Second Isomorphism Theorem/Groups

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


 * $(1): \quad H$ be a subgroup of $G$
 * $(2): \quad N$ be a normal subgroup of $G$.

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

where $\cong$ denotes group isomorphism.

Proof
The fact that $N$ is normal, together with Intersection with Normal Subgroup is Normal, gives us that $N \cap H \lhd H$.

Also, $N \lhd N H = \left \langle {H, N} \right \rangle$ follows from Subset 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.

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

Also see

 * Isomorphism Theorems