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 $\phi$:


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

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

This is true for all $k \in \ker \left({\phi}\right)$ and $x \in G_1$.

From Subgroup is Normal iff Contains Conjugate Elements, it follows that $\ker \left({\phi}\right)$ is a normal subgroup of $G_1$.

Also see

 * Definition:Natural Group Epimorphism