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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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2 Answers

If $h\in H$ and $k\in\ker\varphi$, then $\varphi(hk)=\varphi(h)\varphi(k)=\varphi(h)\in\varphi[H]$, so by definition $hk\in\varphi^{-1}[\varphi[H]]$.

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Because of the details you included in your OP, I'm really surprised you didn't find that inclusion first. I'm wondering if maybe you put up the opposite inclusion that you intended(?).

If that's the case, then keep reading:

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, $x\in \ker(\phi)h\subseteq \ker(\phi)H$.

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