First Isomorphism Theorem

Preface
This theorem applies for Groups, Rings, Modules, Algebras, and any other algebraic structure where you see the word homomorphism.

It is a categorical result i.e. it is independent of the structure used.

Groups
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.

Rings
Let $\phi: R \to S$ be a ring homomorphism.

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

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

where $\cong$ denotes ring isomorphism.

Some authors call this the homomorphism theorem.

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

Proof for Groups
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.

Proof for Rings
In Ring Homomorphism whose Kernel contains Ideal, let $J = \ker \left({\phi}\right)$.

This gives the ring homomorphism $\mu: R / \ker \left({\phi}\right) \to S$ as follows:


 * CommDiagFirstIsomTheorem.png

That is:
 * $\phi = \mu \circ \nu$

Then we have:
 * $\ker \left({\mu}\right) = \ker \left({\phi}\right) / \ker \left({\phi}\right)$

This is the null subring of $R / \ker \left({\phi}\right)$ by Quotient Ring Defined by Ring Itself is Null Ideal.

Then from Kernel of Monomorphism is Trivial it follows that $\mu$ is a monomorphism.

From $\phi = \mu \circ \nu$, we have:
 * $\operatorname {Im} \left({\mu}\right) = \operatorname {Im} \left({\phi}\right)$

It follows that $\mu$ is an isomorphism.

Alternative Names
There is no standard numbering for the Isomorphism Theorems. Different authors use different labellings.

This particular result, for example, is also known as the Homomorphism Theorem.

Result for groups

 * : $\S 7.4$
 * : $\S 52.1$
 * : $\S 8$: Theorem $8.13$

Result for rings

 * : $\S 2.2$: Theorem $2.9$
 * : $\S 60.3$