avewright/proofwiki-math huggingface dataset:
Theorem: Connected Subspace of Linearly Ordered Space

Let $\struct {S, \preceq, \tau}$ be a linearly ordered space.
Let $Y \subseteq S$.
Then $Y$ is connected in $\struct {S, \tau}$ {{iff}} both of the following hold:
:$(1): \quad Y$ is order-convex in $S$
:$(2): \quad \struct {Y, \preceq \restriction_Y}$ is a linear continuum, where $\restriction$ denotes restriction.

Proof:
=== Necessary Conditions ===
Let $Y$ be connected in $\struct {S, \tau}$.
{{AimForCont}} $Y$ is not order-convex in $S$.
Then there exist $a, b, c \in S$ such that:
:$a \prec b \prec c$
:$a, c \in Y$ but $b \notin Y$
Recall that:
:$b^\prec$ denotes the (strict) lower closure of $b$: $b^\prec = \set {u \in S: u \prec b}$
:$b^\succ$ denotes the (strict) upper closure of $b$: $b^\succ = \set {u \in S: b \prec u}$
We have that $Y \cap b^\prec$ and $Y \cap b^\succ$ are separated in $Y$.
{{explain|We probably need a link to a result proving the above statement.}}
That is, by definition, $Y$ is disconnected.
Hence by Proof by Contradiction it follows that $Y$ is order-convex in $S$.
{{qed|lemma}}
{{AimForCont}} $Y$ is order-convex in $S$, but not a linear continuum.
Then by Linearly Ordered Space on Order-Convex Subset is Subspace Topology, the subspace topology on $Y$ is the same as the order topology on $Y$.
Thus by Linearly Ordered Space is Connected iff Linear Continuum, $Y$ is disconnected.
Hence by Proof by Contradiction it follows that $Y$ is a linear continuum.
{{qed|lemma}}
Theorem: Abnormal Subgroup is Self-Normalizing Subgroup

Let $G$ be a group.
Let $H$ be an abnormal subgroup of $G$.
Then $H$ is a self-normalizing subgroup of $G$.

Proof:
We have:
* Abnormal Subgroup is Weakly Abnormal Subgroup
* Weakly Abnormal Subgroup is Self-Normalizing Subgroup
Hence the result.
{{qed}}
Category:Abnormal Subgroups
Category:Self-Normalizing Subgroups
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Theorem: Duality Principle (Category Theory)/Conceptual Duality

Let $\Sigma$ be a statement about metacategories, be it in natural language or otherwise.
Suppose that $\Sigma$ holds for all metacategories.
Then so does its dual statement $\Sigma^*$.

Proof:
From Dual Category of Dual Category, any metacategory $\mathbf C$ may be regarded as a dual category.
Thus it is sufficient to verify $\Sigma^*$ holds in all dual categories.
In any dual category $\mathbf C^{\mathrm {op}}$, we see that:
:$\operatorname{dom} f^{\mathrm {op}} = \paren {\operatorname{cod} f}^{\mathrm {op}}$
:$\operatorname{cod} f^{\mathrm {op}} = \paren {\operatorname{dom} f}^{\mathrm {op}}$
:$f^\mathrm {op} \circ g^\mathrm {op} = \paren {g \circ f}^{\mathrm {op}}$
Thus, the interpretation of $\Sigma^*$ in $\mathbf C^{\mathrm {op}}$ comes down to the interpretation of $\Sigma$ in $\mathbf C$.
That is, except for amending every symbol $\star$ of $\mathbf C$ with a superscript "$\mathrm {op}$", yielding $\star^\mathrm {op}$.
This operation does not change the validity of $\Sigma$ in any possible way, and $\Sigma$ was assumed to hold in $\mathbf C$.
Hence we conclude that $\Sigma^*$ must hold in $\mathbf C^{\mathrm {op}}$ as well.
The result follows.
{{qed}}
Theorem: Product of Complex Number with Conjugate

Let $z = a + i b \in \C$ be a complex number.
Let $\overline z$ denote the complex conjugate of $z$.
Then:
:$z \overline z = a^2 + b^2 = \cmod z^2$
and thus is wholly real.

