permutation|permutation]].
+From [[Symmetric Group is Group]], the [[Definition:Set|set]] of all [[Definition:Permutation|permutations]] on $P$ form the [[Definition:Symmetric Group|symmetric group]] $\struct {\map \Gamma P, \circ}$ on $P$.
+Thus $S_P$ is a [[Definition:Subset|subset]] of $\struct {\map \Gamma P, \circ}$.
+Let $A$ and $B$ be [[Definition:Symmetry Mapping|symmetry mappings]] on $P$.
+From [[Composition of Symmetries is Symmetry]], $A \circ B$ is also a [[Definition:Symmetry Mapping|symmetry mapping]] on $P$.
+Also, by the definition of [[Definition:Symmetry Mapping|symmetry]], $A^{-1}$ is also a [[Definition:Symmetry Mapping|symmetry mapping]] on $P$.
+Thus we have:
+:$A, B \in S_P \implies A \circ B \in S_P$
+:$A \in S_P \implies A^{-1} \in \S_P$
+The result follows from the [[Two-Step Subgroup Test]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Internal Angles of Square}
+Tags: Squares
+
+\begin{theorem}
+The [[Definition:Internal Angle|internal angles]] of a [[Definition:Square (Geometry)|square]] are [[Definition:Right Angle|right angles]].
+\end{theorem}
+
+\begin{proof}
+By definition, a [[Definition:Square (Geometry)|square]] is a [[Definition:Regular Polygon|regular]] [[Definition:Quadrilateral|quadrilateral]].
+From [[Internal Angles of Regular Polygon]], the [[Definition:Internal Angle|internal angles]] of a [[Definition:Square (Geometry)|square]] measure $\dfrac {180 \degrees \paren {4 - 2} } 4 = 90 \degrees$.
+The result follows from [[Measurement of Right Angle]].
+{{qed}}
+[[Category:Squares]]
+n7tjkuonqy6jbbedb9iv3obmyjyrq54
+\end{proof}<|endoftext|>
+\section{Area of Square}
+Tags: Areas of Quadrilaterals, Area of Square
+
+\begin{theorem}
+A [[Definition:Square (Geometry)|square]] has an [[Definition:Area|area]] of $L^2$ where $L$ is the [[Definition:Length of Line|length]] of a [[Definition:Side of Polygon|side]] of the square.
+Thus we have that the [[Definition:Area|area]] is a [[Definition:Real Function|function]] of the [[Definition:Length of Line|length]] of the [[Definition:Side of Polygon|side]]:
+:$\forall L \in \R_{\ge 0}: \map \Area L = L^2$
+where it is noted that the [[Definition:Domain of Mapping|domain]] of $L$ is the [[Definition:Positive Real Number|set of non-negative real numbers]].
+\end{theorem}
+
+\begin{proof}
+Let $\Box ABCD$ be a [[Definition:Square (Geometry)|square]] whose [[Definition:Side of Polygon|side]] $AB$ is of [[Definition:Linear Measure|length]] $L$.
+Let $\Box EFGH$ be a [[Definition:Square (Geometry)|square]] whose [[Definition:Side of Polygon|side]] $EF$ is of [[Definition:Linear Measure|length]] $1$.
+:[[File:AreaOfSquare.png|300px]]
+From the [[Axiom:Area Axioms|Axioms of Area]], the [[Definition:Area|area]] of $\Box EFGH$ is $1$.
+By definition, $AB : EF = L : 1$.
+From [[Similar Polygons are composed of Similar Triangles]], the [[Definition:Ratio|ratio]] of the [[Definition:Area|areas]] of $\Box ABCD$ to $\Box EFGH$ is the [[Definition:Duplicate Ratio|duplicate ratio]] of the [[Definition:Ratio|ratio]] of $AB$ to $EF$.
+Thus by definition of [[Definition:Duplicate Ratio|duplicate ratio]]:
+: $\Box ABCD : \Box EFGH = \left({AB : EF}\right)^2$
+That is:
+: $\dfrac {\Box ABCD} {\Box EFGH} = \left({\dfrac L 1}\right)^2 = L^2$
+That is, the [[Definition:Area|area]] of $\Box ABCD$ has $L^2$ as many units as $\Box EFGH$.
+Hence the result.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+=== Integer Side Length ===
+In the case where $L = 1$, the statement follows from the [[Definition:Area|definition of area]].
+If $L \in \N, L > 1$, then we can divide the square into smaller squares, each of side length one.
+Since there will be $L$ squares of side length one on each side, it follows that there will be $L \cdot L = L^2$ squares of side length one.
+Thus, the area of the square of side length $L$ is $L^2 \cdot 1 = L^2$.
+{{Qed|lemma}}
+=== Rational Side Length ===
+Let $A$ be the [[Definition:Area|area]] of the [[Definition:Square (Geometry)|square]] $S$ with side length $L$.
+If $L$ is a [[Definition:Rational Number|rational number]], then:
+:$\exists p, q \in \N: L = \dfrac p q$
+Create a [[Definition:Square (Geometry)|square]] of [[Definition:Side of Polygon|side]] [[Definition:Length of Line|length]] $p$.
+From the [[Area of Square/Proof 1#Integer Side Length|integer side length case]], its area equals $p^2$.
+Divide the [[Definition:Side of Polygon|sides]] into $q$ equal parts.
+Thus the [[Definition:Square (Geometry)|square]] of [[Definition:Side of Polygon|side]] [[Definition:Length of Line|length]] $p$ is divided into $q^2$ small [[Definition:Square (Geometry)|squares]].
+As they all have [[Definition:Side of Polygon|side]] [[Definition:Length of Line|length]] $\dfrac p q$, each of them equals $S$.
+It follows arithmetically:
+:$A = \dfrac {p^2} {q^2} = \paren {\dfrac p q}^2 = L^2$
+{{Qed|lemma}}
+=== Irrational Side Length ===
+Let $L$ be an [[Definition:Irrational Number|irrational number]].
+Then from [[Rationals are Everywhere Dense in Topological Space of Reals]] we know that within an arbitrarily small distance $\epsilon$ from $L$, we can find a rational number less than $L$ and a rational number greater than $L$.
+In formal terms, we have:
+:$\forall \epsilon > 0: \exists A, B \in \Q_+: A < L < B: \left|{A - L}\right| < \epsilon, \left|{B - L}\right| < \epsilon$
+Thus:
+: $\displaystyle \lim_{\epsilon \to 0^+} A = L$
+: $\displaystyle \lim_{\epsilon \to 0^+} B = L$
+Since a square of side length $B$ can contain a square of side length $L$, which can in turn contain a square of side length $A$, then:
+: $\operatorname {area} \Box B \ge \operatorname {area} \Box L \ge \operatorname {area}\Box A$
+By the [[Area of Square/Proof 1#Rational Side Length|result for rational numbers]]:
+: $\operatorname {area}\Box B = B^2$
+: $\operatorname {area}\Box A = A^2$
+We also note that:
+: $\displaystyle \lim_{B \to L} B^2 = L^2 = \lim_{A \to L} A^2$
+Thus:
+: $\displaystyle \lim_{B \to L} \operatorname {area} \Box B = \lim_{B \to L} B^2 = L^2$
+: $\displaystyle \lim_{A \to L} \operatorname {area} \Box A = \lim_{A \to L} A^2 = L^2$
+Finally:
+: $L^2 \ge \operatorname {area}\Box L \ge L^2$
+so:
+: $\operatorname {area}\Box L = L^2$
+{{Qed}}
+\end{proof}
+
+\begin{proof}
+Let a [[Definition:Square (Geometry)|square]] have a [[Definition:Side of Polygon|side length]] $a \in \R$.
+This square is equivalent to the area under the graph of $\map f x = a$ from $0$ to $a$.
+Thus from [[Definition:Definite Integral/Geometric Interpretation|the geometric interpretation of the definite integral]], the area of the square will be the integral:
+:$\displaystyle A = \int_0^a a \rd l$
+Thus:
+{{begin-eqn}}
+{{eqn | l = A
+      | r = \int_0^a a \rd l
+}}
+{{eqn | r = \bigintlimits {l \cdot a} 0 a
+      | c = [[Integral of Constant]]
+}}
+{{eqn | r = a \cdot a - 0 \cdot a
+}}
+{{eqn | r = a^2
+}}
+{{end-eqn}}
+{{Qed}}
+=== Warning ===
+'''This proof is circular.'''
+The use of the [[Definition:Definite Integral|definite integral]] to represent [[Definition:Area|area]] is based on the fact that the [[Area of Rectangle|area of a rectangle]] is the product of the rectangle's [[Definition:Width|width]] and [[Definition:Height (Linear Measure)|height]].
+That fact is in turn derived from this one.
+However, this demonstration neatly parallels the integration based proofs of the areas of other figures, for example [[Area of Circle]].
+\end{proof}<|endoftext|>
+\section{Order of Symmetric Group}
+Tags: Symmetric Groups
+
+\begin{theorem}
+Let $S$ be a [[Definition:Finite Set|finite set]] of [[Definition:Cardinality|cardinality]] $n$.
+Let $\struct {\map \Gamma S, \circ}$ be the [[Definition:Symmetric Group|symmetric group]] on $S$.
+Then $\struct {\map \Gamma S, \circ}$ has $n!$ elements (see [[Definition:Factorial|factorial]]).
+\end{theorem}
+
+\begin{proof}
+A direct application of [[Cardinality of Set of Bijections]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Powers of Permutation Element}
+Tags: Symmetric Groups
+
+\begin{theorem}
+Let $S_n$ denote the [[Definition:Symmetric Group on n Letters|symmetric group on $n$ letters]].
+Let $\pi \in S_n$, and let $i \in \N^*_n$.
+Let $k \in \Z: k > 0$ be the smallest such that:
+: $\pi^k \left({i}\right) \in \left\{{i, \pi \left({i}\right), \pi^2 \left({i}\right), \ldots, \pi^{k-1} \left({i}\right)}\right\}$
+Then $\pi^k \left({i}\right) = i$.
+\end{theorem}
+
+\begin{proof}
+Suppose $\pi^k \left({i}\right) = \pi^r \left({i}\right)$ for some $r > 0$.
+Then, since $\pi$ has an [[Definition:Inverse Element|inverse]], $\pi^{k - r} \left({i}\right) = i$.
+This contradicts the definition of $k$, so $r = 0$.
+{{qed}}
+[[Category:Symmetric Groups]]
+s6busb2smrfp9a2bew0qxesh1oun8w9
+\end{proof}<|endoftext|>
+\section{Fixed Elements form 1-Cycles}
+Tags: Symmetric Groups
+
+\begin{theorem}
+Let $S_n$ denote the [[Definition:Symmetric Group on n Letters|symmetric group on $n$ letters]].
+Let $\pi \in S_n$.
+Let $\Fix \pi$ be the [[Definition:Fixed Element of Permutation|set of elements fixed by $\pi$]].
+For any $\pi \in S_n$, all the elements of $\Fix \pi$ form [[Definition:Cyclic Permutation|$1$-cycles]].
+\end{theorem}
+
+\begin{proof}
+Let $\pi$ be a permutation, and let $x \in \Fix \pi$.
+From the definition of a [[Definition:Fixed Element of Permutation|fixed element]], $\map \pi x = x$.
+From the definition of a [[Definition:Cyclic Permutation|$k$-cycle]], we see that $1$ is the smallest $k \in \Z: k > 0$ such that $\map {\pi^k} x = x$.
+The result follows.
+{{qed}}
+[[Category:Symmetric Groups]]
+5th1obufnxiwimah16mkfvfzxd09v13
+\end{proof}<|endoftext|>
+\section{Equality of Cycles}
+Tags: Symmetric Groups
+
+\begin{theorem}
+Let $S_n$ denote the [[Definition:Symmetric Group on n Letters|symmetric group on $n$ letters]], realised as the [[Definition:Permutation on n Letters|permutations of $\left\{{1, \ldots, n}\right\}$]].
+Let $\rho = \begin{bmatrix} a_0 & \cdots & a_{k-1} \end{bmatrix}$, $\sigma = \begin{bmatrix} b_0 & \cdots & b_{k-1} \end{bmatrix} \in S_n$ be [[Definition:Cyclic Permutation|$k$-cycles]] of $S_n$.
+For any $d \in \Z$, by [[Integer is Congruent to Integer less than Modulus]] we can associate to $d$ a unique [[Definition:Integer|integer]] $\tilde d \in \left\{{0, \ldots, k-1}\right\}$ such that $d \equiv \tilde d \pmod k$.
+Define $a_d$ and $b_d$ for any $d \in \Z$ by $a_d = a_{\tilde d}$ and $b_d = b_{\tilde d}$
+Choose $i, j \in \left\{{1, \ldots, k}\right\}$ such that:
+:$\displaystyle a_i = \min\left\{ {a_0, \ldots, a_{k-1} }\right\}$
+:$\displaystyle b_j = \min\left\{ {b_0, \ldots, b_{k-1} }\right\}$
+Then $\rho = \sigma$ {{iff}} for all $d \in \Z$, $a_{i + d} = b_{j + d}$.
+That is, $\rho = \sigma$ {{iff}} they are identical when written with the lowest element first.
+\end{theorem}
+
+\begin{proof}
+{{ProofWanted}}
+[[Category:Symmetric Groups]]
+p1lu4rvnq1iznqntvshmssa721dm2ur
+\end{proof}<|endoftext|>
+\section{Identity Permutation is Disjoint from All}
+Tags: Symmetric Groups, Identity Mappings
+
+\begin{theorem}
+Let $S_n$ denote the [[Definition:Symmetric Group on n Letters|symmetric group on $n$ letters]].
+Let $e \in S_n$ be the [[Definition:Identity Mapping|identity permutation]] on $S_n$.
+Then $e$ is [[Definition:Disjoint Permutations|disjoint]] from every permutation $\pi$ on $S_n$ (including itself).
+\end{theorem}
+
+\begin{proof}
+By definition of the [[Definition:Identity Mapping|identity permutation]]:
+: $\forall i \in \N_{>0}: e \left({i}\right) = i$
+Thus $e$ [[Definition:Fixed Element of Permutation|fixes]] all elements of $S_n$.
+Thus each element [[Definition:Moved Element of Permutation|moved]] by a permutation $\pi$ is [[Definition:Fixed Element of Permutation|fixed]] by $e$.
+The set of elements moved by $e$ is $\O$, so the converse is true vacuously.
+{{qed}}
+[[Category:Symmetric Groups]]
+[[Category:Identity Mappings]]
+ej17rkvuch2qhbeqa3wmxc2042jco6v
+\end{proof}<|endoftext|>
+\section{Disjoint Permutations Commute}
+Tags: Symmetric Groups, Permutation Theory
+
+\begin{theorem}
+Let $S_n$ denote the [[Definition:Symmetric Group on n Letters|symmetric group on $n$ letters]].
+Let $\rho, \sigma \in S_n$ such that $\rho$ and $\sigma$ are [[Definition:Disjoint Permutations|disjoint]].
+Then $\rho \sigma = \sigma \rho$.
+\end{theorem}
+
+\begin{proof}
+Let $\rho$ and $\sigma$ be [[Definition:Disjoint Permutations|disjoint permutations]].
+Let $i \in \Fix \rho$.
+Then:
+:$\map {\sigma \rho} i = \map \sigma i$
+whereas:
+:$\map {\rho \sigma} i = \map \rho {\map \sigma i}$
+{{AimForCont}} $\map \sigma i \notin \Fix \rho$.
+Then because $\sigma$ and $\rho$ are [[Definition:Disjoint Permutations|disjoint]] it follows that:
+{{begin-eqn}}
+{{eqn | l = \map \sigma i
+      | o = \in
+      | r = \Fix \sigma
+}}
+{{eqn | ll= \leadsto
+      | l = \map {\sigma^2} i
+      | r = \map \sigma i
+}}
+{{eqn | ll= \leadsto
+      | l = \map {\sigma^{-1} \sigma^2} i
+      | r = \map {\sigma^{-1} \sigma} i
+}}
+{{eqn | ll= \leadsto
+      | l = \map \sigma i
+      | r = i
+}}
+{{end-eqn}}
+But it was previously established that $i \in \Fix \rho$.
+This is a [[Definition:Contradiction|contradiction]].
+Therefore:
+:$\map \sigma i \in \Fix \rho$
+and so:
+:$\map {\rho \sigma} i = \map \sigma i = \map {\sigma \rho} i$
+Let $i \notin \Fix \rho$.
+Then:
+: $i \in \Fix \sigma$
+and the same proof can be performed with $\rho$ and $\sigma$ exchanged.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Permutation Induces Equivalence Relation}
+Tags: Symmetric Groups, Equivalence Relations
+
+\begin{theorem}
+Let $S_n$ denote the [[Definition:Symmetric Group on n Letters|symmetric group on $n$ letters]].
+Let $\pi \in S_n$.
+Let $\mathcal R_\pi$ be the [[Definition:Relation|relation]] defined by:
+:$i \mathrel {\mathcal R_\pi} j \iff \exists k \in \Z: \map {\pi^k} i = j$
+Then $\mathcal R_\pi$ is an [[Definition:Equivalence Relation|equivalence relation]].
+\end{theorem}
+
+\begin{proof}
+Let $\pi \in S_n$.
+First we note that, from [[Element of Finite Group is of Finite Order]], every [[Definition:Element|element]] of a [[Definition:Finite Group|finite group]] has [[Definition:Finite Order Element|finite order]].
+Thus $\pi$ has [[Definition:Finite Order Element|finite order]], so:
+:$\exists r \in \Z: \pi^r = e$
+Checking in turn each of the criteria for [[Definition:Equivalence Relation|equivalence]]:
+=== Reflexive ===
+From above, $\exists r \in \Z: \pi^r = e$.
+Therefore $\exists k \in \Z: \map {\pi^k} i = i$.
+So $\mathcal R_\pi$ is [[Definition:Reflexive Relation|reflexive]].
+{{qed|lemma}}
+=== Symmetric ===
+Let $\map {\pi^k} i = j$.
+Because $\pi^r = e$, we have $\map {\pi^r} j = j$ (from above).
+Thus $\map {\pi^{r - k} } j = i$.
+So $\mathcal R_\pi$ is [[Definition:Symmetric Relation|symmetric]].
+{{qed|lemma}}
+=== Transitive ===
+Let $\map {\pi^{s_1} } i = j, \map {\pi^{s_2} } j = k$.
+Then $\map {\pi^{s_1 + s_2} } i = k$.
+So $\mathcal R_\pi$ is [[Definition:Transitive Relation|transitive]].
+{{qed|lemma}}
+All criteria are met, and so $\mathcal R_\pi$ is an [[Definition:Equivalence Relation|equivalence relation]].
+{{qed}}
+[[Category:Symmetric Groups]]
+[[Category:Equivalence Relations]]
+rjo3omx45c5ldvuigd6uen9pr0jdghw
+\end{proof}<|endoftext|>
+\section{Existence and Uniqueness of Cycle Decomposition}
+Tags: Symmetric Groups, Permutation Theory
+
+\begin{theorem}
+Let $S_n$ denote the [[Definition:Symmetric Group on n Letters|symmetric group on $n$ letters]].
+Every [[Definition:Element|element]] of $S_n$ may be [[Definition:Unique|uniquely expressed]] as a [[Definition:Cycle Decomposition|cycle decomposition]], up to the order of factors.
+\end{theorem}
+
+\begin{proof}
+By definition, a [[Definition:Cycle Decomposition|cycle decomposition]] of an [[Definition:Element|element]] of $S_n$ is a product of [[Definition:Disjoint Permutations|disjoint]] [[Definition:Cyclic Permutation|cycles]].
+=== Construction of Disjoint Permutations ===
+Let $\sigma \in S_n$ be a [[Definition:Permutation|permutation]] on $S_n$.
+Let $\mathcal R_\sigma$ be the [[Definition:Equivalence Relation|equivalence]] defined in [[Permutation Induces Equivalence Relation]].
+Let $\N_n$ be used to denote the [[Definition:Initial Segment of One-Based Natural Numbers|(one-based) initial segment of natural numbers]]:
+:$\N_n = \closedint 1 n = \set {1, 2, 3, \ldots, n}$
+Let $\N_n / \mathcal R_\sigma = \set {E_1, E_2, \ldots, E_m}$ be the [[Definition:Quotient Set|quotient set]] of $\N_n$ determined by $\mathcal R_\sigma$.
+By [[Equivalence Class of Element is Subset]]:
+:$E \in \N_n / \mathcal R_\sigma \implies E \subseteq \N_n$
+For any $E_i \in \N_n / \mathcal R_\sigma$, let $\rho_i: \paren {\N_n \setminus E_i} \to \paren {\N_n \setminus E_i}$ be the [[Definition:Identity Mapping|identity mapping]] on $\N_n \setminus E_i$.
+By [[Identity Mapping is Permutation]], $\rho_i$ is a [[Definition:Permutation|permutation]].
+Also, let $\phi_i = \tuple {E_i, E_i, R}$ be a [[Definition:Relation|relation]] where $R$ is defined as:
+:$\forall x, y \in E_i: \tuple {x, y} \in R \iff \map \sigma x = y$
+It is easily seen that $\phi_i$ is [[Definition:Many-to-One Relation|many to one]].
+For all $x \in E_i$:
+{{begin-eqn}}
+{{eqn | l = x
+      | o = \mathcal R_\sigma
+      | r = \map \sigma x
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \map \sigma x
+      | o = \in
+      | r = E_i
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \sigma \sqbrk {E_i}
+      | o = \subseteq
+      | r = E_i
+      | c = 
+}}
+{{end-eqn}}
+which shows that $\phi_i$ is [[Definition:Left-Total Relation|left-total]].
+It then follows from the definition of a [[Definition:Mapping|mapping]] that $\phi_i: E_i \to E_i$ is a [[Definition:Mapping|mapping]] defined by:
+:$\map {\phi_i} x = \map \sigma x$
+It is seen that $\phi_i$ is an [[Definition:Injection|injection]] because $\sigma$ is an [[Definition:Injection|injection]].
+So by [[Injection from Finite Set to Itself is Surjection/Corollary|Injection from Finite Set to Itself is Permutation]], $\phi_i$ is a [[Definition:Permutation|permutation]] on $E_i$.
+By [[Intersection with Relative Complement is Empty]], $E_i$ and $\N_n \setminus E_i$ are [[Definition:Disjoint Sets|disjoint]].
+By [[Union with Relative Complement]]:
+: $E_i \cup \paren {\N_n \setminus E_i} = \N_n$
+So by [[Union of Bijections with Disjoint Domains and Codomains is Bijection]], let the [[Definition:Permutation|permutation]] $\sigma_i \in S_n$ be defined by:
+:$\map {\sigma_i} x = \map {\paren {\phi_i \cup \rho_i} } x = \begin{cases}
+\map \sigma x & : x \in E_i \\
+x & : x \notin E_i
+\end{cases}$
+By [[Equivalence Classes are Disjoint]], it follows that each of the $\sigma_i$ are [[Definition:Disjoint Permutations|disjoint]].
+{{qed|lemma}}
+=== These Permutations are Cycles ===
+It is now to be shown that all of the $\sigma_i$ are [[Definition:Cyclic Permutation|cycles]].
+From [[Order of Element Divides Order of Finite Group]], there exists $\alpha \in \Z_{\gt 0}$ such that $\sigma_i^\alpha = e$, and so:
+:$\map {\sigma_i^\alpha} x = \map e x = x$
+By the [[Well-Ordering Principle]], let $k = \min \set {\alpha \in \N_{\gt 0}: \map {\sigma_i^\alpha} x = x}$
+Because $\sigma_i$ [[Definition:Fixed Element of Permutation|fixes]] each $y \notin E_i$, it suffices to show that:
+:$E_i = \set {x, \map {\sigma_i} x, \ldots, \map {\sigma_i^{k - 1} } x}$
+for some $x \in E_i$.
+If $x \in E_i$, then for all $t \in \Z$:
+:$x \mathrel {\mathcal R_\sigma} \map {\sigma_i^t} x \implies \map {\sigma_i^t} x \in E_i$
+It has been shown that:
+:$(1) \quad \set {x, \map {\sigma_i} x, \ldots, \map {\sigma_i^{k - 1} } x} \subseteq E_i$
+Let $x, y \in E_i$.
+Then:
+{{begin-eqn}}
+{{eqn | l = x
+      | o = \mathcal R_\sigma
+      | r = y
+}}
+{{eqn | ll= \leadsto
+      | l = \map {\sigma_i^t} x
+      | r = y
+      | c = for some $t \in \Z$, by [[Permutation Induces Equivalence Relation]]
+}}
+{{eqn | ll= \leadsto
+      | l = \map {\sigma_i^{k q + r} } x
+      | r = y
+      | c = for some $q \in \Z$, and $0 \le r \lt k$ by the [[Division Theorem]]
+}}
+{{eqn | ll= \leadsto
+      | l = \map {\sigma_i^r \sigma_i^{k q} } x
+      | r = y
+}}
+{{eqn | ll= \leadsto
+      | l = \map {\sigma_i^r} x
+      | r = y
+      | c = [[Fixed Point of Permutation is Fixed Point of Power]]
+}}
+{{end-eqn}}
+It has been shown that:
+:$(2) \quad E_i \subseteq \set {x, \map {\sigma_i} x, \ldots, \map {\sigma_i^{k - 1} } x}$
+Combining $(1)$ and $(2)$ yields:
+:$E_i = \set {x, \map {\sigma_i} x, \ldots, \map {\sigma_i^{k - 1} } x}$
+{{qed|lemma}}
+=== The Product of These Cycles form the Permutation ===
+Finally, it is now to be shown that $\sigma = \sigma_1 \sigma_2 \cdots \sigma_m$.
+From the [[Fundamental Theorem on Equivalence Relations]]:
+:$x \in \N_n \implies x \in E_j$
+for some $j \in \set {1, 2, \ldots, m}$.
+Therefore:
+{{begin-eqn}}
+{{eqn | l = \map {\sigma_1 \sigma_2 \cdots \sigma_m} x
+      | r = \map {\sigma_1 \sigma_2 \cdots \sigma_j} x
+}}
+{{eqn | r = \map {\sigma_j} x
+      | c = because $\sigma_j \sqbrk {E_j} = E_j$
+}}
+{{eqn | r = \map \sigma x
+      | c = Definition of $\sigma_i$
+}}
+{{end-eqn}}
+and so existence of a [[Definition:Cycle Decomposition|cycle decomposition]] has been shown.
+{{qed|lemma}}
+=== Uniqueness of Cycle Decomposition ===
+Take the [[Definition:Cycle Decomposition|cycle decomposition]] of $\sigma$, which is $\sigma_1 \sigma_2 \cdots \sigma_m$.
+Let $\tau_1 \tau_2 \cdots \tau_s$ be some product of [[Definition:Disjoint Permutations|disjoint]] [[Definition:Cyclic Permutation|cycles]] such that $\sigma = \tau_1 \tau_2 \cdots \tau_s$.
+It is assume that this product describes $\sigma$ completely and does not contain any duplicate [[Definition:Cyclic Permutation|$1$-cycles]].
+Let $x$ be a [[Definition:Moved Element of Permutation|moved element]] of $\sigma$.
+Then there exists a $j \in \set {1, 2, \ldots, s}$ such that $\map {\tau_j} x \ne x$.
+And so:
+{{begin-eqn}}
+{{eqn | l = \map \sigma x
+      | r = \map {\tau_1 \tau_2 \cdots \tau_j} x
+}}
+{{eqn | r = \map {\tau_j} x
+      | c = [[Power of Moved Element is Moved]]
+}}
+{{end-eqn}}
+It has already been shown that $x \in E_i$ for some $i \in \set {1, 2, \ldots, m}$.
+Therefore:
+{{begin-eqn}}
+{{eqn | l = \map {\sigma_i} x
+      | r = \map {\tau_j} x
+}}
+{{eqn | l = \map {\sigma_i^2} x
+      | r = \map {\tau_{j \prime} \tau_j} x
+      | c = by [[Power of Moved Element is Moved]]
+}}
+{{eqn | r = \map {\tau_j^2} x
+      | c = because $\map {\sigma_i^2} x \ne \map {\sigma_i} x$ and this product is [[Definition:Disjoint Permutations|disjoint]]
+}}
+{{eqn | l = \vdots
+      | o = \vdots
+      | r = \vdots
+}}
+{{eqn | l = \map {\sigma_i^{k - 1} } x
+      | r = \map {\tau_j^{k - 1} } x
+}}
+{{end-eqn}}
+This effectively shows that $\sigma_i = \tau_j$.
+Doing this for every $E_i$ implies that $m = s$ and that there exists a $\rho \in S_m$ such that:
+:$\sigma_{\map \rho i} = \tau_i$
+In other words, $\tau_1 \tau_2 \cdots \tau_m$ is just a reordering of $\sigma_1 \sigma_2 \cdots \sigma_m$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Powers of Disjoint Permutations}
+Tags: Symmetric Groups
+
+\begin{theorem}
+Let $S_n$ denote the [[Definition:Symmetric Group on n Letters|symmetric group on $n$ letters]].
+Let $\rho, \sigma$ be [[Definition:Disjoint Permutations|disjoint permutations]].
+Then:
+: $\forall k \in \Z: \paren {\sigma \rho}^k = \sigma^k \rho^k$
+\end{theorem}
+
+\begin{proof}
+A direct application of [[Power of Product of Commutative Elements in Group]], and the fact that [[Disjoint Permutations Commute]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Order of Product of Disjoint Permutations}
+Tags: Permutation Theory, Order of Product of Disjoint Permutations
+
+\begin{theorem}
+Let $S_n$ denote the [[Definition:Symmetric Group on n Letters|symmetric group on $n$ letters]].
+Let $\pi$ be a product of [[Definition:Disjoint Permutations|disjoint permutations]] of [[Definition:Order of Group Element|orders]] $k_1, k_2, \ldots, k_r$.
+Then:
+:$\order \pi = \lcm \set {k_1, k_2, \ldots, k_r}$
+where:
+:$\order \pi$ denotes the [[Definition:Order of Group Element|order]] of $\pi$ in $S_n$
+:$\lcm$ denotes [[Definition:Lowest Common Multiple of Integers|lowest common multiple]].
+\end{theorem}
+
+\begin{proof}
+Suppose $\pi$ is a [[Definition:Cyclic Permutation|cycle]].
+Then from [[Order of Cycle is Length of Cycle]], $\order \pi$ is its [[Definition:Length of Cyclic Permutation|length]].
+As the [[Definition:Lowest Common Multiple of Integers|LCM]] of $n \in \Z$ is $n$ itself, the result follows.
+Let $\pi = \rho_1 \rho_2 \cdots \rho_r$ where:
+: each $\rho_s$ is of [[Definition:Order of Group Element|order]] $k_s$
+: $\rho_1$ to $\rho_r$ are mutually [[Definition:Disjoint Permutations|disjoint permutations]].
+Let $t = \lcm \set {k_1, k_2, \ldots, k_r}$.
+From [[Disjoint Permutations Commute]]:
+:$\pi^t = \rho_1^t \rho_2^t \cdots \rho_r^t$
+We have that:
+:$\forall s: 1 \le s \le r: \exists m \in \Z: t = m s$
+Hence:
+:$\forall s: \rho_s^t = e$
+Thus $\pi^t = e$ and certainly $\order \pi \divides t$.
+Let $\pi^u = e$ for some $u \in \Z_{>0}$.
+We have that $\rho_1, \rho_2, \cdots, \rho_r$ are mutually [[Definition:Disjoint Permutations|disjoint permutations]].
+Thus $\rho_1^u, \rho_2^u, \cdots, \rho_r^u$ are also mutually [[Definition:Disjoint Permutations|disjoint permutations]].
+Then we have that each $\rho_s^u = e$.
+Hence $k_s \divides u$.
+Thus, if $\pi^u = e$, then:
+:$t \divides u$
+Thus, by choosing $u = \order \pi$:
+:$\order \pi = t = \lcm \set {k_1, k_2, \ldots, k_r}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Group Action defines Permutation Representation}
+Tags: Group Actions, Group Homomorphisms
+
+\begin{theorem}
+Let $\map \Gamma X$ be the set of [[Definition:Permutation|permutations]] on a [[Definition:Set|set]] $X$.
+Let $G$ be a [[Definition:Group|group]].
+Let $\phi: G \times X \to X$ be a [[Definition:Group Action|group action]].
