Second Isomorphism Theorem

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


 * $$H$$ be a subgroup of $$G$$.
 * $$N$$ be a normal subgroup of $$G$$.

Then:
 * $$\frac H {H \cap N} \cong \frac {H N} N$$

where $$\cong$$ denotes group isomorphism.

This result is also referred to by some sources as the first isomorphism theorem.

Proof

 * From Quotient Group, for $$G / H$$ to be defined, it is necessary for $$H \triangleleft G$$.

The fact that Intersection with Normal Subgroup is Normal gives us that $$N \cap H \triangleleft H$$.

Also, $$N \triangleleft N H = \left \langle {H, N} \right \rangle$$ follows from Subgroup Product with Normal Subgroup as Generator.


 * Now we define a mapping $$\phi: H \to H N / N$$ by the rule $$\phi \left({h}\right) = h N$$.

Note that $$N$$ need not be a subset of $$H$$. Therefore, the coset $$h N$$ is an element of $$H N / N$$ rather than of $$H / N$$.

Then $$\phi$$ is a homomorphism, as $$\phi \left({x y}\right) = x y N = \left({x N}\right) \left({y N}\right) = \phi \left({x}\right) \phi \left({y}\right)$$.

Then:

$$ $$ $$ $$

Then we see that $$\phi$$ is a surjection because $$h n N = h N \in H N / N$$ is $$\phi \left({h}\right)$$.

The result follows from the First Isomorphism Theorem.