# Image of a Normal Subgroup Under a Surjective Homomorphism is a Normal Subgroup

## Problem 161

Let $f: H \to G$ be a surjective group homomorphism from a group $H$ to a group $G$.
Let $N$ be a normal subgroup of $H$. Show that the image $f(N)$ is normal in $G$.

## Proof.

To show that $f(N)$ is normal, we show that $gf(N)g^{-1}=f(N)$ for any $g \in G$. Equivalently, it suffices to show that $gf(n)g^{-1}\in f(N)$ for all $g\in G$, $n\in N$.
Since $f$ is surjective, for each $g \in G$ there exists $h \in H$ such that $f(h)=g$.

Then we have
\begin{align*}
gf(n)g^{-1}&=f(h)f(n)f(h)^{-1}\\
&=f(hnh^{-1}),
\end{align*}
where the second equality follows since $f$ is a group homomorphism.

Since $N$ is a normal subgroup in $H$, the element $hnh^{-1}$ is in $N$.
Therefore we have $gf(n)g^{-1} \in f(N)$, hence $f(N)$ is a normal subgroup in $G$.

Let $G$ be a finite group and let $H$ be a subset of $G$ such that for any \$a,b \in...