Proof:
By the definition of a complex number, let $z = a + i b$ where $a$ and $b$ are real numbers.
Then:
{{begin-eqn}}
{{eqn | l = z \overline z
      | r = \paren {a + i b} \paren {a - i b}
      | c = {{Defof|Complex Conjugate}}
}}
{{eqn | r = a^2 + a \cdot i b + a \cdot \paren {-i b} + i \cdot \paren {-i} \cdot b^2
      | c = {{Defof|Complex Multiplication}}
}}
{{eqn | r = a^2 + i a b - i a b + b^2
      | c = 
}}
{{eqn | r = a^2 + b^2
      | c = 
}}
{{eqn | r = \paren {\sqrt {a^2 + b^2} }^2
      | c = 
}}
{{eqn | r = \cmod z^2
      | c = {{Defof|Complex Modulus}}
}}
{{end-eqn}}
As $a^2 + b^2$ is wholly real, the result follows.
{{qed}}
Theorem: Compact Subspace of Linearly Ordered Space/Lemma 1

Let $\struct {X, \preceq, \tau}$ be a linearly ordered space.
Let $Y \subseteq X$ be a non-empty subset of $X$.
Let $Y$ be a compact subspace of $\struct {X, \tau}$.
Then $\struct {Y, \preceq \restriction_Y}$ is a complete lattice, where $\restriction$ denotes restriction.

Proof:
{{AimForCont}} $Y$ is not a complete lattice.
Then there is an $S \subseteq Y$ with no least upper bound in $Y$.
Let $U = \set {b \in Y: b \text { is an upper bound of } S}$.
Note that $U$ may be empty.
Let:
:$\UU = \set {u^\succeq: u \in U}$
:$\LL = \set {s^\preceq: s \in S}$.
where:
:$u^\succeq$ is the upper closure of $u$ in $U$
:$s^\preceq$ is the lower closure of $s$ in $S$.
Then $\UU \cup \LL$ covers $Y$:
Let $y \in Y$.
Suppose $y$ is an upper bound for $S$.
Then because $S$ has no least upper bound in $Y$, there exists a $y' \in U$ such that $y' \prec y$.
Thus $y \in y'^\succeq$, so, in the case where $y$ is an upper bound for $S$:
:$\ds y \in \bigcup \UU$
Otherwise, that is if $y$ is ''not''  an upper bound for $S$, there exists an $s \in S$ such that $y \prec s$.
Hence in this case:
:$\ds y \in \bigcup \LL$
$\UU \cup \LL$ has no finite subcover:
Let $F_U$ and $F_L$ be finite subsets of $U$ and $S$, respectively.
Define:
:$\FF_U = \set {u^\succeq: u \in F_S}$
:$\FF_L = \set {s^\preceq: s \in F_L}$
We will show that $\FF_U \cup \FF_L$ does not cover $Y$.
Suppose to the contrary that $\FF_U \cup \FF_L$ is a cover for $Y$.
Since $Y$ is non-empty, $F_L$ or $F_U$ is non-empty.
If $F_L$ is non-empty, it has a maximum $m$.
Since $S$ has no least upper bound, it has no maximum
Therefore there is an $s \in S$ such that $m \prec s$.
Thus $\ds s \notin \bigcup \FF_L$.
By the definition of upper bound, no upper bound of $S$ precedes $m$.
So $\ds m \notin \bigcup \FF_U$.
Thus $\FF_L \cup \FF_U$ does not cover $Y$.
This is a contradiction of our supposition that it does.
Hence it cannot be the case that $\FF_U \cup \FF_L$ is a cover for $Y$.
If $F_U$ is non-empty, it has a minimum $m$.
Then since $S$ has no least upper bound in $Y$, there is a $u \in U$ such that $u \prec m$.
Thus $\ds u \notin \bigcup \FF_U$.
But since $u$ is an upper bound of $S$, it is not succeeded by any element of $S$.
So $\ds u \notin \bigcup \FF_L$.
Thus $\FF_L \cup \FF_U$ does not cover $Y$, a contradiction.
So $\UU \cup \LL$ is an open cover of $Y$ with no finite subcover
So $Y$ is not compact, which contradicts the supposition.
By Proof by Contradiction, it follows that $\struct {Y, \preceq \restriction_Y}$ is a complete lattice.
{{qed}}
Category:Compact Subspace of Linearly Ordered Space
88sn5thyqs2p7g3m4q3okedd2g4srr1
Theorem: Inclusion Mapping on Subgroup is Homomorphism

