23
$\begingroup$

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

$\endgroup$
2
  • 4
    $\begingroup$ 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. $\endgroup$ Apr 2, 2015 at 15:10
  • 3
    $\begingroup$ 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, 2015 at 15:56

1 Answer 1

27
$\begingroup$

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.

$\endgroup$
1
  • 1
    $\begingroup$ Thanks! I was starting to suspect that the answer was no, and it's good to have it confirmed. $\endgroup$
    – Derek Holt
    Apr 2, 2015 at 22:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.