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Let us recall that a group $G$ is called perfect if it coincides with its commutator subgroup $G'$, or equivalently, if its abelianization $G/G'$ is trivial.

Question. Is there any name for a group $G$ whose abelianization $G/G'$ is finite?

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    $\begingroup$ @DanPetersen I don't think that's a good idea. That would mean having a perfect subgroup of finite index, which is not necessarily the case for groups with $G/G'$ finite. $\endgroup$
    – Derek Holt
    Sep 22, 2018 at 11:56
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    $\begingroup$ Yes there are many counterexamples. The infinite dihedral group is solvable and so has no nontrivial perfect subgropus, but it has finite abelianization. $\endgroup$
    – Derek Holt
    Sep 22, 2018 at 12:20
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    $\begingroup$ I suggest "Wilbur". $\endgroup$ Sep 22, 2018 at 13:28
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    $\begingroup$ I suggest “almost perfect” group. A google search indicates this terminology has been used; for instance, here: core.ac.uk/download/pdf/61487184.pdf . $\endgroup$
    – HJRW
    Sep 22, 2018 at 17:56
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    $\begingroup$ Sometimes a group is called indicable if it admits a non-trivial homomorphism to $\mathbb{Z}$. For a finitely-generated group, this is equivalent to having infinite abelianization. So for finitely generated groups, your condition could be called "non-indicable". But I wouldn't recommend this terminology... $\endgroup$
    – Ian Agol
    Sep 22, 2018 at 22:28

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Prompted by the OP, I'm writing my comment as an answer.

I suggest “almost perfect” group. A google search indicates this terminology has been used; for instance, here: core.ac.uk/download/pdf/61487184.pdf .

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