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 $\phi$:


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

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, $\ker \left({\phi}\right) \le \operatorname{Dom} \left({\phi}\right)$.

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

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

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