# Commutator length modulo finite index subgroups

We write $cl$ for the commutator length, i.e. the least number of commutators which multiply to a given element of a group.

Given an element $g$ in the commutator subgroup of the free group $G=F_2$ on two generators, is it true that $$cl_G(g) = \displaystyle \max_{\mbox{H < G finite index normal}} cl_{G/H} (g \mod H)$$ ?

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On the RHS, you mean $\text{cl}_{G/H}$, right? – Andy Putman Jan 14 '13 at 18:49
Indeed, I changed that. – dlbb Jan 14 '13 at 19:37

I think the answer is no and that actually the "finite quotient" commutator length (defined by your formula) is bounded on $[F_2,F_2]$.
Indeed by Nikolov-Segal, in the profinite completion $P$ of $F_2$, the derived subgroup $[P,P]$ is closed; since the set $C$ of commutators is compact, it follows by a Baire argument that $[P,P]$ is boundedly generated by $C$ (first get by Baire that some open neighborhood of the identity of $[P,P]$ has bounded commutator length in $P$ and then use compactness). Observe on the other hand that $[F_2,F_2]=[P,P]\cap F_2$, which is essentially trivial since the abelian group $F_2/[F_2,F_2]$ is residually finite. So in any finite quotient of $F_2$, the commutator length of $w\in [F_2,F_2]$ is bounded by the universal number that arises as upper bound of the commutator length of $[P,P]$ in $P$.
On the other hand, the commutator length of $F_2=\langle x,y\rangle$ is unbounded on $[F_2,F_2]$, as the commutator length of $[x,y]^n$ grows linearly (I think it's due to Bavard).
Nice argument; however, looking at the Nikolov-Segal paper, it appears that the way that they show that $[P,P]$ is closed is to show that every element in it is a product of a bounded number of commutators: ams.org/mathscinet-getitem?mr=2016979 Of course, your argument shows why these things are equivalent. – Ian Agol Jan 15 '13 at 6:48