Third Isomorphism Theorem

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


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

Then: where $$H / N$$ denotes the Quotient Group of $$H$$ by $$N$$;
 * $$H / N$$ is a normal subgroup of $$G / N$$

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

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 $$N \subseteq H, |N| \le |H| \ $$ and so $$|G/N| \ge |G/H| \ $$, indicating $$\phi \ $$ is surjective, so

$$ $$ $$ $$

The result follows from the First Isomorphism Theorem.