hcju/proofwikiqa huggingface dataset:
Follows directly from: :$(1) \quad$ the fundamental theorem of arithmetic :$(2) \quad$ the fact that $\struct {\Z, +, \times}$ is an integral domain :$(3) \quad$ the definitions complete factorization and equivalent factorizations. {{qed}} Category:Integral Domains Category:Integers Category:Factorization Category:Unique Factorization Domains 32n4jwtruzn4meukioxmgtt2arwfda5
That the trivial ordering is compatible with $\circ$ and $*$ follows from Trivial Ordering is Universally Compatible. Conversely, suppose that $\preceq$ is a ordering compatible with $\circ$ and $*$. {{proof wanted}}
The null ring, which contains one element, is ''not'' a field as it is trivial. Therefore any field must contain at least two elements. For $\struct {\set {0_R, 1_R}, +, \circ}$ to be a field: :$\struct {\set {0_R, 1_R}, +}$ must be an abelian group. This is fulfilled as this is the parity group. :$\struct {\set {0_R, 1_R}, \circ}$ must be a commutative division ring. This is fulfilled, as $\struct {\set {0_R, 1_R}^*, \circ} = \struct {\set {1_R}, \circ}$ is the trivial group. :$\circ$ needs to distribute over $+$. This follows directly from Ring Product with Zero and the behaviour of the identity element in a group. {{qed}}
This result follows directly from: :Field has no Proper Zero Divisors :By definition, that a field is a commutative ring. {{qed}}
Let $\struct {F, +, \circ}$ be a field whose zero is $0_F$ and whose unity is $1_F$. Suppose $\exists x, y \in F: x \circ y = 0_F$. Suppose $x \ne 0_F$. Then, by the definition of a field, $x^{-1}$ exists in $F$ and: :$y = 1_F \circ y = x^{-1} \circ x \circ y = x^{-1} \circ 0_F = 0_F$. Otherwise $x = 0_F$. So if $x \circ y = 0_F$, either $x = 0_F$ or $y = 0_F$ as we were to show. {{qed}}
Let $R$ be a finite integral domain whose unity is $1$ and whose zero is $0$. Let $a \in R$ such that $a \ne 0$. We wish to show that $a$ has a product inverse in $R$. So consider the map $f: R \to R$ defined by $f: x \mapsto a x$. We first show that the kernel of $f$ is trivial. Consider that: :$\map \ker f = \set {x \in R: \map f x = 0} = \set {x \in R: a x = 0}$ Since $R$ is an integral domain, it has no proper zero divisors and thus $a x = 0$ means that $a = 0$ or $x = 0$. Since, by definition, $a \ne 0$, then it must be true that $x = 0$. Therefore, $\map \ker f = \set 0$ and so $f$ is injective. Next, the Pigeonhole Principle gives us that an injective mapping from a finite set onto itself is surjective. Since $R$ is finite, the mapping $f$ is surjective. Finally, since $f$ is surjective and $1 \in R$, we have: :$\exists x \in R: \map f x = a x = 1$ So this $x$ is the product inverse of $a$ and the proof is complete. {{qed}}
Let $R$ be a finite integral domain with unity $1$ and zero $0$. Let $R^*$ denote $R \setminus \set 0$, the set $R$ without the zero. As $R$ is finite, we may enumerate the elements of $R$ as: :$x_0 = 0, x_1 = 1, x_2, x_3, \ldots, x_n$ Let $R^*$. Consider the elements: :$x_k x_1, x_k x_2, \ldots, x_k x_n$  which are members of $R$ by closure of the ring product. Since $R$ is an integral domain, by definition it has no proper zero divisors. Therefore all of $x_k x_1, x_k x_2, \ldots, x_k x_n$ are elements of $R^*$. Recall: :$x_k x_p = x_k x_q \implies x_p = x_q$  by the Cancellation Law of Ring Product of Integral Domain. So the list of products consists of $n$ distinct elements and is therefore equal to the whole $R^*$. Since $1 \in R^*$, we have that $x_k x_{k^*} = 1$ for some $k^*$. Therefore, there exists an inverse for every non-zero element, so $R$ is a field. As the underlying set of $R$ is finite, $R$ is by definition a Galois field. {{qed}}
