avewright/proofwiki-math huggingface dataset:
Theorem: Conjugate of Set by Group Product

Let $\struct {G, \circ}$ be a group.
Let $S \subseteq G$.
Let $S^a$ denote the $G$-conjugate of $S$ by $a$ as:
:$S^a := \set {y \in G: \exists x \in S: y = a \circ x \circ a^{-1} } = a \circ S \circ a^{-1}$
Then:
:$\paren {S^a}^b = S^{b \circ a}$

Proof:
$S^a$ 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}}
Theorem: Conjugate of Subgroup is Subgroup

Let $G$ be a group.
Let $H \le G$ be a subgroup of $G$.
Then the conjugate of $H$ by $a$ is a subgroup of $G$:
:$\forall H \le G, a \in G: H^a \le G$

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 = {{Group-axiom|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}}
Theorem: Conjugate of Subgroup is Subgroup

Let $G$ be a group.
Let $H \le G$ be a subgroup of $G$.
Then the conjugate of $H$ by $a$ is a subgroup of $G$:
:$\forall H \le G, a \in G: H^a \le G$

Proof:
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}}
Theorem: Inner Automorphisms form Normal Subgroup of Automorphism Group

Let $G$ be a group.
Then the set $\Inn G$ of all inner automorphisms of $G$ is a normal subgroup of the automorphism group $\Aut G$ of $G$:
:$\Inn G \lhd \Aut G$

Proof:
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 | q = \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}}
Theorem: Subgroup of Abelian Group is Normal

Every subgroup of an abelian group is normal.

Proof:
Let $G$ be an abelian group.
Let $H \le G$ be a subgroup of $G$.
Then for all $a \in G$:
{{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}}
Theorem: Subgroup is Normal iff it contains Product of Inverses

A subgroup $H$ of a group $G$ is normal {{iff}}:
:$\forall a, b \in G: a b \in H \implies a^{-1} b^{-1} \in H$

Proof:
=== 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 $\forall a, b \in G: a b \in H \implies a^{-1} b^{-1} \in H$.
Theorem: Intersection with Normal Subgroup is Normal

Let $G$ be a group.
Let $H$ be a subgroup of $G$, and let $N$ be a normal subgroup of $G$.
Then $H \cap N$ is a normal subgroup of $H$.

Proof:
By Intersection of Subgroups is Subgroup, $H \cap N$ is a subgroup of $N$.
It remains to be shown that $H \cap N$ is normal in $H$.
Let $g \in H$ and $x \in H \cap N$. 
However, $H \le G$ and $ g \in H$ imply:
:$g \in G$
Because $N \lhd G$:
:$\forall n \in N: \forall g \in G: g n g^{-1} \in N$
Because $H \le G$ and therefore closed: 
:$\forall x \in H \cap N: \forall g \in H: g x g^{-1} \in H$
Therefore, from the definition of normal subgroups:
:$N \lhd G \implies g x g^{-1} \in N$
:$x \in N$
:$g \in G$
Thus:
:$g x g^{-1} \in H \cap N$
Because $H$ is a subgroup, $g x g^{-1} \in H \cap N$, $g \in H$, and the definition of normal subgroups:
:$H \cap N \lhd H$
{{qed}}
Theorem: Kernel is Normal Subgroup of Domain

Let $\phi$ be a group homomorphism.
Then the kernel of $\phi$ is a normal subgroup of the domain of $\phi$:
:$\map \ker \phi \lhd \Dom \phi$

Proof:
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}}
Theorem: Coset Product is Well-Defined

Let $\struct {G, \circ}$ be a group.
Let $N$ be a normal subgroup of $G$.
Let $a, b \in G$.
Then the coset product:
:$\paren {a \circ N} \circ \paren {b \circ N} = \paren {a \circ b} \circ N$
is well-defined.

