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, for all $$x, y \in G$$, $$x K = y K \implies \theta \left({x K}\right) = \theta \left({y K}\right)$$.

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.

Comment
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 $$2.9$$
 * : $$\S 60.3$$