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I know that if two finitely presented groups, $G$ and $H$, have non-isomorphic abelianizations, then $G$ and $H$ are not isomorphic.

I am wondering if the converse is also true —- if $G$ and $H$ have isomorphic abelianizations, can we conclude that $G$ and $H$ are isomorphic? My intuition tells me this can’t hold in general, but I’m having a hard time coming up with a counterexample.

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    $\begingroup$ Let $G$ be any non-abelian group and let $H$ be the abelianization of $G$. $\endgroup$
    – lulu
    Oct 17, 2022 at 23:04
  • $\begingroup$ Thanks for the quick response, @lulu $\endgroup$
    – user523692
    Oct 17, 2022 at 23:07
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    $\begingroup$ Another class of example comes from the fact that any nonabelian simple group has trivial abelianization. So any two nonabelian simple groups will in particular have isomorphic abelianizations. $\endgroup$ Oct 18, 2022 at 0:05

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Yes certainly. As @lulu notes just take a non-abelian group. I.e. one which differs from its abelianization.

After all the abelianization of the abelianization is the abelianization (abelianization is sort of idempotent, similar to closure in topology, or projection in linear algebra; doing it twice is the same as doing it once).

So $S_3$?

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