+For $g \in G$, let $\phi_g: X \to X$ be the [[Definition:Mapping|mapping]] defined as:
+:$\map {\phi_g} x = \map \phi {g, x}$
+Let $\tilde \phi: G \to \map \Gamma X$ be the [[Definition:Permutation Representation/Group Action|mapping associated to]] $\phi$, defined by:
+:$\map {\tilde \phi} g := \phi_g$
+Then $\tilde \phi$ is a [[Definition:Group Homomorphism|group homomorphism]].
+\end{theorem}
+
+\begin{proof}
+Note that, by [[Group Action determines Bijection|Group Action Determines Bijection]], $\phi_g \in \map \Gamma X$ for $g \in G$.
+Let $g, h \in G$.
+From the [[Definition:Group Action|definition of group action]]:
+:$\forall \tuple {g, x} \in G \times X: \map \phi {g, x} \in X = g \wedge x \in X$
+First we show that for all $x \in X$:
+:$\map {\phi_g \circ \phi_h} x = \map {\phi_{g h} } x$
+Thus:
+{{begin-eqn}}     
+{{eqn | l = \map {\phi_g \circ \phi_h} x
+      | r = g \wedge \paren {h \wedge x}
+      | c = Definition of $\phi_g$, $\phi_h$
+}}
+{{eqn | r = \paren {g h} \wedge x
+      | c = {{Defof|Group Action}}
+}}
+{{eqn | r = \map {\phi_{g h} } x
+      | c = Definition of $\phi_{g h}$
+}}
+{{end-eqn}}
+Also, we have:
+: $e \wedge x = x \implies \map {\phi_e} x = x$
+where $e$ is the [[Definition:Identity Element|identity]] of $G$.
+Therefore, we have shown that $\tilde \phi: G \to \map \Gamma X: g \mapsto \phi_g$ is a [[Definition:Group Homomorphism|group homomorphism]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Group Action determines Bijection}
+Tags: Group Actions
+
+\begin{theorem}
+Let $*$ be a [[Definition:Group Action|group action]] of $G$ on $X$.
+Then each $g \in G$ determines a [[Definition:Bijection|bijection]] $\phi_g: X \to X$ given by:
+:$\map {\phi_g} x = g * x$
+Its [[Definition:Inverse Mapping|inverse]] is:
+:$\phi_{g^{-1} }: X \to X$.
+These [[Definition:Bijection|bijection]] are sometimes called '''transformations''' of $X$.
+\end{theorem}
+
+\begin{proof}
+=== Proof of Injectivity ===
+Let $x, y \in X$
+Then:
+{{begin-eqn}}
+{{eqn | l = \map {\phi_g} x
+      | r = \map {\phi_g} y
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = g * x
+      | r = g * y
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = g^{-1} * \paren {g * x}
+      | r = g^{-1} * \paren {g * y}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \paren {g^{-1} g} * x
+      | r = \paren {g^{-1} g} * y
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = e * x
+      | r = e * y
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = x
+      | r = y
+      | c = 
+}}
+{{end-eqn}}
+Thus $\phi_g$ is an [[Definition:Injection|injection]].
+{{qed|lemma}}
+=== Proof of Surjectivity ===
+Let $x \in X$.
+Then:
+{{begin-eqn}}
+{{eqn | l = x
+      | r = e * x
+      | c = 
+}}
+{{eqn | r = \paren {g g^{-1} } * x
+      | c = 
+}}
+{{eqn | r = g * \paren {g^{-1} * x}
+      | c = 
+}}
+{{eqn | r = \map {\phi_g} y
+      | c = where $y = g^{-1} * x \in X$
+}}
+{{end-eqn}}
+Thus a group action is a [[Definition:Surjection|surjection]].
+{{qed|lemma}}
+So a group action is an [[Definition:Injection|injection]] and a [[Definition:Surjection|surjection]] and therefore a [[Definition:Bijection|bijection]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Group Action Induces Equivalence Relation}
+Tags: Group Actions, Equivalence Relations
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
+Let $X$ be a [[Definition:Set|set]].
+Let $*: G \times S \to S$ be a [[Definition:Group Action|group action]].
+Let $\mathcal R_G$ be the [[Definition:Equivalence Relation Induced by Group Action|relation induced]] by $G$, that is:
+:$x \mathrel {\mathcal R_G} y \iff y \in \Orb x$
+where:
+:$\Orb x$ denotes the [[Definition:Orbit (Group Theory)|orbit]] of $x \in X$.
+Then:
+:$\mathcal R_G$ is an [[Definition:Equivalence Relation|equivalence relation]].
+:The [[Definition:Equivalence Class|equivalence class]] of an element is its [[Definition:Orbit (Group Theory)|orbit]].
+\end{theorem}
+
+\begin{proof}
+Let $x \mathrel {\mathcal R_G} y \iff y \in \Orb x$.
+Checking in turn each of the critera for [[Definition:Equivalence Relation|equivalence]]:
+=== Reflexivity ===
+$x = e * x \implies x \in \Orb x$ from the definition of [[Definition:Group Action|group action]].
+Thus $\mathcal R_G$ is [[Definition:Reflexive Relation|reflexive]].
+{{qed|lemma}}
+=== Symmetry ===
+{{begin-eqn}}
+{{eqn | l = y
+      | o = \in
+      | r = \Orb x
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \exists g \in G: y
+      | r = g * x
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = g^{-1} * \paren {g * x}
+      | r = g^{-1} * y
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = x
+      | r = g^{-1} * y
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \exists g^{-1} \in G: x
+      | r = g^{-1} * y
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = x
+      | o = \in
+      | r = \Orb y
+      | c = 
+}}
+{{end-eqn}}
+Thus $\mathcal R_G$ is [[Definition:Symmetric Relation|symmetric]].
+{{qed|lemma}}
+=== Transitivity ===
+{{begin-eqn}}
+{{eqn | l = y
+      | o = \in
+      | r = \Orb x, z \in \Orb y
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \exists g_1 \in G: y
+      | r = g_1 * x, \exists g_2 \in G: z = g_2 * y
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = z
+      | r = g_2 * \paren {g_1 * x}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = z
+      | r = \paren {g_2 g_1} * x
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = z
+      | o = \in
+      | r = \Orb x
+      | c = 
+}}
+{{end-eqn}}
+Thus $\mathcal R_G$ is [[Definition:Transitive Relation|transitive]].
+{{qed|lemma}}
+So $\mathcal R_G$ has been shown to be an [[Definition:Equivalence Relation|equivalence relation]]. 
+Hence the result, by definition of an [[Definition:Equivalence Class|equivalence class]].
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Partition Equation}
+Tags: Group Actions, Named Theorems
+
+\begin{theorem}
+Let [[Definition:Group|group]] $G$ [[Definition:Group Action|act on]] a [[Definition:Finite Set|finite set]] $X$.
+Let the distinct [[Definition:Orbit (Group Theory)|orbits]] of $X$ under the [[Definition:Group Action|action]] of $G$ be:
+:$\Orb {x_1}, \Orb {x_2}, \ldots, \Orb {x_s}$
+Then:
+:$\card X = \card {\Orb {x_1} } + \card {\Orb {x_2} } + \cdots + \card {\Orb {x_s} }$
+\end{theorem}
+
+\begin{proof}
+Follows trivially from the fact that the [[Group Action Induces Equivalence Relation]].
+{{explain|cite a result about equiv rels and partitions, and one about cardinality and partitions}}
+{{qed}}
+[[Category:Group Actions]]
+[[Category:Named Theorems]]
+pwvgvk2c6p89r84hq9vudcd3u47firc
+\end{proof}<|endoftext|>
+\section{Definition:Stabilizer}
+Tags: Definitions: Group Actions
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]].
+Let $X$ be a [[Definition:Set|set]].
+Let $*: G \times X \to X$ be a [[Definition:Group Action|group action]].
+For each $x \in X$, the '''stabilizer of $x$ by $G$''' is defined as:
+:$\Stab x := \set {g \in G: g * x = x}$
+where $*$ denotes the [[Definition:Group Action|group action]].
+\end{theorem}<|endoftext|>
+\section{Stabilizer is Subgroup}
+Tags: Group Actions, Stabilizers, Stabilizer is Subgroup
+
+\begin{theorem}
+Let $\struct {G, \circ}$ be a [[Definition:Group|group]] which [[Definition:Group Action|acts on]] a [[Definition:Set|set]] $X$.
+Let $\Stab x$ be the [[Definition:Stabilizer|stabilizer of $x$ by $G$]].
+Then for each $x \in X$, $\Stab x$ is a [[Definition:Subgroup|subgroup]] of $G$.
+\end{theorem}
+
+\begin{proof}
+From the {{GroupActionAxiom|2}}:
+:$e * x = x \implies e \in \Stab x$
+and so $\Stab x$ cannot be [[Definition:Empty Set|empty]].
+Let $g, h \in \Stab x$.
+{{begin-eqn}}
+{{eqn | l = g, h
+      | o = \in
+      | r = \Stab x
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = g * x
+      | r = x
+      | c = {{Defof|Stabilizer}} of $x$ by $G$
+}}
+{{eqn | lo= \land
+      | l = h * x
+      | r = x
+      | c = {{Defof|Stabilizer}} of $x$ by $G$
+}}
+{{eqn | ll= \leadsto
+      | l = g * \paren {h * x}
+      | r = x
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \paren {g \circ h} * x
+      | r = x
+      | c = {{GroupActionAxiom|1}}
+}}
+{{eqn | ll= \leadsto
+      | l = g \circ h
+      | o = \in
+      | r = \Stab x
+      | c = {{Defof|Stabilizer}} of $x$ by $G$
+}}
+{{end-eqn}}
+Let $g \in \Stab x$.
+Then:
+:$x = \paren {g^{-1} \circ g} * x = g^{-1} * \paren {g * x} = g^{-1} * x$
+Hence $g^{-1} \in \Stab x$.
+Thus the conditions for the [[Two-Step Subgroup Test]] are fulfilled, and $\Stab x \le G$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Orbit-Stabilizer Theorem}
+Tags: Group Actions, Stabilizers, Named Theorems, Orbit-Stabilizer Theorem
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]] which [[Definition:Group Action|acts on]] a [[Definition:Finite Set|finite set]] $X$.
+Let $x \in X$.
+Let $\Orb x$ denote  the [[Definition:Orbit (Group Theory)|orbit]] of $x$.
+Let $\Stab x$ denote the [[Definition:Stabilizer|stabilizer of $x$ by $G$]].
+Let $\index G {\Stab x}$ denote the [[Definition:Index of Subgroup|index]] of $\Stab x$ in $G$.
+Then:
+:$\order {\Orb x} = \index G {\Stab x} = \dfrac {\order G} {\order {\Stab x} }$
+\end{theorem}
+
+\begin{proof}
+Let us define the [[Definition:Mapping|mapping]]:
+:$\phi: G \to \Orb x$
+such that:
+:$\map \phi g = g * x$
+where $*$ denotes the [[Definition:Group Action|group action]].
+It is clear that $\phi$ is [[Definition:Surjection|surjective]], because from the definition $x$ was acted on by all the [[Definition:Element|elements]] of $G$.
+Next, from [[Stabilizer is Subgroup/Corollary|Stabilizer is Subgroup: Corollary]]:
+:$\map \phi g = \map \phi h \iff g^{-1} h \in \Stab x$
+This means:
+:$g \equiv h \pmod {\Stab x}$
+Thus there is a well-defined [[Definition:Bijection|bijection]]:
+:$G \mathbin / \Stab x \to \Orb x$
+given by:
+:$g \, \Stab x \mapsto g * x$
+So $\Orb x$ has the same number of elements as $G \mathbin / \Stab x$.
+That is:
+:$\order {\Orb x} = \index G {\Stab x}$
+The result follows.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let $x \in X$.
+Let $\phi: \Orb x \to G \mathbin / \Stab x$ be a [[Definition:Mapping|mapping]] from the [[Definition:Orbit (Group Theory)|orbit]] of $x$ to the [[Definition:Left Coset Space|left coset space]] of $\Stab x$ defined as:
+:$\forall g \in G: \map \phi {g * x} = g \, \Stab x$
+where $*$ is the [[Definition:Group Action|group action]].
+Note: this is ''not'' a [[Definition:Group Homomorphism|homomorphism]] because $\Orb x$ is not a [[Definition:Group|group]].
+Suppose $g * x = h * x$ for some $g, h \in G$.
+Then:
+:$h^{-1} g * x = h^{-1} h * x$
+and so:
+:$h^{-1} g * x = x$
+Thus:
+:$h^{-1} g \in \Stab x$
+so by [[Left Coset Space forms Partition]]:
+:$g \, \Stab x = h \, \Stab x$
+demonstrating that $\phi$ is [[Definition:Well-Defined Mapping|well-defined]].
+Let $\map \phi {g_1 * x} = \map \phi {g_2 * x}$ for some $g_1, g_2 \in G$.
+Then:
+:$g_1 \, \Stab x = g_2 \, \Stab x$
+and so by [[Left Coset Space forms Partition]]:
+:$g_2^{-1} g_1 \in \Stab x$
+So by definition of $\Stab x$:
+:$x = g_2^{-1} g_1 * x$
+Thus:
+{{begin-eqn}}
+{{eqn | l = g_2 * x
+      | r = g_2 * \paren {g_2^{-1} g_1 * x}
+      | c = applying $g_2$ to both sides
+}}
+{{eqn | r = \paren {g_2 g_2^{-1} g_1} x
+      | c = {{Defof|Group Action}}
+}}
+{{eqn | r = g_1 * x
+      | c = 
+}}
+{{end-eqn}}
+thus demonstrating that $\phi$ is [[Definition:Injection|injective]].
+As the [[Definition:Left Coset|left coset]] $g \, \Stab x$ is $\map \phi {g * x}$ by definition of $\phi$, it follows that $\phi$ is a [[Definition:Surjection|surjection]].
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Group Acts on Itself}
+Tags: Group Actions
+
+\begin{theorem}
+Let $\struct {G, \circ}$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
+Then $\struct {G, \circ}$ [[Definition:Group Action|acts on]] itself by the rule:
+:$\forall g, h \in G: g * h = g \circ h$
+\end{theorem}
+
+\begin{proof}
+Follows directly from the [[Definition:Group Axioms|group axioms]] and the definition of a [[Definition:Group Action|group action]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Action of Group on Coset Space is Group Action}
+Tags: Cosets, Group Action on Coset Space
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
+Let $H$ be a [[Definition:Subgroup|subgroup]] of $G$.
+Let $*: G \times G / H \to G / H$ be the [[Definition:Group Action on Coset Space|action on the (left) coset space]]:
+:$\forall g \in G, \forall g' H \in G / H: g * \paren {g' H} := \paren {g g'} H$
+Then $G$ is a [[Definition:Group Action|group action]].
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = a * \paren {b * g' H}
+      | r = a * \paren {\paren {b g'} H}
+      | c = Definition of $*$
+}}
+{{eqn | r = \paren {a \paren {b g'} } H
+      | c = Definition of $*$
+}}
+{{eqn | r = \paren {a b} g' H
+      | c = {{GroupAxiom|1}}
+}}
+{{end-eqn}}
+demonstrating that $*$ satisfies {{GroupActionAxiom|2}}.
+Then:
+{{begin-eqn}}
+{{eqn | l = e * g' H
+      | r = \paren {e g'} H
+      | c = Definition of $*$
+}}
+{{eqn | r = g'H
+      | c = {{GroupAxiom|2}}
+}}
+{{end-eqn}}
+demonstrating that $*$ satisfies {{GroupActionAxiom|1}}.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Group Action on Sets with k Elements}
+Tags: Examples of Group Actions, Group Action on Sets with k Elements
+
+\begin{theorem}
+Let $\struct {G, \circ}$ be a [[Definition:Finite Group|finite group]] whose [[Definition:Identity Element|identity]] is $e$.
+Let $\Bbb S = \set {S \subseteq G: \card S = k}$, that is, the [[Definition:Set|set]] of all of [[Definition:Subset|subsets]] of $G$ which have exactly $k$ [[Definition:Element|elements]].
+Let $G$ act on $\Bbb S$ by the rule:
+:$\forall S \in \Bbb S: g * S = g S = \set {x \in G: x = g s: s \in S}$
+This is a [[Definition:Group Action|group action]], and:
+: $\forall S \in \Bbb S: \order {\Stab S} \divides \card S$
+where $\Stab S$ denotes the [[Definition:Stabilizer|stabilizer of $S$ by $G$]].
+\end{theorem}
+
+\begin{proof}
+First it is necessary to prove that this is a [[Definition:Group Action|group action]].
+The [[Definition:Group Action|action]] is the same as the one defined in [[Group Action on Coset Space]], but this time we are limiting ourselves to the [[Definition:Subset|subsets]] of $G$ which have the same number of [[Definition:Element|elements]].
+From the result in [[Group Action on Coset Space]], we only have to prove that:
+:$\card {g S} = \card S$
+This follows directly from [[Order of Subset Product with Singleton]].
+It remains to be shown that:
+:$\order {\Stab S} \divides \card S$
+By definition of [[Definition:Stabilizer|stabilizer]]:
+:$\Stab S = \set {g \in G: g S = S}$
+It follows that:
+:$\forall s \in S: g s \in S$
+Then:
+:$\Stab S s \subseteq S$
+Thus:
+:$\card {\Stab S s} \le \card S$
+We have from [[Stabilizer is Subgroup/Corollary 1|corollary $1$ to Stabilizer is Subgroup]] that the [[Definition:Stabilizer|stabilizer]] contains the [[Definition:Identity Element|identity]]:
+:$e \in \Stab S \le G$
+It therefore follows that:
+:$s \in S \implies s \in \Stab S s$
+Thus:
+:$\displaystyle S = \bigcup_{t \mathop \in S} \Stab S t$
+By [[Congruence Class Modulo Subgroup is Coset]], distinct [[Definition:Right Coset|right cosets]] are [[Definition:Disjoint Sets|disjoint]].
+Thus $S$ consists of a [[Definition:Set Union|union]] of [[Definition:Disjoint Sets|disjoint]] [[Definition:Right Coset|right cosets]] of $\Stab S$.
+Thus $\displaystyle \bigcup_{t \mathop \in S} \Stab S t$ is a [[Definition:Set Partition|partition]] of $S$.
+Therefore:
+:$\forall s \in S: \card {\Stab S s} \divides \card S$
+As we have:
+:$\forall s \in S: \card {\Stab S s} = \order {\Stab S}$
+the result follows:
+:$\card {\Stab S} \divides \card S$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Group Action on Subgroup of Symmetric Group}
+Tags: Group Actions, Symmetric Groups
+
+\begin{theorem}
+Let $S_n$ be the [[Definition:Symmetric Group on n Letters|symmetric group of $n$ elements]].
+Let $H$ be a [[Definition:Subgroup|subgroup]] of $S_n$.
+Let $X$ be any [[Definition:Set|set]] with $n$ [[Definition:Element|elements]].
+Then $H$ [[Definition:Group Action|acts on]] $X$ as a [[Group Action determines Bijection|group of transformations]] on $X$.
+\end{theorem}
+
+\begin{proof}
+The [[Definition:Identity Mapping|identity permutation]] takes each element of $X$ to itself, thus fulfilling {{GroupActionAxiom|1}}.
+The [[Definition:Group Operation|group operation]] in $S_n$ ensures fulfilment of {{GroupActionAxiom|2}}.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Kernel of Group Action is Normal Subgroup}
+Tags: Group Actions
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
+Let $X$ be a [[Definition:Set|set]].
+Let $\phi: G \times X \to X$ be a [[Definition:Group Action|group action]].
+Let $G_0$ denote the [[Definition:Kernel of Group Action|kernel of $\phi$]].
+Then $G_0$ is a [[Definition:Normal Subgroup|normal subgroup]] of $G$.
+\end{theorem}
+
+\begin{proof}
+Let $h \in G_0$.
+{{begin-eqn}}     
+{{eqn | l = h
+      | o = \in
+      | r = G_0
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = h
+      | o = \in
+      | r = \set {g \in G: \forall x \in X: g \cdot x = x}
+      | c = {{Defof|Kernel of Group Action}}
+}}
+{{eqn | ll= \leadstoandfrom
+      | lo= \forall x \in X:
+      | l = h
+      | o = \in
+      | r = \set {g \in G: g \cdot x = x}
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | lo= \forall x \in X:
+      | l = h
+      | o = \in
+      | r = \Stab x
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = h
+      | o = \in
+      | r = \bigcap_{x \mathop \in X} \Stab x
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = G_0
+      | r = \bigcap_{x \mathop \in X} \Stab x
+      | c = 
+}}
+{{end-eqn}}
+From [[Stabilizer is Subgroup]]:
+:$\Stab x \le G$
+Thus $G_0$ is the [[Definition:Set Intersection|intersection]] of [[Definition:Subgroup|subgroups]].
+By [[Intersection of Subgroups is Subgroup/General Result|Intersection of Subgroups is Subgroup]]:
+:$G_0 \le G$
+To prove [[Definition:Normal Subgroup|normality]] it is sufficient to show:
+:$\forall g \in G: g G_0 g^{-1} = G_0$
+Let $h \in G_0, g \in G$ be arbitrary.
+Then:
+{{begin-eqn}}
+{{eqn | l = \paren {g h g^{-1} } \cdot x
+      | r = g \cdot \paren {h \cdot \paren { g^{-1} \cdot x } }
+      | c = [[Definition:Group Action|Group Action]] [[Definition:Associative Operation|Associates]] with [[Definition:Group Operation|Group Operation]]: {{GroupActionAxiom|1}}
+}}
+{{eqn | r = g \cdot \paren {g^{-1} \cdot x}
+      | c = because $h \in \Stab { g^{-1} \cdot x}$
+}}
+{{eqn | r = \paren {g \cdot  g^{-1} } \cdot x
+      | c = [[Definition:Group Action|Group Action]] [[Definition:Associative Operation|Associates]] with [[Definition:Group Operation|Group Operation]]: {{GroupActionAxiom|1}}
+}}
+{{eqn | r = x
+      | c = $e \cdot x = x$: {{GroupActionAxiom|2}}
+}}
+{{end-eqn}}
+Therefore:
+:$g h g^{-1} \in G_0$
+so:
+:$g G_0 g^{-1} \subseteq G_0$
+Conversely suppose that $h \in G_0$.
+Then by the above:
+:$h' = g^{-1} h g \in G_0$
+Therefore:
+:$h = g h' g^{-1} \in g G_0 g^{-1}$
+and so:
+:$G_0 \subseteq g G_0 g^{-1}$
+This concludes the proof.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Quotient of Transformation Group acts Effectively}
+Tags: Group Actions, Quotient Groups
+
+\begin{theorem}
+Let $G$ be a [[Definition:Transformation Group|transformation group]] (which may or may not be [[Definition:Effective Transformation Group|effective]]) acting on $X$.
+Then the [[Definition:Quotient Group|quotient group]] $G / G_0$, where $G_0$ is the [[Definition:Kernel of Group Action|kernel]], ''does'' act [[Definition:Effective Transformation Group|effectively]] on $X$.
+\end{theorem}<|endoftext|>
+\section{Condition for Group to Act Effectively on Left Coset Space}
+Tags: Group Action on Coset Space, Cosets
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
+Let $H$ be a [[Definition:Subgroup|subgroup]] of $G$.
+Then $G$ [[Definition:Effective Transformation Group|acts effectively]] on the [[Definition:Left Coset Space|left coset space]] $G / H$ {{iff}}:
+:$\ds \bigcap_{a \mathop \in G} H^a = \set e$
+where $H^a$ denotes the [[Definition:Conjugate of Group Subset|conjugate]] of $H$ by $a$.
+\end{theorem}
+
+\begin{proof}
+$G$ [[Definition:Effective Transformation Group|acts effectively]] on the [[Definition:Left Coset Space|left coset space]] $G / H$ {{iff}} $a H \mapsto g a H$ is [[Definition:Faithful Group Action|faithful]], {{iff}}:
+{{begin-eqn}}
+{{eqn | lo= \forall g \in G: \forall a H \in G / H:
+      | l = g a H = a H
+      | o = \implies
+      | r = g = e
+      | c = {{Defof|Faithful Group Action}}
+}}
+{{eqn | ll= \leadstoandfrom
+      | lo= \forall g \in G: \forall a \in G:
+      | l = a^{-1} g a \in H
+      | o = \implies
+      | r = g = e
+      | c = [[Left Cosets are Equal iff Product with Inverse in Subgroup]]
+}}
+{{eqn | ll= \leadstoandfrom
+      | lo= \forall g \in G: \forall a \in G:
+      | l = g \in a H a^{-1}
+      | o = \implies
+      | r = g = e
+}}
+{{eqn | ll= \leadstoandfrom
+      | lo= \forall g \in G: \forall a \in G:
+      | l = g \in H^a
+      | o = \implies
+      | r = g = e
+      | c = {{Defof|Conjugate of Group Subset}}
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \bigcap_{a \mathop \in G} H^a
+      | r  = \set e
+      | c = {{Defof|Intersection of Family}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Set of Permutations is Largest Effective Transformation Group}
+Tags: Group Actions
+
+\begin{theorem}
+The set of [[Definition:Permutation|permutations]] of a [[Definition:Set|set]] $X$ forms the largest [[Definition:Effective Transformation Group|effective transformation group]] of $X$.
+\end{theorem}<|endoftext|>
+\section{Conjugacy Action is Group Action}
+Tags: Conjugacy Action
+
+\begin{theorem}
+Let $\left({G, \circ}\right)$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
+\end{theorem}<|endoftext|>
+\section{Conjugacy Action on Identity}
+Tags: Conjugacy Action
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
+For the [[Definition:Conjugacy Action|conjugacy action]]:
+:$\order {\Orb e} = 1$
+and thus:
+:$\Stab e = G$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = g * e
+      | r = g e g^{-1}
+      | c = 
+}}
+{{eqn | r = g g^{-1}
+      | c = 
+}}
+{{eqn | r = e
+      | c = 
+}}
+{{end-eqn}}
+So the only [[Definition:Conjugate of Group Element|conjugate]] of $e$ is $e$ itself.
+Thus:
+:$\Orb e = \set e$
+and so:
+:$\order {\Orb e} = 1$
+From the [[Orbit-Stabilizer Theorem]], it follows immediately that:
+:$\Stab e = G$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Cauchy's Lemma (Group Theory)}
+Tags: Group Theory, Cauchy's Lemma (Group Theory)
+
+\begin{theorem}
+Let $\struct {G, \circ}$ be a [[Definition:Group|group]] of [[Definition:Order of Structure|finite order]] whose [[Definition:Identity Element|identity]] is $e$.
+Let $p$ be a [[Definition:Prime Number|prime number]] which [[Definition:Divisor of Integer|divides]] the [[Definition:Order of Structure|order]] of $G$.
+Then $\struct {G, \circ}$ has an [[Definition:Element|element]] of [[Definition:Order of Group Element|order]] $p$.
+\end{theorem}
+
+\begin{proof}
+Let $\order G = n$ such that $p \divides n$.
+Let:
+:$X = \set {\tuple {a_1, a_2, \ldots, a_p} \in G^p: a_1 a_2 \cdots a_p = e}$
+where $G^p$ is the [[Definition:Cartesian Space|cartesian product]] $\underbrace {G \times G \times \cdots \times G}_p$.
+The first $p - 1$ [[Definition:Coordinate of Ordered Tuple|coordinates]] of an [[Definition:Element|element]] of $X$ can be chosen arbitrarily.
+The last [[Definition:Coordinate of Ordered Tuple|coordinate]] is determined by the fact that:
+:$a_1 a_2 \cdots a_{p - 1} = a_p^{-1}$
+So from the [[Product Rule for Counting]], it follows that:
+:$\card X = n^{p - 1}$
+Let $C_p$ be a [[Definition:Cyclic Group|cyclic group]] of [[Definition:Order of Structure|order]] $p$ [[Definition:Generator of Cyclic Group|generated]] by the [[Definition:Element|element]] $c$.
+Let $C_p$ [[Definition:Group Action|act on]] the set $X$ by the rule:
+:$c * \tuple {a_1, a_2, \ldots, a_p} = \tuple {a_2, a_3, \ldots, a_p, a_1}$
+By the [[Orbit-Stabilizer Theorem]], the number of [[Definition:Element|elements]] in any [[Definition:Orbit (Group Theory)|orbit]] is a [[Definition:Divisor of Integer|divisor]] of the order of $C_p$, which is $p$.
+As $p$ is [[Definition:Prime Number|prime]], an [[Definition:Orbit (Group Theory)|orbit]] has either $p$ [[Definition:Element|elements]] or $1$ [[Definition:Element|element]] by definition.
+Let $r$ be the number of [[Definition:Orbit (Group Theory)|orbits]] with one [[Definition:Element|element]].
+Let $s$ be the number of [[Definition:Orbit (Group Theory)|orbits]] with $p$ [[Definition:Element|elements]].
+Then by the [[Partition Equation]]:
+:$r + s p = n^{p - 1} = \card X$
+[[Definition:By Hypothesis|By hypothesis]], $p \divides n$, so:
+:$r + s p = n^{p - 1} \implies p \divides r$
+We know that $r \ne 0$ because, for example, the [[Definition:Orbit (Group Theory)|orbit]] of $\tuple {e, e, \ldots, e} \in X$ has only one [[Definition:Element|element]].
+So there must be at least $p$ [[Definition:Orbit (Group Theory)|orbits]] with only one [[Definition:Element|element]].
+Each such [[Definition:Element|element]] has the form $\tuple {a, a, \ldots, a} \in X$ so $a^p = e$.
+So $G$ contains at least $p$ [[Definition:Element|elements]] $x$ satisfying $x^p = e$.
+So $G$ contains an [[Definition:Element|element]] $a \ne e$ such that $a^p = e$.
+That is, $a$ must have [[Definition:Order of Group Element|order]] $p$.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+By the [[First Sylow Theorem/Corollary|corollary to the First Sylow Theorem]], $G$ has [[Definition:Subgroup|subgroups]] of [[Definition:Order of Group|order]] $p^r$ for all $r$ such that $p^r \divides \order G$.
+Thus $G$ has at least one [[Definition:Subgroup|subgroup]] $H$ of [[Definition:Order of Group|order]] $p$.
+As a [[Prime Group is Cyclic]], $H$ is a [[Definition:Cyclic Group|cyclic group]].
+Thus by definition $H$ has an [[Definition:Element|element]] of [[Definition:Order of Group Element|order]] $p$.
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Transposition is Self-Inverse}
+Tags: Symmetric Groups
+
+\begin{theorem}
+All [[Definition:Transposition|transpositions]] are [[Definition:Self-Inverse Element|self-inverse]].
+\end{theorem}
+
+\begin{proof}
+Let $\pi = \begin{bmatrix} k_1 & k_2 \end{bmatrix}$ be a [[Definition:Transposition|transposition]].
+Writing $\pi \pi$ in [[Definition:Permutation on n Letters/Cycle Notation|cycle notation]] gives:
+:$\begin{bmatrix} k_1 & k_2 \end{bmatrix} \begin{bmatrix} k_1 & k_2 \end{bmatrix}$
+from which we see that $k_1 \to k_2 \to k_1$ and $k_2 \to k_1 \to k_2$.
+The result follows from the definition of [[Definition:Self-Inverse Element|self-inverse]].
+{{qed}}
+[[Category:Symmetric Groups]]
+lehv6tztngn5ci5vftba2lh34nrl8ch
+\end{proof}<|endoftext|>
+\section{Conjugates of Transpositions}
+Tags: Symmetric Groups
+
+\begin{theorem}
+Let $k_1, k_2, k_3 \in \left\{{1, 2, \ldots, n}\right\}$.
+Then:
+: $(1): \quad \begin{bmatrix} k_1 & k_2 \end{bmatrix} = \begin{bmatrix} k_3 & k_2 \end{bmatrix} \begin{bmatrix} k_1 & k_3 \end{bmatrix} \begin{bmatrix} k_3 & k_2 \end{bmatrix}$
+: $(2): \quad \begin{bmatrix} k_1 & k_2 \end{bmatrix} = \begin{bmatrix} k_3 & k_1 \end{bmatrix} \begin{bmatrix} k_3 & k_2 \end{bmatrix} \begin{bmatrix} k_3 & k_1 \end{bmatrix}$
+\end{theorem}
+
+\begin{proof}
+$(1):$ Calculating the product of $\begin{bmatrix} k_3 & k_2 \end{bmatrix} \begin{bmatrix} k_1 & k_3 \end{bmatrix} \begin{bmatrix} k_3 & k_2 \end{bmatrix}$:
+:: $k_1 \to k_3 \to k_2$
+:: $k_2 \to k_3 \to k_1$
+:: $k_3 \to k_2 \to k_3$
+hence the result.