Proof:
Let $N \lhd G$ where $G$ is a group.
Let $a, a', b, b' \in G$ such that:
:$a \circ N = a' \circ N$
and:
:$b \circ N = b' \circ N$
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 = {{hypothesis}}
}}
{{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
      | q = \exists n \in N
      | l = b^{-1} \circ a^{-1} \circ a'
      | r = n \circ b^{-1}
      | c = {{Defof|Subset Product}}
}}
{{eqn | ll= \leadsto
      | l = \paren {b^{-1} \circ a^{-1} \circ a'} \circ b'
      | r = \paren {n \circ b^{-1} } \circ b'
      | c = {{Group-axiom|0}}
}}
{{eqn | ll= \leadsto
      | l = \paren {b^{-1} \circ a^{-1} } \circ \paren {a' \circ b'}
      | r = \paren {n \circ b^{-1} } \circ b'
      | c = {{Group-axiom|1}}
}}
{{eqn | ll= \leadsto
      | l = \paren {a \circ b}^{-1} \circ \paren {a' \circ b'}
      | r = \paren {n \circ b^{-1} } \circ b'
      | c = Inverse of Group Product
}}
{{eqn | ll= \leadsto
      | l = \paren {a \circ b}^{-1} \circ \paren {a' \circ b'}
      | r = n \circ \paren{ b^{-1} \circ b' }
      | c = {{Group-axiom|1}}
}}
{{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}}
Theorem: Coset Product is Well-Defined

Let $\struct {G, \circ}$ be a group.
Let $N$ be a normal subgroup of $G$.
Let $a, b \in G$.
Then the coset product:
:$\paren {a \circ N} \circ \paren {b \circ N} = \paren {a \circ b} \circ N$
is well-defined.

Proof:
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:
:$\paren {a \circ N} \circ \paren {b \circ N} \subseteq \paren {a \circ b} \circ N$
Let $n \in N$ be arbitrary.
Then:
{{begin-eqn}}
{{eqn | l = \paren {a \circ b} \circ n
      | o = \in
      | r = \paren {a \circ b} \circ N
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \paren {a \circ e \circ b} \circ n
      | o = \in
      | r = \paren {a \circ b} \circ N
      | c = {{Group-axiom|2}}
}}
{{eqn | ll= \leadsto
      | l = \paren {a \circ e} \circ \paren {b \circ n}
      | o = \in
      | r = \paren {a \circ b} \circ N
      | c = {{Group-axiom|1}}
}}
{{eqn | ll= \leadsto
      | l = \paren {a \circ e} \circ \paren {b \circ n}
      | o = \in
      | r = \paren {a \circ N} \circ \paren {b \circ N}
      | c = {{Defof|Subset Product}}
}}
{{eqn | ll= \leadsto
      | l = a \circ  \paren {b \circ n}
      | o = \in
      | r = \paren {a \circ N} \circ \paren {b \circ N}
      | c = {{Defof|Identity Element}}
}}
{{eqn | ll= \leadsto
      | l = a \circ \paren {b \circ N}
      | o = \subseteq
      | r = \paren {a \circ N} \circ \paren {b \circ N}
      | c = {{hypothesis}}
}}
{{eqn | ll= \leadsto
      | l = \paren {a \circ b} \circ N
      | o = \subseteq
      | r = \paren {a \circ N} \circ \paren {b \circ N}
      | c = {{Corollary|Subset Product within Semigroup is Associative}}
}}
{{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}}
Theorem: Coset Product is Well-Defined

Let $\struct {G, \circ}$ be a group.
Let $N$ be a normal subgroup of $G$.
Let $a, b \in G$.
Then the coset product:
:$\paren {a \circ N} \circ \paren {b \circ N} = \paren {a \circ b} \circ N$
is well-defined.

Proof:
Let $N \lhd G$ where $G$ is a group.
Let $a, a', b, b' \in G$ such that:
:$N \circ a = N \circ a'$
and:
:$N \circ b = N \circ b'$
We need to show that:
:$N \circ \paren {a \circ b} = N \circ \paren {a' \circ b'}$
So:
{{begin-eqn}}
{{eqn | l = N \circ \paren {a \circ b}
      | r = \paren {N \circ a} \circ b
      | c = {{Corollary|Subset Product within Semigroup is Associative}}
}}
{{eqn | r = \paren {N \circ a'} \circ b
      | c = {{hypothesis}}
}}
{{eqn | r = \paren {a' \circ N} \circ b
      | c = {{Defof|Normal Subgroup}}
}}
{{eqn | r = a' \circ \paren {N \circ b}
      | c = {{Corollary|Subset Product within Semigroup is Associative}}
}}
{{eqn | r = a' \circ \paren {N \circ b'}
      | c = {{hypothesis}}
}}
{{eqn | r = \paren {a' \circ N} \circ b'
      | c = {{Corollary|Subset Product within Semigroup is Associative}}
}}
{{eqn | r = \paren {N \circ a'} \circ b'
      | c = {{Defof|Normal Subgroup}}
}}
{{eqn | r = N \circ \paren {a' \circ b'}
      | c = {{Corollary|Subset Product within Semigroup is Associative}}
}}
{{end-eqn}}
{{qed}}
Theorem: Coset Product is Well-Defined

Let $\struct {G, \circ}$ be a group.
Let $N$ be a normal subgroup of $G$.
Let $a, b \in G$.
Then the coset product:
:$\paren {a \circ N} \circ \paren {b \circ N} = \paren {a \circ b} \circ N$
is well-defined.

