While trying to carry out some technical arguments in free groups, I have encountered the following problem, to which I don't know the answer.


Let $F$ be a free group and let $g,a\_1,\ldots,a\_n \in F$. Suppose that $g$ is not equal to a product of conjugates of $a\_1,a\_2,\ldots,a\_n$, in that order. That is, there do not exist $x\_1,x\_2,\ldots,x\_n \in F$ with $g = a\_1^{x\_1}a\_2^{x\_2}\cdots a\_n^{x\_n}$.


Is there necessarily a finite quotient $F/N$ of $F$ in which the same is true of the images of $g,a\_1,\ldots,a\_n$. That is, there do not $x\_1,x\_2,\ldots,x\_n \in F$ such that $g^{-1}a\_1^{x\_1}a\_2^{x\_2}\cdots a\_n^{x\_n} \in N$?



products of conjugates in free groups
The answer to your question is no; there need not be a finite quotient like that. This also answers the question of Lev Glebsky and Luis Manuel Rivera Martinez mentioned in a comment. I learned this argument from Jakub Gismatullin (who presented it in a similar form at a workshop at the Erwin-Schrödinger-Institute in April 2013).


The key is the following deep theorem, proved by Nikolay Nikolov and Dan Segal in [Nikolay Nikolov, Dan Segal, *Generators and commutators in finite groups; abstract quotients of compact groups*, Invent math (2012) 190:513–602].



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> **Theorem (Nikolov-Segal):** There exists a constant $n$ such that the following holds. For every $2$-generator finite group $G$ with generators $g\_1,g\_2$, every element of the subgroup $[G,G]$ is a product of $n$ factors of the form $[g,g\_i^{\pm 1}]$ with $g \in G$, $i =1,2$.
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Let now $F$ be the free group on two generators $g\_1,g\_2$. Observe that $[g,g\_i^{\pm 1}] = (gg\_i^{\pm}g^{-1})g\_i^{\mp}$, so that any commutator with a generator is a product of two conjugates of the generators. Choose $N=2n \cdot 4^n-1$ and find a sequence $a\_0,\dots,a\_N$, where $a\_j \in \{g\_1,g\_1^{-1},g\_2,g\_2^{-1}\}$ and $a\_{2j+1} = a\_{2j}^{-1}$, and such that any possible choice of a sequence of length $n$ appears among sequences $(a\_{2j}, a\_{2j+2}, \dots, a\_{2(j+n-1)})$. (Existence is easy, just concatenate all possible sequence on the even indices and choose appropriate elements for the odd indices.)


Use the well-known fact that the commutator width of $F$ (free group on two generators) is infinite and choose some element $g \in [F,F]$, whose commutator length is strictly larger than $n \cdot 4^n$. It is clear that $g$ is not of the form $a\_0^{x\_0}a\_1^{x\_1} \cdots a\_N^{x\_N}$, since $a\_{2j}^{x\_{2j}} a\_{2j+1}^{x\_{2j+1}}$ is a commutator by construction.


However, in any finite quotient $G=F/N$, we have $gN \in [G,G]$ and thus (by the theorem above) we can find $x\_0,x\_1,\dots,x\_N$ such that $a\_0^{x\_0}a\_1^{x\_1}a\_2^{x\_2} \cdots a\_N^{x\_N} \in gN$. Indeed, we write $gN$ as a product of $n$ commutators with generators, locate a suitable segment of the sequence of the $a\_i$'s, choose appropriate $x\_i$'s there and set all other $x\_i$'s equal to the neutral element.


As a remark, this does not provide an example of a group which is not weakly sofic and it does not disprove Conjecture 2.1 in the paper by Glebsky-Rivera Martinez. It only proves that products of conjugacy classes need not be closed in the pro-finite topology.