+{{qed|lemma}}
+$(2):$ Calculating the product of $\begin{bmatrix} k_3 & k_1 \end{bmatrix} \begin{bmatrix} k_3 & k_2 \end{bmatrix} \begin{bmatrix} k_3 & k_1 \end{bmatrix}$:
+:: $k_1 \to k_3 \to k_2$
+:: $k_2 \to k_3 \to k_1$
+:: $k_3 \to k_1 \to k_3$
+hence the result.
+{{qed}}
+[[Category:Symmetric Groups]]
+psvlecwx3uss76gc4u3nnod6mn6y4ch
+\end{proof}<|endoftext|>
+\section{K-Cycle can be Factored into Transpositions}
+Tags: Cyclic Permutations
+
+\begin{theorem}
+Every [[Definition:Cyclic Permutation|$k$-cycle]] can be factorised into the product of $k - 1$ [[Definition:Transposition|transpositions]].
+This factorisation is not unique.
+\end{theorem}
+
+\begin{proof}
+The cycle $\begin{pmatrix} 1 & 2 & \ldots & k \end{pmatrix}$ has the factorisation:
+:$\begin{pmatrix} 1 & 2 & \ldots & k \end{pmatrix} = \begin{pmatrix} 1 & k \end{pmatrix} \ldots \begin{pmatrix} 1 & 3 \end{pmatrix} \begin{pmatrix} 1 & 2 \end{pmatrix}$
+Therefore, the general [[Definition:Cyclic Permutation|$k$-cycle]] $\begin{pmatrix} i_1 & i_2 & \ldots & i_k \end{pmatrix}$ has the factorisation:
+:$\begin{pmatrix} i_1 & i_2 & \ldots & i_k \end{pmatrix} = \begin{pmatrix} i_1 & i_k \end{pmatrix} \ldots \begin{pmatrix} i_1 & i_3 \end{pmatrix} \begin{pmatrix} i_1 & i_2 \end{pmatrix}$
+The cycle $\begin{pmatrix} 1 & 2 & \ldots & k \end{pmatrix}$ also has the factorisation:
+:$\begin{pmatrix} 1 & 2 & \ldots & k \end{pmatrix} = \begin{pmatrix} 1 & 2 \end{pmatrix} \begin{pmatrix} 2 & 3 \end{pmatrix} \ldots \begin{pmatrix} k - 1 & k \end{pmatrix}$
+Therefore, the general [[Definition:Cyclic Permutation|$k$-cycle]] $\begin{pmatrix} i_1 & i_2 & \ldots & i_k \end{pmatrix}$ also has the factorisation:
+:$\begin{pmatrix} i_1 & i_2 & \ldots & i_k \end{pmatrix} = \begin{pmatrix} i_1 & i_2 \end{pmatrix} \begin{pmatrix} i_2 & i_3 \end{pmatrix} \ldots \begin{pmatrix} i_{k - 1} & i_k \end{pmatrix}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sign of Permutation is Plus or Minus Unity}
+Tags: Symmetric Groups
+
+\begin{theorem}
+Let $n \in \N$ be a [[Definition:Natural Number|natural number]].
+Let $\N_n$ denote the [[Definition:Set|set]] of [[Definition:Natural Numbers|natural numbers]] $\set {1, 2, \ldots, n}$.
+Let $S_n$ denote the [[Definition:Symmetric Group on n Letters|symmetric group on $n$ letters]].
+Let $\sequence {x_k}_{k \mathop \in \N_n}$ be a [[Definition:Finite Sequence|finite]] [[Definition:Real Sequence|sequence in $\R$]].
+Let $\pi \in S_n$.
+Let $\map {\Delta_n} {x_1, x_2, \ldots, x_n}$ be the [[Definition:Product of Differences|product of differences]] of $\tuple {x_1, x_2, \ldots, x_n}$.
+Let $\map \sgn \pi$ be the [[Definition:Sign of Permutation on n Letters|sign of $\pi$]].
+Let $\pi \cdot \map {\Delta_n} {x_1, x_2, \ldots, x_n}$ be defined as:
+:$\pi \cdot \map {\Delta_n} {x_1, x_2, \ldots, x_n} := \map {\Delta_n} {x_{\map \pi 1}, x_{\map \pi 2}, \ldots, x_{\map \pi n} }$
+Then either:
+:$\pi \cdot \Delta_n = \Delta_n$
+or:
+:$\pi \cdot \Delta_n = -\Delta_n$
+That is:
+:$\map \sgn \pi = \begin{cases}
+1 & :\pi \cdot \Delta_n = \Delta_n \\
+-1 & : \pi \cdot \Delta_n = -\Delta_n
+\end{cases}$
+Thus:
+:$\pi \cdot \Delta_n = \map \sgn \pi \Delta_n$
+\end{theorem}
+
+\begin{proof}
+If $\exists i, j \in \N_n$ such that $x_i = x_j$, then $\map {\Delta_n} {x_1, x_2, \ldots, x_n} = 0$ and the result follows trivially.
+So, suppose all the elements $x_k$ are [[Definition:Distinct|distinct]].
+Let us use $\Delta_n$ to denote $\map {\Delta_n} {x_1, x_2, \ldots, x_n}$.
+Let $1 \le a < b \le n$.
+Then $x_a - x_b$ is a [[Definition:Divisor of Integer|divisor]] of $\Delta_n$.
+Then $x_{\map \pi a} - x_{\map \pi b}$ is a factor of $\pi \cdot \Delta_n$.
+There are two possibilities for the ordering of $\map \pi a$ and $\map \pi b$:
+Either $\map \pi a < \map \pi b$ or $\map \pi a > \map \pi b$.
+If the former, then $x_{\map \pi a} - x_{\map \pi b}$ is a factor of $\Delta_n$.
+If the latter, then $-\paren {x_{\map \pi a} - x_{\map \pi b} }$ is a factor of $\Delta_n$.
+The same applies to all factors of $\Delta_n$.
+Thus:
+{{begin-eqn}}
+{{eqn | l = \pi \cdot \Delta_n
+      | r = \pi \cdot \prod_{1 \mathop \le i \mathop < j \mathop \le n} \paren {x_i - x_j}
+      | c = 
+}}
+{{eqn | r = \pm \prod_{1 \mathop \le i \mathop < j \mathop \le n} \paren {x_i - x_j}
+      | c = 
+}}
+{{eqn | r = \pm \Delta_n
+      | c = 
+}}
+{{end-eqn}}
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Parity Function is Homomorphism}
+Tags: Symmetric Groups
+
+\begin{theorem}
+Let $S_n$ denote the [[Definition:Symmetric Group on n Letters|symmetric group on $n$ letters]].
+Let $\pi \in S_n$.
+Let $\map \sgn \pi$ be the [[Definition:Sign of Permutation on n Letters|sign of $\pi$]].
+Let the [[Definition:Parity of Permutation|parity function]] of $\pi$ be defined as:
+:Parity of $\pi = \begin{cases}
+\mathrm {Even} & : \map \sgn \pi = 1 \\
+\mathrm {Odd} & : \map \sgn \pi = -1
+\end{cases}$
+The [[Definition:Mapping|mapping]] $\sgn: S_n \to C_2$, where $C_2$ is the [[Definition:Cyclic Group|cyclic group]] of [[Definition:Order of Structure|order 2]], is a [[Definition:Group Homomorphism|homomorphism]].
+\end{theorem}
+
+\begin{proof}
+We need to show that:
+:$\forall \pi, \rho \in S_n: \map \sgn \pi \, \map \sgn \rho = \map \sgn {\pi \rho}$
+Let $\Delta_n$ be an arbitrary [[Definition:Product of Differences|product of differences]].
+{{begin-eqn}}
+{{eqn | l = \map \sgn {\pi \rho} \Delta_n
+      | r = \pi \rho \cdot \Delta_n
+      | c = {{Defof|Sign of Permutation}}
+}}
+{{eqn | r = \pi \cdot \paren {\rho \cdot \Delta_n}
+      | c = [[Permutation on Polynomial is Group Action]]
+}}
+{{eqn | r = \pi \cdot \paren {\map \sgn \rho \cdot \Delta_n}
+      | c = {{Defof|Sign of Permutation}}
+}}
+{{eqn | r = \pi \, \map \sgn \rho \cdot \Delta_n
+      | c = [[Permutation on Polynomial is Group Action]]
+}}
+{{eqn | r = \map \sgn \pi \, \map \sgn \rho \Delta_n
+      | c = {{Defof|Sign of Permutation}}
+}}
+{{end-eqn}}
+As $\struct {\set {1, -1}, \times}$ is the [[Definition:Parity Group|parity group]], the result follows immediately.
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Permutation on Polynomial is Group Action}
+Tags: Group Actions, Permutation Theory
+
+\begin{theorem}
+Let $n \in \Z: n > 0$.
+Let $F_n$ be the set of all [[Definition:Polynomial|polynomials]] in $n$ variables $x_1, x_2, \ldots, x_n$:
+:$F = \set {\map f {x_1, x_2, \ldots, x_n}: f \text{ is a polynomial in $n$ variables} }$
+Let $S_n$ denote the [[Definition:Symmetric Group on n Letters|symmetric group on $n$ letters]].
+Let $*: S_n \times F \to F$ be the [[Definition:Mapping|mapping]] defined as:
+:$\forall \pi \in S_n, f \in F: \pi * \map f {x_1, x_2, \ldots, x_n} = \map f {x_{\map \pi 1}, x_{\map \pi 2}, \ldots, x_{\map \pi n} }$
+Then $*$ is a [[Definition:Group Action|group action]].
+\end{theorem}
+
+\begin{proof}
+Let $\pi, \rho \in S_n$.
+Let $\pi * f$ be the [[Definition:Permutation on Polynomial|permutation on the polynomial]] $f$ by $\pi$.
+Let $e \in S_n$ be the [[Definition:Identity Element|identity]] of $S_n$.
+From [[Symmetric Group is Group]]:
+:$e * f = f$
+thus fulfilling {{GroupActionAxiom|1}}.
+Then we have that:
+{{begin-eqn}}
+{{eqn | l = \paren {\pi \circ \rho} * f
+      | r = \map \pi {\rho * f}
+      | c = 
+}}
+{{eqn | r = \pi * \paren {\rho * f}
+      | c = 
+}}
+{{end-eqn}}
+thus fulfilling {{GroupActionAxiom|2}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Parity of Inverse of Permutation}
+Tags: Symmetric Groups
+
+\begin{theorem}
+Let $S_n$ denote the [[Definition:Symmetric Group on n Letters|symmetric group on $n$ letters]].
+Then:
+:$\forall \pi \in S_n: \map \sgn \pi = \map \sgn {\pi^{-1} }$
+\end{theorem}
+
+\begin{proof}
+From [[Parity Function is Homomorphism]]:
+:$\map \sgn {I_{S_n} } = 1$
+Thus:
+:$\pi \pi^{-1} = I_{S_n} \implies \map \sgn \pi \, \map \sgn {\pi^{-1} } = 1$
+The result follows immediately.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Parity of Conjugate of Permutation}
+Tags: Symmetric Groups
+
+\begin{theorem}
+Let $S_n$ denote the [[Definition:Symmetric Group on n Letters|symmetric group on $n$ letters]].
+Then:
+:$\forall \pi, \rho \in S_n: \map \sgn {\pi \rho \pi^{-1} } = \map \sgn \rho$
+where $\map \sgn \pi$ is the [[Definition:Sign of Permutation|sign of $\pi$]].
+\end{theorem}
+
+\begin{proof}
+As $\map \sgn \pi = \pm 1$ for any $\pi \in S_n$, we can apply the laws of [[Definition:Commutative Operation|commutativity]] and [[Definition:Associative|associativity]]:
+{{begin-eqn}}
+{{eqn | l = \map \sgn \pi \, \map \sgn \rho
+      | r = \map \sgn \rho \, \map \sgn \pi
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \map \sgn \pi \, \map \sgn \rho \, \map \sgn {\pi^{-1} }
+      | r = \map \sgn \rho \, \map \sgn \pi \, \map \sgn {\pi^{-1} }
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \map \sgn {\pi \rho \pi^{-1} }
+      | r = \map \sgn \rho
+      | c = [[Parity Function is Homomorphism]]
+}}
+{{end-eqn}}
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Parity of K-Cycle}
+Tags: Cyclic Permutations
+
+\begin{theorem}
+Let $\pi$ be a [[Definition:Cyclic Permutation|$k$-cycle]].
+Then:
+:$\map \sgn \pi = \begin{cases}
+1 & : k \ \text {odd} \\
+-1 & : k \ \text {even}
+\end{cases}$
+Thus:
+:$\map \sgn \pi = \paren {-1}^{k - 1}$
+or equivalently:
+:$\map \sgn \pi = \paren {-1}^{k + 1}$
+\end{theorem}
+
+\begin{proof}
+From [[Transposition is of Odd Parity]], any [[Definition:Transposition|transposition]] is of [[Definition:Parity (Permutation)|odd parity]].
+From [[K-Cycle can be Factored into Transpositions]], we see that a [[Definition:Cyclic Permutation|$k$-cycle]] is the product of $k - 1$ [[Definition:Transposition|transposition]]s.
+Thus $\pi$ is even {{iff}} $k$ is [[Definition:Odd Integer|odd]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Alternating Group is Normal Subgroup of Symmetric Group}
+Tags: Symmetric Groups, Alternating Groups
+
+\begin{theorem}
+Let $n \ge 2$ be a [[Definition:Natural Number|natural number]].
+Let $S_n$ denote the [[Definition:Symmetric Group on n Letters|symmetric group on $n$ letters]].
+Let $A_n$ be the [[Definition:Alternating Group|alternating group on $n$ letters]].
+Then $A_n$ is a [[Definition:Normal Subgroup|normal subgroup]] of $S_n$ whose [[Definition:Index of Subgroup|index]] is $2$.
+\end{theorem}
+
+\begin{proof}
+We have that $\map \sgn {S_n}$ is [[Definition:Surjection|onto]] $C_2$.
+Thus from the [[First Isomorphism Theorem for Groups|First Isomorphism Theorem]], $A_n$ consists of the set of [[Definition:Even Permutation|even permutations]] of $S_n$.
+The result follows from [[Subgroup of Index 2 is Normal]].
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Group of Permutations either All or Half Even}
+Tags: Symmetric Groups
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]] of [[Definition:Permutation|permutations]].
+Then either ''exactly half'' of the permutations in $G$ are [[Definition:Even Permutation|even]], or they are ''all'' [[Definition:Even Permutation|even]].
+\end{theorem}
+
+\begin{proof}
+From [[Parity Function is Homomorphism]], the [[Definition:Mapping|mapping]]:
+:$\sgn : G \to \Z_2$
+is a [[Definition:Group Homomorphism|homomorphism]].
+Then:
+:$G / \map \ker \sgn \cong \Img \sgn$
+from the [[First Isomorphism Theorem for Groups|First Isomorphism Theorem]].
+The only possibilities for $\Img \sgn$ are $\set 0$ or $\Z_2$.
+So either:
+:$\order {G / \map \ker \sgn} = 1$
+in which case all the [[Definition:Permutation|permutations]] of $G$ are [[Definition:Even Permutation|even]], or:
+:$\order {G / \map \ker \sgn} = 2$
+in which case exactly half of them are.
+{{qed}}
+[[Category:Symmetric Groups]]
+jktrlc08nm231gxlx8dba3n056hbeqi
+\end{proof}<|endoftext|>
+\section{Cycle Decomposition of Conjugate}
+Tags: Symmetric Groups, Conjugacy, Permutation Theory
+
+\begin{theorem}
+Let $S_n$ denote the [[Definition:Symmetric Group on n Letters|symmetric group on $n$ letters]].
+Let $\pi, \rho \in S_n$.
+The [[Definition:Cycle Decomposition|cycle decomposition]] of the [[Definition:Permutation|permutation]] $\pi \rho \pi^{-1}$ can be obtained from that of $\rho$ by replacing each $i$ in the [[Definition:Cycle Decomposition|cycle decomposition]] of $\rho$ with $\map \pi i$.
+\end{theorem}
+
+\begin{proof}
+Consider the effect of $\pi \rho \pi^{-1}$ on $\map \pi i$:
+:$\map {\pi \rho \pi^{-1} } {\map \pi i} = \map \pi {\map \rho i}$
+That is:
+:$\pi \rho \pi^{-1}$ maps $\map \pi i$ to $\map \pi {\map \rho i}$
+In the [[Definition:Cycle Decomposition|cycle decomposition]] of $\pi \rho \pi^{-1}$, $\map \pi i$ lies to the left of $\map \pi {\map \rho i}$, whereas in the [[Definition:Cycle Decomposition|cycle decomposition]] of $\rho$, $i$ lies to the left of $\map \rho i$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Conjugate Permutations have Same Cycle Type}
+Tags: Conjugacy, Permutation Theory
+
+\begin{theorem}
+Let $n \ge 1$ be a [[Definition:Natural Number|natural number]].
+Let $G$ be a [[Definition:Subgroup|subgroup]] of the [[Definition:Symmetric Group on n Letters|symmetric group on $n$ letters]] $S_n$.
+Let $\sigma, \rho \in G$.
+Then $\sigma$ and $\rho$ are [[Definition:Conjugate of Group Element|conjugate]] {{iff}} they have the same [[Definition:Cycle Type|cycle type]].
+\end{theorem}
+
+\begin{proof}
+Let $\sigma \in G$ have [[Definition:Cycle Type|cycle type]] $\tuple {k_1, k_2, \ldots, k_n}$.
+From [[Existence and Uniqueness of Cycle Decomposition]], $\sigma$ can be expressed uniquely as the [[Definition:Product Element|product]] of [[Definition:Disjoint Permutations|disjoint]] [[Definition:Cyclic Permutation|cycles]]:
+:$\sigma = \alpha_1 \alpha_2 \dotsm \alpha_l$
+where $\alpha_i$ is a [[Definition:Cyclic Permutation|$k_i$-cycle]].
+Let $\tau \in G$ such that $\rho = \tau \sigma \tau^{-1}$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \tau \sigma \tau^{-1}
+      | r = \tau \alpha_1 \alpha_2 \dotsm \alpha_l \tau^{-1}
+      | c = 
+}}
+{{eqn | n = 1
+      | r = \tau \alpha_1 \tau^{-1} \tau \alpha_2 \tau^{-1} \dotsm \tau \alpha_l \tau^{-1}
+      | c = [[Product of Conjugates equals Conjugate of Products]]
+}}
+{{eqn | r = \alpha_{\map \tau 1} \alpha_{\map \tau 2} \dotsm \alpha_{\map \tau l}
+      | c = [[Cycle Decomposition of Conjugate]]
+}}
+{{end-eqn}}
+We have that for $i, j \in \set {1, 2, \ldots, l}$, $\alpha_i$ and $\alpha_j$ are [[Definition:Disjoint Permutations|disjoint]].
+We also have that $\tau$ is a [[Definition:Bijection|bijection]], and so an [[Definition:Injection|injection]].
+It follows that $\alpha_{\map \tau i}$ and $\alpha_{\map \tau j}$ are also [[Definition:Disjoint Permutations|disjoint]] for all $i, j \in \set {1, 2, \ldots, l}$.
+Thus:
+:the [[Definition:Product Element|product]] in $(1)$ is $\rho$ written as the [[Definition:Product Element|product]] of [[Definition:Disjoint Permutations|disjoint]] [[Definition:Cyclic Permutation|cycles]]
+where:
+:$\tau \alpha_i \tau^{-1}$ is a [[Definition:Cyclic Permutation|$k_i$-cycle]].
+Thus $\rho$ has the same [[Definition:Cycle Type|cycle type]] as $\sigma$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Transpositions of Adjacent Elements generate Symmetric Group}
+Tags: Symmetric Groups
+
+\begin{theorem}
+Let $n \in \Z: n > 1$.
+Let $S_n$ denote the [[Definition:Symmetric Group on n Letters|symmetric group on $n$ letters]].
+Then the [[Definition:Transposition|transpositions]] $a_k = \begin{pmatrix} k & k + 1 \end{pmatrix}$ for $1 \le k < n$ are a [[Definition:Generator of Group|set of generators]] for $S_n$.
+They satisfy the relations:
+{{begin-eqn}}
+{{eqn | l = a_k^2
+      | r = e
+      | c = (for $1 \le k < n$)
+}}
+{{eqn | l = \paren {a_k a_{k + 1} }^3
+      | r = e
+      | c = (for $1 \le k < n - 1$)
+}}
+{{eqn | l = \paren {a_i a_j}^2
+      | r = e
+      | c = (for $1 \le i, j < n, \size {i - j} > 1$)
+}}
+{{end-eqn}}
+\end{theorem}<|endoftext|>
+\section{Hermitian Matrix has Real Eigenvalues}
+Tags: Hermitian Matrices, Hermitian Matrix has Real Eigenvalues
+
+\begin{theorem}
+Every [[Definition:Hermitian Matrix|Hermitian matrix]] has [[Definition:Eigenvalue|eigenvalues]] which are all [[Definition:Real Number|real numbers]].
+\end{theorem}
+
+\begin{proof}
+Let $\mathbf A$ be a [[Definition:Hermitian Matrix|Hermitian matrix]].
+Then, by definition:
+: $\mathbf A = \mathbf A^*$
+where $\mathbf A^*$ denotes the [[Definition:Conjugate Transpose of Matrix|conjugate transpose]] of $\mathbf A$.
+Let $\lambda$ be an [[Definition:Eigenvalue|eigenvalue]] of $\mathbf A$.
+Let $\mathbf v$ be an [[Definition:Eigenvector|eigenvector]] corresponding to the [[Definition:Eigenvalue|eigenvalue]] $\lambda$.
+By definition of [[Definition:Eigenvector|eigenvector]]:
+: $\mathbf{A v} = \lambda \mathbf v$
+[[Definition:Matrix Product (Conventional)|Left-multiplying]] both sides by $\mathbf v^*$, we obtain:
+:$(1):\quad \mathbf v^* \mathbf{A v} = \mathbf v^* \lambda \mathbf v = \lambda \mathbf v^* \mathbf v$
+Firstly, note that both $\mathbf v^* \mathbf{A v}$ and $\mathbf v^* \mathbf v$ are $1 \times 1$-matrices.
+{{explain|Links to the proof of this fact}}
+Now observe that, using [[Conjugate Transpose of Matrix Product/General Case|Conjugate Transpose of Matrix Product: General Case]]:
+:$\left({\mathbf v^* \mathbf{A v}}\right)^* = \mathbf v^* \mathbf A^* \left({\mathbf v^*}\right)^*$
+As $\mathbf A$ is [[Definition:Hermitian Matrix|Hermitian]], and $\left({\mathbf v^*}\right)^* = \mathbf v$ by [[Conjugate Transpose is Involution]], it follows that:
+:$\mathbf v^* \mathbf A^* \left({\mathbf v^*}\right)^* = \mathbf v^* \mathbf{A v}$
+That is, $\mathbf v^* \mathbf{A v}$ is also [[Definition:Hermitian Matrix|Hermitian]].
+By [[Product with Conjugate Transpose Matrix is Hermitian]], $\mathbf v^* \mathbf v$ is [[Definition:Hermitian Matrix|Hermitian]]. 
+So both $\mathbf v^* \mathbf{A v}$ and $\mathbf v^* \mathbf v$ are [[Definition:Hermitian Matrix|Hermitian]] $1 \times 1$ matrices. 
+Now suppose that we have for some $a,b \in \C$:
+:$\mathbf v^* \mathbf{A v} = \begin{bmatrix}a\end{bmatrix}$
+:$\mathbf v^* \mathbf v = \begin{bmatrix}b\end{bmatrix}$
+Note that $b \ne 0$ as an [[Definition:Eigenvector|eigenvector]] is by definition [[Definition:Zero Vector|non-zero]].
+By definition of a [[Definition:Hermitian Matrix|Hermitian matrix]]:
+: $a = \bar a$ and $b = \bar b$
+where $\bar a$ denotes the [[Definition:Complex Conjugate|complex conjugate]] of $a$.
+By [[Complex Number equals Conjugate iff Wholly Real]], it follows that $a, b \in \R$, that is, are [[Definition:Real Number|real]].
+From equation $(1)$, it follows that $\begin{bmatrix}a\end{bmatrix} = \lambda \begin{bmatrix}b\end{bmatrix}$. 
+Thus, $a = \lambda b$, i.e. $\lambda = \dfrac a b$ (recall that $b \ne 0$).
+Hence $\lambda$, being a [[Definition:Quotient (Algebra)|quotient]] of [[Definition:Real Number|real numbers]], is [[Definition:Real Number|real]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let $\mathbf A$ be a [[Definition:Hermitian Matrix|Hermitian matrix]].
+Then, by definition, $\mathbf A = \mathbf A^\dagger$, where $^\dagger$ designates the [[Definition:Conjugate Transpose of Matrix|conjugate transpose]].
+Let $\lambda$ be an [[Definition:Eigenvalue|eigenvalue]] of $\mathbf A$.
+Let $\mathbf v$ be an [[Definition:Eigenvector|eigenvector]] corresponding to the [[Definition:Eigenvalue|eigenvalue]] $\lambda$ of $\mathbf A$.
+Denote with $\left\langle{\cdot, \cdot}\right\rangle$ the [[Definition:Inner Product|inner product]] on $\C$. 
+{{begin-eqn}}
+{{eqn | l = \lambda * \left\langle v, v\right\rangle
+      | r = \left\langle \lambda*v, v\right\rangle
+      | c = [[Properties of Complex Inner Product]]
+}}
+{{eqn | r = \left\langle \mathbf A*v, v\right\rangle
+      | c = {{Defof|Eigenvector}}: $\lambda*v = \mathbf A*v$
+}}
+{{eqn | r = \left\langle v, \mathbf A^\dagger * v\right\rangle
+      | c = [[Properties of Adjugate]]
+}}
+{{eqn | r = \left\langle v, \mathbf A*v\right\rangle
+      | c = $\mathbf A$ is [[Definition:Hermitian Matrix|Hermitian]], so $\mathbf A^\dagger = \mathbf A$
+}}
+{{eqn | r = \left\langle v, \lambda*v\right\rangle
+      | c = {{Defof|Eigenvector}}: $\lambda*v = \mathbf A*v$
+}}
+{{eqn | r = \overline{\lambda}*\left\langle v, v\right\rangle
+      | c = [[Properties of Complex Inner Product]]
+}}
+{{end-eqn}}
+We have that $v \ne 0$, and because of the [[Definition:Positive Definite|positive definiteness]], it must be that:
+: $\left\langle v, v\right\rangle \ne 0$
+{{explain|A link is needed to whatever it is ($\left\langle v, v\right\rangle$, presumably) is [[Definition:Positive Definite|positive definite]].}}
+Thus:
+: $\left\langle v, v\right\rangle \ne 0$
+So we can divide both sides by $\left\langle v, v\right\rangle$.
+Thus $\lambda = \overline{\lambda}$.
+By [[Complex Number equals Conjugate iff Wholly Real]], $\lambda$ is a [[Definition:Real Number|real number]].
+$\lambda$ was arbitrary, so it follows that every [[Definition:Eigenvalue|eigenvalue]] is a [[Definition:Real Number|real number]].
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Permutation of Cosets}
+Tags: Symmetric Groups, Cosets, Permutation Theory
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]] and let $H \le G$.
+Let $\mathbb S$ be the set of all distinct [[Definition:Left Coset|left cosets]] of $H$ in $G$.
+Then:
+:$(1): \quad$ For any $g \in G$, the [[Definition:Mapping|mapping]] $\theta_g: \mathbb S \to \mathbb S$ defined by $\map {\theta_g} {x H} = g x H$ is a [[Definition:Permutation|permutation]] of $\mathbb S$.
+:$(2): \quad$ The mapping $\theta$ defined by $\map \theta g = \theta_g$ is a [[Definition:Group Homomorphism|homomorphism]] from $G$ into the [[Definition:Symmetric Group|symmetric group]] on $\mathbb S$.
+:$(3): \quad$ The [[Definition:Kernel of Group Homomorphism|kernel]] of $\theta$ is the [[Definition:Subgroup|subgroup]] $\displaystyle \bigcap_{x \mathop \in G} x H x^{-1}$.
+\end{theorem}
+
+\begin{proof}
+First we need to show that $\theta_g$ is [[Definition:Well-Defined Mapping|well-defined]] and [[Definition:Injection|injective]].
+{{begin-eqn}}
+{{eqn | l = x H
+      | r = y H
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = y^{-1} x
+      | o = \in
+      | r = H
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \paren {g y}^{-1} g x
+      | r = y^{-1} x \in H
+      | c = 
+}}
+{{eqn | ll= \leadstoandfrom
+      | l = \map {\theta_g} {y H}
+      | r = \map {\theta_g} {x H}
+      | c = 
+}}
+{{end-eqn}}
+Thus $\theta_g$ is [[Definition:Well-Defined Mapping|well-defined]] and [[Definition:Injection|injective]].
+Then we see that $\forall x H \in \mathbb S: \map {\theta_g} {g^{-1} x H} = x H$, so $\theta_g$ is [[Definition:Surjection|surjective]].
+Thus $\theta_g$ is a [[Definition:Well-Defined Mapping|well-defined]] [[Definition:Bijection|bijection]] on $\mathbb S$, and therefore a [[Definition:Permutation|permutation]] on $\mathbb S$.
+Next we see:
+{{begin-eqn}}
+{{eqn | l = \map {\theta_{u v} } {x H}
+      | r = u v x H
+      | c = 
+}}
+{{eqn | r = \map {\theta_u} {v x H}
+      | c = 
+}}
+{{eqn | r = \map {\theta_u} {\map {\theta_v }{x H} }=
+      | c = 
+}}
+{{end-eqn}}
+This shows that $\theta_{u v} = \theta_u \theta_v$, and thus:
+:$\map \theta {u v} = \map \theta u \, \map \theta v$
+Thus $\theta$ is a [[Definition:Group Homomorphism|homomorphism]].
+Now to calculate $\map \ker \theta$:
+{{begin-eqn}}
+{{eqn | l = \map \ker \theta
+      | r = \set {g \in G: \theta_g = I_\mathbb S}
+      | c = 
+}}
+{{eqn | r = \set {g \in G: \forall x \in G: \map {\theta_g} {x H} = x H}
+      | c = 
+}}
+{{eqn | r = \set {g \in G: \forall x \in G: g x h = x H}
+      | c = 
+}}
+{{eqn | r = \set {g \in G: \forall x \in G: x^{-1} g x h = H}
+      | c = 
+}}
+{{eqn | r = \set {g \in G: \forall x \in G: x^{-1} g x \in H}
+      | c = 
+}}
+{{eqn | r = \set {g \in G: \forall x \in G: g \in x H x^{-1} }
+      | c = 
+}}
+{{eqn | r = \bigcap_{x \mathop \in G} x H x^{-1}
+      | c = 
+}}
+{{end-eqn}}
+as required.
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Cayley's Representation Theorem}
+Tags: Symmetric Groups, Group Theory, Subgroups, Cayley's Representation Theorem
+
+\begin{theorem}
+Let $S_n$ denote the [[Definition:Symmetric Group on n Letters|symmetric group on $n$ letters]].
+Every finite [[Definition:Group|group]] is [[Definition:Group Isomorphism|isomorphic]] to a [[Definition:Subgroup|subgroup]] of $S_n$ for some $n \in \Z$.
+\end{theorem}
+
+\begin{proof}
+Let $H = \set e$.
+We can apply [[Permutation of Cosets]] to $H$ so that:
+:$\mathbb S = G$
+and:
+:$\map \ker \theta = \set e$
+The result follows by the [[First Isomorphism Theorem for Groups]].
+{{Qed}}
+\end{proof}
+
+\begin{proof}
+Let $G$ be any arbitrary [[Definition:Finite Group|finite group]] whose [[Definition:Identity Element|identity]] is $e_G$.
+Let $S$ be the [[Definition:Symmetric Group|symmetric group]] on the elements of $G$, where $e_S$ is the [[Definition:Identity Element|identity]] of $S$.
+For any $x \in G$, let $\lambda_x$ be the [[Definition:Left Regular Representation|left regular representation of $G$ with respect to $x$]].
