First Isomorphism Theorem/Groups

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

Let $\ker \left({\phi}\right)$ be the kernel of $\phi$.

Then:
 * $\operatorname {Im} \left({\phi}\right) \cong G_1 / \ker \left({\phi}\right)$

where $\cong$ denotes group isomorphism.

Some authors call this the homomorphism theorem.

Others combine this result with Kernel is Subgroup and Kernel is Normal Subgroup of Domain.

Proof
Let $K = \ker \left({\phi}\right)$.

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:
 * $\forall x \in G_1: \theta \left({x K}\right) = \phi \left({x}\right)$

is well-defined.

That is, we need to ensure that:
 * $\forall x, y \in G: x K = y K \implies \theta \left({x K}\right) = \theta \left({y K}\right)$

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

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

Since we also have that $\phi \left({x}\right) = \phi \left({y}\right) \implies x K = y K$, it follows that $\theta \left({x K}\right) = \theta \left({y K}\right) \implies x K = y K$.

So $\theta$ is injective.

We also note that $\operatorname {Im} \left({\theta}\right) = \left\{{\theta \left({x K}\right): x \in G}\right\}$.

So:

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

Thus $\theta$ is a monomorphism whose image equals $\operatorname {Im} \left({\phi}\right)$.

The result follows.