Let $F$ be the free group on (say) two generators, $a$ and $b$. Let $A$ and $B$ be (freely reduced) elements of $F$. Let $W(X, Y)$ denote a word on the words $X, Y$.


-Is it ever true that the equation $W(a, b) = W(A, B)$, has finitely many non-conjugate solutions (by conjugate solution I mean there exists a word $V$ such that $V^{-1}AV = A^{\prime}$ and $V^{-1}BV=B^{\prime}$)?


For example, take $W(a, b) = a^{-1}b^nab^m$. We therefore want to find $A$, $B$ such that $a^{-1}b^nab^m = A^{-1}B^nAB^m$. We can take $A=b^ia$ and $B=b$ for all $i$, and so this equation has infinitely many (non-conjugate) solutions.


In fact, $a\mapsto b^ia$, $b \mapsto b$ defines an automorphism of $F$ (as free groups are Hopfian). Further, different $i$'s give different coset representatives of Out(F), and so a related question would be,


-Does there exist a word $W \in F$ such that there are only finitely many outer automorphisms $\phi$ such that $W\phi = W$?


I cannot seem to get anywhere with this. The only examples I can find are, essentially, trivial. For example, $W(a, b) = a$. However, this doesn't quite work, as then $b$ can be whatever we want (essentially, exclude this because its boring).


Any help/ideas of papers to look at would be greatly appreciated.



Solutions to some equations in a free group
Let $G=\langle x,y,c\mid w(x,y)=c\rangle$ (which, in this case, happens to be a free group). The set of solutions to the system that you are looking for corresponds precisely to the set of homomorphisms $f:G\to F$ satisfying $f(c)=w(a,b)$. The basic idea of Sela's approach to the study of equations over $F$ (see [here](http://www.ams.org/mathscinet/search/publdoc.html?arg3=&co4=AND&co5=AND&co6=AND&co7=AND&dr=all&pg4=AUCN&pg5=TI&pg6=PC&pg7=ALLF&pg8=ET&review_format=html&s4=sela%252C%2520z%2a&s5=&s6=&s7=&s8=All&vfpref=html&yearRangeFirst=&yearRangeSecond=&yrop=eq&r=9&mx-pid=1863735), *et seq.*) is to study the splittings of $G$ as an amalgamated free product or HNN extension. Specifically, we have the following. (I'm not sure this exact statement is written down anywhere, but I think I've got it correct. To really do this right, you should probably intone the magic words 'graded limit group' somewhere.)


**Theorem (Sela):** There are infinitely many conjugacy classes of solutions if and only if $G$ splits (as an amalgamated free product or HNN extension) over a cyclic group, with infinite centraliser, relative to $c$. (That is, $c$ should be conjugate into one of the vertex groups of the splitting.)


*Sketch proof:* If there are infinitely many solutions then one can take a Gromov--Hausdorff limit and get an action on a real tree. The Rips machine promotes this to an action on a simplicial tree.


Conversely, if $G$ splits as hypothesised then a Dehn twist in this splitting gives one infinitely many outer automorphisms that preserve $w$, as required. *QEDish*


So, in order to write down an example of the form you want, you simply need a $w$ such that $F$ has no cyclic splittings relative to $w$. This is easier said than done, though in principle any `sufficiently generic' $w$ should do. In [a recent paper](http://arxiv.org/abs/1006.2123), Cashen and Macura described an algorithm for finding cyclic splittings of $F$ relative to $w$. (In fact, they describe their algorithm as finding cut pairs in a certain decomposition space. But these turn out to correspond to cyclic splittings, as Cashen has [shown](http://arxiv.org/abs/1009.2492).)


According to Theorem 6.3 of Cashen and Macura, any word $w$ for which the Whitehead graph is complete will be `rigid', ie $F$ does not split cyclically relative to $w$. For instance, I think the word


$w=aba^{-1}b^{-2}ab^{-1}a^{-2}$


works.


