# If $N \leq G$ is the kernel of some group homomorphism then it is normal

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?

• 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 $x\in gNg^{-1}$...".
– Dave
Jul 9, 2022 at 19:05
• good point @Dave Jul 9, 2022 at 19:06

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.