# 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

\[f^{-1}(f(H))=HN.\]

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## 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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