Let $\struct {G, \circ}$ be a group.
Let $\struct {H, \circ_{\restriction H} }$ be a subgroup of $G$.
Let $i: H \to G$ be the inclusion mapping from $H$ to $G$.
Then $i$ is a group homomorphism.

Proof:
Let $x, y \in H$.
From {{Group-axiom|0}}, $x \circ_{\restriction H} y \in H$.
Then:
{{begin-eqn}}
{{eqn | l = \map i {x \circ_{\restriction H} y}
      | r = x \circ_{\restriction H} y
      | c = {{Defof|Inclusion Mapping}}
}}
{{eqn | r = x \circ y
      | c = {{Defof|Restriction of Operation}}
}}
{{eqn | r = \map i x \circ \map i y
      | c = {{Defof|Inclusion Mapping}}
}}
{{end-eqn}}
Hence the result by definition of group homomorphism.
{{qed}}
Theorem: Mapping to Identity is Unique Constant Homomorphism

Let $\struct {G, \circ}$ and $\struct {H, *}$ be groups whose identities are $e_G$ and $e_H$ respectively.
Then there exists a unique constant mapping from $G$ to $H$ which is a homomorphism:
:$\phi_{e_H}: G \to H: \forall g \in G: \map {\phi_{e_H} } g = e_H$

Proof:
Let $h \in H$ such that $\phi_h: G \to H$ is a (group) homomorphism, where $\phi_h$ is defined as:
:$\forall g \in G: \map {\phi_h} g = h$
Then from Group Homomorphism Preserves Identity:
:$\map {\phi_h} {e_G} = e_H$
and so $h = e_H$.
Hence the result by definition of constant mapping.
It remains to prove that such a constant mapping is in fact a homomorphism.
Let $x, y \in G$.
{{begin-eqn}}
{{eqn | l = \map {\phi_{e_H} } {x \circ y}
      | r = e_H
      | c = Definition of $\phi_{e_H}$
}}
{{eqn | r = e_H * e_H
      | c = {{Defof|Identity Element}}
}}
{{eqn | r = \map {\phi_{e_H} } x \circ \map {\phi_{e_H} } y
      | c = by definition of $\phi_{e_H}$
}}
{{end-eqn}}
Thus the morphism property is demonstrated, and $\phi_{e_H}$ is seen to be a homomorphism.
{{qed}}
Theorem: Set Union is Self-Distributive/General Result

Let $\family {\mathbb S_i}_{i \mathop \in I}$ be an $I$-indexed family of sets of sets.
Then:
:$\ds \bigcup_{i \mathop \in I} \bigcup \mathbb S_i = \bigcup \bigcup_{i \mathop \in I} \mathbb S_i$

