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It is an old result of Schützenberger that in a free group, a basic commutator cannot be a proper power. A look at the original reference

M.-P. Schützenberger, Sur l'équation $a^{2+n} = b^{2+m}c^{2+p}$ dans un groupe libre, C. R. Acad. Sci. Paris 248 (1959), 2435–2436 (French).

quickly reveals that a lot of details are missing and some claims appear to be wrong.

Question: Let $F$ be a free group and $a,b,c \in F$ with $c \neq 1$ and $n \geq 2$. Why is $[a,b] \neq c^n$?

In particular, it would be nice to have a somewhat geometric proof of this apparently fundamental fact. There is an algebraic proof in

G. Baumslag, Some aspects of groups with unique roots, Acta Math 104(3) (1960), 217–303.

as Lemma 36.4 but it is relies on various technical computations and is hard to grasp.

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3 Answers 3

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The case $n=2$ (originally due to Lyndon) admits a very nice geometric argument: one notes that elements $a,b,c$ with $[a,b]=c^2$ lead to a map from the surface $\Sigma_{-1}$ of Euler characteristic -1 to a graph. Pulling back midpoints of edges, one obtains essential, two-sided, simple closed curves on $\Sigma_{-1}$. Pinching these, Euler characteristic shows that one of the resulting components must be a projective plane, and the result follows.

For $n> 2$, Duncan--Howie proved the stronger theorem that, for any non-trivial element $c$ of $F_2$, the stable commutator length of $c$ is at least $1/2$. (In fact, by passing to covers, one can also deduce the $n=2$ case from the Duncan--Howie theorem.) When you decode it, their proof is quite geometric, relying on defining a sort of combinatorial vector field on surfaces.

Louder and I gave a sort of generalization of Duncan--Howie's proof in our work on one-relator groups. Again, the idea is to count Euler characteristic using some sort of geometric data (we call it a stacking). We don't spell it out explicitly, but you can use our ideas to give a geometric proof of the Duncan--Howie theorem, and in particular of the result that you want.

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  • $\begingroup$ How does the result for $n>2$ follow from the Duncan-Howie theorem? the latter a priori says nothing for fixed $n$, at least in the way you state it (which is something asymptotic on $c^n$ for $n\to\infty$)? $\endgroup$
    – YCor
    Dec 9, 2016 at 4:24
  • $\begingroup$ @Yves, ${scl}(w)$ is the infimum of the commutator lengths of $w^n$ over $n$. $\endgroup$
    – HJRW
    Dec 9, 2016 at 6:00
  • $\begingroup$ Ah, thanks, I thought of it as a limit and not an infimum, so now I see why $n=2$ is special. $\endgroup$
    – YCor
    Dec 9, 2016 at 6:06
  • $\begingroup$ Actually, I do not understand all this about $n=2$. Duncan and Howie do not use the term scl at all. Their Theorem 5.1 implies, in particular, that, a non-identity product of $k$ commutators in a free group cannot be an $n$th power for $n\geqslant2k$. What is special for $n=2$? $\endgroup$ Aug 12, 2020 at 9:22
  • $\begingroup$ @AntonKlyachko, you’re quite right: the Duncan—Howie theorem also handles the $n=2$ case. I confused myself by rephrasing the result in terms of scl, and then using the “wrong” definition of scl, namely as the stablised commutator length. (Since this was four years ago, I don’t remember why I wanted to do it this way!) If you use the correct definition of scl, namely as the stabilised Euler characteristic, then indeed there’s nothing special about the $n=2$ case. Thanks for pointing this out. Perhaps I’ll rewrite the answer to clarify. $\endgroup$
    – HJRW
    Aug 12, 2020 at 10:26
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(a) On the free 2-step nilpotent group $L$ on $(u,v)$, the commutator $z=[u,v]$ is not a proper power.

Indeed, since $L/[L,L]$ is torsion-free, any root of $z$ should lie in $[L,L]$ and the latter is readily seen to be the cyclic group generated by $z$. (Remark: $L$ is isomorphic to the integral 3-dimensional Heisenberg group.)$\qquad\Box$

I'll use three other very classical facts, without proof: the hardest (b) has both geometric and combinatorial proofs (the simplest using that $\pi_1$ of graphs are free), while (c),(d) are elementary exercises ((c) follows from Malcev's more general results but is very easy by hand here).

