# 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, $N \lhd N H = \gen {H, N}$ follows from Subset Product with Normal Subgroup as Generator.

Now we define a mapping $\phi: H \to H N / N$ by the rule:

$\map \phi h = h N$

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:

$\map \phi {x y} = x y N = \paren {x N} \paren {y N} = \map \phi x \map \phi y$

Then:

 $\displaystyle \map \ker \phi$ $=$ $\displaystyle \set {h \in H: \map \phi h = e_{H N / N} }$ $\displaystyle$ $=$ $\displaystyle \set {h \in H: h N = N}$ $\displaystyle$ $=$ $\displaystyle \set {h \in H: h \in N}$ $\displaystyle$ $=$ $\displaystyle H \cap N$

Then we see that $\phi$ is a surjection because $h n N = h N \in H N / N$ is $\map \phi h$.

The result follows from the First Isomorphism Theorem.

$\blacksquare$

## Also known as

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