# Group Homomorphism, Preimage, and Product of Groups

## Problem 208

Let $G, G’$ be groups and let $f:G \to G’$ be a group homomorphism.
Put $N=\ker(f)$. Then show that we have
$f^{-1}(f(H))=HN.$

## Proof.

$(\subset)$ Take an arbitrary element $g\in f^{-1}(f(H))$. Then we have $f(g)\in f(H)$.
It follows that there exists $h\in H$ such that $f(g)=f(h)$.
Since $f$ is a group homomorphism, we obtain
$f(h^{-1}g)=e’,$ where $e’$ is the identity element of the group $G’$.

This implies that $h^{-1}g\in \ker(f)=N$, hence we have
\begin{align*}
g\in hN\subset HN.
\end{align*}
Therefore we have $f^{-1}(f(H)) \subset HN$.

$(\supset)$ On the other hand, let $g\in HN$ be an arbitrary element.
Then we can write $g=hn$ with $h \in H$ and $n\in N$.
We have
\begin{align*}
f(g)&=f(hn)=f(h)f(n)\\
&=f(h)e’=f(h)\in f(H)
\end{align*}
since $f$ is a group homomorphism and $f(n)=e’$.
Thus we have
$g\in f^{-1}(f(H))$ and $f^{-1}(f(H)) \supset HN$.

Therefore, putting the two continents together gives
$f^{-1}(f(H))=HN$ as required.

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