# A question about composition of the inverse image of a group homomorphism and the homomorphism itself

Suppose $\phi:G_1 \rightarrow G_2$ is a group homomorphism and $H \leq G_1$. Show that $\phi^{-1}(\phi(H))=H \cdot \ker(\phi)$.

Attempt at a solution:

I was easily able to show that $\phi^{-1}(\phi(H))\subseteq H\cdot \ker(\phi)$ due to the fact that if $h \in H$ and $k\in \ker(\phi)$ then $\phi(hk)=\phi(h)$.

However I'm having trouble with the reverse inclusion. That is, showing that if $hk\in H\cdot \ker(\phi)$ then $hk\in \phi^{-1}(\phi(H))$.

Any help would be greatly appreciated.

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If $h\in H$ and $k\in\ker\varphi$, then $\varphi(hk)=\varphi(h)\varphi(k)=\varphi(h)\in\varphi[H]$, so by definition $hk\in\varphi^{-1}[\varphi[H]]$.
Suppose $x\in \phi^{-1}(\phi(H))$. Then $\phi(x)=\phi(h)$ for some $h\in H$. It follows that $\phi(xh^{-1})=1$, hence $xh^{-1}\in \ker(\phi)$. Thus, $x\in \ker(\phi)h\subseteq \ker(\phi)H$.