# Second Isomorphism Theorem/Groups

## Contents

## 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**.

## Also see

## Sources

- 1967: John D. Dixon:
*Problems in Group Theory*... (previous) ... (next): $1$: Subgroups: $1.\text{T}.5$ - 1967: George McCarty:
*Topology: An Introduction with Application to Topological Groups*... (previous) ... (next): $\text{II}$: Problem $\text{HH}$ - 1971: Allan Clark:
*Elements of Abstract Algebra*... (previous) ... (next): Chapter $2$: Group Homomorphism and Isomorphism: $\S 69$. The Second Isomorphism Theorem - 1996: John F. Humphreys:
*A Course in Group Theory*... (previous) ... (next): Chapter $8$: The Homomorphism Theorem: Theorem $8.15$