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.

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

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 $\left|{N}\right| \le \left|{H}\right|$ and so $\left|{G / N}\right| \ge \left|{G / H}\right|$, indicating $\phi$ is surjective, so:

The result follows from the First Isomorphism Theorem.