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.