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**](http://dx.doi.org/10.1016/0021-8693(81)90265-9), J. Algebra 68, 276-297 (1981). [ZBL0526.20024](https://zbmath.org/?q=an:0526.20024).



Commutator problem vs conjugacy/word problem
Denis Osin [*Osin, Denis*, [**Small cancellations over relatively hyperbolic groups and embedding theorems**](http://dx.doi.org/10.4007/annals.2010.172.1), Ann. Math. (2) 172, No. 1, 1-39 (2010). [ZBL1203.20031](https://zbmath.org/?q=an:1203.20031).] 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.


