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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?

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    $\begingroup$ 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 $\endgroup$
    – Benjamin Steinberg
    Commented Mar 24, 2021 at 14:42
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    $\begingroup$ 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. $\endgroup$
    – YCor
    Commented Mar 24, 2021 at 17:32
  • $\begingroup$ @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. $\endgroup$
    – David Roberson
    Commented Mar 24, 2021 at 17:43
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    $\begingroup$ @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! :)) $\endgroup$
    – HJRW
    Commented Mar 27, 2021 at 11:33
  • $\begingroup$ @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"!) $\endgroup$
    – YCor
    Commented Mar 28, 2021 at 23:33

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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.

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