+From [[Regular Representations in Group are Permutations]], $\forall x \in G: \lambda_x \in S$.
+So, we can define a [[Definition:Mapping|mapping]] $\theta: G \to S$ as:
+:$\forall x \in G: \map \theta x = \lambda_x$
+From [[Composition of Regular Representations]], we have:
+:$\forall x, y \in G: \lambda_x \circ \lambda_y = \lambda_{x y}$
+where in this context $\circ$ denotes [[Definition:Composition of Mappings|composition of mappings]].
+Thus by definition of $\theta$:
+:$\map \theta x \circ \map \theta y = \map \theta {x y}$
+demonstrating that $\theta$ is a [[Definition:Group Homomorphism|homomorphism]].
+Having established that fact, we can now consider $\map \ker \theta$, where $\ker$ denotes the [[Definition:Kernel of Group Homomorphism|kernel]] of $\theta$.
+Let $x \in G$.
+We have that:
+{{begin-eqn}}
+{{eqn | l = x
+      | o = \in
+      | r = \map \ker \theta
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \map \theta x
+      | r = e_S
+      | c = {{Defof|Kernel of Group Homomorphism}}
+}}
+{{eqn | ll= \leadsto
+      | l = \lambda_x
+      | r = I_G
+      | c = where $I_G$ is the [[Definition:Identity Mapping|identity mapping]] of $G$
+}}
+{{eqn | ll= \leadsto
+      | l = \map {\lambda_x} {e_G}
+      | r = \map {I_G} {e_G}
+      | c = {{Defof|Identity Mapping}}
+}}
+{{eqn | ll= \leadsto
+      | l = x
+      | r = e_G
+      | c = Definitions of $\lambda_x$ and $I_G$
+}}
+{{end-eqn}}
+So $\map \ker \theta$ can contain no element other than $e_G$.
+So, since clearly $e_G \in \map \ker \theta$, it follows that:
+:$\map \ker \theta = \set {e_G}$
+By [[Kernel is Trivial iff Monomorphism]], it follows that $\theta$ is a [[Definition:Group Monomorphism|monomorphism]].
+By [[Monomorphism Image is Isomorphic to Domain]], we have that:
+:$\theta \sqbrk G \cong \Img \theta$
+that is, $\theta$ is [[Definition:Group Isomorphism|isomorphic]] to its image.
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Basic Results about Modules}
+Tags: Module Theory
+
+\begin{theorem}
+Let $\struct {G, +_G}$ be an [[Definition:Abelian Group|abelian group]] whose [[Definition:Identity Element|identity]] is $e$.
+Let $\struct {R, +_R, \times_R}$ be a [[Definition:Ring (Abstract Algebra)|ring]] whose [[Definition:Ring Zero|zero]] is $0_R$.
+Let $\struct {G, +_G, \circ}_R$ be an [[Definition:Module|$R$-module]].
+Let $x \in G, \lambda \in R, n \in \Z$.
+Let $\sequence {x_m}$ be a [[Definition:Sequence|sequence of elements of $G$]].
+Let $\sequence {\lambda_m}$ be a [[Definition:Sequence|sequence of elements of $R$]] that is, [[Definition:Scalar (Module)|scalars]].
+Then:
+\end{theorem}<|endoftext|>
+\section{Basic Results about Unitary Modules}
+Tags: Unitary Modules
+
+\begin{theorem}
+Let $\struct {G, +_G}$ be an [[Definition:Abelian Group|abelian group]] whose [[Definition:Identity Element|identity]] is $e$.
+Let $\struct {R, +_R, \times_R}$ be a [[Definition:Ring with Unity|ring with unity]] whose [[Definition:Ring Zero|zero]] is $0_R$ and whose [[Definition:Unity of Ring|unity]] is $1_R$.
+Let $\struct {G, +_G, \circ}_R$ be an [[Definition:Unitary Module|unitary $R$-module]].
+Let $x \in G, n \in \Z$.
+Then:
+\end{theorem}<|endoftext|>
+\section{Epimorphism preserves Modules}
+Tags: Epimorphisms, Module Theory
+
+\begin{theorem}
+Let $\left({G, +_G, \circ}\right)_R$ be an [[Definition:Module|$R$-module]].
+Let $\left({H, +_H, \circ}\right)_R$ be an [[Definition:R-Algebraic Structure|$R$-algebraic structure]].
+Let $\phi: G \to H$ be an [[Definition:R-Algebraic Structure Epimorphism|epimorphism]].
+Then $H$ is an [[Definition:Module|$R$-module]].
+\end{theorem}
+
+\begin{proof}
+If $\left({G, +_G: \circ}\right)_R$ is an [[Definition:Module|$R$-module]], then:
+$\forall x, y, \in G, \forall \lambda, \mu \in R$:
+: $(1): \quad \lambda \circ \left({x +_G y}\right) = \left({\lambda \circ x}\right) +_G \left({\lambda \circ y}\right)$
+: $(2): \quad \left({\lambda +_R \mu}\right) \circ x = \left({\lambda \circ x}\right) +_G \left({\mu \circ x}\right)$
+: $(3): \quad \left({\lambda \times_R \mu}\right) \circ x = \lambda \circ \left({\mu \circ x}\right)$
+If $\phi: G \to H$ is an [[Definition:R-Algebraic Structure Epimorphism|epimorphism]], then:
+: $\forall x, y \in G: \phi \left({x +_G y}\right) = \phi \left({x}\right) +_H \phi \left({y}\right)$
+: $\forall x \in S: \forall \lambda \in R: \phi \left({\lambda \circ x}\right) = \lambda \circ \phi \left({x}\right)$
+: $\forall y \in H: \exists x \in G: y = \phi \left({x}\right)$
+As $\phi$ is an epimorphism, we can accurately specify the behaviour of all elements of $H$, as they are the images of elements of $G$. If $\phi$ were not an epimorphism, i.e. not [[Definition:Surjection|surjective]], we would have no way of knowing the behaviour of elements of $H$ outside of the [[Definition:Image of Subset under Mapping|image]] of $G$. Hence the specification that $\phi$ needs to be an epimorphism.
+Now we check the [[Definition:Module|criteria for $H$ being a module]], in turn.
+=== Module $(1)$ ===
+{{begin-eqn}}
+{{eqn | l=\lambda \circ \left({\phi \left({x}\right) +_H \phi \left({y}\right)}\right)
+      | r=\lambda \circ \left({\phi \left({x +_G y}\right)}\right)
+      | c=
+}}
+{{eqn | r=\phi \left({\lambda \circ \left({x +_G y}\right)}\right)
+      | c=
+}}
+{{eqn | r=\phi \left({\left({\lambda \circ x}\right) +_G \left({\lambda \circ y}\right)}\right)
+      | c=
+}}
+{{eqn | r=\phi \left({\lambda \circ x}\right) +_H \phi \left({\lambda \circ y}\right)
+      | c=
+}}
+{{end-eqn}}
+Thus $(1)$ is shown to hold for $H$.
+=== Module $(2)$ ===
+{{begin-eqn}}
+{{eqn | l=\left({\lambda +_R \mu}\right) \circ \phi \left({x}\right)
+      | r=\phi \left({\left({\lambda +_R \mu}\right) \circ x}\right)
+      | c=
+}}
+{{eqn | r=\phi \left({\left({\lambda \circ x}\right) +_G \left({\mu \circ x}\right)}\right)
+      | c=
+}}
+{{eqn | r=\phi \left({\lambda \circ x}\right) +_H \phi \left({\mu \circ x}\right)
+      | c=
+}}
+{{eqn | r=\lambda \circ \phi \left({x}\right) +_H \mu \circ \phi \left({x}\right)
+      | c=
+}}
+{{end-eqn}}
+Thus $(2)$ is shown to hold for $H$.
+=== Module $(3)$ ===
+{{begin-eqn}}
+{{eqn | l=\left({\lambda \times_R \mu}\right) \circ \phi \left({x}\right)
+      | r=\phi \left({\left({\lambda \times_R \mu}\right) \circ x}\right)
+      | c=
+}}
+{{eqn | r=\phi \left({\lambda \circ \left({\mu \circ x}\right)}\right)
+      | c=
+}}
+{{eqn | r=\lambda \circ \left({\phi \left({\mu \circ x}\right)}\right)
+      | c=
+}}
+{{eqn | r=\lambda \circ \left({\mu \circ \phi \left({x}\right)}\right)
+      | c=
+}}
+{{end-eqn}}
+Thus $(3)$ is shown to hold for $H$.
+So all the criteria for $H$ to be an [[Definition:Module|$R$-module]] are fulfilled.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Condition for Linear Transformation}
+Tags: Linear Transformations
+
+\begin{theorem}
+Let $G$ be a [[Definition:Unitary Module|unitary $R$-module]], and let $H$ be an [[Definition:Module|$R$-module]].
+Let $\phi: G \to H$ be a [[Definition:Mapping|mapping]].
+Then $\phi$ is a [[Definition:Linear Transformation|linear transformation]] {{iff}}:
+:$\forall x, y \in G: \forall \lambda, \mu \in R: \map \phi {\lambda x + \mu y} = \lambda \map \phi x + \mu \map \phi y$
+\end{theorem}
+
+\begin{proof}
+Any [[Definition:Linear Transformation|linear transformation]] clearly satisfies the condition.
+Let $\phi$ be such that the condition is satisfied.
+Let $\lambda = \mu = 1_R$.
+Then $\map \phi {x + y} = \map \phi x + \map \phi y$.
+Now let $\mu = 0_R$.
+Then $\map \phi {\lambda x} = \lambda \map \phi x$.
+Thus the conditions are fulfilled for $\phi$ to be a [[Definition:R-Algebraic Structure Homomorphism|homomorphism]], that is, a [[Definition:Linear Transformation|linear transformation]].
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Module of All Mappings is Module}
+Tags: Examples of Modules
+
+\begin{theorem}
+Let $\struct {R, +_R, \times_R}$ be a [[Definition:Ring (Abstract Algebra)|ring]].
+Let $\struct {G, +_G, \circ}_R$ be an [[Definition:Module|$R$-module]].
+Let $S$ be a [[Definition:Set|set]].
+Let $\struct {G^S, +_G', \circ}_R$ be the [[Definition:Module of All Mappings|module of all mappings]] from $S$ to $G$.
+Then $\struct {G^S, +_G', \circ}_R$ is an [[Definition:Module|$R$-module]].
+\end{theorem}
+
+\begin{proof}
+To show that $\struct {G^S, +_G', \circ}_R$ is an [[Definition:Module|$R$-module]], we verify the following:
+$\forall f, g, \in G^S, \forall \lambda, \mu \in R$:
+:$(1): \quad \lambda \circ \paren {f +_G' g} = \paren {\lambda \circ f} +_G' \paren {\lambda \circ g}$
+:$(2): \quad \paren {\lambda +_R \mu} \circ f = \paren {\lambda \circ f} +_G \paren {\mu \circ f}$
+:$(3): \quad \paren {\lambda \times_R \mu} \circ f = \lambda \circ \paren {\mu \circ f}$
+=== Criterion 1 ===
+:$(1): \quad \lambda \circ \paren {f +_G' g} = \paren {\lambda \circ f} +_G' \paren {\lambda \circ g}$
+Let $x \in S$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \lambda \circ \paren {\map {\paren {f +_G' g} } x}
+      | r = \lambda \circ \paren {\map f x +_G \map g x}
+      | c = 
+}}
+{{eqn | r = \paren {\lambda \circ \map f x} +_G \paren {\lambda \circ \map g x}
+      | c = {{Defof|Module}}
+}}
+{{eqn | r = \paren {\map {\paren {\lambda \circ f} } x} +_G \paren {\map {\paren {\lambda \circ g} } x}
+      | c = 
+}}
+{{eqn | r = \map {\paren {\paren {\lambda \circ f} +_G' \paren {\lambda \circ g} } } x
+      | c = 
+}}
+{{end-eqn}}
+Thus $(1)$ holds.
+{{qed|lemma}}
+=== Criterion 2 ===
+: $(2): \quad \paren {\lambda +_R \mu} \circ f = \paren {\lambda \circ f} +_G \paren {\mu \circ f}$
+Let $x \in S$.
+{{begin-eqn}}
+{{eqn | l = \map {\paren {\paren {\lambda +_R \mu} \circ f} } x
+      | r = \paren {\lambda +_R \mu} \circ \paren {\map f x}
+      | c = 
+}}
+{{eqn | r = \paren {\lambda \circ \map f x} +_G \paren {\mu \circ \map f x}
+      | c = 
+}}
+{{eqn | r = \map {\paren {\lambda \circ f} } x +_G \map {\paren {\mu \circ f} } x
+      | c = 
+}}
+{{end-eqn}}
+Thus $(2)$ holds.
+{{qed|lemma}}
+=== Criterion 3 ===
+:$(3): \quad \paren {\lambda \times_R \mu} \circ f = \lambda \circ \paren {\mu \circ f}$
+{{begin-eqn}}
+{{eqn | l = \map {\paren {\paren {\lambda \times_R \mu} \circ f} } x
+      | r = \paren {\lambda \times_R \mu} \circ \paren {\map f x}
+      | c = 
+}}
+{{eqn | r = \lambda \circ \paren {\mu \circ \map f x}
+      | c = 
+}}
+{{eqn | r = \lambda \circ \paren {\map {\paren {\mu \circ f} } x}
+      | c = 
+}}
+{{eqn | r = \map {\paren {\lambda \circ \paren {\mu \circ f} } } x
+      | c = 
+}}
+{{end-eqn}}
+Thus $(3)$ holds.
+{{qed|lemma}}
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Kuratowski's Theorem}
+Tags: Graph Theory
+
+\begin{theorem}
+The following conditions on a [[Definition:Graph (Graph Theory)|graph]] $\Gamma$ are equivalent:
+:$(1): \quad \Gamma$ is [[Definition:Planar Graph|planar]]
+:$(2): \quad \Gamma$ contains no [[Definition:Graph Subdivision|subdivision]] of either the [[Definition:Complete Graph|complete graph $K_5$]] or the [[Definition:Complete Bipartite Graph|complete bipartite graph $K_{3, 3}$]].
+\end{theorem}
+
+\begin{proof}
+Let $\Gamma$ be a [[Definition:Graph (Graph Theory)|graph]] with $V$ [[Definition:Vertex of Graph|vertices]], $E$ [[Definition:Edge of Graph|edges]] and $F$ [[Definition:Face of Graph|faces]].
+The proof proceeds in two parts: $(1) \implies (2)$ and $(2) \implies (1)$.
+=== $(1)$ implies $(2)$ ===
+First we consider $K_5$.  
+$K_5$ has $5$ [[Definition:Vertex of Graph|vertices]] and $10$ [[Definition:Edge of Graph|edge]]s.
+By the [[Euler Polyhedron Formula]], a [[Definition:Planar Embedding|planar embedding]] would have $7$ [[Definition:Face of Graph|faces]].  
+But each [[Definition:Face of Graph|face]] has at least $3$ [[Definition:Edge of Graph|edge]]s, while each [[Definition:Edge of Graph|edge]] bounds at most two [[Definition:Face of Graph|face]]s.
+Suppose we count the incident [[Definition:Edge of Graph|edge]]-[[Definition:Face of Graph|face]] pairs.
+Then the number of [[Definition:Face of Graph|face]]s is at most:
+:$E \dfrac 2 3 = 6 \dfrac 2 3$
+which [[Definition:Contradiction|contradicts]] the deduction that $K_5$, if [[Definition:Planar Graph|planar]], should have $7$ [[Definition:Face of Graph|face]]s.
+Hence $K_5$ is [[Definition:Non-Planar Graph|non-planar]].
+Now consider $K_{3, 3}$.
+$K_{3, 3}$ has $6$ [[Definition:Vertex of Graph|vertices]] and $9$ [[Definition:Edge of Graph|edge]]s.
+Hence $K_{3, 3}$ has $5$ [[Definition:Face of Graph|face]]s if it is [[Definition:Planar Graph|planar]].
+Since $K_{3, 3}$ is [[Definition:Bipartite Graph|bipartite]], from [[Graph is Bipartite iff No Odd Cycles]], each [[Definition:Cycle (Graph Theory)|cycle]] in $K_{3,3}$ has an [[Definition:Even Integer|even number]] of [[Definition:Edge of Graph|edge]]s.
+Hence any [[Definition:Face of Graph|face]] must have at least $4$ [[Definition:Edge of Graph|edge]]s.
+So the number of [[Definition:Face of Graph|face]]s is at most
+:$E \dfrac 2 4 = 4 \dfrac 1 2$
+which [[Definition:Contradiction|contradicts]] the deduction that $K_{3, 3}$, if [[Definition:Planar Graph|planar]], should have $5$ [[Definition:Face of Graph|face]]s.
+Hence $K_{3, 3}$ is [[Definition:Non-Planar Graph|non-planar]].
+Now consider a [[Definition:Graph Subdivision|subdivision]] of either graph.
+<!--
+A subdivision of a graph is a replacement of any [[Definition:Edge of Graph|edge]] $[x,y]$ with two [[Definition:Edge of Graph|edge]]s, $[x,z]$ and $[z,y]$, where $z$ is a new vertex.
+-->
+The arguments as above continue to work as before, since both $V$ and $E$ have increased by one in the [[Euler Polyhedron Formula|Euler formula]].
+So the expected value of $F$ is unchanged.
+Hence if a [[Definition:Graph (Graph Theory)|graph]] $\Gamma$ contains a [[Definition:Graph Subdivision|subdivision]] of $K_5$ or $K_{3,3}$, then $\Gamma$ is not [[Definition:Planar Graph|planar]].
+Taking the [[Definition:Contrapositive Statement|contrapositive]]:
+:$\Gamma$ is a [[Definition:Planar Graph|planar graph]] $\implies \Gamma$ does not contain a [[Definition:Graph Subdivision|subdivision]] of either $K_5$ or $K_{3, 3}$.
+=== $(2)$ implies $(1)$ ===
+{{ProofWanted}}
+{{namedfor|Kazimierz Kuratowski|cat = Kuratowski}}
+[[Category:Graph Theory]]
+gedzpex6s6czckd0wgmg0gqzs6smt6d
+\end{proof}<|endoftext|>
+\section{Fundamental Theorem of Finite Abelian Groups}
+Tags: Fundamental Theorems, Abelian Groups, Proofs by Induction
+
+\begin{theorem}
+Every [[Definition:Finite Group|finite]] [[Definition:Abelian Group|abelian group]] is an [[Definition:Internal Group Direct Product|internal group direct product]] of [[Definition:Cyclic Group|cyclic groups]] whose [[Definition:Order of Group|orders]] are [[Definition:Prime Power|prime powers]].
+The number of terms in the [[Definition:Internal Group Direct Product|product]] and the [[Definition:Order of Group|orders]] of the [[Definition:Cyclic Group|cyclic groups]] are [[Definition:Unique|uniquely determined]] by the group.
+\end{theorem}
+
+\begin{proof}
+Let $G$ be a [[Definition:Finite Group|finite]] [[Definition:Abelian Group|abelian group]].
+By means of [[Abelian Group is Product of Prime-power Order Groups]], we factor it uniquely into groups of [[Definition:Prime Power|prime-power]] [[Definition:Order of Group|order]].
+Then, [[Abelian Group of Prime-power Order is Product of Cyclic Groups]] applies to each of these factors.
+Hence we conclude $G$ factors into [[Definition:Prime Power|prime-power]] [[Definition:Order of Group|order]] [[Definition:Cyclic Group|cyclic groups]].
+The factorisation of $G$ into [[Definition:Prime Power|prime-power]] [[Definition:Order of Group|order]] factors is already [[Definition:Unique|unique]].
+Therefore, a demonstration of the uniqueness of the secondary factorisation suffices:
+Suppose $\order G = p^k$ with $p$ a [[Definition:Prime Number|prime]].
+Let $G$ have the following two factorisations:
+:$G = H_1 \times H_2 \times \cdots \times H_m = K_1 \times K_2 \times \dotsb \times K_n$
+where the $H_i$'s and $K_i$'s are [[Definition:Non-Trivial Group|non-trivial]] [[Definition:Cyclic Group|cyclic]] [[Definition:Subgroup|subgroups]] with
+:$\order {H_1} \ge \order {H_2} \ge \dotsb \ge \order {H_m}|$
+:$\order {K_1} \ge \order {K_2} \ge \dotsb \ge \order {K_n}$
+We proceed through [[Second Principle of Mathematical Induction|complete induction]] on $k$.
+=== Basis for the induction ===
+For $k = 1$, the statement follows from [[Prime Group is Cyclic]].
+=== Induction Hypothesis ===
+Now we assume the theorem is true for all [[Definition:Abelian Group|abelian groups]] of [[Definition:Order of Structure|order]] $p^l$, where $l < k$.
+=== Induction Step ===
+By [[Power of Elements is Subgroup]], $G^p = \set {x^p: x \in G}$ is a [[Definition:Proper Subgroup|proper subgroup]] of $G$.
+It follows that:
+:$G^p = H_1^p \times \cdots \times H_{m'}^p = K_1^p \times \cdots \times K_{n'}^p$
+where:
+:$m'$ is the largest integer $i$ such that $\order {H_i} > p$
+:$n'$ is the largest integer $j$ such that $\order {K_j} > p$
+This $m'$ and $n'$ ensure that the [[Definition:Internal Group Direct Product|direct products]] above do not have trivial factors.
+Also, by [[Cauchy's Group Theorem]], we have:
+:$\order {G^p} < \order G$
+This means that we can apply the induction hypothesis.
+It follows that:
+:$m' = n'$
+and:
+:$\order {H_i^p} = \order {K_i^p}$
+for $i = 1, 2, \ldots, m'$.
+We know that $\order {H_i} = p \order {H_i^p}$, and the same for $K_i$ and $K_i^p$.
+It follows that $\order {H_i} = \order {K_i}$ for all $i = 1, 2, \ldots, m'$.  
+Now by construction of $m'$:
+:$i > m' \implies \order {H_i} = p = \order {K_i}$
+It follows that we have:
+:$\order {H_1} \order {H_2} \dotsm \order {H_{m'} } p^{m - m'} = \order G = \order {K_1} \order {K_2} \dotsm \order {K_{n'} } p^{n - n'}$
+Therefore:
+:$m - m' = n - n'$
+and as $m' = n'$, it must follow that $m = n$.
+The result follows by [[Second Principle of Mathematical Induction|induction]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Finite Direct Product of Modules is Module}
+Tags: Module Theory, Direct Products, Finite Direct Product of Modules is Module
+
+\begin{theorem}
+Let $\struct {R, +_R, \times_R}$ be a [[Definition:Ring (Abstract Algebra)|ring]].
+Let $\struct {G_1, +_1, \circ_1}_R, \struct {G_2, +_2, \circ_2}_R, \ldots, \struct {G_n, +_n, \circ_n}_R$ be [[Definition:Module|$R$-modules]].
+Let:
+:$\ds G = \prod_{k \mathop = 1}^n G_k$
+be their [[Definition:Module Direct Product|direct product]].
+Then $G$ is a [[Definition:Module|module]].
+\end{theorem}
+
+\begin{proof}
+This is a special case of [[Direct Product of Modules is Module]].
+\end{proof}
+
+\begin{proof}
+We need to show that:
+$\forall x, y, \in G, \forall \lambda, \mu \in R$:
+:$(1): \quad \lambda \circ \paren {x + y} = \paren {\lambda \circ x} + \paren {\lambda \circ y}$
+:$(2): \quad \paren {\lambda +_R \mu} \circ x = \paren {\lambda \circ x} + \paren {\mu \circ x}$
+:$(3): \quad \paren {\lambda \times_R \mu} \circ x = \lambda \circ \paren {\mu \circ x}$
+Checking the criteria in order:
+=== Criterion 1 ===
+:$(1): \quad \lambda \circ \paren {x + y} = \paren {\lambda \circ x} + \paren {\lambda \circ y}$
+Let $x = \tuple {x_1, x_2, \ldots, x_n}, y = \tuple {y_1, y_2, \ldots, y_n} \in G$.
+{{begin-eqn}}
+{{eqn | l = \lambda \circ \paren {x + y}
+      | r = \lambda \circ \paren {\tuple {x_1, x_2, \ldots, x_n} + \tuple {y_1, y_2, \ldots, y_n} }
+      | c = 
+}}
+{{eqn | r = \lambda \circ \tuple {x_1 +_1 y_1, x_2 +_2 y_2, \ldots, x_n +_n y_n}
+      | c = 
+}}
+{{eqn | r = \tuple {\lambda \circ_1 \paren {x_1 + y_1}, \lambda \circ_2 \paren {x_2 + y_2}, \ldots, \lambda \circ_n \paren {x_n + y_n} }
+      | c = 
+}}
+{{eqn | r = \tuple {\lambda \circ_1 x_1, \lambda \circ_2 x_2, \ldots, \lambda \circ_n x_n} + \tuple {\lambda \circ_1 y_1, \lambda \circ_2 y_2, \ldots, \lambda \circ_n y_n}
+      | c = 
+}}
+{{eqn | r = \paren {\lambda \circ \tuple {x_1, x_2, \ldots, x_n} } + \paren {\lambda \circ \tuple {y_1, y_2, \ldots, y_n} }
+      | c = 
+}}
+{{eqn | r = \paren {\lambda \circ x} + \paren {\lambda \circ y}
+      | c = 
+}}
+{{end-eqn}}
+So $(1)$ holds.
+{{qed|lemma}}
+=== Criterion 2 ===
+:$(2): \quad \paren {\lambda +_R \mu} \circ x = \paren {\lambda \circ x} + \paren {\mu \circ x}$
+Let $x = \tuple {x_1, x_2, \ldots, x_n} \in G$.
+{{begin-eqn}}
+{{eqn | l = \paren {\lambda +_R \mu} \circ x
+      | r = \paren {\lambda +_R \mu} \circ \tuple {x_1, x_2, \ldots, x_n}
+      | c = 
+}}
+{{eqn | r = \tuple {\paren {\lambda +_R \mu} \circ_1 x_1, \paren {\lambda +_R \mu} \circ_2 x_2, \ldots, \paren {\lambda +_R \mu} \circ_n x_n}
+      | c = 
+}}
+{{eqn | r = \tuple {\paren {\lambda \circ_1 x_1} +_1 \paren {\mu \circ_1 x_1}, \paren {\lambda \circ_2 x_2} +_2 \paren {\mu \circ_2 x_2}, \ldots, \paren {\lambda \circ_n x_n} +_n \paren {\mu \circ_n x_n} }
+      | c = 
+}}
+{{eqn | r = \tuple {\lambda \circ_1 x_1, \lambda \circ_2 x_2, \lambda \circ_n x_n} + \tuple {\mu \circ_1 x_1, \mu \circ_2 x_2, \ldots, \mu \circ_n x_n}
+      | c = 
+}}
+{{eqn | r = \paren {\lambda \circ_1 \tuple {x_1, x_2, \ldots, x_n} } + \paren {\mu \circ_1 \tuple {x_1, x_2, \ldots, x_n} }
+      | c = 
+}}
+{{eqn | r = \paren {\lambda \circ x} + \paren {\mu \circ x}
+      | c = 
+}}
+{{end-eqn}}
+So $(2)$ holds.
+{{qed|lemma}}
+=== Criterion 3 ===
+:$(3): \quad \paren {\lambda \times_R \mu} \circ x = \lambda \circ \paren {\mu \circ x}$
+Let $x = \tuple {x_1, x_2, \ldots, x_n} \in G$.
+{{begin-eqn}}
+{{eqn | l = \paren {\lambda \times_R \mu} \circ x
+      | r = \paren {\lambda \times_R \mu} \circ \tuple {x_1, x_2, \ldots, x_n}
+      | c = 
+}}
+{{eqn | r = \tuple {\paren {\lambda \times_R \mu} \circ_1 x_1, \paren {\lambda \times_R \mu} \circ_2 x_2, \ldots, \paren {\lambda \times_R \mu} \circ_n x_n}
+      | c = 
+}}
+{{eqn | r = \tuple {\lambda \circ_1 \paren {\mu \circ_1 x_1}, \lambda \circ_2 \paren {\mu \circ_2 x_2}, \ldots, \lambda \circ_n \paren {\mu \circ_n x_n} }
+      | c = 
+}}
+{{eqn | r = \lambda \circ \tuple {\mu \circ_1 x_1, \mu \circ_2 x_2, \ldots, \mu \circ_n x_n}
+      | c = 
+}}
+{{eqn | r = \lambda \circ \paren {\mu \circ \tuple {x_1, x_2, \ldots, x_n} }
+      | c = 
+}}
+{{eqn | r = \lambda \circ \paren {\mu \circ x}
+      | c = 
+}}
+{{end-eqn}}
+So $(3)$ holds.
+{{qed|lemma}}
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Subring Module}
+Tags: Subrings, Module Theory
+
+\begin{theorem}
+Let $\struct {R, +, \times}$ be a [[Definition:Ring (Abstract Algebra)|ring]].
+Let $\struct {S, +_S, \times_S}$ be a [[Definition:Subring|subring]] of $R$.
+Let $\struct {G, +_G, \circ}_R$ be an [[Definition:Module|$R$-module]].
+Let $\circ_S$ be the [[Definition:Restriction of Operation|restriction]] of $\circ$ to $S \times G$.
+Then $\struct {G, +_G, \circ_S}_S$ is an [[Definition:Module|$S$-module]].
+The module $\struct {G, +_G, \circ_S}_S$ is called the '''$S$-module obtained from $\struct {G, +_G, \circ}_R$ by restricting scalar multiplication'''.
+{{refactor|Extract the below into its own page}}
+If $\struct {G, +_G, \circ}_R$ is a [[Definition:Unitary Module|unitary $R$-module]] and $1_R \in S$, then $\struct{G, +_G, \circ_S}_S$ is also [[Definition:Unitary Module|unitary]].
+\end{theorem}
+
+\begin{proof}
+We have that:
+:$\forall a, b \in S: a +_S b = a + b$
+:$\forall a, b \in S: a \times_S b = a \times b$
+:$\forall a \in S: \forall x \in G = a \circ_S x = a \circ x$
+as $+_S$, $\times_S$ and $\circ_S$ are [[Definition:Restriction of Operation|restrictions]].
+Let us verify the [[Definition:Module Axioms|module axioms]].
+=== {{Module-axiom|1}} ===
+We need to show that:
+:$\forall a \in S: \forall x, y \in G: a \circ_S \paren {x +_G y} = a \circ_S x +_G a \circ_S y$
+We have:
+{{begin-eqn}}
+{{eqn | l = a \circ_S \paren {x +_G y}
+      | r = a \circ \paren {x +_G y}
+}}
+{{eqn | r = a \circ x +_G a \circ y
+      | c = {{Module-axiom|1}} on $\struct {G, +_G, \circ}_R$
+}}
+{{eqn | r = a \circ_S x +_G a \circ_S y
+}}
+{{end-eqn}}
+{{qed|lemma}}
+=== {{Module-axiom|2}} ===
+We need to show that:
+:$\forall a,b \in S: \forall x \in G: \paren {a +_S b} \circ_S x = a \circ_S x +_G b \circ_S y$
+We have:
+{{begin-eqn}}
+{{eqn | l = \paren {a +_S b} \circ_S x
+      | r = \paren {a + b} \circ x
+}}
+{{eqn | r = a \circ x + b \circ x
+      | c = {{Module-axiom|2}} on $\struct {G, +_G, \circ}_R$
+}}
+{{eqn | r = a \circ_S x +_G b \circ_S x
+}}
+{{end-eqn}}
+{{qed|lemma}}
+=== {{Module-axiom|3}} ===
+We need to show that:
+:$\forall a, b \in S: \forall x \in G: \paren {a \times_S b} \circ_S x = a \circ_S \paren {b \circ_S x}$
+We have:
+{{begin-eqn}}
+{{eqn | l = \paren {a \times_S b} \circ_S x
+      | r = \paren {a \times b} \circ x
+}}
+{{eqn | r = a \circ \paren {b \circ x}
+      | c = {{Module-axiom|3}} on $\struct {G, +_G, \circ}_R$
+}}
+{{eqn | r = a \circ_S \paren {b \circ_S x}
+}}
+{{end-eqn}}
+{{qed|lemma}}
+Thus $\struct {G, +_G, \circ_S}_S$ is an [[Definition:Module|$S$-module]].
