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.


By Kernel of Group Homomorphism is Subgroup:

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


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

Then:

\(\displaystyle \map \phi {x k x^{-1} }\) \(=\) \(\displaystyle \map \phi x \map \phi k \paren {\map \phi x}^{-1}\) Homomorphism to Group Preserves Inverses
\(\displaystyle \) \(=\) \(\displaystyle \map \phi x e_{G_2} \paren {\map \phi x}^{-1}\) Definition of Kernel of Group Homomorphism
\(\displaystyle \) \(=\) \(\displaystyle \map \phi x \paren {\map \phi x}^{-1}\) Definition of Identity Element
\(\displaystyle \) \(=\) \(\displaystyle e_{G_2}\) 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 $G_1$.

$\blacksquare$


Also see


Sources