Proof:
By the definition of set union, we have:
:$\ds \forall i \in I: \forall S \in \mathbb S_i: S \in \bigcup_{j \mathop \in I} \mathbb S_j$
Hence, by Set is Subset of Union: General Result:
:$\ds \forall i \in I: \forall S \in \mathbb S_i: S \subseteq \bigcup \bigcup_{j \mathop \in I} \mathbb S_j$
By Union is Smallest Superset: General Result, it follows that:
:$\ds \forall i \in I: \bigcup \mathbb S_i \subseteq \bigcup \bigcup_{j \mathop \in I} \mathbb S_j$
Therefore, by Union is Smallest Superset: Family of Sets, we conclude that:
:$\ds \bigcup_{i \mathop \in I} \bigcup \mathbb S_i \subseteq \bigcup \bigcup_{j \mathop \in I} \mathbb S_j$
By Set is Subset of Union General Result and then Set is Subset of Union of Family, we have:
:$\ds \forall i \in I: \forall S \in \mathbb S_i: S \subseteq \bigcup \mathbb S_i \subseteq \bigcup_{j \mathop \in I} \bigcup \mathbb S_j$
Because $\subseteq$ is transitive, we can apply the definition set union to conclude that:
:$\ds \forall S \in \bigcup_{i \mathop \in I} \mathbb S_i: S \subseteq \bigcup_{j \mathop \in I} \bigcup \mathbb S_j$
Hence, by Union is Smallest Superset: General Result:
:$\ds \bigcup \bigcup_{i \mathop \in I} \mathbb S_i \subseteq \bigcup_{j \mathop \in I} \bigcup \mathbb S_j$
By definition of set equality, the result follows.
{{qed}}
Category:Set Union is Self-Distributive
3mbjhuuxw75qsxd5urj4liu61ryredf
Theorem: Existence and Uniqueness of Generated Topology

Let $X$ be a set.
Let $\SS \subseteq \powerset X$ be a subset of the power set of $X$.
Then there exists a unique topology $\map \tau \SS$ on $X$ such that:
:$(1): \quad \SS \subseteq \map \tau \SS$.
:$(2): \quad$ For any topology $\TT$ on $X$, the implication $\SS \subseteq \TT \implies \map \tau \SS \subseteq \TT$ holds.

Proof:
=== Existence ===
Define:
:$\mathbb E = \leftset {\TT \subseteq \powerset X: \SS \subseteq \TT}$ and $\TT$ is a topology on $\rightset X$
Since $\powerset X$ is a topology on $X$, it follows that $\mathbb E$ is non-empty.
Hence, we can define:
:$\ds \map \tau \SS = \bigcap \mathbb E$
It follows that $\map \tau \SS$ is a topology on $X$.
By Intersection is Largest Subset: General Result, it follows that $\SS \subseteq \map \tau \SS$.
By Intersection is Subset: General Result, it follows that if $\SS \subseteq \TT$ and $\TT$ is a topology on $X$, then $\map \tau \SS \subseteq \TT$.
{{qed|lemma}}
Theorem: Equivalent Conditions for Cover by Collection of Subsets

Let $X$ be a set.
Then the following conditions are equivalent for a subset $\CC \subseteq \powerset X$ of the power set of $X$:
:$(1): \quad \CC$ is a cover for $X$.
:$(2): \quad \ds X = \bigcup \CC$.
:$(3): \quad \ds \exists \SS \subseteq \CC: X = \bigcup \SS$.

Proof:
=== $(1)$ implies $(2)$ ===
By definition, $\CC$ covers $X$ {{iff}} $X \subseteq \ds \bigcup \CC$.
By Union of Subsets is Subset, we have that:
:$\ds \bigcup \CC \subseteq X$
since $\CC \subseteq \powerset X$.
By definition of set equality, it follows that $X = \ds \bigcup \CC$.
{{qed|lemma}}
Theorem: Subgroup equals Conjugate iff Normal

:$\forall g \in G: g \circ N \circ g^{-1} = N$
:$\forall g \in G: g^{-1} \circ N \circ g = N$