+{{qed|lemma}}
+It remains to prove the final statement:
+If $\struct {G, +_G, \circ}_R$ is a [[Definition:Unitary Module|unitary $R$-module]] and $1_R \in S$, then $\struct {G, +_G, \circ_S}_S$ is also [[Definition:Unitary Module|unitary]].
+To show that $\struct {G, +_G, \circ_S}_S$ is [[Definition:Unitary Module|unitary]], we must prove that:
+:$\forall x \in G: 1_R \circ_S x = x$
+Since $1_R \in S$ by assumption, the [[Product Element|product]] $1_R \circ_S x$ is defined. 
+We now have:
+{{begin-eqn}}
+{{eqn | l = 1_R \circ_S x
+      | r = 1_R \circ x
+}}
+{{eqn | r = x
+      | c = [[Definition:Unitary Module Axioms|Unitary Module Axiom $(4)$]] on $\struct {G, +_G, \circ}_R$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Z-Module Associated with Abelian Group is Unitary Z-Module}
+Tags: Examples of Unitary Modules, Z-Module Associated with Abelian Group
+
+\begin{theorem}
+Let $\struct {G, *}$ be an [[Definition:Abelian Group|abelian group]] with [[Definition:Identity Element|identity]] $e$.
+Let $\struct {G, *, \circ}_\Z$ be the [[Definition:Z-Module Associated with Abelian Group|$Z$-module associated with $G$]].
+Then $\struct {G, *, \circ}_\Z$ is a [[Definition:Unitary Module|unitary $Z$-module]].
+\end{theorem}
+
+\begin{proof}
+The notation $*^n x$ can be written as $x^n$.
+Let us verify that $\struct {G, *, \circ}_\Z$ is a [[Definition:Unitary Module|unitary $\Z$-module]] by verifying the axioms in turn.
+=== Axiom $(\text M 1)$ ===
+We need to show that $n \circ \paren {x * y} = \paren {n \circ x} * \paren {n \circ y}$.
+From the definition, $n \circ x = x^n$ and so $n \circ \paren {x * y} = \paren {x * y}^n$
+From [[Power of Product in Abelian Group]]:
+:$\paren {x * y}^n = x^n * y^n = \paren {n \circ x} * \paren {n \circ y}$
+{{qed|lemma}}
+=== Axiom $(\text M 2)$ ===
+We need to show that $\paren {n + m} \circ x = \paren {n \circ x} * \paren {m \circ x}$.
+That is, that $x^{n + m} = x^n * x^m$.
+This is an instance of [[Powers of Group Elements/Sum of Indices|Powers of Group Elements: Sum of Indices]].
+{{qed|lemma}}
+=== Axiom $(\text M 3)$ ===
+We need to show that $\paren {n \times m} \circ x = n \circ \paren {m \circ x}$.
+That is, that $x^{n m} = \paren {x^m}^n$.
+This follows directly from [[Powers of Group Elements/Product of Indices|Powers of Group Elements: Product of Indices]].
+{{qed|lemma}}
+=== Axiom $(\text M 4)$ ===
+We need to show that $\forall x \in G: 1 \circ x = x$.
+That is, that $x^1 = x$.
+This follows from the definition of [[Definition:Power of Group Element|Power of Group Element]].
+{{qed|lemma}}
+Having verified all four axioms, we have shown that $\struct {G, *, \circ}_\Z$ is a [[Definition:Unitary Module|unitary $\Z$-module]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Projection on Cartesian Product of Modules}
+Tags: Module Theory, Cartesian Product, Epimorphisms, Projections
+
+\begin{theorem}
+Let $G$ be the [[Definition:Cartesian Product|cartesian product]] of a [[Definition:Sequence|sequence]] $\sequence {G_n}$ of [[Definition:Module|$R$-modules]].
+Then for each $j \in \closedint 1 n$, the [[Definition:Projection (Mapping Theory)|projection]] $\pr_j$ on the $j$th co-ordinate is an [[Definition:R-Algebraic Structure Epimorphism|epimorphism]] from $G$ onto $G_j$.
+\end{theorem}<|endoftext|>
+\section{Canonical Injection into Cartesian Product of Modules}
+Tags: Module Theory, Cartesian Product, Monomorphisms
+
+\begin{theorem}
+Let $G$ be the [[Definition:Cartesian Product|cartesian product]] of a [[Definition:Sequence|sequence]] $\sequence {G_n}$ of [[Definition:Module|$R$-modules]].
+Then for each $j \in \closedint 1 n$, the [[Definition:Canonical Injection (Abstract Algebra)|canonical injection]] $\inj_j$ from $G_j$ into $G$ is a [[Definition:R-Algebraic Structure Monomorphism|monomorphism]].
+\end{theorem}
+
+\begin{proof}
+$G$ can be seen as functions:
+:$\displaystyle f: A \to \bigcup_{a \mathop \in A} G_a$
+{{explain|Clarify the above sentence.}}
+Let $a \in A$.
+Let $x, y \in G_a$.
+Let $r \in R$.
+So both $x + y \in G_a$ and $r x \in G_a$.
+Let $b \in A$.
+=== Case 1 ===
+{{explain|Explain and clarify the notation}}
+Let $b = a$.
+Then:
+:$\map {\map {\inj_a} {x + y} } b = x + y = \map {\map {\inj_a} x} b + \map {\map {\inj_a} y} b$
+and:
+:$\map {\map {\inj_a} {r x} } b = r x = r \, \map {\map {\inj_a} x} b$
+{{qed|lemma}}
+=== Case 2 ===
+Let $b \ne a$.
+Then:
+:$\map {\map {\inj_a} {x + y} } b = 0 + 0 = \map {\map {\inj_a} x} b + \map {\map {\inj_a} y} b$
+and:
+:$\map {\map {\inj_a} {r x} } b = 0 = \map r 0 = r \, \map {\map {\inj_a} x} b$
+Therefore:
+:$\map {\inj_a} {x + y} = \map {\inj_a} x + \map {\inj_a} y$
+and:
+:$\map {\inj_a} {r x} = r \, \map {\inj_a} x$
+{{qed|lemma}}
+So $\inj_a$ is a [[Definition:R-Algebraic Structure Homomorphism|homomorphism]].
+Combined with [[Canonical Injection is Injection]] gives that $\inj_a$ is a [[Definition:R-Algebraic Structure Monomorphism|monomorphism]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Submodule Test}
+Tags: Submodules
+
+\begin{theorem}
+Let $\left({G, +, \circ}\right)_R$ be a [[Definition:Unitary Module|unitary $R$-module]].
+Let $H$ be a [[Definition:Empty Set|non-empty]] [[Definition:Subset|subset]] of $G$.
+Then $\left({H, +, \circ}\right)_R$ is a [[Definition:Submodule|submodule]] of $G$ iff:
+:$\forall x, y \in H: \forall \lambda \in R: x + y \in H, \lambda \circ x \in H$
+\end{theorem}
+
+\begin{proof}
+If the conditions are fulfilled, then:
+:$x \in H \implies -x = \left({- 1_R}\right) \circ x \in H$
+Thus $H$ is a [[Definition:Subgroup|subgroup]] of $\left({G, +}\right)$ by the [[Two-Step Subgroup Test]], and hence a [[Definition:Submodule|submodule]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Module is Submodule of Itself}
+Tags: Submodules
+
+\begin{theorem}
+Let $\left({G, +_G, \circ}\right)_R$ be an [[Definition:Module|$R$-module]].
+Then $\left({G, +_G, \circ}\right)_R$ is a [[Definition:Submodule|submodule]] of itself.
+\end{theorem}
+
+\begin{proof}
+Follows directly from the fact that a [[Group is Subgroup of Itself|group is a subgroup of itself]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Null Module Submodule of All}
+Tags: Submodules, Module Theory
+
+\begin{theorem}
+Let $\left({G, +_G, \circ}\right)_R$ be an [[Definition:Module|$R$-module]].
+Then the [[Definition:Null Module|null module]]:
+:$\left({\left\{{e_G}\right\}, +_G, \circ}\right)_R$
+is a [[Definition:Submodule|submodule]] of $\left({G, +_G, \circ}\right)_R$.
+\end{theorem}
+
+\begin{proof}
+Follows directly from the fact that the [[Trivial Subgroup is Subgroup|trivial subgroup is a subgroup of $\left({G, +_G}\right)$]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{First Sylow Theorem}
+Tags: Sylow Theorems, First Sylow Theorem
+
+\begin{theorem}
+Let $p$ be a [[Definition:Prime Number|prime number]].
+Let $G$ be a [[Definition:Group|group]] such that:
+:$\order G = k p^n$
+where:
+:$\order G$ denotes the [[Definition:Order of Structure|order]] of $G$
+:$p$ is not a [[Definition:Divisor of Integer|divisor]] of $k$.
+Then $G$ has at least one [[Definition:Sylow p-Subgroup|Sylow $p$-subgroup]].
+\end{theorem}
+
+\begin{proof}
+Let $\order G = k p^n$ such that $p \nmid k$.
+Let $\mathbb S = \set {S \subseteq G: \order S = p^n}$, that is, the set of all of [[Definition:Subset|subsets]] of $G$ which have exactly $p^n$ [[Definition:Element|elements]].
+Let $N = \order {\mathbb S}$.
+Now $N$ is the number of ways $p^n$ [[Definition:Element|elements]] can be chosen from a set containing $p^n k$ [[Definition:Element|elements]].
+ 
+From [[Cardinality of Set of Subsets]], this is given by:
+:$N = \dbinom {p^n k} {p^n} = \dfrac {\paren {p^n k} \paren {p^n k - 1} \cdots \paren {p^n k - i} \cdots \paren {p^n k - p^n + 1} } {\paren {p^n} \paren {p^n - 1} \cdots \paren {p^n - i} \cdots \paren 1}$
+From [[Binomial Coefficient involving Power of Prime]]:
+:$\dbinom {p^n k} {p^n} \equiv k \pmod p$
+Thus:
+:$N \equiv k \pmod p$
+Now let $G$ [[Definition:Group Action|act]] on $\mathbb S$ by the rule:
+:$\forall S \in \mathbb S: g * S = g S = \set {x \in G: x = g s: s \in S}$
+That is, $g * S$ is the [[Definition:Left Coset|left coset]] of $S$ by $g$.
+From [[Group Action on Sets with k Elements]], this is a [[Definition:Group Action|group action]].
+Now, let $\mathbb S$ have $r$ [[Definition:Orbit (Group Theory)|orbits]] under this [[Definition:Group Action|action]].
+From [[Set of Orbits forms Partition]], the [[Definition:Orbit (Group Theory)|orbits]] [[Definition:Set Partition|partition]] $\mathbb S$.
+Let these [[Definition:Orbit (Group Theory)|orbits]] be represented by $\set {S_1, S_2, \ldots, S_r}$, so that:
+{{begin-eqn}}
+{{eqn | l = \mathbb S
+      | r = \Orb {S_1} \cup \Orb {S_2} \cup \ldots \cup \Orb {S_r}
+      | c = 
+}}
+{{eqn | l = \size {\mathbb S}
+      | r = \size {\Orb {S_1} } + \card {\Orb {S_2} } + \ldots + \size {\Orb {S_r} }
+      | c = 
+}}
+{{end-eqn}}
+If each [[Definition:Orbit (Group Theory)|orbit]] had [[Definition:Length of Orbit|length]] [[Definition:Divisor of Integer|divisible]] by $p$, then $p \divides N$.
+But this can not be the case, because, as we have seen:
+:$N \equiv k \pmod p$
+So at least one [[Definition:Orbit (Group Theory)|orbit]] has [[Definition:Length of Orbit|length]] which is ''not'' divisible by $p$.
+Let $S \in \set {S_1, S_2, \ldots, S_r}$ be such that $\size {\Orb S)} = m: p \nmid m$.
+Let $s \in S$.
+It follows from [[Group Action on Prime Power Order Subset]] that:
+:$\Stab S s = S$
+and so:
+:$\size {\Stab S} = \size S = p^n$
+From [[Stabilizer is Subgroup]]:
+:$\Stab S \le G$
+Thus $\Stab S$ is the [[Definition:Subgroup|subgroup]] of $G$ with $p^n$ [[Definition:Element|elements]] of which we wanted to prove the existence.
+{{Qed}}
+\end{proof}
+
+\begin{proof}
+Let $\Bbb S = \set {S \subseteq G: \order S = p^n}$, that is, the set of all of [[Definition:Subset|subsets]] of $G$ which have exactly $p^n$ [[Definition:Element|elements]].
+Now let $G$ [[Definition:Group Action|act]] on $\Bbb S$ by the rule:
+:$\forall S \in \Bbb S: g * S = g S = \set {x \in G: x = g s: s \in S}$
+From [[Set of Orbits forms Partition]], the [[Definition:Orbit (Group Theory)|orbits]] [[Definition:Set Partition|partition]] $\mathbb S$.
+Let these [[Definition:Orbit (Group Theory)|orbits]] be represented by $\set {S_1, S_2, \ldots, S_n}$, so that:
+:$(1): \quad \size {\mathbb S} = \size {\Orb {S_1} } + \card {\Orb {S_2} } + \ldots + \size {\Orb {S_r} }$
+Thus each $\Orb {S_i}$ is the [[Definition:Orbit (Group Theory)|orbit]] under $*$ of some $S_i$ whose [[Definition:Order of Group|order]] is $p^n$.
+By the [[Orbit-Stabilizer Theorem]]:
+:$(2): \quad \order {\Orb {S_i} } = \dfrac {\order G} {\order {\Stab {S_i} } }$ for all $i \in \set {1, 2, \ldots, n}$
+where $\Stab {S_i}$ is the [[Definition:Stabilizer|stabilizer]] of $S_i$ under $*$.
+Let $s \in S_i$ and $x \in \Stab {S_i}$.
+Then $s x \in S_i$ because $S_i x = S_i$.
+Let $s$ be fixed while $x$ runs over $\Stab {S_i}$.
+Then:
+:$\forall s \in S_i: s \Stab {S_i} \subseteq S_i$
+Therefore $S_i$ is the [[Definition:Set Union|union]] of all the [[Definition:Left Coset|left cosets]] $s \Stab {S_i}$.
+These [[Definition:Left Coset|left cosets]] may not all be [[Definition:Distinct Elements|distinct]].
+Let $r_i$ be the number of [[Definition:Distinct Elements|distinct]] [[Definition:Left Coset|left cosets]] of $s \Stab {S_i}$.
+Then:
+:$\card {S_i} = r_i \card {\Stab {S_i} }$
+because:
+: [[Definition:Distinct Elements|distinct]] [[Definition:Left Coset|left cosets]] are [[Definition:Disjoint Sets|disjoint]]
+: all have the same [[Definition:Cardinality|cardinality]] $\card {\Stab {S_i} }$.
+Thus:
+:$(3): \quad \card {\Stab {S_i} } \divides \card {S_i} = p^n$
+Because the only [[Definition:Divisor of Integer|divisors]] of $p^n$ are smaller [[Definition:Integer Power|powers]] of $p$:
+:$\card {\Stab {S_i} } = p^{n_i}$
+for some $n_i \le n$, for $i = 1, 2, \ldots, r$.
+We can write:
+:$g = \order G = k p^n$
+where $p$ does not [[Definition:Divisor of Integer|divide]] $k$.
+From $(2)$ and $(3)$:
+:$\card {S_i} = k p^{d_i}$
+where $d_i = n - n_i$ for $i = 1, 2, \ldots, r$.
+From [[Cardinality of Set of Subsets]]:
+:$\card {\Bbb S} = \dbinom {k p^n} {p_n}$
+We have from [[Binomial Coefficient involving Power of Prime]] that:
+:$\dbinom {k p^n} {p_n} \equiv k \pmod p$
+But $p$ does not [[Definition:Divisor of Integer|divide]] $k$.
+Hence $p$ does not [[Definition:Divisor of Integer|divide]] $\card {\Bbb S}$.
+Hence by $(1)$ and $(4)$:
+:$p \nmid k \paren {p^{d_1} + p^{d_2} + \cdots p^{d_r} }$
+If $d_i > 0$ for all $i = 1, 2, \ldots, r$, then $p$ would [[Definition:Divisor of Integer|divide]] this [[Definition:Integer|integer]].
+So there must be some $i$ such that $d_i = 0$.
+Then, for this $i$:
+:$n_i = n$
+That is:
+:$\card {\Stab {S_i} } = p^n$
+Thus $\Stab {S_i}$ is the [[Definition:Subgroup|subgroup]] of $G$ of [[Definition:Order of Group|order]] $p_n$ that was asserted.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{H-Cobordism Theorem}
+Tags: Topology
+
+\begin{theorem}
+Let $X^n, Y^n$ be two [[Definition:Simply Connected|simply connected]] manifolds.
+Let $n \in \N: n \ge 5$ and $\exists W$ such that $W$ is an [[Definition:Cobordism|h-cobordism]] between $X$ and $Y$.
+Then $\exists \psi: W \to X \times \closedint 0 1$ such that $\psi$ is a [[Definition:Diffeomorphism|diffeomorphism]].
+In particular, $X$ and $Y$ are [[Definition:Diffeomorphism|diffeomorphic]].
+\end{theorem}
+
+\begin{proof}
+{{ProofWanted|In progress}}
+[[Category:Topology]]
+2guptkt0e7yhebs6rugk4vfd5rl5e1l
+\end{proof}<|endoftext|>
+\section{Group has Subgroups of All Prime Power Factors}
+Tags: P-Groups, Subgroups
+
+\begin{theorem}
+Let $p$ be a [[Definition:Prime Number|prime]].
+Let $G$ be a [[Definition:Finite Group|finite group]] of [[Definition:Order of Structure|order]] $n$.
+If $p^k \divides n$ then $G$ has at least one [[Definition:Subgroup|subgroup]] of [[Definition:Order of Structure|order]] $p^k$.
+\end{theorem}
+
+\begin{proof}
+From [[Composition Series of Group of Prime Power Order]], a [[Definition:P-Group|$p$-group]] has [[Definition:Subgroup|subgroups]] corresponding to every [[Definition:Divisor of Integer|divisor]] of its [[Definition:Order of Structure|order]].
+Thus, taken with the [[First Sylow Theorem]], a [[Definition:Finite Group|finite group]] has a [[Definition:Subgroup|subgroup]] corresponding to every [[Definition:Prime Power|prime power]] [[Definition:Divisor of Integer|divisor]] of its [[Definition:Order of Structure|order]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Normalizer of Sylow p-Subgroup}
+Tags: P-Groups, Subgroups, Normalizers, Sylow p-Subgroups
+
+\begin{theorem}
+Let $P$ be a [[Definition:Sylow p-Subgroup|Sylow $p$-subgroup]] of a [[Definition:Finite Group|finite group]] $G$.
+Let $\map {N_G} P$ be the [[Definition:Normalizer|normalizer]] of $P$.
+Then any [[Definition:P-Subgroup|$p$-subgroup]] of $\map {N_G} P$ is contained in $P$.
+In particular, $P$ is the unique [[Definition:Sylow p-Subgroup|Sylow $p$-subgroup]] of $\map {N_G} P$.
+\end{theorem}
+
+\begin{proof}
+Let $Q$ be a [[Definition:P-Subgroup|$p$-subgroup]] of $N = \map {N_G} P$.
+Let $\order Q = p^m, \order P = p^n$.
+By [[Normalizer of Subgroup is Largest Subgroup containing that Subgroup as Normal Subgroup]]:
+:$P \lhd \map {N_G} P$
+thus by [[Subset Product with Normal Subgroup as Generator]]:
+:$\gen {P, Q} = P Q$
+Thus by [[Order of Subgroup Product]]:
+:$P Q \le G: \order {P Q} = p^{n + m - s}$
+where $\order {P \cap Q} = p^s$.
+Since $n$ is the highest power of $p$ dividing $\order G$, this is possible only when $m \le s$.
+Since $P \cap Q \le Q, s \le m$ thus we conclude that $m = s$ and therefore $P \cap Q = Q$.
+Thus from [[Intersection with Subset is Subset]]:
+:$Q \subseteq P$
+In particular, if $Q$ is a [[Definition:Sylow p-Subgroup|Sylow $p$-subgroup]] of $\map {N_G} P$, then $Q = P$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Second Sylow Theorem}
+Tags: Sylow Theorems
+
+\begin{theorem}
+Let $P$ be a [[Definition:Sylow p-Subgroup|Sylow $p$-subgroup]] of the [[Definition:Finite Group|finite group]] $G$.
+Let $Q$ be any [[Definition:P-Subgroup|$p$-subgroup]] of $G$.
+Then $Q$ is a [[Definition:Subset|subset]] of a [[Definition:Conjugate of Group Subset|conjugate]] of $P$.
+\end{theorem}
+
+\begin{proof}
+Let $P$ be a [[Definition:Sylow p-Subgroup|Sylow $p$-subgroup]] of $G$.
+Let $\mathbb S$ be the [[Definition:Set|set]] of all [[Definition:Distinct|distinct]] [[Definition:Conjugate of Group Subset|$G$-conjugates]] of $P$:
+:$\mathbb S = \set {g P g^{-1}: g \in G}$
+Let $h * S$ be the [[Definition:Conjugacy Action|conjugacy action]]:
+:$\forall h \in P, S \in \mathbb S: h * S = h S h^{-1}$
+From [[Conjugacy Action on Subgroups is Group Action]], this is a [[Definition:Group Action|group action]] for $S \le G$.
+To show it is [[Definition:Closed Algebraic Structure|closed]] for $S \in \mathbb S$:
+{{begin-eqn}}
+{{eqn | l = S
+      | o = \in
+      | r = \mathbb S
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \exists g \in G: S
+      | r = g P g^{-1}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = h * S
+      | r = h \paren {g P g^{-1} } h^{-1}
+      | c = 
+}}
+{{eqn | r = \paren {h g} P \paren {h g}^{-1}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = h * S
+      | o = \in
+      | r = \mathbb S
+      | c = 
+}}
+{{end-eqn}}
+So, consider the [[Definition:Orbit (Group Theory)|orbits]] and [[Definition:Stabilizer|stabilizers]] of $\mathbb S$ under this [[Definition:Group Action|group action]].
+Since $\forall S \in \mathbb S: \Stab S \le P$, we have that:
+:$\size {\Stab S} \divides \order P$
+Therefore, by the [[Orbit-Stabilizer Theorem]], these [[Definition:Length of Orbit|orbit lengths]] are all [[Definition:Congruence Modulo Integer|congruent]] to either $0$ or $1$ modulo $p$, since $P$ is a [[Definition:Sylow p-Subgroup|Sylow $p$-subgroup]] of $G$.
+Note that this will imply, as we shall mark later on:
+:$\size {\mathbb S} \equiv 1 \pmod p$
+Now, $h * P = h P h^{-1} = P$, so:
+:$\Orb P = \set P$
+We now show that $P$ is the only [[Definition:Element|element]] of $\mathbb S$ such that $\size {\Orb S} = 1$.
+If $g P g^{-1}$ has one [[Definition:Element|element]] in its [[Definition:Orbit (Group Theory)|orbit]], then:
+:$\forall x \in P: x \paren {g P g^{-1} } x^{-1} = g P g^{-1}$
+Thus $\forall x \in P$ we have that:
+:$g^{-1} x g \in \map {N_G} P$
+From [[Order of Conjugate Element equals Order of Element]], we have that:
+:$\order {g^{-1} x g} = \order x$
+Thus $P_1 = g^{-1} P g$ is a [[Definition:P-Subgroup|$p$-subgroup]] of $\map {N_G} P$.
+As $P$ and $P_1$ have the same number of [[Definition:Element|element]]s, $P_1$ is a [[Definition:Sylow p-Subgroup|Sylow $p$-subgroup]] of $\map {N_G} P$.
+Hence $P_1 = P$ by [[Normalizer of Sylow p-Subgroup]], so $g P g^{-1} = P$.
+Thus $P$ is the only [[Definition:Element|element]] of $\mathbb S$ whose [[Definition:Orbit (Group Theory)|orbit]] has [[Definition:Length of Orbit|length]] $1$.
+From [[Stabilizer of Coset Action on Set of Subgroups]], $P = \map {N_G} P$.
+Thus, for any $g \notin P$, $\size {\Orb {g P g^{-1} } }$ under conjugation by [[Definition:Element|element]]s of $P$ has [[Definition:Orbit (Group Theory)|orbit]] greater than $1$.
+Hence:
+: $\size {\mathbb S} \equiv 1 \pmod p$
+as promised.
+Next we consider [[Definition:Orbit (Group Theory)|orbit]]s of $\mathbb S$ under conjugation by [[Definition:Element|element]]s of $Q$.
+Since every [[Definition:Orbit (Group Theory)|orbit]] has [[Definition:Length of Orbit|length]] a power of $p$, the above conclusion shows there is at least one [[Definition:Orbit (Group Theory)|orbit]] of [[Definition:Length of Orbit|length]] $1$.
+So there is an [[Definition:Element|element]] $g$ such that:
+:$\forall x \in Q: x \paren {g P g^{-1} } x^{-1} = g P g^{-1}$
+As previously:
+:$g^{-1} Q g \subseteq \map {N_G} P$
+So by [[Normalizer of Sylow p-Subgroup]]:
+:$g^{-1} Q g \subseteq P$
+Thus $Q \subseteq g P g^{-1}$ as required.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Third Sylow Theorem}
+Tags: Sylow Theorems, Third Sylow Theorem
+
+\begin{theorem}
+All the [[Definition:Sylow p-Subgroup|Sylow $p$-subgroups]] of a [[Definition:Finite Group|finite group]] are [[Definition:Conjugate of Group Subset|conjugate]].
+\end{theorem}
+
+\begin{proof}
+Suppose $P$ and $Q$ are [[Definition:Sylow p-Subgroup|Sylow $p$-subgroups]] of $G$.
+By the [[Second Sylow Theorem]], $Q$ is a [[Definition:Subset|subset]] of a [[Definition:Conjugate of Group Subset|conjugate]] of $P$.
+But since $\order P = \order Q$, it follows that $Q$ must equal a [[Definition:Conjugate of Group Subset|conjugate]] of $P$.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let $G$ be a [[Definition:Finite Group|finite group]] of [[Definition:Order of Group|order]] $p^n m$, where $p \nmid m$ and $n > 0$.
+Let $H$ be a [[Definition:Sylow p-Subgroup|Sylow $p$-subgroup]] of $G$.
+We have that:
+:$\order H = p^n$
+:$\index G H = m$
+Let $S_1, S_2, \ldots, S_m$ denote the [[Definition:Left Coset|left cosets]] of $G \pmod H$.
+We have that $G$ [[Definition:Group Action|acts on]] $G / H$ by the rule:
+:$g * S_i = g S_i$.
+Let $H_i$ denote the [[Definition:Stabilizer|stabilizer]] of $S_i$.
+By the [[Orbit-Stabilizer Theorem]]:
+:$\order {H_i} = p^n$
+while:
+:$S_i = g H \implies g H g^{-1} \subseteq H_i$
+Because $\order {g H g^{-1} } = \order H = \order {H_i}$, we have:
+:$g H g^{-1} \subseteq H_i$
+Let $H'$ be a second [[Definition:Sylow p-Subgroup|Sylow $p$-subgroup]] of $G$.
+Then $H'$ acts on $G / H$ by the same rule as $G$.
+Since $p \nmid m$, there exists at least one [[Definition:Orbit (Group Theory)|orbit]] under $H'$ whose [[Definition:Cardinality|cardinality]] is not [[Definition:Divisor of Integer|divisible]] by $p$.
+Suppose that $S_1, S_2, \ldots, S_r$ are the [[Definition:Element|elements]] of an [[Definition:Orbit (Group Theory)|orbit]] where $p \nmid r$.
+Let $K = H' \cap H_1$.
+Then $K$ is the [[Definition:Stabilizer|stabilizer]] of $S_1$ under the action of $H'$.
+Therefore:
+:$\index {H'} K = r$
+However:
+:$\order {H'} = p^n$
+and:
+:$p \nmid r$
+from which it follows that:
+:$r = 1$
+and:
+:$K = H'$
+Therefore:
+:$\order K = \order {H'} = \order {H_1} = p^n$
+and:
+:$H' = K = H_1$
+Thus $H'$ and $H$ are [[Definition:Conjugate of Group Subset|conjugates]].
+\end{proof}<|endoftext|>
+\section{Fourth Sylow Theorem}
+Tags: Sylow Theorems, Fourth Sylow Theorem
+
+\begin{theorem}
+The number of [[Definition:Sylow p-Subgroup|Sylow $p$-subgroups]] of a [[Definition:Finite Group|finite group]] is [[Definition:Congruence (Number Theory)|congruent]] to $1 \pmod p$.
+\end{theorem}
+
+\begin{proof}
+Let $G$ be a [[Definition:Finite Group|finite group]] of [[Definition:Order of Group|order]] $p^n m$, where $p \nmid m$ and $n > 0$.
+Let $r$ be the number of [[Definition:Sylow p-Subgroup|Sylow $p$-subgroups]] of $G$.
+Let $H$ be a [[Definition:Sylow p-Subgroup|Sylow $p$-subgroup]] of $G$.
+We have that:
+:$\order H = p^n$
+:$\index G H = m$
+Let $S_1, S_2, \ldots, S_m$ denote the [[Definition:Element|elements]] of the [[Definition:Left Coset Space|left coset space]] of $G / H$.
+We have that $H$ [[Definition:Group Action|acts on]] $G / H$ by the rule:
+:$g * S_i = g S_i$
+for $S_i \in G / H$.
+Unless $H = G$ and $r = 1$, there is more than $1$ [[Definition:Orbit (Group Theory)|orbit]].
+{{explain|Why? See [[Orbits of Group Action on Sets with Power of Prime Size]]}}
+We have that $H$ is the [[Definition:Stabilizer|stabilizer]] of the [[Definition:Left Coset|coset]] $H$, which must be one of $S_1, S_2, \ldots, S_m$.
+Let $S_1, S_2, \ldots, S_k$ be the [[Definition:Element|elements]] of $G / H$ whose [[Definition:Stabilizer|stabilizer]] is $H$.
+From the [[Orbit-Stabilizer Theorem]] and from $\order H = p^n$ we see there are $2$ cases:
+:$(1): \quad$ The [[Definition:Orbit (Group Theory)|orbit]] of $S_i$ contains $p^t$ [[Definition:Element|elements]] where $0 < t < n$
+:$(2): \quad$ The [[Definition:Orbit (Group Theory)|orbit]] of $S_i$ contains only the [[Definition:Element|element]] $S_i$.
+$(2)$ occurs {{iff}} $S_i$ is one of the [[Definition:Left Coset|cosets]] $S_1, S_2, \ldots, S_k$ whose [[Definition:Stabilizer|stabilizer]] is $H$.
+So counting the [[Definition:Element|elements]] of $G / H$, we see that:
+:$m = k + u p$
+or:
+:$m \equiv k \pmod p$
+From the [[Fifth Sylow Theorem]], we have:
+:$m \equiv k r \pmod p$
+and so:
+:$k r \equiv k \pmod p$
+from which it follows:
+:$r \equiv 1 \pmod p$
+because $k \not \equiv 0 \pmod p$.
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sylow p-Subgroup is Unique iff Normal}
+Tags: Normal Subgroups, Sylow p-Subgroups
+
+\begin{theorem}
+A [[Definition:Group|group]] $G$ has exactly one [[Definition:Sylow p-Subgroup|Sylow $p$-subgroup]] $P$ {{iff}} $P$ is [[Definition:Normal Subgroup|normal]].
+\end{theorem}
+
+\begin{proof}
+If $G$ has precisely one [[Definition:Sylow p-Subgroup|Sylow $p$-subgroup]], it must be [[Definition:Normal Subgroup|normal]] from [[Unique Subgroup of a Given Order is Normal]].
+Suppose a [[Definition:Sylow p-Subgroup|Sylow $p$-subgroup]] $P$ is [[Definition:Normal Subgroup|normal]].
+Then it equals its [[Definition:Conjugate of Group Subset|conjugates]].
+Thus, by the [[Third Sylow Theorem]], there can be only one such [[Definition:Sylow p-Subgroup|Sylow $p$-subgroup]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Intersection of Normal Subgroup with Sylow P-Subgroup}
+Tags: Normal Subgroups, Set Intersection, Sylow p-Subgroups
+
+\begin{theorem}
+Let $P$ be a [[Definition:Sylow p-Subgroup|Sylow $p$-subgroup]] of a [[Definition:Finite Group|finite group]] $G$.
