# How to solve this one-variable equation in a free group?

Let $F_n$ be a free group, $u, v \in F_n$, where $[u,v] \neq 1$. Trying to show that the following equation does not have a solution in $F_n$ :

$$[v,x] = [u,v] .$$

Any ideas are appreciated. I seem to have a proof but it's ugly (combinatorial, cancellation theory etc.) and not short (so, more likely to have a silly mistake :) )

Immediate thing to notice is that if $x=w$ is a solution, then so is the "line" $x=C(v)w$, where $C(v)$ is a centralizer of $v$.

Also, not difficult to show that there are no solution in a free group of rank two generated by $u$ and $v$.

thanks!

-
One remark, the subgroup $\langle u,v,x\rangle$ must be rank 2, since it is the image of a genus 2 surface group (it is a standard fact that the image of a genus $g$ surface group in a free group must have rank $\leq g$, by considering preimages of midpoints of edges). –  Ian Agol Jan 25 '13 at 18:37
@Ian: The fact that $\langle u,v,x\rangle$ is of rank 2 is obvious: it is of rank at most 3, but cannot be free of rank 3 since it satisfies a non-trivial relation. The equation is both quadratic and has one variable. There are lots of papers about solving quadratic equations and equations with only one variable in the free group. These papers provide algorithms, but I do not see how to apply those results here, except by using Wicks forms (which people do when they solve quadratic equations) as I suggested in my answer. –  Mark Sapir Jan 25 '13 at 19:26

This is not a complete solution, but it may help. Suppose the equation has a solution. Then use the Wicks forms (Lyndon, R. C.; Wicks, M. J. Commutators in free groups. Canad. Math. Bull. 24 (1981), no. 1, 101–106.): there exist reduced decompositions $u\equiv a\bar c, x\equiv cb$, $v\equiv p\bar q, u\equiv qr$ such that $[u,x]\equiv abc\bar a\bar b\bar c, [v,u]\equiv pqr\bar p\bar q\bar r$ where $\equiv$ means there are no cancellations in these words, and $\bar z=z^{-1}$. Since $abc\bar a\bar b\bar c=pqr\bar p\bar q\bar r$, we have that either $\bar c$ is a suffix of $\bar r$ or vice versa, etc. The point is that with Wicks forms you can work as in the free semigroup (no cancellations).

-
thanks guys. Mark, sure I am familiar with those papers (Miasnikov, Remeslennikov-Chiswell, and old ones - Appel, etc.). From classification of solutions, all that is clear is that solution (if it exists) is a "line". From algorithmical perspective, since this one is not a specific equation (in a sense that u and v are arbitrary), those methods do not help here (at least directly). I will definitely look at Wicks forms. Seems like a more efficient way to go than the direct (cancellations) one. –  Alexey Jan 26 '13 at 3:26
It might be that this is "more" a quadratic equation than an equation in one variable. So you may want to look at papers by Alina Vdovina and co-authors. –  Mark Sapir Jan 26 '13 at 4:01

Hi Mark, just wanted to share the solution to this problem as well as a method used. I found quite a powerful tool buried at the end of this paper

"A Classification of Fully Residually Free Groups of Rank Three or Less", Benjamin Fine, Anthony M Gaglione, Alexei Myasnikov, Gerhard Rosenberger, Dennis Spellman.

Here is the result that follows from the proofs/discussion of that paper and stated in the Section 8, Theorem 6:

Let $F_{u,v} = \langle u ,v \rangle$ be a free group of rank two and suppose that $a(u,v)$ and $b(u,v)$ are non-trivial and not proper powers in $F_{u,v}$. If $a(u,v)$ and $b(u,v)$ are conjugate in some overgroup $F_{u,v} < F$ then either $a(u,v)$ and $b(u,v)$ are already conjugate in $F_{u,v}$ or the element $ta(u,v)t^{-1}b(u,v)^{-1}$ is a primitive in a free group of rank three $F_{3} = \langle t,u,v \rangle$.

Let me now demonstrate that the equation $[x,u]=[u,v]$ has no solutions in $F$. Suppose the above equation has a solution $x=\omega$. We have then $\omega^{-1}u^{-1}\omega =[u,v]u^{-1}$, so that elements $u^{-1}$ and $[u,v]u^{-1}$ are conjugated in $F$. First, notice that $u^{-1}$ and $[u,v]u^{-1}$ are neither proper powers nor conjugated in $F_2= \langle u,v \rangle$. Hence, by the above result, element $e=t^{-1}u^{-1}tu[v,u]$ must be primitive in $F_3 = \langle t,u,v \rangle$. However element $e$ belongs to the derived subgroup $F_3 '$ and as such can not be primitive.

-