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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 $g^{-1}a_1^{x_1}a_2^{x_2}\cdots a_n^{x_n} \in N$?

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    $\begingroup$ If I'm not mistaken, here is another way to phrase this question: Is the product of finitely many conjugacy classes in $F$ closed with respect to the profinite topology on $F$? According to arxiv.org/abs/0709.0026, (last paragraph on page 2), this was open in 2008. $\endgroup$ – Anton Malyshev Apr 2 '15 at 15:10
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    $\begingroup$ Ah that's interesting because what I was trying to do was connected with weakly sofic groups. It looks as though I may have just rediscovered the same open problem, which is probably very difficult. $\endgroup$ – Derek Holt Apr 2 '15 at 15:56
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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].

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 $g \in G$, $i =1,2$.

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.

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    $\begingroup$ Thanks! I was starting to suspect that the answer was no, and it's good to have it confirmed. $\endgroup$ – Derek Holt Apr 2 '15 at 22:59

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