+Let $N$ be a [[Definition:Normal Subgroup|normal subgroup]] of $G$.
+Then $P \cap N$ is a [[Definition:Sylow p-Subgroup|Sylow $p$-subgroup]] of $N$.
+\end{theorem}
+
+\begin{proof}
+Since $N \lhd G$, we see that:
+:$\gen {P, N} = P N$
+from [[Subset Product with Normal Subgroup as Generator]].
+Since $P \cap N \le P$, it follows that:
+:$\order {P \cap N} = p^k$
+where $k > 0$.
+By [[Order of Subgroup Product]]:
+:$\order {P N} \order {P \cap N} = \order P \order N$
+Hence from [[Lagrange's Theorem (Group Theory)|Lagrange's Theorem]]:
+:$\index N {P \cap N} = \index {P N} P$
+By [[Tower Law for Subgroups]]:
+:$\index G P = \index G {P N} \index {P N} P$
+Thus:
+:$\index {P N} P \divides \index G P$
+where $\divides$ denotes [[Definition:Divisor of Integer|divisibility]].
+Thus:
+:$p \nmid \index {P N} P$
+so:
+:$p \nmid \index N {P \cap N}$
+Thus $P \cap N$ is a [[Definition:Sylow p-Subgroup|Sylow $p$-subgroup]] of $N$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Quotient of Sylow P-Subgroup}
+Tags: Quotient Groups, Sylow p-Subgroups
+
+\begin{theorem}
+Let $P$ be a [[Definition:Sylow p-Subgroup|Sylow $p$-subgroup]] of a [[Definition:Finite Group|finite group]] $G$.
+Let $N$ be a [[Definition:Normal Subgroup|normal subgroup]] of $G$.
+Then $P N / N$ is a [[Definition:Sylow p-Subgroup|Sylow $p$-subgroup]] of $G / N$.
+\end{theorem}
+
+\begin{proof}
+We have that $P \le G$ and $ N \lhd G$.
+So by the [[Second Isomorphism Theorem for Groups]]:
+:$P N / N \cong P / \paren {P \cap N}$
+We have that:
+:$P N / N = \set {p N : p \in P}$
+and so every [[Definition:Element|element]] of $P N / N$ has [[Definition:Order of Group Element|order]] a [[Definition:Power of Group Element|power]] of $p$.
+Hence $P N / N$ is a [[Definition:P-Subgroup|$p$-subgroup]] of $G / N$.
+From [[Intersection of Normal Subgroup with Sylow P-Subgroup]], we have that:
+:$p \nmid \index G {P N}$
+So $P N / N$ is a [[Definition:Sylow p-Subgroup|Sylow $p$-subgroup]] of $G / N$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Fifth Sylow Theorem}
+Tags: Sylow Theorems, Fifth Sylow Theorem
+
+\begin{theorem}
+The number of [[Definition:Sylow p-Subgroup|Sylow $p$-subgroups]] of a [[Definition:Finite Group|finite group]] is a [[Definition:Divisor of Integer|divisor]] of their common [[Definition:Index of Subgroup|index]].
+\end{theorem}
+
+\begin{proof}
+Let $G$ be a [[Definition:Finite Group|finite group]] of [[Definition:Order of Group|order]] $p^n m$, where $p \nmid m$ and $n > 0$.
+Let $r$ be the number of [[Definition:Sylow p-Subgroup|Sylow $p$-subgroups]] of $G$.
+It is to be shown that $r \divides m$.
+Let $H$ be a [[Definition:Sylow p-Subgroup|Sylow $p$-subgroup]] of $G$.
+We have that:
+:$\order H = p^n$
+:$\index G H = m$
+Let $S_1, S_2, \ldots, S_m$ denote the [[Definition:Element|elements]] of the [[Definition:Left Coset Space|left coset space]] of $G / H$.
+We have that $G$ [[Definition:Group Action|acts on]] $G / H$ by the rule:
+:$g * S_i = g S_i$
+for $S_i \in G / H$.
+There is only one [[Definition:Orbit (Group Theory)|orbit]] under this action, namely the whole of $G / H$.
+{{explain|Why? See [[Orbits of Group Action on Sets with Power of Prime Size]]}}
+Therefore the [[Definition:Stabilizer|stabilizer]] of each $S_i$ is a [[Definition:Subgroup|subgroup]] of $G$ of [[Definition:Index of Subgroup|index]] $m$ and [[Definition:Order of Group|order]] $p^n$.
+In other words, each $S_i$ has a [[Definition:Sylow p-Subgroup|Sylow $p$-subgroup]] as a [[Definition:Stabilizer|stabilizer]].
+Now it is shown that each [[Definition:Sylow p-Subgroup|Sylow $p$-subgroup]] is the [[Definition:Stabilizer|stabilizer]] of one or more of the [[Definition:Left Coset|cosets]] $S_1, S_2, \ldots, S_m$.
+We have that $H$ is the [[Definition:Stabilizer|stabilizer]] of the [[Definition:Left Coset|coset]] $H$, which must be one of $S_1, S_2, \ldots, S_m$.
+Let $S_1, S_2, \ldots, S_k$ be the [[Definition:Element|elements]] of $G / H$ whose [[Definition:Stabilizer|stabilizer]] is $H$.
+By the [[Third Sylow Theorem]], any other [[Definition:Sylow p-Subgroup|Sylow $p$-subgroup]] of $H$ is a [[Definition:Conjugate of Group Subset|conjugates]] $g H g^{-1}$ of $H$.
+Thus it is seen that $g H g^{-1}$ is a [[Definition:Stabilizer|stabilizer]] of the [[Definition:Left Coset|cosets]] $g S_1, g S_2, \ldots, g S_k$.
+So each of the $r$ distinct [[Definition:Sylow p-Subgroup|Sylow $p$-subgroups]] of $G$ is the [[Definition:Stabilizer|stabilizer]] of exactly $k$ [[Definition:Element|elements]] of $G / H$.
+Thus:
+:$m = k r$
+and so:
+:$r \divides m$
+as required.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Finite Submodule of Function Space}
+Tags: Submodules, Function Spaces
+
+\begin{theorem}
+Let $\left({G, +}\right)$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
+Let $R$ be a [[Definition:Ring (Abstract Algebra)|ring]].
+Let $\left({G, +, \circ}\right)_R$ be an [[Definition:Module|$R$-module]].
+Let $S$ be a set.
+Let $G^S$ the set of all [[Definition:Mapping|mappings]] $f: S \to G$.
+Let $G^{\left({S}\right)}$ be the set of all [[Definition:Mapping|mappings]] $f: S \to G$ such that $f \left({x}\right) = e$ for all but [[Definition:Finite|finitely]] many elements $x$ of $S$.
+Then:
+: $\left({G^{\left({S}\right)}, +', \circ}\right)_R$ is a [[Definition:Submodule|submodule]] of $\left({G^S, +, \circ}\right)_R$
+where $+'$ is the [[Definition:Induced Structure|operation induced]] on $G^{\left({S}\right)}$ by $+$.
+\end{theorem}
+
+\begin{proof}
+Let $\left({G, +, \circ}\right)_R$ be an [[Definition:Module|$R$-module]] and $S$ be a [[Definition:Set|set]].
+We need to show that $\left({G^{\left({S}\right)}, +'}\right)$ is a [[Definition:Group|group]].
+Let $f, g \in G^{\left({S}\right)}$.
+Let:
+: $F = \left\{{x \in S: f \left({x}\right) \ne e}\right\}$
+: $G = \left\{{x \in S: g \left({x}\right) \ne e}\right\}$
+From the definition of $f$ and $g$, both $F$ and $G$ are [[Definition:Finite|finite]].
+Since $e + e = e$ by definition of [[Definition:Identity Element|identity element]], it follows that if:
+: $f \left({x}\right) + g \left({x}\right) \ne e$
+then necessarily $f \left({x}\right) \ne e$ or $g \left({x}\right) \ne e$.
+That is:
+: $\left({f +' g}\right) \left({x}\right) = f \left({x}\right) + g \left({x}\right) \ne e \implies x \in F \cup G$
+But as $F$ and $G$ are both [[Definition:Finite|finite]], it follows that $F \cup G$ is also [[Definition:Finite|finite]].
+Hence $f +' g \in G^{\left({S}\right)}$ and $\left({G^{\left({S}\right)}, +'}\right)$ is [[Definition:Closed Algebraic Structure|closed]].
+Now let $f \in G^{\left({S}\right)}$.
+Let $f^*$ be the [[Induced Structure Inverse]] of $f$.
+Thus $f^* \left({x}\right) = - \left({f \left({x}\right)}\right)$.
+Again, let $F = \left\{{x \in S: f \left({x}\right) \ne e}\right\}$.
+It follows directly that $x \in S \setminus F \implies f^* \left({x}\right) = e$.
+Hence $f^* \left({x}\right) \ne e \implies x \in F$ and hence $f^* \in G^{\left({S}\right)}$.
+So by the [[Two-Step Subgroup Test]], it follows that $\left({G^{\left({S}\right)}, +'}\right)$ is a [[Definition:Subgroup|subgroup]] of $\left({G^S, +}\right)$.
+Hence $\left({G^{\left({S}\right)}, +', \circ}\right)_R$ is an [[Definition:Module|$R$-module]].
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Polynomial Functions form Submodule of All Functions}
+Tags: Module Theory, Polynomial Theory
+
+\begin{theorem}
+Let $K$ be a [[Definition:Commutative and Unitary Ring|commutative ring with unity]].
+Let $K^K$ be the [[Definition:Module of All Mappings|$K$-module mappings]] $f: K \to K$.
+Let $P \left({K}\right) \subseteq K^K$ be the [[Definition:Set|set]] of all [[Definition:Polynomial Function (Abstract Algebra)|polynomial functions]] on $K$.
+Then $P \left({K}\right)$ is a $K$-[[Definition:Submodule|submodule]] of $K^K$.
+\end{theorem}<|endoftext|>
+\section{Intersection and Sum of Submodules}
+Tags: Submodules
+
+\begin{theorem}
+Let $\left({G, +, \circ}\right)_R$ be an [[Definition:Module|$R$-module]].
+Let $H$ and $K$ be [[Definition:Submodule|submodules]] of $G$.
+Then $H + K$ and $H \cap K$ are also [[Definition:Submodule|submodules]] of $G$.
+The [[Definition:Set Intersection|intersection]] of any [[Definition:Set|set]] of submodules of $G$ is a [[Definition:Submodule|submodule]].
+Thus if $S \subseteq G$, the [[Definition:Set Intersection|intersection]] of all submodules of $G$ containing $S$ is the smallest submodule of $G$ containing $S$.
+\end{theorem}<|endoftext|>
+\section{Linear Transformation of Submodule}
+Tags: Linear Transformations
+
+\begin{theorem}
+Let $G$ and $H$ be [[Definition:Module|$R$-modules]].
+Let $\phi: G \to H$ be a [[Definition:Linear Transformation|linear transformation]].
+Then:
+:$(1): \quad$ If $M$ is a [[Definition:Submodule|submodule]] of $G$, $\phi \sqbrk M$ is a [[Definition:Submodule|submodule]] of $H$
+:$(2): \quad$ If $N$ is a [[Definition:Submodule|submodule]] of $H$, $\phi^{-1} \sqbrk N$ is a [[Definition:Submodule|submodule]] of $G$
+:$(3): \quad$ The [[Definition:Codomain of Mapping|codomain]] of $\phi$ is a [[Definition:Submodule|submodule]] of $H$
+:$(4): \quad$ The [[Definition:Kernel of Linear Transformation|kernel]] of $\phi$ is a [[Definition:Submodule|submodule]] of $G$.
+\end{theorem}
+
+\begin{proof}
+Since a [[Definition:Linear Transformation|linear transformation]] $\phi: G \to H$ is, in particular, a [[Definition:Group Homomorphism|homomorphism]] from the [[Definition:Group|group]] $G$ to the [[Definition:Group|group]] $H$, it follows that:
+:$(1): \quad$ By [[Homomorphism with Cancellable Codomain Preserves Identity]], $\map \phi {e_G} = e_H$
+:$(2): \quad$ By [[Homomorphism with Identity Preserves Inverses]], $\map \phi {-x} = -\map \phi x$.
+From [[Epimorphism preserves Modules]] and definition of [[Definition:Surjection|surjection]], it follows that as $M$ is a [[Definition:Submodule|submodule]] of $G$, then $\phi \sqbrk M$ is a [[Definition:Submodule|submodule]] of $H$.
+The result follows ...
+{{stub}}
+\end{proof}<|endoftext|>
+\section{Elements of Module with Equal Images under Linear Transformations form Submodule}
+Tags: Linear Transformations
+
+\begin{theorem}
+Let $G$ and $H$ be [[Definition:Module|$R$-modules]].
+Let $\phi$ and $\psi$ be [[Definition:Linear Transformation|linear transformations]] from $G$ into $H$.
+Then the set $S = \set {x \in G: \map \phi x = \map \psi x}$ is a [[Definition:Submodule|submodule]] of $G$.
+\end{theorem}
+
+\begin{proof}
+Let $x, y \in S$.
+Let $\lambda \in R$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \map \phi {x + y}
+      | r = \map \phi x + \map \phi y
+      | c = {{Defof|Linear Transformation}}
+}}
+{{eqn | r = \map \psi x + \map \psi y
+      | c = $x, y \in S$
+}}
+{{eqn | r = \map \psi {x + y}
+      | c = {{Defof|Linear Transformation}}
+}}
+{{eqn | l = \map \phi {\lambda \circ x}
+      | r = \lambda \circ \map \phi x
+      | c = {{Defof|Linear Transformation}}
+}}
+{{eqn | r = \lambda \circ \map \psi x
+      | c = $x \in S$
+}}
+{{eqn | r = \map \psi {\lambda \circ x}
+      | c = {{Defof|Linear Transformation}}
+}}
+{{end-eqn}}
+Hence $x + y, \lambda \circ x \in S$.
+By [[Submodule Test]], $S$ is a [[Definition:Submodule|submodule]] of $G$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Poincaré Conjecture}
+Tags: Topology, Millennium Problems
+
+\begin{theorem}
+Let $\Sigma^m$ be a smooth $m$-manifold.
+Let $\Sigma^m$ satisfy:
+:$H_0 \struct {\Sigma; \Z} = 0$
+and:
+:$H_m \struct {\Sigma; \Z} = \Z$
+{{explain|Definition of the notation $H_0 \struct {\Sigma; \Z}$, nature of $H_0$ and $H_m$}}
+Then $\Sigma^m$ is homeomorphic to the $m$-sphere $\Bbb S^m$.
+\end{theorem}
+
+\begin{proof}
+The proof proceeds on several dimensional-cases. Note that the case $m = 3$ is incredibly intricate, and that a full proof would be impractical to produce here.  An outline of the $m = 3$ case will be given instead.
+=== Dimension $m = 1$ ===
+Follows from the [[Classification of Compact One-Manifolds]].
+{{qed|lemma}}
+=== Dimension $m = 2$ ===
+Follows from the [[Classification of Compact Two-Manifolds]].
+{{qed|lemma}}
+=== Dimension $m = 3$ ===
+Follows from [[Thurston's Geometrization Conjecture]], proved by {{AuthorRef|Grigori Perelman}}.
+{{ProofWanted}}
+{{qed|lemma}}
+=== Dimension $m = 4$ ===
+Follows from $4$-dimensional [[Topological h-Cobordism Theorem]] of {{AuthorRef|Andrew John Casson}} and {{AuthorRef|Michael Hartley Freedman}}.
+{{ProofWanted}}
+{{qed|lemma}}
+=== Dimension $m = 5$ ===
+{{ProofWanted}}
+Summary:
+Any $\Sigma^5$ bounds a contractible $6$-manifold $Z$.
+Let $\Bbb D^6$ be a $6$-disk (AKA 6-ball).
+{{explain|Open or closed disk / ball?}}
+Then $Z - \Bbb D^6$ is an $h$-cobordism between $\Sigma$ and $\partial \Bbb D^6 = \Bbb S^5$.
+Hence $\Sigma$ is differomorphic to $\Bbb S^5$ by the [[h-Cobordism Theorem]].
+{{qed|lemma}}
+=== Dimension $m \ge 6$ ===
+{{ProofWanted}}
+We can cut two small $m$-disks $D', D''$ from $\Sigma$.
+The remaining manifold, $\Sigma \setminus \paren {D' \cup D''}$ is an h-cobordism between $\partial D'$ and $\partial D''$.
+These are just two copies of $\Bbb S^{m-1}$.
+By the $h$-cobordism theorem, there exists a diffeomorphism:
+:$\phi: \Sigma \setminus \paren {D' \cup D''} \to \Bbb S^{m - 1} \times \closedint 0 1$
+which can be chosen to restrict to the identity on one of the $\Bbb S^{m - 1}$.
+Let $\Xi$ denote this $\Bbb S^{m - 1}$ such that $\phi$ restricts to the identity.
+Since $\psi \vert_\Xi = Id$, we can extend $\psi$ across $D''$, the interior of $\Xi$ to obtain a diffeomorphism $\phi': \Sigma \setminus D'' \to \Bbb S^{m - 1} \cup D'$.
+{{explain|What does $\psi \vert_\Xi$ mean?}}
+Let $\Bbb D^m$ denote this latter manifold, which is merely an $m$-disk.
+Our diffeomorphism $\phi': \Sigma \setminus D'' \to \Bbb D^m$ induces a diffeomorphism on the boundary spheres $\Bbb S^{m - 1}$.
+Any diffeomorphism of the boundary sphere $\Bbb S^{m - 1}$ can be extended radially to the whole disk:
+:$\map {\operatorname {int} } {\Bbb S^{m - 1} } = D''$
+but only as a homeomorphism of D''.  
+Hence the extended function $\phi'': \Sigma \to \Bbb S^m$ is a homeomorphism.
+{{qed}}
+{{Namedfor|Jules Henri Poincaré|cat = Poincaré}}
+\end{proof}<|endoftext|>
+\section{Linear Transformation of Generated Module}
+Tags: Linear Transformations
+
+\begin{theorem}
+Let $G$ and $H$ be [[Definition:Module|$R$-modules]].
+Let $\phi$ and $\psi$ be [[Definition:Linear Transformation|linear transformations]] $G$ into $H$. 
+Let $S$ be a [[Definition:Generator of Module|generator]] for $G$.
+Suppose that $\forall x \in S: \map \phi x = \map \psi x$.
+Then $\phi = \psi$.
+\end{theorem}<|endoftext|>
+\section{Generated Submodule is Linear Combinations}
+Tags: Unitary Modules, Linear Algebra
+
+\begin{theorem}
+Let $G$ be a [[Definition:Unitary Module|unitary $R$-module]].
+Let $S \subseteq G$.
+Then the [[Definition:Generator of Module|submodule $H$ generated by $S$]] is the [[Definition:Set|set]] of all [[Definition:Linear Combination|linear combinations]] of $S$.
+\end{theorem}
+
+\begin{proof}
+First the extreme case:
+The smallest submodule of $G$ containing $\O$ is $\set {e_G}$.
+[[Definition:Linear Combination of Empty Set|By definition]], $\set {e_G}$ is the set of all linear combinations of $\varnothing$.
+Now the general case:
+Let $\O \subset S \subseteq G$.
+Let $L$ be the set of all [[Definition:Linear Combination of Subset|linear combinations]] of $S$.
+Since $G$ is a [[Definition:Unitary Module|unitary $R$-module]], every element $x \in S$ is the linear combination $1_R x$, so $S \subseteq L$.
+But $L$ is closed for addition and scalar multiplication, so [[Submodule Test|is a submodule]].
+Thus $H \subseteq L$.
+But as every linear combination of $S$ clearly belongs to any submodule of $G$ which contains $S$, we also have $L \subseteq H$.
+\end{proof}<|endoftext|>
+\section{Empty Set is Linearly Independent}
+Tags: Empty Set, Linear Algebra
+
+\begin{theorem}
+The [[Definition:Empty Set|empty set]] is a [[Definition:Linearly Independent Set|linearly independent set]].
+\end{theorem}
+
+\begin{proof}
+There are no [[Definition:Sequence|sequences]] at all of $n$ terms of the [[Definition:Empty Set|empty set]] for any $n > 0$.
+Hence the result holds [[Definition:Vacuous Truth|vacuously]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Subset of Module Containing Identity is Linearly Dependent}
+Tags: Module Theory, Linear Algebra
+
+\begin{theorem}
+Let $G$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
+Let $\struct {R, +, \circ}$ be a [[Definition:Ring (Abstract Algebra)|ring]] whose [[Definition:Ring Zero|zero]] is $0_R$.
+Let $\struct {G, +_G, \circ}_R$ be an [[Definition:Module|$R$-module]].
+Let $H \subseteq G$ such that $e \in H$.
+Then $H$ is a [[Definition:Linearly Dependent Set|linearly dependent set]].
+\end{theorem}
+
+\begin{proof}
+From [[Scalar Product with Identity]], $\forall \lambda: \lambda \circ e = e$.
+Let $H \subseteq G$ such that $e \in H$.
+Consider any [[Definition:Sequence|sequence]] $\sequence {a_k}_{1 \mathop \le k \mathop \le n}$ in $H$ which includes $e$.
+So, let $a_j = e$ for some $j \in \closedint 1 n$.
+Let $c \in R \ne 0_R$.
+Consider the [[Definition:Sequence|sequence]] $\sequence {\lambda_k}_{1 \mathop \le k \mathop \le n}$ of [[Definition:Element|elements]] of $R$ defined as:
+:$\lambda_k = \begin{cases}
+c & : k \ne j \\
+0_R & : k= j
+\end{cases}$
+Then:
+{{begin-eqn}}
+{{eqn | l = \sum_{k \mathop = 1}^n \lambda_k \circ a_k
+      | r = \lambda_1 \circ a_1 + \lambda_2 \circ a_2 + \cdots + \lambda_j \circ a_j + \cdots + \lambda_n \circ a_n
+      | c = 
+}}
+{{eqn | r = 0_R \circ a_1 + 0_R \circ a_2 + \cdots + c \circ e +  \cdots + 0_R \circ a_n
+      | c = 
+}}
+{{eqn | r = e + e + \cdots + e + \cdots + e
+      | c = 
+}}
+{{eqn | r = e
+      | c = 
+}}
+{{end-eqn}}
+Thus there exists a [[Definition:Sequence|sequence]] $\sequence {\lambda_k}_{1 \mathop \le k \mathop \le n}$ in which not all $\lambda_k = 0_R$ such that:
+:$\displaystyle \sum_{k \mathop = 1}^n \lambda_k \circ a_k = e$
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Subset of Linearly Independent Set is Linearly Independent}
+Tags: Unitary Modules, Linear Algebra
+
+\begin{theorem}
+A [[Definition:Subset|subset]] of a [[Definition:Linearly Independent Set|linearly independent set]] is also [[Definition:Linearly Independent Set|linearly independent]].
+\end{theorem}
+
+\begin{proof}
+Let $G$ be an [[Definition:Unitary Module|unitary $R$-module]].
+Then $\sequence {a_n}$ is a [[Definition:Linearly Independent Sequence|linearly independent sequence]] {{iff}} $\set {a_1, a_2, \ldots, a_n}$ is a [[Definition:Linearly Independent Set|linearly independent set]] of $G$.
+So suppose that $\set {a_1, a_2, \ldots, a_n}$ is a [[Definition:Linearly Independent Set|linearly independent set]] of $G$.
+Then clearly $\sequence {a_n}$ is a [[Definition:Linearly Independent Sequence|linearly independent sequence]] of $G$.
+Conversely, let $\sequence {a_n}$ be a [[Definition:Linearly Independent Sequence|linearly independent sequence]] of $G$.
+Let $\sequence {b_m}$ be a [[Definition:Sequence of Distinct Terms|sequence of distinct terms]] of $\set {a_1, a_2, \ldots, a_n}$.
+Let $\sequence {\mu_m}$ be a sequence of [[Definition:Scalar (Vector Space)|scalars]] such that $\displaystyle \sum_{j \mathop = 1}^m \mu_j b_j = 0$.
+For each $k \in \closedint 1 n$, let:
+:$\lambda_k = \begin{cases}
+\mu_j & : j \text { is the unique index such that } a_k = b_j \\
+0 & : a_k \notin \set {b_1, b_2, \ldots, b_m}
+\end{cases}$
+Then:
+:$\displaystyle 0 = \sum_{j \mathop = 1}^m \mu_j b_j = \sum_{k \mathop = 1}^n \lambda_k a_k$
+Thus:
+:$\forall k \in \closedint 1 n: \lambda_k = 0$
+As $\set {\mu_1, \ldots, \mu_m} \subseteq \set {\lambda_1, \ldots, \lambda_n}$, it follows that:
+:$\forall j \in \closedint 1 m: \mu_j = 0$
+and so $\sequence {b_m}$ has been shown to be a [[Definition:Linearly Independent Sequence|linearly independent sequence]].
+Hence the result.
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Superset of Linearly Dependent Set}
+Tags: Unitary Modules, Linear Algebra
+
+\begin{theorem}
+Any [[Definition:Set|set]] containing a [[Definition:Linearly Dependent Set|linearly dependent set]] is also [[Definition:Linearly Dependent Set|linearly dependent]].
+\end{theorem}
+
+\begin{proof}
+Suppose $\left\{{a_1, a_2, \ldots, a_n}\right\}$ is a [[Definition:Linearly Independent Set|linearly independent set]].
+Then by [[Subset of Linearly Independent Set]], any [[Definition:Subset|subset]] $\left\{{b_1, b_2, \ldots, b_m}\right\}$ of $\left\{{a_1, a_2, \ldots, a_n}\right\}$ must itself be [[Definition:Linearly Independent Set|linearly independent]].
+Thus if $\left\{{b_1, b_2, \ldots, b_m}\right\}$ is [[Definition:Linearly Dependent Set|linearly dependent]], then so must $\left\{{a_1, a_2, \ldots, a_n}\right\}$ be.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Number of Ordered Bases from a Basis}
+Tags: Module Theory
+
+\begin{theorem}
+Each [[Definition:Basis (Linear Algebra)|basis]] of $n$ elements determines $n!$ [[Definition:Ordered Basis|ordered bases]].
+\end{theorem}
+
+\begin{proof}
+Follows directly from [[Cardinality of Set of Bijections]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Classification of Compact Two-Manifolds}
+Tags: Algebraic Topology
+
+\begin{theorem}
+Any [[Definition:Smooth Manifold|smooth]], [[Definition:Compact Space|compact]], [[Definition:Path-Connected|path-connected]] [[Definition:Topological Manifold|manifold]] of [[Definition:Dimension (Topology)|dimension $2$]] is [[Definition:Diffeomorphism|diffeomorphic]] to [[Definition:Sphere (Topology)|the sphere $\mathbb S^2$]], a [[Definition:Connected Sum|connected sum]] of [[Definition:Torus (Topology)|tori $\mathbb T^2$]], or a [[Definition:Connected Sum|connected sum]] of [[Definition:Projective Space|projective spaces $\mathbb{RP}^2$]].
+Any such [[Definition:Topological Manifold|$2$-manifold]] with [[Definition:Boundary of Manifold|boundary]] is [[Definition:Diffeomorphism|diffeomorphic]] to [[Definition:Sphere (Topology)|the sphere $\mathbb S^2$]], a [[Definition:Connected Sum|connected sum]] of [[Definition:Torus (Topology)|tori $\mathbb T^2$]], or a [[Definition:Connected Sum|connected sum]] of [[Definition:Projective Space|projective spaces $\mathbb{RP}^2$]], with a number of [[Definition:Open Disk|open disks]] removed.
+The [[Definition:Euler Characteristic|Euler characteristic]], [[Definition:Orientability|orientability]], and number of [[Definition:Boundary Curve|boundary curves]] suffice to describe a [[Definition:Topological Surface|surface]].
+\end{theorem}
+
+\begin{proof}
+It is known that the [[Definition:Connected Sum|connected sum]] of $g$ [[Definition:Torus (Topology)|tori]], $\mathbb T_1^2 \# \mathbb T_2^2 \# \ldots \# \mathbb T_g^2$, which we denote $g \mathbb T^2$, is [[Definition:Orientable|orientable]] and has [[Definition:Euler Characteristic|Euler characteristic]] $2 - 2 g - b$, where $b$ is the number of [[Definition:Boundary Curve|boundary curves]].
+It is also known that the [[Definition:Connected Sum|connected sum]] of $p$ [[Definition:Projective Space|projective spaces]] $p \mathbb{RP}^2$ is [[Definition:Non-Orientable|non-orientable]] and has [[Definition:Euler Characteristic|Euler characteristic]] $2 - p - b$.
+Thus the [[Definition:Euler Characteristic|Euler characteristic]], number of [[Definition:Boundary Curve|boundary curves]] and [[Definition:Orientability|orientability]] distinguish any closed, [[Definition:Path-Connected Space|path-connected]] [[Definition:Topological Manifold|$2$-manifold]].  
+=== Lemma ===
+A [[Definition:Compact Space|compact]], [[Definition:Boundaryless Manifold|boundaryless]] [[Definition:Topological Manifold|$2$-manifold]] $S$ is [[Definition:Diffeomorphism|diffeomorphic]] to a [[Definition:Polyhedral Disk|polyhedral disk]] $P$ with [[Definition:Edge of Polyhedron|edges]] identified pairwise.
+That is, for any [[Definition:Closed Set (Topology)|closed]], [[Definition:Connected Topological Space|connected]] [[Definition:Topological Manifold|$2$-manifold]], $\exists$ a [[Definition:Polyhedral Disk|polyhedral disk]] $P$ and an [[Definition:Equivalence Relation|equivalence relation]] $\sim$ such that $S \cong P \setminus \sim$.
+=== Proof of Lemma ===
+{{ProofWanted}}
+{{qed|lemma}}
+With this lemma, the classification can be completed.
+Throughout the proof, the term [[Definition:Topological Surface|surface]] is used in its [[Definition:Topology (Mathematical Branch)|topological]] meaning of [[Definition:Smooth Manifold|smooth]] [[Definition:Topological Manifold|$2$-manifold]].
+Consider the [[Definition:Polygonal Representation|polygonal representation]] guaranteed to exist by the lemma above.
+Suppose there is only one pair of [[Definition:Edge of Polyhedron|edges]] on $P$.
+Then they are either identified in an [[Definition:Orientable|orientable]] or [[Definition:Non-Orientable|non-orientable]] manner.
+Thus they yield either $\mathbb S^2$ or $\mathbb{RP}^2$, respectively.
+Now suppose there is more than one pair of [[Definition:Edge of Polyhedron|edges]] in $P$ identified by $\sim$.
+If it can be shown that such a [[Definition:Topological Surface|surface]] can always be [[Definition:Decomposable Set|decomposed]] into the [[Definition:Connected Sum|connected sum]] of:
+: either a $\mathbb T^2$ a $\mathbb{RP}^2$
+and:
+: a [[Definition:Topological Surface|surface]] described by a [[Definition:Polyhedral Disk|polyhedral disk]] with fewer [[Definition:Edge of Polyhedron|edges]] than $P$
+the classification of [[Definition:Path-Connected|path-connected]] [[Definition:Topological Surface|surface]] will be complete.
+There are $5$ cases to be examined:
+=== Case 1 ===
+There are two adjacent [[Definition:Edge of Polyhedron|edges]] identified with opposite [[Definition:Orientation (Topology)|orientations]].
+Then the identification can be performed to obtain a new [[Definition:Polyhedral Disk|polyhedral disk]] with one less pair of [[Definition:Edge of Polyhedron|edges]].
+{{qed|lemma}}
+=== Case 2 ===
+There are two adjacent [[Definition:Edge of Polyhedron|edges]] identified with the same [[Definition:Orientation (Topology)|orientations]].
+Then there is a [[Definition:Curve (Topology)|curve]] in $P$ from the unshared points of these two [[Definition:Edge of Polyhedron|edges]] which, under $\sim$, becomes a [[Definition:Simple Closed Curve|simple closed curve]] in $S$.
+{{explain|Is "curve" in this context synonymous with [[Definition:Arc (Topology)]]?}}
+The [[Definition:Polyhedral Disk|triangular disk]] $\Delta$ created by this [[Definition:Curve (Topology)|curve]] and the two [[Definition:Edge of Polyhedron|edges]] is just two [[Definition:Edge of Polyhedron|edges]] identified with the same [[Definition:Orientation (Topology)|orientations]], with a [[Definition:Boundary of Manifold|boundary]].