An integral domain is by definition a ring which has no proper zero divisors. By definition, a Galois field is a field whose underlying set is finite. The result follows from Finite Ring with No Proper Zero Divisors is Field. {{qed}}
{{AimForCont}} $\struct {D, +, \circ}$ is a finite integral domain which is not a field. From Non-Field Integral Domain has Infinite Number of Ideals, $\struct {D, +, \circ}$ has an infinite number of distinct ideals. But this contradicts the assertion that $\struct {D, +, \circ}$ is finite. Hence the result by Proof by Contradiction. {{qed}}
$F$ is a field and $F \subseteq F$ from Set is Subset of Itself. {{Qed}}
For $\map Z K$ to be a subfield of $K$, it needs to be a division ring that is commutative. Thus the result follows directly from Center of Ring is Commutative Subring. {{qed}}
=== Necessary Condition === Let $\struct {R, +, \circ}$ be a field. The result follows from Field has 2 Ideals. {{qed|lemma}} === Sufficient Condition === Suppose that the only ideals of $\struct {R, +, \circ}$ are $\struct {R, +, \circ}$ and $\set {0_R}$. The result follows from Commutative and Unitary Ring with 2 Ideals is Field {{Qed}}
We have that the kernel of $K$ is an ideal. From Ideals of Division Ring, $\ker \left({K}\right)$ must therefore either be $0_K$ or $K$. * If $\ker \left({K}\right) = 0_K$, then by Ring Epimorphism with Trivial Kernel is Isomorphism $\phi$ is an ring isomorphism, thus making $R$ a division ring like $K$. * If $\ker \left({K}\right) = K$, then $\forall x \in K: \phi \left({K}\right) = 0_R$. As $\phi$ is an epimorphism, it is surjective and therefore $R = \left\{{0_R}\right\}$. {{qed}}
Let us set up mappings $\theta: H \to H x$ and $\phi: H x \to H$ as follows: :$\forall u \in H: \map \theta u = u x$ :$\forall v \in H x: \map \phi v = v x^{-1}$ From Element in Right Coset iff Product with Inverse in Subgroup: :$v \in H x \implies v x^{-1} \in H$ Now: :$\forall v \in H x: \theta \circ \map \phi v = v x^{-1} x = v$ :$\forall u \in H: \phi \circ \map \theta u = u x x^{-1} = u$ Thus $\theta \circ \phi = I_{H x}$ and $\phi \circ \theta = I_H$ are identity mappings. So $\theta = \phi^{-1}$: both are bijections and one is the inverse of the other. This establishes, for each $x \in G$, the set equivalence between $H$ and $H x$: :$H \simeq H x$ In particular, for any $x, y \in G$, it follows from $H x \simeq H$ and $H y \simeq H$ that: :$H x \simeq H y$ by Set Equivalence is Equivalence Relation. Similarly, we can set up mappings $\alpha: H \to x H$ and $\beta: x H \to H$ as follows: :$\forall u \in H: \map \alpha u = x u$ :$\forall v \in x H: \map \beta v = x^{-1} v$ Analogous to above reasoning gives $\alpha = \beta^{-1}$ which establishes $H \simeq x H$.  Also similarly, $x H \simeq y H$ for all $x, y \in G$. Hence the result. {{qed}}
Let there be exactly $r$ different left cosets of $H$ in $G$. Let a complete repetition-free list of these left cosets be: :$a_1 H, a_2 H, a_3 H \ldots, a_r H: a_1, a_2, \ldots, a_r \in G$ From Left Coset Space forms Partition, every element of $G$ is contained in exactly one of the left cosets. Let $x \in G$. Then, for $1 \le i \le r$: {{begin-eqn}} {{eqn | l = x       | o = \in       | r = H a_i^{-1}       | c =  }} {{eqn | ll= \iff       | l = x \paren {a_i^{-1} }^{-1}       | o = \in       | r = H       | c = Element in Right Coset iff Product with Inverse in Subgroup }} {{eqn | ll= \iff       | l = x a_i       | o = \in       | r = H       | c = Inverse of Group Inverse }} {{eqn | ll= \iff       | l = \paren {x^{-1} }^{-1} a_i       | o = \in       | r = H       | c = Inverse of Group Inverse }} {{eqn | ll= \iff       | l = x^{-1}       | o = \in       | r = a_i H       | c = Element in Left Coset iff Product with Inverse in Subgroup }} {{end-eqn}} Since $x^{-1} \in a_i H$ is true for precisely one value of $i$, it follows that $x \in H a_i^{-1}$ is also true for precisely that value of $i$. So there are exactly $r$ different right cosets of $H$ in $G$, and a complete repetition-free list of these is: :$H a_1^{-1}, H a_2^{-1}, H a_3^{-1}, \ldots, H a_r^{-1}$ The result follows. {{qed}}
