Third Isomorphism Theorem

Theorem
Let $$H \triangleleft G, N \triangleleft G$$, $$N \subseteq H$$.

Then $$H / N \triangleleft G / N$$ and:

$$\frac {G / N} {H / N} \cong \frac G H$$.

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 $$N \subseteq H, |N| \leq |H| \ $$ and so $$|G/N| \geq |G/H| \ $$, indicating $$\phi \ $$ is surjective, so

$$ $$ $$ $$

The result follows from the First Isomorphism Theorem.

Comment
There is no standard numbering for the Isomorphism Theorems. Different authors use different labellings.

This particular result, for example, is also known as the First Isomorphism Theorem.