Proof:
By definition, a subgroup is normal in $G$ {{iff}}:
:$\forall g \in G: g \circ N = N \circ g$
First note that:
:$(1): \quad \paren {\forall g \in G: g \circ N \circ g^{-1} = N} \iff \paren {\forall g \in G: g^{-1} \circ N \circ g = N}$
which is shown by, for example, setting $h := g^{-1}$ and substituting.
=== Necessary Condition ===
Suppose that $N$ is normal in $G$.
Then:
{{begin-eqn}}
{{eqn | q = \forall g \in G
      | l = g \circ N
      | r = N \circ g
      | c = {{Defof|Normal Subgroup}} of $G$
}}
{{eqn | ll= \leadsto
      | l = \paren {g \circ N} \circ g^{-1}
      | r = \paren {N \circ g} \circ g^{-1}
      | c = {{Defof|Subset Product}}
}}
{{eqn | ll= \leadsto
      | l = g \circ N \circ g^{-1}
      | r = N \circ \paren {g \circ g^{-1} }
      | c = Subset Product within Semigroup is Associative: Corollary
}}
{{eqn | ll= \leadsto
      | l = g \circ N \circ g^{-1}
      | r = N \circ e
      | c = {{Defof|Inverse Element}}
}}
{{eqn | ll= \leadsto
      | l = g \circ N \circ g^{-1}
      | r = N
      | c = Coset by Identity
}}
{{end-eqn}}
Similarly:
{{begin-eqn}}
{{eqn | q = \forall g \in G
      | l = N \circ g
      | r = g \circ N
      | c = {{Defof|Normal Subgroup}}
}}
{{eqn | ll= \leadsto
      | r = g^{-1} \circ \paren {N \circ g}
      | l = g^{-1} \circ \paren {g \circ N}
      | c = {{Defof|Subset Product}}
}}
{{eqn | ll= \leadsto
      | l = g^{-1} \circ N \circ g
      | r = \paren {g^{-1} \circ g} \circ N
      | c = Subset Product within Semigroup is Associative: Corollary
}}
{{eqn | ll= \leadsto
      | l = g^{-1} \circ N \circ g
      | r = e \circ N
      | c = {{Defof|Inverse Element}}
}}
{{eqn | ll= \leadsto
      | l = g^{-1} \circ N \circ g
      | r = N
      | c = Coset by Identity
}}
{{end-eqn}}
{{qed|lemma}}
=== Sufficient Condition ===
Let $N$ be a subgroup of $G$ such that:
: $\forall g \in G: g \circ N \circ g^{-1} = N$
and so from $(1)$ above:
: $\forall g \in G: g^{-1} \circ N \circ g = N$
Then:
{{begin-eqn}}
{{eqn | q = \forall g \in G
      | l = g \circ N \circ g^{-1}
      | r = N
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \paren {g \circ N \circ g^{-1} } \circ g
      | r = N \circ g
      | c = {{Defof|Subset Product}}
}}
{{eqn | ll= \leadsto
      | l = \paren {g \circ N} \circ \paren {g^{-1} \circ g}
      | r = N \circ g
      | c = Subset Product within Semigroup is Associative: Corollary
}}
{{eqn | ll= \leadsto
      | l = \paren {g \circ N} \circ e
      | r = N \circ g
      | c = {{Defof|Inverse Element}}
}}
{{eqn | ll= \leadsto
      | l = g \circ N
      | r = N \circ g
      | c = Coset by Identity
}}
{{end-eqn}}
{{qed}}
Theorem: Subgroup equals Conjugate iff Normal

:$\forall g \in G: g \circ N \circ g^{-1} = N$
:$\forall g \in G: g^{-1} \circ N \circ g = N$

Proof:
From Subgroup is Superset of Conjugate iff Normal, $N$ is normal in $G$ {{iff}}:
:$\forall g \in G: N \supseteq g \circ N \circ g^{-1}$
:$\forall g \in G: N \supseteq g^{-1} \circ N \circ g$
From Subgroup is Subset of Conjugate iff Normal, $N$ is normal in $G$ {{iff}}:
:$\forall g \in G: N \subseteq g \circ N \circ g^{-1}$
:$\forall g \in G: N \subseteq g^{-1} \circ N \circ g$
The result follows by definition of set equality.
{{qed}}
Theorem: Subgroup is Normal iff Contains Conjugate Elements

Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $N$ be a subgroup of $G$.
Then $N$ is normal in $G$ {{iff}}:
{{begin-eqn}}
{{eqn | n = 1
      | q = \forall g \in G
      | l = n \in N
      | o = \iff
      | r = g \circ n \circ g^{-1} \in N
}}
{{eqn | n = 2
      | q = \forall g \in G
      | l = n \in N
      | o = \iff
      | r = g^{-1} \circ n \circ g \in N
}}
{{end-eqn}}

Proof:
By definition, a subgroup is normal in $G$ {{iff}}:
:$\forall g \in G: g \circ N = N \circ g$
=== Necessary Condition ===
Suppose that $g \circ N = N \circ g$, by definition 1 of normality in $G$.
Let $n \in N$.
Then:
{{begin-eqn}}
{{eqn | l = g \circ n
      | o = \in
      | r = N \circ g
      | c = {{Defof|Coset}}
}}
{{eqn | ll= \leadstoandfrom
      | q = \exists n_1 \in N
      | l = g \circ n
      | r = n_1 \circ g
      | c = {{Defof|Coset}}
}}
{{eqn | ll= \leadstoandfrom
      | l = g \circ n \circ g^{-1}
      | r = n_1
      | c = Division Laws for Groups
}}
{{eqn | ll= \leadstoandfrom
      | l = g \circ n \circ g^{-1}
      | o = \in
      | r = N
      | c = Definition of $n_1$
}}
{{end-eqn}}
{{qed|lemma}}
=== Sufficient Condition ===
Suppose that:
: $\forall g \in G: \paren {n \in N \iff g \circ n \circ g^{-1} \in N}$
Let $g \circ n \circ g^{-1} \in N$.
{{begin-eqn}}
{{eqn | q = \exists n_1 \in N
      | l = g \circ n \circ g^{-1}
      | r = n_1
      | c = 
}}
{{eqn | ll= \leadsto
      | l = g \circ n
      | r = n_1 \circ g
      | c = Division Laws for Groups
}}
{{eqn | ll= \leadsto
      | l = g \circ n
      | o = \in
      | r = N \circ g
      | c = {{Defof|Coset}}
}}
{{eqn | ll= \leadsto
      | l = g \circ N
      | o = \subseteq
      | r = N \circ g
      | c = {{Defof|Subset}}
}}
{{end-eqn}}
Similarly:
{{begin-eqn}}
{{eqn | q = \exists n_2 \in N
      | l = g \circ n \circ g^{-1}
      | r = n_2
      | c = 
}}
{{eqn | ll= \leadsto
      | l = n \circ g^{-1}
      | r = g^{-1} \circ n_2
      | c = Division Laws for Groups
}}
{{eqn | ll= \leadsto
      | l = n \circ g^{-1}
      | o = \in
      | r = g^{-1} \circ N
      | c = {{Defof|Coset}}
}}
{{eqn | ll= \leadsto
      | l = N \circ g^{-1}
      | o = \subseteq
      | r = g^{-1} \circ N
      | c = {{Defof|Subset}}
}}
{{end-eqn}}
As $g$ is arbitrary, then so is $g^{-1}$.
Thus:
:$N \circ g \subseteq g \circ N$
By definition of set equality:
:$g \circ N = N \circ g$
Theorem: Subgroup is Normal iff Left Cosets are Right Cosets

Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $N$ be a subgroup of $G$.
Then $N$ is normal in $G$ (by definition 1) {{iff}}:
:Every right coset of $N$ in $G$ is a left coset
or equivalently:
:The right coset space of $N$ in $G$ equals its left coset space.