Let $G$ be a group and let $H \le G$. Consider the mapping $\phi$ from the left coset space to the right coset space defined as: :$\forall g \in G: \map \phi {g H} = H g^{-1}$ We need to show that $\phi$ is a bijection. First we need to show that $\phi$ is well-defined. That is, that $a H = b H \implies \map \phi {a H} = \map \phi {b H}$. Suppose $a H = b H$. {{begin-eqn}} {{eqn | l = a H = b H       | o = \iff       | r = a^{-1} b \in H       | c = Left Cosets are Equal iff Product with Inverse in Subgroup }} {{eqn | l = H a^{-1} = H b^{-1}       | o = \iff       | r = a^{-1} \paren {b^{-1} }^{-1} \in H       | c = Right Cosets are Equal iff Product with Inverse in Subgroup }} {{end-eqn}} But $a^{-1} \paren {b^{-1} }^{-1} = a^{-1} b \in H$ as $a H = b H$. So $H a^{-1} = H b^{-1}$ and $\phi$ is well-defined. Next we show that $\phi$ is injective: Suppose $\exists x, y \in G: \map \phi {x H} = \map \phi {y H}$. Then $H x^{-1} = H y^{-1}$, so $x^{-1} = e_G x^{-1} = h y^{-1}$ for some $h \in H$. Thus $h = x^{-1} y \implies h^{-1} = y^{-1} x$. As $H$ is a subgroup, $h^{-1} \in H$. Thus: : $y^{-1} x \in H$ So by Left Cosets are Equal iff Product with Inverse in Subgroup: :$x H = y H$ Thus $\phi$ is injective. Next we show that $\phi$ is surjective: Let $H x$ be a right coset of $H$ in $G$. Since $x = \paren {x^{-1} }^{-1}$, $H x = \map \phi {x^{-1} H}$ and so $\phi$ is surjective. Thus $\phi$ constitutes a bijection from the left coset space to the right coset space, and the result follows. {{qed}}
{{begin-eqn}} {{eqn | o =        | r = \forall x, y \in G: x^{-1} y = y x^{-1}       | c =  }} {{eqn | o = \leadsto       | r = \paren {x \equiv^l y \pmod H \iff y \equiv^r x \pmod H}       | c = {{Defof|Congruence Modulo Subgroup}} }} {{eqn | o = \leadsto       | r = \paren {x \equiv^l y \pmod H \iff x \equiv^r y \pmod H}       | c = Congruence Modulo Subgroup is Equivalence Relation, therefore Symmetric }} {{end-eqn}} {{qed}}
$S^a$ is defined as $a \circ S \circ a^{-1}$ from the definition of the conjugate of a set. From the definition of subset product with a singleton, this can be seen to be the same thing as: :$S^a = \set a \circ S \circ \set {a^{-1} }$. Thus we can express $\paren {S^a}^b$ as $b \circ \paren {a \circ S \circ a^{-1} } \circ b^{-1}$, and understand that the {{RHS}} refers to subset products. From Subset Product within Semigroup is Associative (which applies because $\circ$ is associative), it then follows directly that: {{begin-eqn}} {{eqn | l = \paren {S^a}^b       | r = b \circ \paren {a \circ S \circ a^{-1} } \circ b^{-1}       | c = from above }} {{eqn | r = \paren {b \circ a} \circ S \circ \paren {a^{-1} \circ b^{-1} }       | c = Subset Product within Semigroup is Associative }} {{eqn | r = \paren {b \circ a} \circ S \circ \paren {b \circ a}^{-1}       | c = Inverse of Group Product }} {{eqn | r = S^{b \circ a}       | c = {{Defof|Conjugate of Group Subset}} }} {{end-eqn}} {{Qed}} === Proof for Alternative Definition === Using the same preliminary argument as above, we then follow: {{begin-eqn}} {{eqn | l = \paren {S^a}^b       | r = b^{-1} \circ \paren {a^{-1} \circ S \circ a} \circ b       | c = from above }} {{eqn | r = \paren {b^{-1} \circ a^{-1} } \circ S \circ \paren {a \circ b}       | c = Subset Product within Semigroup is Associative }} {{eqn | r = \paren {a \circ b}^{-1} \circ S \circ \paren {a \circ b}       | c = Inverse of Group Product }} {{eqn | r = S^{a \circ b}       | c = {{Defof|Conjugate (Group Theory)/Subset/Also defined as|Conjugate of Group Subset}} }} {{end-eqn}} {{Qed}}
