First Isomorphism Theorem/Groups

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:

 $\ds x K$ $=$ $\ds y K$ $\ds \leadstoandfrom \ \$ $\ds x^{-1} y$ $\in$ $\ds K = \map \ker \phi$ Left Cosets are Equal iff Product with Inverse in Subgroupâ€Ž $\ds \leadstoandfrom \ \$ $\ds \map \phi {x^{-1} y}$ $=$ $\ds e_{G_2}$ $\ds \leadstoandfrom \ \$ $\ds \paren {\map \phi x}^{-1} \map \phi y$ $=$ $\ds e_{G_2}$ $\ds \leadstoandfrom \ \$ $\ds \map \phi x$ $=$ $\ds \map \phi y$

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:

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

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

 $\ds \forall x, y \in G: \,$ $\ds$  $\ds \map \theta {x K y K}$ $\ds$ $=$ $\ds \map \theta {x y K}$ $\ds$ $=$ $\ds \map \phi {x y}$ $\ds$ $=$ $\ds \map \phi x \map \phi y$ $\ds$ $=$ $\ds \map \theta {x K} \map \theta {y K}$

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.