(b) Subgroups of free groups are free.

(c) $L$ is Hopfian (all its surjective group endomorphisms are automorphisms).

(d) If $N$ is a nilpotent group and $M$ a subgroup generating $N$ modulo $[N,N]$ (that is, $M[N,N]=N$), then $M=N$.

This allows to prove the result.

In a free group $F$, if $a,b,c$ are two elements and $[a,b]\neq 1$ then $[a,b]$ is not a proper power $c^n$ of $c$.

By (a), we can suppose that $F$ is generated by $a,b,c$. If $d$ is the rank of $F$, it is also the rank of its abelianization, and the relation $[a,b]=c^n$ then forces $d\le 2$. Also $d\ge 2$ since $a,b$ don't commute. So $F$ is free on 2 generators (up to now this is exactly Baumslag's argument).

Hence $L=F/[F,[F,F]]$ is free 2-step nilpotent on 2 generators. Denote by $A,B,C$ the images of $a,b,c$ in $L$. Since $L/[L,L]$ is torsion-free and $C^n\in [L,L]$, we have $C\in [L,L]$. By (d), we deduce that $L$ is generated by $A,B$. So the endomorphism of $L$ mapping the two free generators to $A$ and $B$ is surjective. By (c) deduce that $L$ is free 2-step nilpotent on $(A,B)$. Since $[A,B]=C^n$, we can use (a) to get a contradiction. (Baumslag's conclusion of the argument also uses a 2-step nilpotent group, but I haven't followed as it's more technical)

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  • $\begingroup$ Iirc, this is very similar to Baumslag's argument (though he uses the Hopf-ness of the pro-p completion). $\endgroup$
    – HJRW
    Dec 8, 2016 at 19:55
  • $\begingroup$ @HJRW, Andreas said that Baumslag's proof relies "on various technical computations and is hard to grasp". I hope you don't find this about this proof (even if it's the same in spirit). $\endgroup$
    – YCor
    Dec 8, 2016 at 22:11
  • $\begingroup$ I agree that Baumslag's proof may not look like this at first glance. But I think it's essentially the same. $\endgroup$
    – HJRW
    Dec 8, 2016 at 22:21
  • $\begingroup$ @HJRW no problem. I claim no originality, it's an exercise anyway. $\endgroup$
    – YCor
    Dec 8, 2016 at 22:21
  • $\begingroup$ It's very nice! I just thought it might be good to connect the dots. $\endgroup$
    – HJRW
    Dec 8, 2016 at 22:26
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There is a visualisation of Schützenberger's observation based on a variant of the car-crash lemma that says that

for any multiple motion on a map on a closed oriented surface $S$, the number of points of complete collision is at least $\chi(S)+\sum\limits_D(d_D-1)$, where the summation runs over all faces $D$ of the map and $d_D$ is the number of cars moving around a face $D$.

See [this self-advertisement] for exact definitions.

So, if a non-identity commutator in a free group $F(x,y,\dots)$ is, e.g., a cube of a word $c$, we obtain a one-face map on a torus, where the label of the face is $c^3$. Let three cars move around this face counter-clockwise with a constant speed one edge per a minute; at a moment $t$ each car moves along an edge labelled by the $[t]$th letter of the word $c$ (where the integer part $[t]$ of $t\in\mathbb R$ is counted modulo the length of $c$).

3 cars on a torus

The car-crash lemma asserts that $$ \small \pmatrix{ \hbox{the number of points} \\\\ \hbox{of complete collision} } \geqslant \begin{pmatrix} \hbox{the Euler characteristic} \\\\ \hbox{of the torus,} \\\\ \hbox{i.e. 0} \end{pmatrix} + \pmatrix{ \hbox{$d_D$, i.e. the number of cars} \\\\ \hbox{moving around the unique face $D$,} \\\\ \hbox{i.e. 3} } -1=2. %} $$ On the other hand, a collision inside an edge is impossible:

Cars colliding inside an edge

such a collision would imply that some letter of the word $c$ is $x$ and $x^{-1}$ simultaneously. A similar argument shows that a collision at a vertex cannot occur too. This contradiction proves that a commutator cannot be a cube.

This approach was used in [this self-advertisement], [this self-advertisement], and [this self-advertisement] to obtain generalisations of Schützenberger's observation and Duncan--Howie's theorem (mentioned by Henry Wilton).

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