Does $H_1/\ker(\phi_{|H_1})=H_2/\ker(\phi_{|H_2})$ imply that $H_1=H_2$? Let $\phi: G \rightarrow G'$ be a group homomorphism that is onto and $H_1 \subseteq H_2$ are subgroups of $G$. In addition, $\phi(H_1)=\phi(H_2)$ and $\ker(\phi_{|H_1})=\ker(\phi_{|H_2})$.
I suspect that if $H_1/\ker(\phi_{|H_1})=H_2/\ker(\phi_{|H_2})$ then  $H_1=H_2$. How should I prove my claim? (Without assuming finite order.)

You do not even need your claim. You need the following three ingredients (being onto is obviously also completely irrelevant for the claim, $H_1$ and $H_2$ do not care for $G^\prime$ being large, do they?):


*

*$H_1 \subset H_2$

*$\phi(H_1)=\phi(H_2)$

*$\operatorname{ker}(\phi_{|H_1})=\operatorname{ker}(\phi_{|H_2})$
Then $H_1=H_2$ holds. The proof is pretty straightforward. Take an element $x \in H_2$, then we have $\phi(x)=\phi(y)$ with some $y \in H_1$ by the second ingredient. Now consider $x-y$ and use the third ingredient to finally deduce $x \in H_1$.
