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:

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:

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

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

The result follows.

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

 * Isomorphism Theorems