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Suppose $N \leq G$ is a subgroup of some group and it is the kernel of some group homomorphism $\phi$. Prove $N \unlhd G$

We want to show that $gNg^{-1}=N$ for every $g \in G$, given that $N:=$ ker $\phi$ some $\phi:G \rightarrow H$ a homomorphism. To show $gNg^{-1} \subset N$, its enough to show for every $x \in gNg^{-1}$ is in the kernel of $\phi$. We have that

\begin{align} \phi(x)&= \phi(gng^{-1}) && \text{for some $n \in N$}\\ &=\phi(g)\phi(n)\phi(g)^{-1} && \text{as $\phi$ is a homomorphism}\\ &=\phi(g)e_H\phi(g)^{-1} && \text{as $n \in N:=$ ker $\phi$}\\ &=\phi(g)\phi(g)^{-1} && \text{definition of mult by identity}\\ &=e_H && \text{definition of inverses} \end{align}

thus $x \in$ ker $\phi$ and thus $x \in N$. Is this enough of a direct proof that $N$ is normal if its the kernel of some group homomorphism?

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    $\begingroup$ This is a good proof, and just showing $gNg^{-1}\subseteq N$ is enough (since it's for all $g\in G$). One small note: we should say "...for all $x\in gNg^{-1}$...". $\endgroup$
    – Dave
    Jul 9, 2022 at 19:05
  • $\begingroup$ good point @Dave $\endgroup$
    – anonymous
    Jul 9, 2022 at 19:06

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Or, if you like, you could use the first isomorphism theorem: since the quotient $G/N\cong \phi(G)$ is a group, $N$ is normal.

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