+Hence:
+:$\Delta \cong \mathbb{RP}^2 - \mathbb D_\Delta^2$
+and therefore if we remove the two [[Definition:Edge of Polyhedron|edges]] in question from $P$, we construct a [[Definition:Polygonal Disk|polygonal disk]] $P - \Delta$ such that:
+:$\left({\left({P - \Delta}\right) \setminus \sim}\right) \# \mathbb{RP}^2 \cong S$
+{{Explain|"polygonal disk" or "polyhedral disk"? Terminology is confused here.}}
+{{qed|lemma}}
+=== Case 3 ===
+There are two non-adjacent [[Definition:Edge of Polyhedron|edges]] in $P$, identified with the same [[Definition:Orientation (Topology)|orientations]].
+There exists a [[Definition:Curve (Topology)|curve]] $C$ in $P$ from the endpoint of one of these [[Definition:Edge of Polyhedron|edges]] to the identified point in the other.
+A new [[Definition:Polygonal Disk|polygonal disk]], and a new [[Definition:Equivalence Relation|equivalence relation]], can be defined as follows:
+Let $P'$ be the identification of the [[Definition:Edge of Polyhedron|edges]] in question and the separation of the disk along $C$.
+Let $\sim'$ be all the equivalences of $\sim$ together with the new equivalence taking one of the copies of $C$ in $P'$ to the other, with appropriate [[Definition:Orientation (Topology)|orientation]].
+$P'$ and $\sim'$ are constructed so that $P' \setminus \sim' \cong S$.
+But now $P'$ satisfies [[Classification of Compact Two-Manifolds#Case 2|Case 2]] because of the [[Definition:Orientation (Topology)|orientations]] imposed on the copies of $C$.
+{{qed|lemma}}
+=== Case 4 ===
+There are two non-adjacent edges in $P$, identified with opposite [[Definition:Orientation (Topology)|orientations]], such that neither edge is between any other pair of identified edges on the perimeter of $P$.
+If we identify the two edges in question, a cylinder $Y$ is obtained.
+There is a curve $C$ in $Y$ such that if $Y$ is separated along $C$, two cylinders are obtained.
+If the separation is maintained, but the remaining identifications are performed, then two [[Definition:Topological Surface|surfaces]], each with [[Definition:Boundary of Manifold|boundary]] $C$, are obtained.
+So the original [[Definition:Topological Surface|surface]] $S$ was the [[Definition:Connected Sum|connected sum]] of two [[Definition:Topological Surface|surfaces]], each with fewer edges in their respective polyhedral disks than $S$ had. 
+{{qed|lemma}}
+=== Case 5 ===
+There are two non-adjacent edges identified with opposing [[Definition:Orientation (Topology)|orientations]], such that some other pair of identified edges are interlaced with the edges in question on the perimeter of $P$.
+By the preceding cases, it is possible to decompose this [[Definition:Topological Surface|surface]] through the removal of all edges with the same [[Definition:Orientation (Topology)|orientation]], and forming a [[Definition:Connected Sum|connected sum]] with a number of projective planes.
+Hence we regard $\sim$ as identifying any two edges with opposing [[Definition:Orientation (Topology)|orientations]] only.
+Let the pairs of identified edges that are interlaced be $a_1, a_2, b_1, b_2$, such that $a_1 \sim a_2$ and $b_1 \sim b_2$.
+By performing the identification on $a$, a cylinder is obtained.
+By further performing the identification on $b$, the remaining edges of $P$ form the [[Definition:Boundary of Manifold|boundary]] of a disk on a [[Definition:Torus (Topology)|torus]] $\mathbb T^2$.
+Hence the original [[Definition:Topological Surface|surface]] was the [[Definition:Connected Sum|connected sum]]:
+:$\mathbb T^2 \# \left({P - a_1 - a_2 - b_1 - b_2}\right) \setminus \sim$
+{{qed|lemma}}
+Any polyhedral disk with more than one pair of identified edges must satisfy at least one of the above $5$ cases.
+So it can be decomposed into the [[Definition:Connected Sum|connected sum]] of either a projective plane or a [[Definition:Torus (Topology)|torus]] with a [[Definition:Topological Surface|surface]] described by a polyhedral disk with fewer pairs of identified edges.
+Since the polyhedral disk with only one pair of identified edges is either a sphere or a projective plane, every [[Definition:Topological Surface|surface]] without [[Definition:Boundary of Manifold|boundary]] $S$ is either a sphere or the [[Definition:Connected Sum|connected sum]] of a collection of [[Definition:Torus (Topology)|tori]] and projective planes.
+Since:
+:$\mathbb T^2 \# \mathbb{RP}^2 = \mathbb{RP}^2 \# \mathbb{RP}^2 \# \mathbb{RP}^2$
+any [[Definition:Compact Space|compact]] [[Definition:Topological Surface|surface]] without [[Definition:Boundary of Manifold|boundary]] is [[Definition:Diffeomorphism|diffeomorphic]] to either $\mathbb S^2, g \mathbb T^2$, or $p \mathbb{RP}^2$.
+The case for [[Definition:Topological Surface|surfaces]] with [[Definition:Boundary of Manifold|boundary]] is obtained through the removal and insertion of $\mathbb D^2$s to the [[Definition:Topological Surface|surface]].
+{{qed}}
+[[Category:Algebraic Topology]]
+piqrmyefe4oz6jpyvzf2snkoitr6rys
+\end{proof}<|endoftext|>
+\section{Standard Ordered Basis is Basis}
+Tags: Module Theory, Linear Algebra
+
+\begin{theorem}
+Let $\struct {R, +, \circ}$ be a [[Definition:Ring with Unity|ring with unity]] whose [[Definition:Ring Zero|zero]] is $0_R$ and whose [[Definition:Unity of Ring|unity]] is $1_R$.
+Let $n$ be a [[Definition:Positive Integer|positive integer]].
+For each $j \in \closedint 1 n$, let $e_j$ be the [[Definition:Ordered Tuple|ordered $n$-tuple]] of [[Definition:Element|elements]] of $R$ whose $j$th entry is $1_R$ and all of whose other entries is $0_R$.
+Then $\sequence {e_n}$ is an [[Definition:Ordered Basis|ordered basis]] of the [[Definition:Module on Cartesian Product|$R$-module $R^n$]].
+This [[Definition:Ordered Basis|ordered basis]] is called the '''[[Definition:Standard Ordered Basis|standard ordered basis of $R^n$]]'''.
+The corresponding [[Definition:Set|set]] $\set {e_1, e_2, \ldots, e_n}$ is called the '''[[Definition:Standard Basis|standard basis of $R^n$]]'''.
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \sum_{k \mathop = 1}^n \lambda_k e_k
+      | r = \lambda_1 \tuple {1_R, 0_R, 0_R, \ldots, 0_R}
+      | c = 
+}}
+{{eqn | o = +
+      | r = \lambda_2 \tuple {0_R, 1_R, 0_R, \ldots, 0_R}
+      | c = 
+}}
+{{eqn | o = +
+      | r = \ldots
+      | c = 
+}}
+{{eqn | o = +
+      | r = \lambda_n \tuple {0_R, 0_R, 0_R, \ldots, 1_R}
+      | c = 
+}}
+{{eqn | r = \tuple {\lambda_1, \lambda_2, \lambda_3, \ldots, \lambda_n}
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Basis for Finite Submodule of Function Space}
+Tags: Submodules, Function Spaces, Linear Algebra
+
+\begin{theorem}
+Let $\struct {R, +, \circ}$ be a [[Definition:Ring with Unity|ring with unity]] whose [[Definition:Ring Zero|zero]] is $0_R$ and whose [[Definition:Unity of Ring|unity]] is $1_R$.
+Let $A$ be a [[Definition:Set|set]].
+For each $a \in A$, let $f_a: A \to R$ be defined as:
+:$\forall x \in A: \map {f_a} x = \begin{cases}
+1 & : x = a \\
+0 & : x \ne a
+\end{cases}$
+Then $B = \set {f_a: a \in A}$ is a [[Definition:Basis (Linear Algebra)|basis]] of the [[Finite Submodule of Function Space]] $R^{\paren A}$.
+\end{theorem}
+
+\begin{proof}
+Let $\sequence {a_k}_{1 \mathop \le k \mathop \le n}$ be a [[Definition:Sequence of Distinct Terms|sequence of distinct terms]] of $A$.
+Let $\sequence {\lambda_k}_{1 \mathop \le k \mathop \le n}$ a [[Definition:Sequence|sequence]] of [[Definition:Scalar (Module)|scalars]].
+Then: $\displaystyle \sum_{k \mathop = 1}^n \lambda_k f_{a_k}$ is the [[Definition:Mapping|mapping]] whose value at $a_k$ is $\lambda_k$ and whose value at any $x$ not in $\set {a_1, a_2, \ldots, a_n}$ is zero.
+Hence $B$ is a [[Definition:Generator of Module|generator]] of $R^{\left({A}\right)}$ which is [[Definition:Linearly Independent Set|linearly independent]].
+Thus, by definition, $B$ is a [[Definition:Basis (Linear Algebra)|basis]] of $R^{\paren A}$.
+If $A = \closedint 1 n$, then $B$ is the [[Definition:Standard Basis|standard basis]] of $R^n$.
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Unique Representation by Ordered Basis}
+Tags: Unitary Modules, Linear Algebra
+
+\begin{theorem}
+Let $G$ be a [[Definition:Unitary Module|unitary $R$-module]].
+Then $\sequence {a_n}$ is an [[Definition:Ordered Basis|ordered basis]] of $G$ {{Iff}}:
+:For every $x \in G$ there exists [[Definition:Exactly One|one and only one]] [[Definition:Sequence|sequence]] $\sequence {\lambda_n}$ of [[Definition:Scalar (Module)|scalars]] such that $\displaystyle x = \sum_{k \mathop = 1}^n \lambda_k a_k$.
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+Let $\sequence {a_n}$ be an [[Definition:Ordered Basis|ordered basis]] of $G$.
+Then every element of $G$ is a [[Definition:Linear Combination|linear combination]] of $\set {a_1, \ldots, a_n}$, which is a [[Definition:Generator of Module|generator]] of $G$, by [[Generated Submodule is Linear Combinations]].
+Thus there exists at least one such sequence of [[Definition:Scalar (Module)|scalar]].
+Now suppose there were two such sequences of [[Definition:Scalar (Module)|scalars]]: $\sequence {\lambda_n}$ and $\sequence {\mu_n}$.
+That is, suppose $\displaystyle \sum_{k \mathop = 1}^n \lambda_k a_k = \sum_{k \mathop = 1}^n \mu_k a_k$.
+Then:
+{{begin-eqn}}
+{{eqn | l =\ sum_{k \mathop = 1}^n \paren {\lambda_k - \mu_k} a_k
+      | r = \sum_{k \mathop = 1}^n \paren {\lambda_k a_k - \mu_k a_k}
+      | c = 
+}}
+{{eqn | r = \sum_{k \mathop = 1}^n \lambda_k a_k - \sum_{k \mathop = 1}^n \mu_k a_k
+      | c = 
+}}
+{{eqn | r = 0
+      | c = 
+}}
+{{end-eqn}}
+So $\lambda_k = \mu_k$ for all $k \in \closedint 1 n$ as $\sequence {a_n}$ is a [[Definition:Linearly Independent Sequence|linearly independent sequence]].
+{{qed|lemma}}
+=== Sufficient Condition ===
+Now suppose there is [[Definition:Exactly One|one and only one]] [[Definition:Sequence|sequence]] $\sequence {\lambda_n}$ such that the condition holds.
+It is clear that $\set {a_1, \ldots, a_n}$ [[Definition:Generator of Module|generates]] $G$.
+Suppose $\displaystyle \sum_{k \mathop = 1}^n \lambda_k a_k = 0$.
+Then, since also $\displaystyle \sum_{k \mathop = 1}^n 0 a_k = 0$, we have, by hypothesis:
+:$\forall k \in \closedint 1 n: \lambda_k = 0$
+Therefore $\sequence {a_n}$ is a [[Definition:Linearly Independent Sequence|linearly independent sequence]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Isomorphism from R^n via n-Term Sequence}
+Tags: Module Theory, Unitary Modules, Isomorphisms
+
+\begin{theorem}
+Let $G$ be a [[Definition:Unitary Module|unitary $R$-module]].
+Let $\sequence {a_k}_{1 \mathop \le k \mathop \le n}$ be an [[Definition:Ordered Basis|ordered basis]] of $G$.
+Let $R^n$ be the [[Definition:Module on Cartesian Product|$R$-module $R^n$]].
+Let $\psi: R^n \to G$ be defined as:
+:$\displaystyle \map \psi {\sequence {\lambda_k}_{1 \mathop \le k \mathop \le n} } = \sum_{k \mathop = 1}^n \lambda_k a_k$
+Then $\psi$ is an [[Definition:R-Algebraic Structure Isomorphism|isomorphism]].
+\end{theorem}
+
+\begin{proof}
+By [[Unique Representation by Ordered Basis]], $\psi$ is a [[Definition:Bijection|bijection]].
+We have:
+{{begin-eqn}}
+{{eqn | l = \sum_{k \mathop = 1}^n \lambda_k a_k + \sum_{k \mathop = 1}^n \mu_k a_k
+      | r = \sum_{k \mathop = 1}^n \paren {\lambda_k a_k + \mu_k a_k}
+      | c = 
+}}
+{{eqn | r = \sum_{k \mathop = 1}^n \paren {\lambda_k + \mu_k} a_k
+      | c = 
+}}
+{{end-eqn}}
+and we have:
+{{begin-eqn}}
+{{eqn | l = \beta \sum_{k \mathop = 1}^n \lambda_k a_k
+      | r = \sum_{k \mathop = 1}^n \beta \paren {\lambda_k a_k}
+      | c = 
+}}
+{{eqn | r = \sum_{k \mathop = 1}^n \paren {\beta \lambda_k} a_k
+      | c = 
+}}
+{{end-eqn}}
+thus proving that $\psi$ is also a [[Definition:R-Algebraic Structure Homomorphism|homomorphism]].
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Unitary R-Modules with n-Element Bases Isomorphic}
+Tags: Unitary Modules
+
+\begin{theorem}
+Any two [[Definition:Unitary Module|unitary $R$-modules]] having [[Definition:Basis (Linear Algebra)|bases]] of $n$ [[Definition:Element|elements]] are [[Definition:R-Algebraic Structure Isomorphism |isomorphic]].
+\end{theorem}
+
+\begin{proof}
+From [[Isomorphism from R^n via n-Term Sequence]], they are both [[Definition:R-Algebraic Structure Isomorphism|isomorphic]] to the [[Definition:Module on Cartesian Product|$R$-module $R^n$]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{R-Module R^n is n-Dimensional}
+Tags: Module on Cartesian Product
+
+\begin{theorem}
+The [[Definition:Module on Cartesian Product|$R$-module $R^n$]] is [[Definition:Dimension (Linear Algebra)|$n$-dimensional]].
+\end{theorem}
+
+\begin{proof}
+Follows directly from [[Standard Ordered Basis is Basis]].
+{{qed}}
+[[Category:Module on Cartesian Product]]
+4e2u0vlwjpxp8kx7zj4kipknso5ie65
+\end{proof}<|endoftext|>
+\section{Unique Linear Transformation Between Modules}
+Tags: Linear Transformations
+
+\begin{theorem}
+Let $G$ and $H$ be [[Definition:Unitary Module|unitary $R$-modules]].
+Let $\left \langle {a_n} \right \rangle$ be an [[Definition:Ordered Basis|ordered basis]] of $G$.
+Let $\left \langle {b_n} \right \rangle$ be a [[Definition:Sequence|sequence]] of elements of $H$.
+Then there is a unique [[Definition:Linear Transformation|linear transformation]] $\phi: G \to H$ satisfying $\forall k \in \left[{1 \,.\,.\, n}\right]: \phi \left({a_k}\right) = b_k$
+\end{theorem}
+
+\begin{proof}
+By [[Isomorphism from R^n via n-Term Sequence]], the mapping $\phi: G \to H$ defined as:
+: $\displaystyle \phi \left({\sum_{k \mathop = 1}^n \lambda_k a_k}\right) = \sum_{k \mathop = 1}^n \lambda_k b_k$
+is well-defined.
+Thus:
+: $\forall k \in \left[{1 \,.\,.\, n}\right]: \phi \left({a_k}\right) = b_k$
+{{wtd|Verification that $\phi$ is linear needs to be done.}}
+By [[Linear Transformation of Generated Module]], $\phi$ is the only [[Definition:Linear Transformation|linear transformation]] whose value at $a_k$ is $b_k$ for all $k \in \left[{1 \,.\,.\, n}\right]$.
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Linear Transformation from Ordered Basis less Kernel}
+Tags: Linear Transformations
+
+\begin{theorem}
+Let $G$ and $H$ be [[Definition:Unitary Module|unitary $R$-modules]].
+Let $\phi: G \to H$ be a non-zero [[Definition:Linear Transformation|linear transformation]].
+Let $G$ be [[Definition:Dimension of Module|$n$-dimensional]].
+Let $\left \langle {a_n} \right \rangle$ be any [[Definition:Ordered Basis|ordered basis]] of $G$ such that $\left\{{a_k: r + 1 \le k \le n}\right\}$ is the [[Definition:Basis (Linear Algebra)|basis]] of the [[Definition:Kernel of Linear Transformation|kernel]] of $\phi$.
+Then $\left \langle {\phi \left({a_r}\right)} \right \rangle$ is an [[Definition:Ordered Basis|ordered basis]] of the [[Definition:Image of Mapping|image]] of $\phi$.
+\end{theorem}
+
+\begin{proof}
+Suppose:
+:$\displaystyle \sum_{k \mathop = 1}^r \lambda_k \phi \left({a_k}\right) = 0$
+Then:
+:$\displaystyle \phi \left({\sum_{k \mathop = 1}^r \lambda_k a_k}\right) = 0$
+So $\displaystyle \sum_{k \mathop = 1}^r \lambda_k \phi \left({a_k}\right)$ belongs to the [[Definition:Kernel of Linear Transformation|kernel]] of $\phi$ and hence is also a [[Definition:Linear Combination|linear combination]] of $\left\{{a_k: r + 1 \le k \le n}\right\}$.
+Thus $\forall k \in \left[{1 \,.\,.\, r}\right]: \lambda_k = 0$ since $\left \langle {a_n} \right \rangle$ is [[Definition:Linearly Independent Sequence|linearly independent]].
+Thus the sequence $\left \langle {\phi \left({a_r}\right)} \right \rangle$ is linearly independent.
+We have $\forall k \in \left[{r + 1 \,.\,.\, n}\right]: \phi \left({a_k}\right) = 0$.
+So let $x \in G$.
+Let $\displaystyle x = \sum_{k \mathop = 1}^n \mu_k a_k$.
+Then:
+: $\displaystyle \phi \left({x}\right) = \sum_{k \mathop = 1}^n \mu_k \phi \left({a_k}\right) = \sum_{k \mathop = 1}^r \mu_k \phi \left({a_k}\right)$
+Therefore $\left \langle {\phi \left({a_r}\right)} \right \rangle$ is an [[Definition:Ordered Basis|ordered basis]] of the [[Definition:Image of Mapping|image]] of $\phi \left({G}\right)$.
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Addition of Linear Transformations}
+Tags: Linear Transformations
+
+\begin{theorem}
+Let $\left({G, +_G, \circ}\right)_R$ and $\left({H, +_H, \circ}\right)_R$ be [[Definition:Module|$R$-modules]].
+Let $\phi: G \to H$ and $\psi: G \to H$ be [[Definition:Linear Transformation|linear transformations]].
+Let $\phi +_H \psi$ be the operation on $H^G$ induced by $+_H$ as defined in [[Definition:Induced Structure|Induced Structure]].
+Then $\phi +_H \psi: G \to H$ is a [[Definition:Linear Transformation|linear transformation]].
+\end{theorem}<|endoftext|>
+\section{Negative Linear Transformation}
+Tags: Linear Transformations
+
+\begin{theorem}
+Let $\left({G, +_G, \circ}\right)_R$ and $\left({H, +_H, \circ}\right)_R$ be [[Definition:Module|$R$-modules]].
+Let $\phi: G \to H$ be a [[Definition:Linear Transformation|linear transformation]].
+Let $- \phi$ be the negative of $\phi$ as defined in [[Induced Structure Inverse]].
+Then $- \phi: G \to H$ is also a [[Definition:Linear Transformation|linear transformation]].
+\end{theorem}
+
+\begin{proof}
+From the definition of a [[Definition:Module|module]], the [[Definition:Group|group]] $\left({H, +_H}\right)$ is [[Definition:Abelian Group|abelian]].
+Therefore we can apply [[Inverse Mapping in Induced Structure]] to show that $- \phi: G \to H$ is a [[Definition:R-Algebraic Structure Homomorphism|homomorphism]].
+Then:
+{{begin-eqn}}
+{{eqn | l=\left({- \phi}\right) \left({\lambda \circ x}\right)
+      | r=-\phi \left({\lambda \circ x}\right)
+      | c=
+}}
+{{eqn | r=- \lambda \circ \phi \left({x}\right)
+      | c=
+}}
+{{eqn | r=\lambda \circ \left({- \phi \left({x}\right)}\right)
+      | c=
+}}
+{{eqn | r=\lambda \circ \left({- \phi}\right) \left({x}\right)
+      | c=
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Linear Transformations form Abelian Group}
+Tags: Linear Transformations, Abelian Groups
+
+\begin{theorem}
+Let $\struct {G, +_G}$ and $\struct {H, +_H}$ be [[Definition:Group|groups]].
+Let $\struct {R, +_R, \times_R}$ be a [[Definition:Ring (Abstract Algebra)|ring]].
+Let $\struct {G, +_G, \circ}_R$ and $\struct {H, +_H, \circ}_R$ be [[Definition:Module|$R$-modules]].
+Let $\map {\mathcal L_R} {G, H}$ be [[Definition:Set of All Linear Transformations|the set of all linear transformations]] from $G$ to $H$.
+Let $\oplus_H$ be the operation on $H^G$ as defined in [[Addition of Linear Transformations]].
+Then $\struct {\map {\LL_R} {G, H}, \oplus_H}$ is an [[Definition:Abelian Group|abelian group]].
+\end{theorem}
+
+\begin{proof}
+From [[Structure Induced by Group Operation is Group]], $\struct {H^G, \oplus_H}$ is a [[Definition:Group|group]]
+Let $\phi, \psi \in \map {\LL_R} {G, H}$.
+From [[Addition of Linear Transformations]], $\phi \oplus_H \psi \in \map {\LL_R} {G, H}$.
+From [[Negative Linear Transformation]], $-\phi \in \map {\LL_R} {G, H}$.
+Thus, from the [[Two-Step Subgroup Test]], $\struct {\map {\LL_R} {G, H}, \oplus_H}$ is a [[Definition:Subgroup|subgroup]] of $\struct {H^G, \oplus_H}$.
+{{finish|Still need to show that it is abelian}}
+\end{proof}<|endoftext|>
+\section{Morse-Sard Theorem}
+Tags: Topology
+
+\begin{theorem}
+Let $f: X \to Y$ be any [[Definition:Smooth Mapping|smooth map]] of [[Definition:Topological Manifold|manifolds]].
+Then [[Definition:Almost Every|almost every]] point in $Y$ is a [[Definition:Regular Value|regular value]] of $f$.
+\end{theorem}
+
+\begin{proof}
+{{ProofWanted|In progress}}
+\end{proof}<|endoftext|>
+\section{Linear Transformation from Center of Scalar Ring}
+Tags: Linear Transformations
+
+\begin{theorem}
+Let $\struct {G, +_G, \circ}_R$ and $\struct {H, +_H, \circ}_R$ be [[Definition:Module|$R$-modules]].
+Let $\phi: G \to H$ be a [[Definition:Linear Transformation|linear transformation]].
+Let $\map Z R$ be the [[Definition:Center of Ring|center]] of the [[Definition:Scalar Ring of Module|scalar ring]] $R$.
+Let $\lambda \in \map Z R$.
+Then $\lambda \circ \phi$ is a [[Definition:Linear Transformation|linear transformation]].
+\end{theorem}
+
+\begin{proof}
+Let $\lambda \in \lambda \in \map Z R$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \map {\paren {\lambda \circ \phi} } {x +_G y}
+      | r = \lambda \circ \map \phi {x +_G y}
+      | c = 
+}}
+{{eqn | r = \lambda \circ \paren {\map \phi x +_H \map \phi y}
+      | c = 
+}}
+{{eqn | r = \lambda \circ \map \phi x +_H \lambda \circ \map \phi y
+      | c = 
+}}
+{{eqn | r = \map {\paren {\lambda \circ \phi} } x +_H \map {\paren {\lambda \circ \phi} } y
+      | c = 
+}}
+{{end-eqn}}
+Because $\lambda \in \map Z R$, $\lambda$ [[Definition:Commute|commutes]] with all elements of $R$.
+So $\forall \mu \in R: \lambda \circ \mu = \mu \circ \lambda$.
+Thus:
+{{begin-eqn}}
+{{eqn | l = \map {\paren {\lambda \circ \phi} } {\mu \circ x}
+      | r = \lambda \circ \map \phi {\mu \circ x}
+      | c = 
+}}
+{{eqn | r = \lambda \circ \mu \circ \map \phi x
+      | c = 
+}}
+{{eqn | r = \mu \circ \lambda \circ \map \phi x
+      | c = as $\lambda \in \map Z R$
+}}
+{{eqn | r = \mu \circ \map {\paren {\lambda \circ \phi} } x
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Linear Transformations of Commutative Scalar Ring}
+Tags: Linear Transformations
+
+\begin{theorem}
+Let $R$ be a [[Definition:Commutative Ring|commutative ring]].
+Let $\left({G, +_G, \circ}\right)_R$ and $\left({H, +_H, \circ}\right)_R$ be [[Definition:Module|$R$-modules]].
+Let $\mathcal L_R \left({G, H}\right)$ be [[Definition:Set of All Linear Transformations|the set of all linear transformations]] from $G$ to $H$.
+Then $\mathcal L_R \left({G, H}\right)$ is a [[Definition:Submodule|submodule]] of the [[Definition:Module of All Mappings|$R$-module $H^G$]].
+If $H$ is a [[Definition:Unitary Module|unitary module]], then so is $\mathcal L_R \left({G, H}\right)$.
+\end{theorem}
+
+\begin{proof}
+From [[Group equals Center iff Abelian]], the [[Definition:Center of Ring|center]] of a [[Definition:Commutative Ring|commutative ring]] is the entire ring.
+The result follows from [[Linear Transformation from Center of Scalar Ring]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Product of Linear Transformations}
+Tags: Linear Transformations
+
+\begin{theorem}
+Let $R$ be a [[Definition:Commutative and Unitary Ring|commutative ring with unity]] whose [[Definition:Ring Zero|zero]] is $0_R$ and whose [[Definition:Unity of Ring|unity]] is $1_R$.
+Let $\struct {G, +_G, \circ}_R$ be a [[Definition:Unitary Module|unitary $R$-module]] such that $\map \dim G = n$.
+Let $\struct {H, +_H, \circ}_R$ be a [[Definition:Unitary Module|unitary $R$-module]] such that $\map \dim H = m$.
+Let $\map {\LL_R} {G, H}$ be [[Definition:Set of All Linear Transformations|the set of all linear transformations]] from $G$ to $H$.
+Then:
+:$\map \dim {\map {\LL_R} {G, H} } = n m$
+Let $\sequence {a_n}$ be an [[Definition:Ordered Basis|ordered basis]] for $G$.
+Let $\sequence {b_m}$ be an [[Definition:Ordered Basis|ordered basis]] for $H$.
+Let $\phi_{i j}: G \to H$ be the unique [[Definition:Linear Transformation|linear transformation]] defined for each $i \in \closedint 1 n, j \in \closedint 1 m$ which satisfies:
+:$\forall k \in \closedint 1 n: \map {\phi_{i j} } {a_k} = \delta_{i k} b_j$
+where $\delta$ is the [[Definition:Kronecker Delta|Kronecker delta]].
+Then:
+:$\set {\phi_{i j}: i \in \closedint 1 n, j \in \closedint 1 m}$
+is a [[Definition:Basis (Linear Algebra)|basis]] for $\map \dim {\map {\LL_R} {G, H} }$.
+\end{theorem}
+
+\begin{proof}
+Let $B = \set {\phi_{i j}: i \in \closedint 1 n, j \in \closedint 1 m}$.
+Let $\displaystyle \sum_{j \mathop = 1}^m \sum_{i \mathop = 1}^n \lambda_{i j} \phi_{i j} = 0$.
+Then:
+:$\displaystyle \forall k \in \closedint 1 n: 0 = \sum_{j \mathop = 1}^m \sum_{i \mathop = 1}^n \lambda_{i j} \map {\phi_{i j} } {a_k} = \sum_{j \mathop = 1}^m \lambda_{k j} b_j$
+So:
+:$\forall j \in \closedint 1 n: \lambda_{k j} = 0$
+Hence $B$ is [[Definition:Linearly Independent Set|linearly independent]].
+Now let $\phi \in \map {\LL_R} {G, H}$.
+Let $\tuple {\alpha_{i 1}, \alpha_{i 2}, \ldots, \alpha_{i m} }$ be the [[Definition:Sequence|sequence]] of [[Definition:Scalar (Module)|scalars]] that satisfies:
+:$\displaystyle \forall i \in \closedint 1 n: \map \phi {a_i} = \sum_{j \mathop = 1}^m \alpha_{i j} b_j$
+Then:
+:$\displaystyle \forall k \in \closedint 1 n: \map \phi {a_k} = \map {\paren {\sum_{j \mathop = 1}^m \sum_{i \mathop = 1}^n \alpha_{i j} u_{i j} } } {a_k}$
+by a calculation similar to the preceding.
+So, by [[Linear Transformation of Generated Module]]:
+:$\displaystyle u = \sum_{j \mathop = 1}^m \sum_{i \mathop = 1}^n \alpha_{i j} u_{i j}$
+Thus $B$ is a [[Definition:Generator of Module|generator]] for $\phi \in \map {\LL_R} {G, H}$.
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Ring of Linear Operators}
+Tags: Linear Operators
+
+\begin{theorem}
+Let $\map {\LL_R} G$ be [[Definition:Set of All Linear Transformations|the set of all linear operators]] on $G$.
+{{explain|the precise nature of $G$}}
+Let $\phi \circ \psi$ denote the [[Definition:Composition of Mappings|composition]] of the two [[Definition:Linear Operator|linear operators]] $\phi$ and $\psi$.
+Then $\struct {\map {\LL_R} G, +, \circ}$ is a [[Definition:Ring (Abstract Algebra)|ring]].
+\end{theorem}
+
+\begin{proof}
+Follows from [[Composite of Homomorphisms is Homomorphism/R-Algebraic Structure|Composite of R-Algebraic Structure Homomorphisms is Homomorphism]], as it is a [[Definition:Subring|subring]] of the ring of all endomorphisms of the [[Definition:Abelian Group|abelian group]] $\struct {G, +}$.
+{{ProofWanted}}
+[[Category:Linear Operators]]
+eda5k9ojbrvu9i00azjlutffjtyrjib
+\end{proof}<|endoftext|>
+\section{Dimension of Algebraic Dual}
+Tags: Linear Transformations
+
+\begin{theorem}
+Let $G$ be an [[Definition:Dimension (Linear Algebra)|$n$-dimensional]] [[Definition:Module|$R$-module]].
+Let $G^*$ be the [[Definition:Algebraic Dual|algebraic dual]] of $G$.
+Let $G^{**}$ be the [[Definition:Algebraic Dual|algebraic dual]] of $G^*$.
+Then $G^*$ and $G^{**}$ are also [[Definition:Dimension (Linear Algebra)|$n$-dimensional]].
+\end{theorem}
+
+\begin{proof}
+Follows directly from [[Product of Linear Transformations]].
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Annihilator is Submodule of Algebraic Dual}
+Tags: Linear Transformations
+
+\begin{theorem}
+Let $R$ be a [[Definition:Commutative Ring|commutative ring]].
+Let $G$ be a [[Definition:Module|module]] over $R$.
+Let $M$ be a [[Definition:Submodule|submodule]] of $G$.
+Let $G^*$ be the [[Definition:Algebraic Dual|algebraic dual]] of $G$.
