Third Isomorphism Theorem/Groups/Proof 2

Theorem

Let $G$ be a group, and let:

$H, N$ be normal subgroups of $G$
$N$ be a subset of $H$.

Then:

$(1): \quad N$ is a normal subgroup of $H$
$(2): \quad H / N$ is a normal subgroup of $G / N$
where $H / N$ denotes the quotient group of $H$ by $N$
$(3): \quad \dfrac {G / N} {H / N} \cong \dfrac G H$
where $\cong$ denotes group isomorphism.

Proof

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 $\phi \circ q_{H / N} = \psi$
$H / N \subseteq \map \ker \psi$

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.