Proof:
=== Necessary Condition ===
Let $N$ be a normal subgroup of $G$ by Definition 1.
Then the equality of the coset spaces follows directly from definition of normal subgroup and coset.
{{qed|lemma}}
=== Sufficient Condition ===
Suppose that every right coset of $N$ in $G$ is a left coset of $N$ in $G$.
Let $g \in G$.
Since every right coset of $N$ in $G$ is a left coset, there exists an $h \in G$ such that $N \circ g = h \circ N$.
By Element of Group is in its own Coset:
:$g \in N \circ g = h \circ N$
From Element in Left Coset iff Product with Inverse in Subgroup:
: $g^{-1} \circ h \in N$
Then:
{{begin-eqn}}
{{eqn | l = N \circ g
      | r = \paren {g \circ g^{-1} } \circ \paren {h \circ N}
      | c = {{Defof|Inverse Element}} and Coset by Identity
}}
{{eqn | r = g \circ \paren {\paren { g^{-1} \circ h } \circ N}
      | c = Subset Product within Semigroup is Associative: Corollary
}}
{{eqn | r = g \circ N
      | c = $g^{-1} \circ h \in N$ and Left Coset Equals Subgroup iff Element in Subgroup
}}
{{end-eqn}}
Since this holds for all $g \in G$, $N$ is normal in $G$ (by definition 1).
{{Qed}}
Theorem: Symmetric Group has Non-Normal Subgroup

Let $S_n$ be the (full) symmetric group on $n$ elements, where $n \ge 3$.
Then $S_n$ contains at least one subgroup which is not normal.

Proof:
Let $S_n$ act on the set $S$.
Let $e$ be the identity of $S_n$, by definition the identity mapping $I_S$ on $S$.
As $S$ has at least three elements, three can be arbitrary selected and called $a$, $b$ and $c$.
Let $\rho$ be a transposition of $S_n$, transposing elements $a$ and $b$.
$\rho$ can be described in cycle notation as $\paren {a \ b}$.
From Transposition is Self-Inverse it follows that $\set {e, \rho}$ is a subgroup of $S_n$.
Let $\pi$ be the permutation on $S$ described in cycle notation as $\paren {a \ b \ c}$.
By inspection it is found that $\pi^{-1} = \paren {a \ c \ b}$.
Then we have:
{{begin-eqn}}
{{eqn | l = \pi^{-1} \rho \pi
      | r = \paren {a \ c \ b} \paren {a \ b} \paren {a \ b \ c}
      | c = 
}}
{{eqn | r = \paren {a \ c}
      | c = by evaluation
}}
{{eqn | o = \notin
      | r = \set {e, \rho}
      | c = 
}}
{{end-eqn}}
So, by definition, $\set {e, \rho}$ is not a normal subgroup.
{{qed}}
Theorem: Disjoint Union is Coproduct in Category of Sets

Let $\mathbf{Set}$ be the category of sets.
Let $S$ and $T$ be sets.
Then their disjoint union $S \sqcup T$ is a coproduct in $\mathbf{Set}$.

Proof:
We have the implicit mappings $i_1: S \to S \sqcup T$ and $i_2: T \to S \sqcup T$ defined by:
:$i_1 \left({s}\right) = \left({s, 1}\right)$
:$i_2 \left({t}\right) = \left({t, 2}\right)$
Now given a set $V$ and mappings $f: S \to V$ and $g: T \to V$, there is to be a unique $\left[{f, g}\right]: S \sqcup T \to V$ such that:
:$f = \left[{f, g}\right] \circ i_1$
:$g = \left[{f, g}\right] \circ i_2$
Define $\left[{f, g}\right]$ by:
:$\left[{f, g}\right] \left({x, \delta}\right) := \begin{cases}
 f \left({x}\right) & \text{if $\delta = 1$} \\
 g \left({x}\right) & \text{if $\delta = 2$}\end{cases}$
Now it is immediate that $\left[{f, g}\right]$ so defined satisfies the two conditions above.
Furthermore, these conditions fix $\left[{f, g}\right]$ uniquely, since every $\left({x, \delta}\right) \in S \sqcup T$ has $\delta = 1$ or $\delta = 2$.
Hence the result, by definition of coproduct.
{{qed}}
Theorem: Inclusion Mapping on Subgroup is Monomorphism

Let $\struct {G, \circ}$ be a group.
Let $\struct {H, \circ {\restriction_H} }$ be a subgroup of $G$.
Let $i: H \to G$ be the inclusion mapping from $H$ to $G$.
Then $i$ is a group monomorphism.