Proof:
Let $N \lhd G$ where $G$ is a group.
By Left Congruence Class Modulo Subgroup is Left Coset, it can be shown that the left congruence modulo $N$ is an equivalence relation.
Let $\RR^l_N$ denote the equivalence relation on $G$
:$\RR^l_N := \set {\tuple {x, y} \in G \times G: x^{-1} y \in N}$
Let $\struct {G / N, \circ_N}$ be the quotient group of $G$ by $N$, where $\circ_N$ denotes the operation induced on $G / N$ by $\circ$.
{{proofread|How to use the definition of operation induced on quotient set to complete the proof}}
By Left Congruence Class Modulo Subgroup is Left Coset again, the equivalence class $\eqclass g {\RR^l_N} \in \struct {G / N, \circ_N}$ of an element $g \in G$ is the left coset $g N$.
Let $a, a', b, b' \in G$, such that:
:$\eqclass a {\RR^l_N} = \eqclass {a'} {\RR^l_N}$
and:
:$\eqclass b {\RR^l_N} = \eqclass {b'} {\RR^l_N}$
It follows that if:
:$a, a', b, b' \in G$ 
then:
:$ a \circ N = a' \circ N$
and:
:$ b \circ N = b' \circ N$.
We need to show that:
:$\eqclass {a \circ b} {\RR^l_N} = \eqclass {a' \circ b'} {\RR^l_N}$.
That is:
:$ \paren { a \circ b } \circ N = \paren {a' \circ b'} \circ N$
We have:
{{begin-eqn}}
{{eqn | l = \paren {a \circ b} \circ N 
      | r = a \circ  b \circ \paren{N \circ N}
      | c = Product of Subgroup with Itself
}}
{{eqn | r = a \circ \paren{ b \circ N} \circ N
      | c = Subset Product within Semigroup is Associative
}}
{{eqn | r = a \circ \paren{ N \circ b} \circ N
      | c = {{Defof|Normal Subgroup}}
}}
{{eqn | r = \paren {a \circ N} \circ \paren {b \circ N}
      | c = Subset Product within Semigroup is Associative
}}
{{eqn | r = \paren {a' \circ N} \circ \paren {b' \circ N}
      | c = By Hypothesis
}}
{{eqn | r = a' \circ \paren{ N \circ b'} \circ N
      | c = Subset Product within Semigroup is Associative
}}
{{eqn | r = a' \circ \paren{ b' \circ N} \circ N
      | c = {{Defof|Normal Subgroup}}
}}
{{eqn | r = \paren{a' \circ  b'} \circ \paren{N \circ N}
      | c = Subset Product within Semigroup is Associative
}}
{{eqn | r = \paren {a' \circ b'} \circ N 
      | c = Product of Subgroup with Itself
}}
{{end-eqn}}
Hence:
:$\eqclass {a \circ b} {\RR^l_N} = \eqclass {a' \circ b'} {\RR^l_N}$.
{{qed}}
Theorem: Coset Product is Well-Defined

Let $\struct {G, \circ}$ be a group.
Let $N$ be a normal subgroup of $G$.
Let $a, b \in G$.
Then the coset product:
:$\paren {a \circ N} \circ \paren {b \circ N} = \paren {a \circ b} \circ N$
is well-defined.

Proof:
Let $N \lhd G$ where $G$ is a group.
Let $a, a', b, b' \in G$ such that:
:$a \circ N = a' \circ N$
and:
:$b \circ N = b' \circ N$
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 = {{hypothesis}}
}}
{{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 = \paren{ a \circ b } \circ N
      | c = as $n_3 \circ n_2 \in N$
}}
{{eqn | ll= \leadsto
      | l = \paren{ a' \circ b' } \circ N
      | r = \paren{ a \circ b }\circ N
      | c = {{Defof|Left Coset}}
}}
{{end-eqn}}
{{qed}}
Theorem: Quotient Group of Abelian Group is Abelian

Let $G$ be an abelian group.
Let $N \le G$.
Then the quotient group $G / N$ is abelian.

