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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.)

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    $\begingroup$ With $\operatorname{ker}(\phi(H_1))$, you actually mean $\operatorname{ker}(\phi_{|H_1})$, right? $\endgroup$ – MooS Nov 23 '15 at 7:20
  • $\begingroup$ That is correct. $\endgroup$ – user112358 Nov 23 '15 at 7:40
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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 $x \in H_1$.

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  • $\begingroup$ When you say "$x-y$", what do you mean exactly? Is "$-$" a group operation? $\endgroup$ – user112358 Nov 23 '15 at 7:46
  • $\begingroup$ He means $ x+ (-y) $. There is no harm in writing this as $ x-y $. $\endgroup$ – user175531 Nov 23 '15 at 7:56
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    $\begingroup$ @Lewis MooS is writting additively. If you prefer replace $x-y$ with $xy^{-1}$ which is what it would be written multiplicatively. $\endgroup$ – Nex Nov 23 '15 at 7:58

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