Third Isomorphism Theorem/Groups/Proof 2

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

Let $q_H$ and $q_N$ be the quotient mappings from $G$ to $\dfrac G H$ and $G$ to $\dfrac G N$ respectively.

Hence:
 * $N \subseteq \map \ker {q_H}$

From Quotient Theorem for Group Homomorphisms: Corollary 2, it therefore follows that:
 * there exists a group epimorphism $\psi: \dfrac G N \to \dfrac G H$ such that $\psi \circ q_N = q_H$

Then we have that :
 * there exists a group epimorphism $\phi: \dfrac {G / N} {H / N} \to \dfrac G N$ such that $\phi \circ q_{H / N} = \psi$


 * $H / N \subseteq \map \ker \psi$
 * $H / N \subseteq \map \ker \psi$

Thus we form the composite:
 * $\phi \circ q_{H / N} \circ q_N = q_H$

and it remains to be shown that $\phi$ is an isomorphism.