# Homomorphism preserves normality

let $\phi:G\rightarrow G'$ be a homomorphism, and let N' be a normal subgroup of G'. I want to show that $\phi^{-1}[N']$ is also normal subgroup of G.

My work :

since homomorphism preserves subgroup, I only need to show that the normality of $\phi^{-1}[N']$.

let $N=\phi^{-1}[N']$.

first, I will show that $gN$ is a subset of $Ng$.

take $gn$ in $gN$ s.t $g$ is in $G$, $n$ is in $N$.

then $\phi(gn)=\phi(g)\phi(n)$ is in $\phi(g)N'$

and since $N'$ is normal, for some $n'$ in $N'$, $\phi(g)\phi(n)=n'\phi(g)$ and since $N=\phi^{-1}[N']$, there exists some $n_{0}$ in $N$ s.t $\phi(n_{0})=n'$.

then $\phi(gn)=\phi(g)\phi(n)=n'\phi(g)=\phi(n_{0})\phi(g)=\phi(n_{0}g)$

and at here, I want to say that $gn=n_{0}g$ and finish, but I can't guarantee the homomorphism is injective. and it seems I have to do this without the injectiveness. any advice?

From $\phi(gn)=\phi(n_{0}g)$, you get $gn=n_{0}gk$, with $k \in \ker\phi$.

The key point you're missing is that $N$ contains $\ker\phi$.

idea: $\phi^{-1}(N')$ contains the kernel of $\phi$

You might as well note that $\phi^{-1}(N')$ is the kernel of $G\rightarrow G' \rightarrow G'/N'$, and kernels are normal.