Determining whether two free groups are conjugates Let $A= \langle a,b \rangle$, i.e. the free group on two generators. Also, let $B=\langle b^{-1}a^{-1}bab,b^{-1}a^2b, b^2, a^2, ab^2a^{-1}, abab^{-1}a^{-1} \rangle$, $C= \langle a^{-1}b^{-1}aba,a^{-1}b^2a, a^2, b^2, ba^2b^{-1}, baba^{-1}b^{-1} \rangle$.
I'm trying to determine whether $B$ and $C$ are conjugate in $A$. From a glance, they don't look like they are but I'm not entirely sure how to formally write it. I know that the abelinations of $B$ and $C$ are the same in $C$, but that doesn't tell me whether they are actually conjugate or not.
 A: One way to solve this by hand with pencil and paper (or e-tablet and e-pencil) is to compute the Stallings graphs of $B$ and $C$.
To describe the Stallings graph, start with the rose graph $R_A$ with one vertex $V$, and with two edges that are oriented and labelled $a$ and $b$ respectively. Any finite rank subgroup $H < A$ has a unique Stallings graph, which is a finite connected graph $\Gamma_H$ without valence 1 vertices, and which is labelled by assigning to each edge an orientation and one of the letters $a$ or $b$. The topological meaning of this is that the induced graph map $f : \Gamma_H \to R_A$ (taking each oriented, labelled edge of $\Gamma_H$ to the corresponding oriented labelled edge of $R_A$) is locally injective, and that $H$ is conjugate to the image of the induced homomorphism $f_*: \pi_1(\Gamma_H,w) \to \pi_1(R_A,v) = \langle a,b \rangle$ for any vertex $w \in \Gamma_H$. The uniqueness condition means that the label preserving isomorphism class of $\Gamma_H$ is uniquely determined by the conjugacy class of $H$.
Thus, if you can compute the Stallings graphs of $B$ and $C$ then you can examine the graphs to determine whether $B$ and $C$ are conjugate.
To compute the Stallings graph of $B$, for example, start with a rank 6 rose graph having one loop for each of the 6 words listed in the generating set for $B$. Subdivide the first loop into five edges labelled $b^{-1} a^{-1} bab$; subdivide  the second loop subdivided into 4 edges labelled $b^{-1} a a b$; and subdivide similarly for the remaining four edges.
Now repeat the following step inductively: look for a vertex having two directions coming out of that vertex with identical labels, and compute the quotient graph by identifying those two edges. For example, the first loop has first edge labelled $b^{-1}$ and the second loop has first edge labelled $b^{-1}$; so identify those edges.
The number of edges will decrease strictly with each repetition, and therefore eventually you will arrive at a graph such that at each vertex, any two directions coming out of that vertex have distinct labels. The graph you end up with might have valence 1 vertices; inductively remove each such vertex and its incident edge. At the end of that process, the resulting labelled graph is the Stallings graph of $B$.
You can find a detailed description with examples in my book.
A: To prove that $B=C$, we just have to prove that all of the generators of $B$ are in $C$ and vice versa. Many of these are clear. Here is a proof that the first generator of $C$ is in $B$ - the others are similar. (Of course this was all guided by computer calculations.)
$$a^{-1}b^{-1}aba=(a^{-2})(ab^2a^{-1})^{-1}(abab^{-1}a^{-1})(ab^2a^{-1})(a^2).$$
