Determining whether an element of a free product of cyclic groups is a commutator.

Question

Let $G=C_{n_1}*\cdots*C_{n_k}=\langle a_1,\cdots,a_k\mid a_1^{n_1}=\cdots=a_k^{n_k}=1\rangle$ be a free product of finitely many finite cyclic groups. Given a word $g=g_1\dots g_n\in G$, is there an algorithm to determine whether $g$ is a commutator in $G$, i.e. whether there exist $a,b\in G$ such that $g=aba^{-1}b^{-1}$?

A specific example that I am interested in is $G=C_4*C_4$ and $g=(a_1a_2a_1a_2^{-1})^2$. I believe that this element is not a commutator, but I don't see a simple proof.

The only "test" I currently know of is that $g$ must reduce to the identity in the Abelianization of $G$, but of course, this only says that $g$ is a product of commutators and not necessarily a single commutator.

Answer 1

For your example, the following (very naive) computation in Magma seems to show that the image of your element $g$ in a finite quotient of $G = C_4 * C_4$ of order $256$ is not a commutator, so it cannot be a commutator in $G$ itself.

> G  := Group<x,y | x^4,y^4>;
> P<a,b>:=pQuotient(G,2,3);                                        
> #P;
256
> S:={ (x,y) : x in P, y in P};
>  (a*b*a*b^-1)^2 in S;
false

In fact $g$ is the unique element in $[P,P]$ that is not a commutator.

> #S;
15
> #DerivedGroup(P);
16
> { x : x in DerivedGroup(P) | not x in S} eq { (a*b*a*b^-1)^2 };
true