+Then the [[Definition:Annihilator on Algebraic Dual|annihilator]] $M^\circ$ of $M$ is a [[Definition:Submodule|submodule]] of $G^*$.
+Similarly, let $N$ be a [[Definition:Submodule|submodule]] of $G^*$.
+Let $G^{**}$ be the [[Definition:Algebraic Dual|algebraic dual]] of $G^*$.
+Then the [[Definition:Annihilator on Algebraic Dual|annihilator]] $N^\circ$ of $N$ is a [[Definition:Submodule|submodule]] of $G^{**}$.
+\end{theorem}<|endoftext|>
+\section{Inverse Evaluation Isomorphism of Annihilator}
+Tags: Linear Transformations
+
+\begin{theorem}
+Let $R$ be a [[Definition:Commutative Ring|commutative ring]]. 
+Let $G$ be a [[Definition:Module|module]] over $R$ whose [[Definition:Dimension of Module|dimension]] is [[Definition:Finite|finite]].
+Let $G^*$ be the [[Definition:Algebraic Dual|algebraic dual]] of $G$.
+Let $G^{**}$ be the [[Definition:Algebraic Dual|algebraic dual]] of $G^*$.
+Let $N$ be a [[Definition:Submodule|submodule]] of $G^*$.
+Let $J$ be the [[Definition:Evaluation Isomorphism|evaluation isomorphism]] from $G$ onto $G^{**}$.
+Let $N^\circ$ be the [[Definition:Annihilator on Algebraic Dual|annihilator]] of $N$.
+Then:
+:$J^{-1} \left({N^\circ}\right) = \left\{{x \in G: \forall t' \in N: t' \left({x}\right) = 0}\right\}$
+\end{theorem}
+
+\begin{proof}
+As $G$ is finite-dimensional, then by [[Evaluation Isomorphism is Isomorphism]] $J: G \to G^{**}$ is an [[Definition:Module Isomorphism|isomorphism]], and therefore a [[Definition:Surjection|surjection]].
+Thus:
+:$N^\circ = \left\{{x^\wedge \in G^{**}: \forall t' \in N: x^\wedge \left({t'}\right) = 0}\right\}$
+where $x^\wedge$ is as defined in the definition of the [[Definition:Evaluation Linear Transformation|evaluation linear transformation]].
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Properties of Evaluation Linear Transformation}
+Tags: Linear Transformations
+
+\begin{theorem}
+Let $R$ be a [[Definition:Commutative Ring|commutative ring]].
+Let $G$ be an [[Definition:Module|$R$-module]].
+Let $G^*$ be the [[Definition:Algebraic Dual|algebraic dual]] of $G$.
+Let $\left \langle {x, t'} \right \rangle$ be the [[Definition:Evaluation Linear Transformation|evaluation linear transformation]] from $G$ to $G^{**}$.
+Then the [[Definition:Mapping|mapping]] $\phi: G \times G^* \to R$ defined as $\forall \left({x, t'}\right) \in G \times G^*: \phi \left({x, t'}\right) = \left \langle {x, t'} \right \rangle$ satisfies the following properties:
+: $(1): \quad \forall x, y \in G: \forall t' \in G^*: \left \langle {x + y, t'} \right \rangle = \left \langle {x, t'} \right \rangle + \left \langle {y, t'} \right \rangle$
+: $(2): \quad \forall x \in G: \forall s', t' \in G^*: \left \langle {x, s' + t'} \right \rangle = \left \langle {x, s'} \right \rangle + \left \langle {x, t'} \right \rangle$
+: $(3): \quad \forall x \in G: \forall s', t' \in G^*: \forall \lambda \in R: \left \langle {\lambda x, t'} \right \rangle = \lambda \left \langle {x, t'} \right \rangle = \left \langle {x, \lambda t'} \right \rangle$
+\end{theorem}<|endoftext|>
+\section{Zero Vector Space Product iff Factor is Zero}
+Tags: Linear Algebra, Vector Algebra, Zero Vector Space Product iff Factor is Zero, Zero Vectors
+
+\begin{theorem}
+Let $F$ be a [[Definition:Field (Abstract Algebra)|field]] whose [[Definition:Field Zero|zero]] is $0_F$ and whose [[Definition:Unity of Field|unity]] is $1_F$.
+Let $\struct {\mathbf V, +, \circ}_F$ be a [[Definition:Vector Space|vector space]] over $F$, as defined by the [[Definition:Vector Space Axioms|vector space axioms]].
+Let $\mathbf v \in \mathbf V, \lambda \in F$.
+Then:
+:$\lambda \circ \mathbf v = \bszero \iff \paren {\lambda = 0_F \lor x = \bszero}$
+\end{theorem}
+
+\begin{proof}
+A [[Definition:Vector Space|vector space]] is a [[Definition:Module|module]], so all results about modules also apply to vector spaces.
+So from [[Scalar Product with Identity]] it follows directly that:
+:$\lambda = 0_F \lor \mathbf v = e \implies \lambda \circ \mathbf v = \bszero$
+Next, suppose $\lambda \circ \mathbf v = \bszero$ but $\lambda \ne 0_F$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \bszero
+      | r = \lambda^{-1} \circ \bszero
+      | c = [[Zero Vector Scaled is Zero Vector]]
+}}
+{{eqn | r = \lambda^{-1} \circ \paren {\lambda \circ \mathbf v}
+      | c = as $\lambda \circ \mathbf v = \bszero$
+}}
+{{eqn | r = \paren {\lambda^{-1} \circ \lambda} \circ \mathbf v
+      | c = {{Vector-space-axiom|7}}
+}}
+{{eqn | r = 1 \circ \mathbf v
+      | c = {{Field-axiom|M4}}
+}}
+{{eqn | r = \mathbf v
+      | c = {{Vector-space-axiom|8}}
+}}
+{{end-eqn}}
+{{Qed}}
+\end{proof}
+
+\begin{proof}
+The [[Definition:Sufficient Condition|sufficient condition]] is proved in [[Vector Scaled by Zero is Zero Vector]], and in [[Zero Vector Scaled is Zero Vector]].
+The [[Definition:Necessary Condition|necessary condition]] is proved in [[Vector Product is Zero only if Factor is Zero]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Homomorphic Image of Vector Space}
+Tags: Homomorphisms, Linear Algebra
+
+\begin{theorem}
+Let $\struct {K, +_K, \times_K}$ be a [[Definition:Division Ring|division ring]].
+Let $\struct {V, +_V, \circ_V}_K$ be a [[Definition:Vector Space|$K$-vector space]].
+Let $\struct {W, +_W, \circ_W}_K$ be a [[Definition:R-Algebraic Structure|$K$-algebraic structure]].
+Let $\phi: V \to W$ be a [[Definition:R-Algebraic Structure Homomorphism|homomorphism]], i.e. a [[Definition:Linear Transformation on Vector Space|linear transformation]].
+Then the [[Definition:Homomorphic Image|homomorphic image]] of $\phi$ is a [[Definition:Vector Space|$K$-vector space]].
+\end{theorem}
+
+\begin{proof}
+Let us write $\phi \sqbrk V$ for the [[Definition:Homomorphic Image|homomorphic image]] of $\phi$.
+From [[Homomorphic Image of R-Module is R-Module]], $\phi \sqbrk V$ is a [[Definition:Module|$K$-module]].
+It thus suffices to show that $\phi \sqbrk V$ is [[Definition:Unitary Module|unitary]], since then it will be a [[Definition:Vector Space|$K$-vector space]].
+To this end, let $1_K$ be the [[Definition:Unity of Ring|unity]] of $K$.
+Then for any $\map \phi {\mathbf v} \in \phi \sqbrk V$, compute:
+{{begin-eqn}}
+{{eqn | l = 1_K \circ_W \map \phi {\mathbf v}
+      | r = \map \phi {1_K \circ_V \mathbf v}
+      | c = $\phi$ is a [[Definition:Linear Transformation on Vector Space|linear transformation]]
+}}
+{{eqn | r = \map \phi {\mathbf v}
+      | c = $V$ is a [[Definition:Vector Space|$K$-vector space]]
+}}
+{{end-eqn}}
+Hence $\phi \sqbrk V$ is [[Definition:Unitary Module|unitary]], and so a [[Definition:Vector Space|$K$-vector space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Direct Product of Vector Spaces is Vector Space}
+Tags: Linear Algebra
+
+\begin{theorem}
+Let $K$ be a [[Definition:Field (Abstract Algebra)|field]].
+Let $V_1, V_2, \ldots, V_n$ be [[Definition:Vector Space|$K$-vector spaces]].
+Let $\struct {V, + , \circ}_K$ be their [[Definition:Direct Product of Vector Spaces|direct product]].
+Then $\struct {V, + , \circ}_K$ is a [[Definition:Vector Space|$K$-vector space]].
+\end{theorem}
+
+\begin{proof}
+This follows directly from [[Finite Direct Product of Modules is Module]] and the definition of [[Definition:Vector Space|vector space]].
+\end{proof}<|endoftext|>
+\section{Definition:Vector Space on Cartesian Product}
+Tags: Definitions: Examples of Vector Spaces
+
+\begin{theorem}
+Let $\struct {K, +, \circ}$ be a [[Definition:Division Ring|division ring]].
+Let $n \in \N_{>0}$.
+Let $+: K^n \times K^n \to K^n$ be defined as:
+:$\tuple {\alpha_1, \ldots, \alpha_n} + \tuple {\beta_1, \ldots, \beta_n} = \tuple {\alpha_1 +_R \beta_1, \ldots, \alpha_n +_R \beta_n}$
+Let $\times: K \times K^n \to K^n$ be defined as:
+:$\lambda \times \tuple {\alpha_1, \ldots, \alpha_n} = \tuple {\lambda \times_R \alpha_1, \ldots, \lambda \times_R \alpha_n}$
+Then $\struct {K^n, +, \times}_K$ is '''the $K$-vector space $K^n$'''.
+\end{theorem}<|endoftext|>
+\section{Definition:Vector Space over Division Subring}
+Tags: Definitions: Examples of Vector Spaces
+
+\begin{theorem}
+Let $\struct {R, +, \circ}$ be a [[Definition:Ring with Unity|ring with unity]] whose [[Definition:Unity of Ring|unity]] is $1_R$.
+Let $S$ be a [[Definition:Division Subring|division subring]] of $R$, such that $1_R \in S$.
+Then $\struct {R, +, \circ_S}_S$, where $\circ_S$ is the [[Definition:Restriction of Operation|restriction]] of $\circ$ to $S \times R$, is the '''vector space on $R$ over the [[Definition:Division Subring|division subring]] $S$'''.
+\end{theorem}<|endoftext|>
+\section{Complex Numbers form Vector Space over Reals}
+Tags: Linear Algebra, Complex Analysis, Examples of Vector Spaces
+
+\begin{theorem}
+Let $\R$ be the set of [[Definition:Real Number|real numbers]].
+Let $\C$ be the set of [[Definition:Complex Number|complex numbers]].
+Then the [[Definition:Module|$\R$-module]] $\C$ is a [[Definition:Vector Space|vector space]].
+\end{theorem}
+
+\begin{proof}
+Recall that [[Real Numbers form Field]].
+Thus by definition, $\R$ is also a [[Definition:Division Ring|division ring]].
+Thus we only need to show that [[Definition:Module|$\R$-module]] $\C$ is a [[Definition:Unitary Module|unitary module]], by demonstrating the module properties:
+$\forall x, y, \in \C, \forall \lambda, \mu \in \R$:
+:$(1): \quad \lambda \paren {x + y} = \paren {\lambda x} + \paren {\lambda y}$
+:$(2): \quad \paren {\lambda + \mu} x = \paren {\lambda x} + \paren {\mu x}$
+:$(3): \quad \paren {\lambda \mu} x = \lambda \paren {\mu x}$
+:$(4): \quad 1 x = x$
+As $\lambda, \mu \in \R$ it follows that $\lambda, \mu \in \C$.
+Thus from [[Complex Multiplication Distributes over Addition]], $(1)$ and $(2)$ immediately follow.
+$(3)$ follows from [[Complex Multiplication is Associative]].
+$(4)$ follows from [[Complex Multiplication Identity is One]], as $1 + 0 i$ is the [[Definition:Unity of Field|unity]] of $\C$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Division Ring is Vector Space over Prime Subfield}
+Tags: Division Rings, Subfields, Linear Algebra
+
+\begin{theorem}
+Let $\struct {K, +, \times}$ be a [[Definition:Division Ring|division ring]].
+Let $\struct {S, +, \times}$ be the [[Definition:Prime Subfield|prime subfield]] of $K$
+Then $\struct {K, +, \times_S}_S$ is an [[Definition:Vector Space|$S$-vector space]], where $\times_S$ is the [[Definition:Restriction of Operation|restriction]] of $\times$ to $S \times K$.
+\end{theorem}<|endoftext|>
+\section{Vector Space over Subring}
+Tags: Linear Algebra
+
+\begin{theorem}
+Let $K$ be a [[Definition:Division Subring|division subring]] of the [[Definition:Division Ring|division ring]] $\struct {L, +_L, \times_L}$.
+Let $\struct {G, +_G, \circ}_L$ be a [[Definition:Vector Space|$L$-vector space]].
+Then $\struct {G, +_G, \circ_K}_K$ is a [[Definition:Vector Space|$K$-vector space]], where $\circ_K$ is the [[Definition:Restriction of Operation|restriction]] of $\circ$ to $K \times G$.
+The [[Definition:Vector Space|$K$-vector space]] $\struct {G, +_G, \circ_K}_K$ is called the '''$K$-vector space obtained from $\struct {L, +_L, \times_L}$ by restricting scalar multiplication'''.
+\end{theorem}<|endoftext|>
+\section{Subspaces of Dimension 2 Real Vector Space}
+Tags: Linear Algebra, Subspaces of Dimension 2 Real Vector Space
+
+\begin{theorem}
+Take the [[Definition:Real Vector Space|$\R$-vector space]] $\left({\R^2, +, \times}\right)_\R$.
+Let $S$ be a [[Definition:Vector Subspace|subspace]] of $\left({\R^2, +, \times}\right)_\R$.
+Then $S$ is one of:
+: $(1): \quad \left({\R^2, +, \times}\right)_\R$
+: $(2): \quad \left\{{0}\right\}$
+: $(3): \quad$ A line through the origin.
+\end{theorem}
+
+\begin{proof}
+Let $S$ be a non-zero [[Definition:Vector Subspace|subspace]] of $\left({\R^2, +, \times}\right)_\R$.
+Then $S$ contains a [[Definition:Zero Vector|non-zero vector]] $\left({\alpha_1, \alpha_2}\right)$.
+Hence $S$ also contains $\left\{{\lambda \times \left({\alpha_1, \alpha_2}\right), \lambda \in \R}\right\}$.
+From [[Equation of Straight Line in Plane]], this set may be described as a line through the origin.
+Suppose $S$ also contains a [[Definition:Zero Vector|non-zero vector]] $\left({\beta_1, \beta_2}\right)$ which is not on that line.
+Then $\alpha_1 \times \beta_2 - \alpha_2 \times \beta_1 \ne 0$.
+Otherwise $\left({\beta_1, \beta_2}\right)$ would be $\zeta \times \left({\alpha_1, \alpha_2}\right)$, where either $\zeta = \beta_1 / \alpha_1$ or $\zeta = \beta_2 / \alpha_2$ according to whether $\alpha_1 \ne 0$ or $\alpha_2 \ne 0$.
+But then $S = \left({\R^2, +, \times}\right)_\R$.
+Because, if $\left({\gamma_1, \gamma_2}\right)$ is any vector at all, then:
+: $\left({\gamma_1, \gamma_2}\right) = \lambda \times \left({\alpha_1, \alpha_2}\right) + \mu \times \left({\beta_1, \beta_2}\right)$
+where $\lambda = \dfrac {\gamma_1 \times \beta_2 - \gamma_2 \times \beta_1} {\alpha_1 \times \beta_2 - \alpha_2 \times \beta_1}, \mu = \dfrac {\alpha_1 \times \gamma_2 - \alpha_2 \times \gamma_1} {\alpha_1 \times \beta_2 - \alpha_2 \times \beta_1}$
+which we get by solving the simultaneous eqns:
+{{begin-eqn}}
+{{eqn | l=\alpha_1 \times \lambda + \beta_1 \times \mu
+      | r=0
+      | c=
+}}
+{{eqn | l=\alpha_2 \times \lambda + \beta_2 \times \mu
+      | r=0
+      | c=
+}}
+{{end-eqn}}
+The result follows.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Follows directly from [[Dimension of Proper Subspace is Less Than its Superspace]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Subspace of Real Continuous Functions}
+Tags: Vector Subspaces, Analysis
+
+\begin{theorem}
+Let $\mathbb J = \set {x \in \R: a \le x \le b}$ be a [[Definition:Closed Real Interval|closed interval]] of the [[Definition:Real Number Line|real number line]] $\R$.
+Let $\map \CC {\mathbb J}$ be the set of all [[Definition:Continuous Real Function|continuous real functions]] on $\mathbb J$.
+Then $\struct {\map \CC {\mathbb J}, +, \times}_\R$ is a [[Definition:Vector Subspace|subspace]] of the [[Definition:Vector Space|$\R$-vector space]] $\struct {\R^{\mathbb J}, +, \times}_\R$.
+\end{theorem}
+
+\begin{proof}
+By definition, $\map \CC {\mathbb J} \subseteq \R^{\mathbb J}$.
+Let $f, g \in \map \CC {\mathbb J}$.
+By [[Two-Step Vector Subspace Test]], it needs to be shown that:
+:$\paren 1: \quad f + g \in \map \CC {\mathbb J}$
+:$\paren 2: \quad \lambda f \in \map \CC {\mathbb J}$ for any $\lambda \in \R$
+$\paren 1$ follows by [[Sum Rule for Continuous Functions]].
+$\paren 2$ follows by [[Multiple Rule for Continuous Functions]].
+Hence $\struct {\map \CC {\mathbb J}, +, \times}_\R$ is a [[Definition:Vector Subspace|subspace]] of the [[Definition:Vector Space|$\R$-vector space]] $\struct {\R^{\mathbb J}, +, \times}_\R$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Singleton is Linearly Independent}
+Tags: Linear Algebra
+
+\begin{theorem}
+Let $K$ be a [[Definition:Division Ring|division ring]].
+Let $\struct {G, +_G}$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
+Let $\struct {G, +_G, \circ}_K$ be a [[Definition:Vector Space|$K$-vector space]] whose [[Definition:Zero Scalar|zero]] is $0_K$.
+Let $x \in G: x \ne e$.
+Then $\set x$ is a [[Definition:Linearly Independent Set|linearly independent subset]] of $G$.
+\end{theorem}
+
+\begin{proof}
+The only [[Definition:Sequence of Distinct Terms|sequence of distinct terms]] in $\set x$ is the one that goes: $x$.
+Suppose $\exists \lambda \in K: \lambda \circ x = e$.
+From [[Zero Vector Space Product iff Factor is Zero]] it follows that $\lambda = 0$.
+Hence the result from definition of [[Definition:Linearly Independent Set|linearly independent set]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Linearly Dependent Sequence of Vector Space}
+Tags: Linear Algebra
+
+\begin{theorem}
+Let $\struct {G, +}$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $\mathbf 0$.
+Let $\struct {G, +, \circ}_K$ be a [[Definition:Vector Space|$K$-vector space]].
+Let $\sequence {a_k}_{1 \mathop \le k \mathop \le n}$ be a [[Definition:Sequence|sequence]] of distinct [[Definition:Zero Vector|non-zero vectors]] of $G$.
+Then $\sequence {a_k}_{1 \mathop \le k \mathop \le n}$ is [[Definition:Linearly Dependent Sequence|linearly dependent]] {{iff}}:
+:$\exists p \in \closedint 2 n: a_p$ is a [[Definition:Linear Combination|linear combination]] of $\sequence {a_k}_{1 \mathop \le k \mathop \le p - 1}$
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+Suppose $\sequence {a_k}_{1 \mathop \le k \mathop \le n}$ is [[Definition:Linearly Dependent Sequence|linearly dependent]].
+[[Definition:By Hypothesis|By hypothesis]], the [[Definition:Set|set]] of all [[Definition:Integer|integers]] $r \in \closedint 1 n$ such that $\sequence {a_k}_{1 \mathop \le k \mathop \le r}$ is [[Definition:Linearly Independent Sequence|linearly dependent]] is not [[Definition:Empty Set|empty]].
+Let $p$ be its [[Definition:Smallest Element|smallest element]].
+Then from [[Singleton is Linearly Independent]], $p \ge 2$, as $a_1 \ne \mathbf 0$ and hence $\set {a_1}$ is [[Definition:Linearly Independent Sequence|linearly independent]].
+Also, there exist [[Definition:Scalar (Vector Space)|scalars]] $\lambda_1, \ldots, \lambda_p$, not all of which are [[Definition:Zero Scalar|zero]], such that $\displaystyle \sum_{k \mathop = 1}^p \lambda_k \circ a_k = \mathbf 0$.
+Suppose $\lambda_p = 0$.
+Then not all of $\lambda_1, \ldots, \lambda_{p-1}$ can be [[Definition:Zero Scalar|zero]].
+Then $\sequence {a_k}_{1 \mathop \le k \mathop \le p-1}$ is [[Definition:Linearly Independent Sequence|linearly dependent]].
+That contradicts the definition of $p$, so $\lambda_p \ne 0$.
+So, because:
+:$\displaystyle \lambda_p \circ a_p = - \sum_{k \mathop = 1}^{p - 1} \lambda_k \circ a_k$
+we must have:
+:$\displaystyle a_p = \sum_{k \mathop = 1}^{p - 1} \paren {-{\lambda_p}^{-1} \lambda_k} \circ a_k$
+and thus $a_p$ is a [[Definition:Linear Combination|linear combination]] of $\sequence {a_k}_{1 \mathop \le k \mathop \le p - 1}$.
+{{qed|lemma}}
+=== Sufficient Condition ===
+Now suppose that $a_p$ is a [[Definition:Linear Combination|linear combination]] of $\sequence {a_k}_{1 \mathop \le k \mathop \le p - 1}$.
+Then:
+:$\displaystyle a_p = \sum_{k \mathop = 1}^{p - 1} \mu_k \circ a_k$
+So we can assign values to $\lambda_k$ as follows:
+:$\forall k \in \closedint 1 n: \lambda_k = \begin{cases}
+\mu_k & : k < p \\
+-1 & : k = p \\
+0 & : k > p \\
+\end{cases}$
+Then:
+:$\displaystyle \sum_{k \mathop = 1}^n \lambda_k \circ a_k = \mathbf 0$
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Vector Space has Basis Between Linearly Independent Set and Finite Spanning Set}
+Tags: Bases of Vector Spaces, Generators of Vector Spaces
+
+\begin{theorem}
+Let $K$ be a [[Definition:Division Ring|division ring]].
+Let $G$ be a [[Definition:Finitely Generated Module|finitely generated]] [[Definition:Vector Space|$K$-vector space]].
+Let $H$ be a [[Definition:Linearly Independent Set|linearly independent subset]] of $G$.
+Let $F$ be a [[Definition:Finite Set|finite]] [[Definition:Generator of Vector Space|generator]] for $G$ such that $H \subseteq F$.
+Then there is a [[Definition:Basis of Vector Space|basis]] $B$ for $G$ such that $H \subseteq B \subseteq F$.
+\end{theorem}
+
+\begin{proof}
+Let $\mathbb S$ be the [[Definition:Set|set]] of all $S \subseteq G$ such that $S$ is a [[Definition:Generator of Vector Space|generator]] for $G$ and that $H \subseteq S \subseteq F$.
+Because $F \in \mathbb S$, it follows that $\mathbb S \ne \O$.
+Because $F$ is [[Definition:Finite Set|finite]], then so is every [[Definition:Element|element]] of $\mathbb S$.
+Let $R = \set {r \in \Z: r = \card S \in \mathbb S}$.
+That is, $R$ is the set of all the integers which are the number of [[Definition:Element|elements]] in [[Definition:Generator of Vector Space|generators]] for $G$ that are subsets of $F$.
+Let $n$ be the smallest [[Definition:Element|element]] of $R$.
+Let $B$ be an [[Definition:Element|element]] of $\mathbb S$ such that $\card B = n$.
+We note that as $H$ is a [[Definition:Linearly Independent Set|linearly independent set]], it does not contain $0$ by [[Subset of Module Containing Identity is Linearly Dependent]].
+Then $0 \notin B$, or $B \setminus \set 0$ would be a [[Definition:Generator of Vector Space|generator]] for $G$ with $n - 1$ [[Definition:Element|elements]].
+This would contradict the definition of $n$.
+Let $m = \card H$.
+Let $\sequence {a_n}$ be a [[Definition:Sequence of Distinct Terms|sequence of distinct vectors]] such that $H = \set {a_1, \ldots, a_m}$ and $B = \set {a_1, \ldots, a_n}$.
+Suppose $B$ were [[Definition:Linearly Dependent Set|linearly dependent]].
+By [[Linearly Dependent Sequence of Vector Space]], there would exist $p \in \closedint 2 n$ and [[Definition:Scalar (Vector Space)|scalars]] $\mu_1, \ldots, \mu_{p - 1}$ such that $\displaystyle a_p = \sum_{k \mathop = 1}^{p - 1} \mu_k a_k$.
+As $H$ is [[Definition:Linearly Independent Set|linearly independent]], $p > m$ and therefore $B' = B \setminus \set {a_p}$ would contain $H$.
+{{explain|why $H$ being L.I. implies $p > m$}}
+Now if $\displaystyle x = \sum_{k \mathop = 1}^n \lambda_k a_k$, then:
+:$\displaystyle x = \sum_{k \mathop = 1}^{p - 1} \paren {\lambda_k + \lambda_p \mu_k} a_k + \sum_{k \mathop = p + 1}^n \lambda_k a_k$
+Hence $B'$ would be a [[Definition:Generator of Vector Space|generator]] for $G$ containing $n - 1$ [[Definition:Element|elements]], which contradicts the definition of $n$.
+Thus $B$ must be [[Definition:Linearly Independent Set|linearly independent]] and hence is a [[Definition:Basis of Vector Space|basis]].
+{{qed}}
+{{Proofread}}
+\end{proof}<|endoftext|>
+\section{Finitely Generated Vector Space has Basis}
+Tags: Bases of Vector Spaces
+
+\begin{theorem}
+Let $K$ be a [[Definition:Division Ring|division ring]].
+Let $V$ be a [[Definition:Finitely Generated Module|finitely generated]] [[Definition:Vector Space|vector space]] over $K$.
+Then $V$ has a [[Definition:Finite Set|finite]] [[Definition:Basis of Vector Space|basis]].
+\end{theorem}
+
+\begin{proof}
+This follows from [[Vector Space has Basis Between Linearly Independent Set and Finite Spanning Set]].
+It suffices to find:
+:A [[Definition:Linearly Independent Set|linearly independent]] [[Definition:Subset|subset]] $L\subset V$
+:A [[Definition:Finite Set|finite]] [[Definition:Generator of Vector Space|generator]] $S\subset V$
+with $L\subset S$.
+By [[Empty Set is Linearly Independent]], we make take $L = \O$ and $S$ any [[Definition:Finite Set|finite]] [[Definition:Generator of Vector Space|generator]], which exists because $V$ is [[Definition:Finitely Generated Module|finitely generated]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Bases of Finitely Generated Vector Space have Equal Cardinality}
+Tags: Bases of Vector Spaces
+
+\begin{theorem}
+Let $K$ be a [[Definition:Division Ring|division ring]].
+Let $G$ be a [[Definition:Finitely Generated Module|finitely generated]] [[Definition:Vector Space|$K$-vector space]].
+Then any two [[Definition:Basis of Vector Space|bases]] of $G$ are [[Definition:Finite Set|finite]] and [[Definition:Set Equivalence|equivalent]].
+{{improve|I still think there's a simpler way of saying "they have the same number of elements" than bringing in all that top-heavy technical set theoretic language of "set equivalence" and "cardinality" and so on. Since we are talking about a finite set, the complexities which arise with regard to transfinites do not arise, so there is no direct need to go into such depth.}}
+\end{theorem}
+
+\begin{proof}
+Since a [[Definition:BasisBasis of Vector Space|basis]] is, by definition, both [[Definition:Linearly Independent Set|linearly independent]] and a [[Definition:Generator of Vector Space|generator]], the result follows directly from [[Size of Linearly Independent Subset is at Most Size of Finite Generator]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Linearly Independent Subset also Independent in Generated Subspace}
+Tags: Linear Algebra
+
+\begin{theorem}
+Let $G$ be a [[Definition:Finitely Generated Module|finitely generated]] [[Definition:Vector Space|$K$-vector space]].
+Let $S$ be a [[Definition:Linearly Independent Set|linearly independent subset]] of $G$.
+Let $M$ be the [[Definition:Vector Subspace|subspace]] of $G$ [[Definition:Generator of Module|generated]] by $S$.
+If $M \ne G$, then $\forall b \in G: b \notin M$, the set $S \cup \set b$ is [[Definition:Linearly Independent Set|linearly independent]].
+\end{theorem}
+
+\begin{proof}
+Suppose that:
+:$\displaystyle \sum_{k \mathop = 1}^n \lambda_k x_k + \lambda b = 0$
+where $\sequence {x_n}$ is a [[Definition:Sequence of Distinct Terms|sequence of distinct vectors]] of $S$.
+If $\displaystyle \lambda \ne 0$, then $\displaystyle b = -\lambda^{-1} \paren {\sum_{k \mathop = 1}^n \lambda_k x_k} \in M$ which contradicts the definition of $b$.
+Hence $\lambda = 0$, and so:
+:$\displaystyle \sum_{k \mathop = 1}^n \lambda_k x_k = 0$
+Therefore $\lambda_1 = \lambda_2 = \cdots = \lambda_n = \lambda = 0$ as $S$ is [[Definition:Linearly Independent Set|linearly independent]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sufficient Conditions for Basis of Finite Dimensional Vector Space}
+Tags: Generators of Vector Spaces, Bases of Vector Spaces, Sufficient Conditions for Basis of Finite Dimensional Vector Space
+
+\begin{theorem}
+Let $K$ be a [[Definition:Division Ring|division ring]].
+Let $n \ge 0$ be a [[Definition:Natural Number|natural number]].
+Let $E$ be an [[Definition:Dimension of Vector Space|$n$-dimensional]] [[Definition:Vector Space|vector space]] over $K$.
+Let $B \subseteq E$ be a [[Definition:Subset|subset]] such that $\card B = n$.
+{{TFAE}}
+: $(1): \quad$ $B$ is a [[Definition:Basis of Vector Space|basis]] of $E$.
+: $(2): \quad$ $B$ is [[Definition:Linearly Independent Set|linearly independent]].
+: $(3): \quad$ $B$ is a [[Definition:Generator of Module|generator]] for $E$.
+\end{theorem}
+
+\begin{proof}
+=== 1 implies 2 and 3 ===
+Let $B$ be a [[Definition:Basis of Vector Space|basis]] of $E$.
+Then conditions $(2)$ and $(3)$ follow directly by the definition of [[Definition:Basis of Vector Space|basis]].
+{{qed|lemma}}
+=== 2 implies 1 ===
+Let $B$ be [[Definition:Linearly Independent Set|linearly independent]].
+Suppose $B$ does not [[Definition:Generator of Vector Space|generate]] $E$.
+Then, because $\card B = n$, by [[Linearly Independent Subset also Independent in Generated Subspace]] there would be a [[Definition:Linearly Independent Set|linearly independent subset]] of $n + 1$ [[Definition:Vector (Linear Algebra)|vectors]] of $E$.
+But this would contradict [[Size of Linearly Independent Subset is at Most Size of Finite Generator]].
+Thus condition $(2)$ implies $(1)$.
+{{qed|lemma}}
+=== 3 implies 1 ===
+Let $B$ be a [[Definition:Generator of Vector Space|generator]] for $E$.
+By [[Vector Space has Basis Between Linearly Independent Set and Finite Spanning Set]], $B$ contains a [[Definition:Basis of Vector Space|basis]] $B'$ of $E$.
+But $B'$ has $n$ elements and hence $B' = B$ by [[Bases of Finitely Generated Vector Space have Equal Cardinality]].
+Thus condition $(3)$ implies $(1)$.
+{{Qed}}
+\end{proof}
\ No newline at end of file