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}}
Let $*: G \times G / H \to G / H$ be the 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 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 stabilizer of $a H$ under $*$. The result follows from Stabilizer is Subgroup. {{qed}}
Let $G$ be a group whose identity is $e$. From Inner Automorphisms form Subgroup of Automorphism Group, $\Inn G$ forms a subgroup of $\Aut G$. It remains to be shown that $\Inn G$ is normal in $\Aut G$. Let $\kappa_x: G \to G$ be the 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 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}}
Let $H \le G$ where $G$ is abelian. Then: {{begin-eqn}} {{eqn | l = y       | o = \in       | r = H^a       | c = where $H^a$ is the 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 = because $a y a^{-1} = y$ as $G$ is abelian }} {{end-eqn}} {{Qed}}
=== Necessary Condition === Suppose $H$ is 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 = Group properties }} {{end-eqn}} As $H$ is a 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}}
By Intersection of Subgroups is Subgroup, $H \cap N$ is a subgroup of $N$. It remains to be shown that $H$ is 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 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}}
Let $\phi: G_1 \to G_2$ be a group homomorphism, where the 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 normal subgroup of $G_1$. {{qed}}
Let $N \lhd G$ where $G$ is a 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 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 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 = 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}}
Let $N \lhd G$ where $G$ is a group. Consider $\paren {a \circ N} \circ \paren {b \circ N}$ as a 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 normal, each 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 set equality. {{qed}}
Let $N \lhd G$ where $G$ is a 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 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 normal }} {{end-eqn}} {{qed}}
Let $N \lhd G$ where $G$ is a 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}}
Let $N \lhd G$ where $G$ is a 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 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 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}}
First we note that because $G$ is abelian, from Subgroup of Abelian Group is Normal we have $N \lhd G$. Thus $G / N$ exists for ''all'' subgroups of $G$. Let $X = x N, Y = y N$ where $x, y \in G$. From the definition of 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 abelian. {{qed}} Category:Quotient Groups meo3u1osmjyzur71i3fnrzfyink4s68
Let $\mathcal R_\phi$ be the 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 equivalence defined by $K$. Thus, again by Quotient Theorem for Epimorphisms, there is a unique epimorphism $\psi: G / K \to H$ satisfying $\psi \circ q_K = \phi$. {{qed}}
=== Proof of Normal Subgroup === {{:Congruence Relation on Group induces Normal Subgroup}} === Proof of Equality of Relations === {{:Normal Subgroup induced by Congruence Relation defines that Congruence}} === Proof that Quotient Structure is Quotient Group === {{:Quotient Structure on Group defined by Congruence equals Quotient Group}}