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EDIT: I originally insisted that the perfect group in question be finite, however I now realize that I do not need this condition, only that the generators used in the presentation have finite order. Perhaps this makes things easier.

A group $G$ is said to be perfect if its abelianization is trivial. Equivalently, a group is perfect if the subgroup generated by its commutators (elements of the form $g^{-1}h^{-1}gh$ for $g,h \in G$) is the whole group.

We say a presentation $G = \langle S | R\rangle$ is a locally commuting presentation if $S$ is a set of generators and $R$ is a set of relations (words in $S$ set equal to identity) such that whenever two generators $a$ and $b$ appear in a word in $R$, the word $a^{-1}b^{-1}ab$ is also in $R$ (this is the relation requiring that $a$ and $b$ commute). See also this question: Nonabelian finite groups with "locally commuting" presentation

My question is whether there is some nontrivial perfect group $G$ which has a locally commuting presentation using generators of finite order (and using only finitely many generators and relations), or is there some reason why this is not possible?

David Roberson
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    I don't know if this is a useful observation but this local commuting relations seems to mean you are looking at quotients of right angled artin groups by elements which are products of vertices from a clique – Benjamin Steinberg Mar 24 '21 at 14:42
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    A slightly categorical view: If $G$ is a group we can consider the group $W_G$ obtained by amalgamating all abelian subgroups of $G$ along their intersections. There is a canonical surjection $W_G$. Say that $G$ is the amalgamation of its abelian subgroups if $W_G\to G$ is an isomorphism. I guess the question is equivalent to whether this is plausible for a nontrivial finite perfect group. It's also (more obviously) equivalent to being an amalgam of abelian groups. – YCor Mar 24 '21 at 17:32
  • @BenjaminSteinberg I'm sorry to say that I don't really know anything about right angled artin groups. So your comment may be very useful but unfortunately not very useful to me. – David Roberson Mar 24 '21 at 17:43
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    @YCor: "amalgam" isn't quite right -- and if it were, the answer would be trivially "no", since non-trivial amalgams are always infinite. I think the correct statement is that $G$ should be the pushout of the diagram of inclusions of its abelian subgroups. Phrased like this -- "Can a pushout of a finite diagram of finite abelian groups be a finite perfect group?" -- the question is a very nice one, and should have a well-known answer. (Unfortunately it's not well known to me! :)) – HJRW Mar 27 '21 at 11:33
  • @HJRW I'm obviously using a more general notion of amalgam than the one from Bass-Serre theory. Of course it's a particular case of colimit / pushout, and Bass-Serre theory essentially describes amalgams along a tree. (In addition to being intuitive, the word amalgam is also practical because there's a verb "amalgamate"!) – YCor Mar 28 '21 at 23:33
  • @YCor: It wasn't obvious to me! I strongly suggest using "pushout" (which after all isn't a very long word) for pushouts, and reserving "amalgam" for the context of Bass--Serre theory, where edge maps are assumed to be injective and there are no higher intersections. The distinction is very important, because the very first lemma of Bass--Serre theory -- that vertex groups embed -- fails for arbitrary pushouts... – HJRW Mar 29 '21 at 10:14
  • (cont'd) ... If you like, we can discuss the merits or otherwise of using the term "amalgam" for developable complexes of groups. But that's not what this question, or at least its first version, is about. – HJRW Mar 29 '21 at 10:15

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I guess I should point out that with Josse van Dobben de Bruyn and Simon Schmidt we have answered this question here (see Section 5): https://arxiv.org/abs/2311.04889

We show that if one uses all the relations arising from pairwise commuting elements of order 2 in the alternating group $A_n$, then the resulting group is perfect and nontrivial for $n \ge 7$. For $n=7$, the resulting group is the triple cover of $A_7$. This construction also allows us to construct the first example of a graph whose automorphism group is trivial but whose quantum automorphism group is not trivial.

David Roberson
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