# First Isomorphism Theorem/Groups

## Contents

## Theorem

Let $\phi: G_1 \to G_2$ be a group homomorphism.

Let $\map \ker \phi$ be the kernel of $\phi$.

Then:

- $\Img \phi \cong G_1 / \map \ker \phi$

where $\cong$ denotes group isomorphism.

## Proof

Let $K = \map \ker \phi$.

By Kernel is Normal Subgroup of Domain, $G_1 / K$ exists.

We need to establish that the mapping $\theta: G_1 / K \to G_2$ defined as:

- $\forall x \in G_1: \map \theta {x K} = \map \phi x$

is well-defined.

That is, we need to ensure that:

- $\forall x, y \in G: x K = y K \implies \map \theta {x K} = \map \theta {y K}$

Let $x, y \in G: x K = y K$.

Then:

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

\(\displaystyle \leadstoandfrom \ \ \) | \(\displaystyle x^{-1} y\) | \(\in\) | \(\displaystyle K = \map \ker \phi\) | Left Cosets are Equal iff Product with Inverse in Subgroup | |||||||||

\(\displaystyle \leadstoandfrom \ \ \) | \(\displaystyle \map \phi {x^{-1} y}\) | \(=\) | \(\displaystyle e_{G_2}\) | ||||||||||

\(\displaystyle \leadstoandfrom \ \ \) | \(\displaystyle \paren {\map \phi x}^{-1} \map \phi y\) | \(=\) | \(\displaystyle e_{G_2}\) | ||||||||||

\(\displaystyle \leadstoandfrom \ \ \) | \(\displaystyle \map \phi x\) | \(=\) | \(\displaystyle \map \phi y\) |

Thus we see that $\theta$ is well-defined.

Since we also have that:

- $\map \phi x = \map \phi y \implies x K = y K$

it follows that:

- $\map \theta {x K} = \map \theta {y K} \implies x K = y K$

So $\theta$ is injective.

We also note that:

- $\Img \theta = \set {\map \theta {x K}: x \in G}$

So:

\(\displaystyle \Img \theta\) | \(=\) | \(\displaystyle \set {\map \theta {x K}: x \in G}\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \set {\map \phi x: x \in G}\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \Img \phi\) |

We also note that $\theta$ is a homomorphism:

\(\displaystyle \forall x, y \in G: \ \ \) | \(\displaystyle \) | \(\) | \(\displaystyle \map \theta {x K y K}\) | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \map \theta {x y K}\) | |||||||||||

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

\(\displaystyle \) | \(=\) | \(\displaystyle \map \phi x \map \phi y\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \map \theta {x K} \map \theta {y K}\) |

Thus $\theta$ is a monomorphism whose image equals $\Img \phi$.

The result follows.

$\blacksquare$

## Also known as

Some sources call this **the homomorphism theorem**.

Others combine this result with Group Homomorphism Preserves Subgroups, Kernel of Group Homomorphism is Subgroup and Kernel is Normal Subgroup of Domain.

Still others do not assign a special name to this theorem at all.

## Also see

## Sources

- 1965: J.A. Green:
*Sets and Groups*... (previous) ... (next): $\S 7.4$. Kernel and image: The Homomorphism Theorem - 1966: Richard A. Dean:
*Elements of Abstract Algebra*... (previous) ... (next): $\S 1.10$: Theorem $24$: Corollary - 1971: Allan Clark:
*Elements of Abstract Algebra*... (previous) ... (next): Chapter $2$: Group Homomorphism and Isomorphism: $\S 66$ - 1978: Thomas A. Whitelaw:
*An Introduction to Abstract Algebra*... (previous) ... (next): $\S 52.1$ The first isomorphism theorem - 1996: John F. Humphreys:
*A Course in Group Theory*... (previous) ... (next): Chapter $8$: The Homomorphism Theorem: Theorem $8.13: \ (3)$