# Third Isomorphism Theorem/Groups

## Theorem

Let $G$ be a group, and let:

- $H, N$ be normal subgroups of $G$
- $N$ be a subset of $H$.

Then:

- $(1): \quad H / N$ is a normal subgroup of $G / N$
- where $H / N$ denotes the quotient group of $H$ by $N$

- $(2): \quad \dfrac {G / N} {H / N} \cong \dfrac G H$
- where $\cong$ denotes group isomorphism.

### Corollary 1

Let $G$ be a group.

Let $N$ be a normal subgroup of $G$.

Let $q: G \to \dfrac G N$ be the quotient epimorphism from $G$ to the quotient group $\dfrac G N$.

Let $K$ be the kernel of $q$.

Then:

- $\dfrac G N \cong \dfrac {G / K} {N / K}$

### Corollary 2

Let $G$ and $H$ be groups whose identities are $e_G$ and $e_H$ respectively.

Let $\phi: G \to H$ be a group homomorphism.

Let $K$ be the kernel of $\phi$.

Let $N$ be a normal subgroup of $G$.

Let $q: G \to \dfrac G N$ be the quotient epimorphism from $G$ to the quotient group $\dfrac G N$.

Then $N \subseteq K$ if and only if there exists a group homomorphism $\psi: \dfrac G N \to H$ such that:

- $\phi = \psi \circ q$

and:

- $\dfrac G K \cong \dfrac {G / N} {K / N}$

## Proof

We define a mapping:

- $\phi: G / N \to G / H$ by $\map \phi {g N} = g H$

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 $\map \phi {x N} = \map \phi {y N}$ and $\phi$ is indeed well-defined.

Now $\phi$ is a homomorphism, from:

\(\displaystyle \map \phi {x N} \map \phi {y N}\) | \(=\) | \(\displaystyle \paren {x H} \paren {y H}\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle x y H\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \map \phi {x y N}\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \map \phi {x N y N}\) |

Also, since $N \subseteq H$, it follows that:

- $\order N \le \order H$

So:

- $\order {G / N} \ge \order {G / H}$, indicating $\phi$ is surjective.

So:

\(\displaystyle \map \ker \phi\) | \(=\) | \(\displaystyle \set {g N \in G / N: \map \phi {g N} = e_{G / H} }\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \set {g N \in G / N: g H = H}\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \set {g N \in G / N: g \in H}\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle H / N\) |

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**, and by others as the **second isomorphism theorem**.

## Also see

## Sources

- 1967: John D. Dixon:
*Problems in Group Theory*... (previous) ... (next): $1$: Subgroups: $1.\text{T}.4$ - 1971: Allan Clark:
*Elements of Abstract Algebra*... (previous) ... (next): Chapter $2$: Group Homomorphism and Isomorphism: $\S 68$. The First Isomorphism Theorem - 1996: John F. Humphreys:
*A Course in Group Theory*... (previous) ... (next): Chapter $8$: The Homomorphism Theorem: Theorem $8.16$