Third Isomorphism Theorem/Groups/Proof 1

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

We define a mapping:
 * $\phi: G / N \to G / H$ by $\map \phi {g N} = g H$

Since $\phi$ is defined on cosets, we need to check that $\phi$ is well-defined.

Suppose $x N = y N$.

Then:
 * $y^{-1} x \in N$

Then:
 * $N \le H \implies y^{-1} x \in H$

and so:
 * $x H = y H$

So:
 * $\map \phi {x N} = \map \phi {y N}$

and $\phi$ is indeed well-defined.

Now $\phi$ is a homomorphism, from:

Also, since $N \subseteq H$, it follows that:
 * $\order N \le \order H$

So:
 * $\order {G / N} \ge \order {G / H}$, indicating $\phi$ is surjective.

So:

The result follows from the First Isomorphism Theorem.