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I have the following problem:

Let $f \colon G_1 \rightarrow G_2$ be an epimorphism, $H_2/G_2$ and $H_1=f^{-1}(H_2)$. Prove that $G_1/H_1 \cong G_2/H_2$. Is this still true if $f$ isn't surjective?

What does $H_2/G_2$ mean? I would assume it means "is a subgroup of" but I'm not sure. Also, which groups are supposed to be isomorphic? Again, I assume it's the quotient groups but I'm not sure. Thank you.

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    $\begingroup$ Seems like there's a typo. $\endgroup$ Commented Nov 21, 2014 at 15:57
  • $\begingroup$ @AmitaiYuval Possibly, though it's used multiple times in the problem set. This is really weird notation. $\endgroup$
    – Mike
    Commented Nov 21, 2014 at 15:59

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Since you want to show an isomorphism of groups you do need that $H_2$ is normal in $G_2$. Hence I guess that $H_2/G_2$ means $H_2\trianglelefteq G_2$, i.e. $H_2$ is a normal subgroup of $G_2$. I have never seen this notation thou

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  • $\begingroup$ Yeah, I assumed so too. Thanks for the quick answer. $\endgroup$
    – Mike
    Commented Nov 21, 2014 at 16:00

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