Third Isomorphism Theorem/Groups/Proof 3

Proof
From Normal Subgroup which is Subset of Normal Subgroup is Normal in Subgroup, $N$ is a normal subgroup of $H$.

Let $q_H$ denote the quotient mapping from $G$ to $\dfrac G H$.

Let $q_N$ denote the quotient mapping from $G$ to $\dfrac G N$.

Let $\RR$ be the congruence relation defined by $N$ in $G$.

Let $\TT$ be the congruence relation defined by $H$ in $G$.

Thus from Congruence Relation induces Normal Subgroup:


 * $q_H = q_\TT$

and:
 * $q_N = q_\RR$

where $q_\RR$ and $q_\TT$ denote the quotient epimorphisms induced by $\RR$ and $\TT$ respectively.

We have that:

That is:
 * $\RR \subseteq \TT$

Let $\SS$ be the relation on the quotient group $G / N$ which satisfies:
 * $\forall X, Y \in G / N: X \mathrel \SS Y \iff \exists x \in X, y \in Y: x \mathrel \TT y$

That is, by definition of $\TT$:
 * $\forall X, Y \in G / N: X \mathrel \SS Y \iff \exists x \in X, y \in Y: x H = y H$

Then by Equivalence Relation induced by Congruence Relation on Quotient Structure is Congruence: Corollary:
 * $\SS$ is a congruence relation on $G / N$.

Hence by Congruence Relation on Group induces Normal Subgroup:
 * $\SS$ induces a normal subgroup of $G / N$.

Again from Equivalence Relation induced by Congruence Relation on Quotient Structure is Congruence: Corollary:
 * there exists a unique isomorphism $\phi$ from $G / N$ to $G / H$ which satisfies:
 * $\phi \circ q_\SS \circ q_\RR = q_\TT$
 * where $q_\SS$, $q_\RR$ and $q_\TT$ denote the quotient epimorphisms as appropriate.

That is:


 * $\phi \circ q_{H / N} \circ q_N = q_H$

and the result follows.