Kernel is Normal Subgroup of Domain

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Theorem

Let $\phi$ be a group homomorphism.

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

$\map \ker \phi \lhd \Dom \phi$

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.

$\map \ker \phi \le \Dom \phi$

Let $k \in \map \ker \phi, x \in G_1$.

Then:

 $\ds \map \phi {x k x^{-1} }$ $=$ $\ds \map \phi x \map \phi k \paren {\map \phi x}^{-1}$ Homomorphism to Group Preserves Inverses $\ds$ $=$ $\ds \map \phi x e_{G_2} \paren {\map \phi x}^{-1}$ Definition of Kernel of Group Homomorphism $\ds$ $=$ $\ds \map \phi x \paren {\map \phi x}^{-1}$ Definition of Identity Element $\ds$ $=$ $\ds e_{G_2}$ Definition of Inverse Element

So:

$x k x^{-1} \in \map \ker \phi$

This is true for all $k \in \map \ker \phi$ and $x \in G_1$.

From Subgroup is Normal iff Contains Conjugate Elements, it follows that $\map \ker \phi$ is a normal subgroup of $G_1$.

$\blacksquare$