Third Isomorphism Theorem/Groups

Theorem
Let $G$ be a group, and let:


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

Then:
 * $H / N$ is a normal subgroup of $G / N$
 * where $H / N$ denotes the quotient group of $H$ by $N$


 * $\dfrac {G / N} {H / N} \cong \dfrac G H$
 * where $\cong$ denotes group isomorphism.

Proof
We define a mapping:
 * $\phi: G / N \to G / H$ by $\phi \left({g N}\right) = g H$

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

Suppose $x N = y N \implies y^{-1} x \in N$.

Then $N \le H \implies y^{-1} x \in H$ and so $x H = y H$.

So $\phi \left({x N}\right) = \phi \left({y N}\right)$ and $\phi$ is indeed well-defined.

Now $\phi$ is a homomorphism, from:

Also, since $N \subseteq H$, it follows that:
 * $\left|{N}\right| \le \left|{H}\right|$

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

So:

The result follows from the First Isomorphism Theorem.

Also known as
This result is also referred to by some sources as the first isomorphism theorem, and by others as the second isomorphism theorem.