Proof:
We have:
* Inclusion Mapping on Subgroup is Homomorphism
* Inclusion Mapping is Injection
The result follows by definition of (group) monomorphism.
{{qed}}
Theorem: Inverse of Group Isomorphism is Isomorphism

Let $\struct {G, \circ}$ and $\struct {H, *}$ be groups.
Let $\phi: \struct {G, \circ} \to \struct {H, *}$ be a mapping.
Then $\phi$ is an isomorphism {{iff}} $\phi^{-1}: \struct {H, *} \to \struct {G, \circ}$ is also an isomorphism.

Proof:
A specific instance of Inverse of Algebraic Structure Isomorphism is Isomorphism.
{{qed}}
Theorem: Inverse of Group Isomorphism is Isomorphism

Let $\struct {G, \circ}$ and $\struct {H, *}$ be groups.
Let $\phi: \struct {G, \circ} \to \struct {H, *}$ be a mapping.
Then $\phi$ is an isomorphism {{iff}} $\phi^{-1}: \struct {H, *} \to \struct {G, \circ}$ is also an isomorphism.

Proof:
=== Necessary Condition ===
Let $\phi: G \to H$ be an isomorphism.
Then by definition $\phi$ is a bijection.
From Bijection iff Inverse is Bijection it follows that:
:$\exists \phi^{-1}: \struct {H, *} \to \struct {G, \circ}$
such that $\phi^{-1}$ is also a bijection.
Thus:
{{begin-eqn}}
{{eqn | q = \forall g \in G, h \in H
      | l = \map \phi g = h
      | o = \iff
      | r = \map {\phi^{-1} } h = g
      | c = Inverse Element of Bijection
}}
{{eqn | ll= \leadsto
      | l = \map \phi {g_1 \circ g_2}
      | r = h_1 * h_2
      | c = {{Defof|Morphism Property}}
}}
{{eqn | ll= \leadsto
      | l = \map {\phi^{-1} } {h_1 * h_2}
      | r = g_1 \circ g_2
      | c = Inverse Element of Bijection
}}
{{eqn | r = \map {\phi^{-1} } {h_1} \circ \map {\phi^{-1} } {h_2}
}}
{{end-eqn}}
So $\phi^{-1}: \struct {H, *} \to \struct {G, \circ}$ is a homomorphism.
$\phi^{-1}$ is also (from above) a bijection.
Thus, by definition, $\phi^{-1}$ is an isomorphism.
{{qed|lemma}}
=== Sufficient Condition ===
Let $\phi^{-1}: \struct {H, *} \to \struct {G, \circ}$ be an isomorphism.
Applying the same result as above in reverse, we have that $\paren {\phi^{-1} }^{-1}: \struct {G, \circ} \to \struct {H, *}$ is also an isomorphism.
But by Inverse of Inverse of Bijection:
: $\paren {\phi^{-1} }^{-1} = \phi$
and hence the result.
{{qed}}
Theorem: Topology as Magma of Sets

The concept of a topology is an instance of a magma of sets.

Proof:
It will suffice to define partial mappings such that the axiom for a magma of sets crystallises into the axioms for a topology.
Let $X$ be any set, and let $\powerset X$ be its power set.
Define:
:$\phi_1: \powerset X \to \powerset X: \map {\phi_1} S := X$
:$\phi_2: \powerset X^2 \to \powerset X: \map {\phi_2} {S, T} := S \cap T$
For each index set $I$, define:
:$\ds \phi_I: \powerset X^I \to \powerset X: \map {\phi_I} {\family {S_i}_{i \mathop \in I} } := \bigcup_{i \mathop \in I} S_i$
It is blatantly obvious that these partial mappings capture the axioms for a topology.
{{finish}}
{{qed}}
Category:Topology
Category:Magmas of Sets
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