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For a finitely presented group $G$, generated by a finite set $A$, the commutator problem is the decision problem: given a word $w$ over the alphabet $A \cup A^{-1}$, can one decide if $w$ is a commutator, i.e. whether there exist words $x, y$ such that $w = [x, y]$ in $G$. Here $[x, y] = x^{-1}y^{-1}xy$ is the commutator. The commutator problem was solved for free groups by Wicks in 1962.

In 1981, Comerford & Edmunds [1] asked whether decidability of the commutator problem for $G$ implies decidability of the conjugacy problem, or even the word problem, for $G$. Has there been any recent progress on this question since then, or any results in a similar direction?

[1] Comerford, Leo P. jun.; Edmunds, Charles C., Quadratic equations over free groups and free products, J. Algebra 68, 276-297 (1981). ZBL0526.20024.

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    $\begingroup$ This would imply that every f.p. group in which every element is a commutator, has solvable conjugacy problem. This sounds suspicious. $\endgroup$
    – YCor
    Commented May 12, 2021 at 10:38
  • $\begingroup$ @YCor That's a good point. As far as I can tell it gives no obvious algorithm for solving the conjugacy problem in finite simple groups, for example (even though it is of course decidable). I can't right now think of a candidate f.p. group in which every element is a commutator but in which not every element is conjugate. $\endgroup$
    – Carl-Fredrik Nyberg Brodda
    Commented May 12, 2021 at 13:03

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Denis Osin [Osin, Denis, Small cancellations over relatively hyperbolic groups and embedding theorems, Ann. Math. (2) 172, No. 1, 1-39 (2010). ZBL1203.20031.] proved that every torsion-free countable group can be embedded into a $2$-generated group with exactly two conjugacy classes of elements, i.e., any two non-trivial elements will be conjugate.

So, let $H$ be a finitely generated torsion-free countable group with unsolvable word problem. Let's embed $H$ into a $2$-generated group $G$ with $2$ conjugacy classes using Osin's theorem. Then every element in $G$ is a commutator and $G$ has unsolvable word problem. Thus the commutator problem in $G$ is solvable but neither the word problem nor the conjugacy problem are decidable.

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    $\begingroup$ Excellent, thanks! This is very nice. I wonder if an example can be constructed where the word problem is decidable but not the conjugacy problem. This trick won't work, as the conjugacy problem and word problem are of course equivalent in a group with only $2$ conjugacy classes. $\endgroup$
    – Carl-Fredrik Nyberg Brodda
    Commented May 12, 2021 at 18:16
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    $\begingroup$ From my experience, almost any reasonable combination of decision properties can be achieved in a finitely generated group obtained using small cancellation methods. $\endgroup$
    – Ashot Minasyan
    Commented May 12, 2021 at 21:07

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