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 $\phi$:

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


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:

\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \phi \left({x k x^{-1} }\right)\) \(=\) \(\displaystyle \phi \left({x}\right) \phi \left({k}\right) \left({\phi \left({x}\right)}\right)^{-1}\) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \)          Homomorphism to Group Preserves Inverses          
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(=\) \(\displaystyle \phi \left({x}\right) \left({\phi \left({x}\right)}\right)^{-1}\) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \)          as $k \in \ker \left({\phi}\right)$          
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(=\) \(\displaystyle \phi \left({x}\right) \left({\phi \left({x}\right)}\right)^{-1}\) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \)          Property of Identity          
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(=\) \(\displaystyle e_{G_2}\) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \)          Property of Inverse          

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


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 $G_1$.

$\blacksquare$

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