Kernel is Normal Subgroup of Domain

Theorem
Let $$\phi$$ be a group homomorphism.

Then the kernel of $$\phi$$ is a normal subgroup of the domain of $$\phi$$:


 * $$\mathrm{ker} \left({\phi}\right) \triangleleft \mathrm{Dom} \left({\phi}\right)$$.

Proof
Let $$\phi: G_1 \to G_2$$ be a group homomorphism, where the identities of $$G_1$$ and $$G_2$$ are $$e_{G_1}$$ and $$e_{G_2}$$ respectively.

By Kernel is Subgroup, $$\mathrm{ker} \left({\phi}\right) \le \mathrm{Dom} \left({\phi}\right)$$.

Let $$k \in \mathrm{ker} \left({\phi}\right), x \in G_1$$. Then:

$$ $$ $$ $$

So $$\phi \left({x k x^{-1}}\right) \in \mathrm{ker} \left({\phi}\right)$$.

As this is true for all $$x \in G_1$$, then from Normal Subgroup Equivalent Definitions, $$\mathrm{ker} \left({\phi}\right)$$ is a normal subgroup of $$G_1$$.