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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asked Sep 22, 2018 at 10:53
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7$\begingroup$ @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 HoltCommented Sep 22, 2018 at 11:56
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7$\begingroup$ 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 HoltCommented Sep 22, 2018 at 12:20
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2$\begingroup$ I suggest "Wilbur". $\endgroup$– Keith KearnesCommented Sep 22, 2018 at 13:28
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2$\begingroup$ I suggest “almost perfect” group. A google search indicates this terminology has been used; for instance, here: core.ac.uk/download/pdf/61487184.pdf . $\endgroup$– HJRWCommented Sep 22, 2018 at 17:56
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2$\begingroup$ 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 AgolCommented Sep 22, 2018 at 22:28
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1 Answer
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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 .
