TITLE: Decidability of conjugacy problem for finitely generated subgroups of free groups
QUESTION [10 upvotes]: The conjugacy problem for a free group $F_n$ on $n$ letters has an easy solution.  Each element of $F_n$ is conjugate to a unique and easily computable "cyclically reduced element" (this means that if you arrange the word around a circle, then there are no cancellations), so two elements of $F_n$ are conjugate if and only if they have the same cyclically reduced conjugates.
I've been trying unsuccessfully to generalize this to solve the following problem.  Let $\{x_1,\ldots,x_k\}$ and $\{y_1,\ldots,y_{k'}\}$ be two finite sets of elements of $F_n$.  Let $G_x$ and $G_y$ be the subgroups of $F_n$ generated by the $x_i$ and the $y_i$, respectively.  Is there an algorithm to decide if $G_x$ and $G_y$ are conjugate?  Does anyone know how to do this?  Thank you very much!

REPLY [7 votes]: There is an algorithm to do this.  I would have thought that it was classical, but in any case an algorithm is given in: I. Kapovich and A. Myasnikov "Stallings foldings and the subgroup structure of free groups", J. Algebra 248 (2002), no 2,   pp. 608-668.  In the online version I found here, it is Corollary 7.8 on page 18.