Proof:
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
Theorem: Quotient Theorem for Group Epimorphisms

Let $\struct {G, \oplus}$ and $\struct {H, \odot}$ be groups.
Let $\phi: \struct {G, \oplus} \to \struct {H, \odot}$ be a group epimorphism.
Let $e_G$ and $e_H$ be the identities of $G$ and $H$ respectively.
Let $K = \map \ker \phi$ be the kernel of $\phi$.
There exists one and only one group isomorphism $\psi: G / K \to H$ satisfying:
:$\psi \circ q_K = \phi$
where $q_K$ is the quotient epimorphism from $G$ to $G / K$.

Proof:
Let $\RR_\phi$ be the equivalence on $G$ defined by $\phi$.
{{begin-eqn}}
{{eqn | q = \forall x \in G
      | l = e_G
      | o = \RR_\phi
      | r = x
      | c = 
}}
{{eqn | ll= \leadstoandfrom
      | l = \map \phi x
      | r = \map \phi {e_G}
      | c = Definition of $\RR_\phi$
}}
{{eqn | ll= \leadstoandfrom
      | l = \map \phi x
      | r = e_H
      | c = Homomorphism to Group Preserves Identity
}}
{{end-eqn}}
Thus:
:$K = \eqclass {e_G} {\RR_\phi}$
From the Quotient Theorem for Epimorphisms:
:$\RR_\phi$ is compatible with $\oplus$
Thus from Kernel is Normal Subgroup of Domain:
:$K \lhd G$
From Congruence Relation induces Normal Subgroup, $\RR_\phi$ 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}}
Theorem: Congruence Relation induces Normal Subgroup

Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $\RR$ be a congruence relation for $\circ$.
Let $H = \eqclass e \RR$, where $\eqclass e \RR$ is the equivalence class of $e$ under $\RR$.
Then:
:$(1): \quad \struct {H, \circ \restriction_H}$ is a normal subgroup of $G$
:$(2): \quad \RR$ is the equivalence relation $\RR_H$ defined by $H$
:$(3): \quad \struct {G / \RR, \circ_\RR}$ is the subgroup $\struct {G / H, \circ_H}$ of the semigroup $\struct {\powerset G, \circ_\PP}$.

Proof:
=== Proof of Normal Subgroup ===
{{:Congruence Relation on Group induces Normal Subgroup}}
Theorem: Preimage of Normal Subgroup of Quotient Group under Quotient Epimorphism is Normal

Let $G$ be a group.
Let $H \lhd G$ where $\lhd$ denotes that $H$ is a normal subgroup of $G$.
Let $K \lhd G / H$.
Let $L = q_H^{-1} \sqbrk K$, where:
:$q_H: G \to G / H$ is the quotient epimorphism from $G$ to the quotient group $G / H$
:$q_H^{-1} \sqbrk K$ is the preimage of $K$ under $q_H$.
Then:
:$L \lhd G$
and there exists a group isomorphism $\phi: \paren {G / H} / K \to G / L$ defined as:
:$\phi \circ q_K \circ q_H = q_L$

Proof:
By Quotient Mapping on Structure is Epimorphism, both $q_K$ and $q_H$ are epimorphisms.
From Composite of Group Epimorphisms is Epimorphism we have that $q_K \circ q_H: G \to \paren {G / H} / K$ is also an epimorphism.
Now:
:$\forall x \in G: x \in \map \ker {q_K \circ q_H} \iff \map {q_K} {\map {q_H} x} = K = e_{G / H}$
This means the same as:
:$\map {q_H} x \in \map \ker {q_K} = K$
But:
:$\map {q_H} x \in K \iff x \in \map {q_H^{-1} } K = L$
Thus:
:$L = \map \ker {q_K \circ q_H}$
By Kernel is Normal Subgroup of Domain:
:$L \lhd G$
{{qed|lemma}}
By Quotient Mapping on Structure is Epimorphism, both $q_K$ and $q_H$ are epimorphisms.
From Composite of Group Epimorphisms is Epimorphism we have that $q_K \circ q_H: G \to \paren {G / H} / K$ is also an epimorphism.
Then {{apriori}}:
:$L \lhd G$
Hence by Quotient Theorem for Group Epimorphisms:
:there exists a group isomorphism $\psi: G / L \to \paren {G / H} / K$ satisfying:
::$\psi \circ q_L = q_K \circ q_L$
Let $\phi = \psi^{-1}$.
Then $\phi$ is a group isomorphism from $\paren {G / H} / K$ to $G / L$:
:$\phi \circ q_k \circ q_H = \phi \circ \psi \circ q_L = q_L$
{{qed}}