Proving $yxy$ and $xyxyx$ generate a free subgroup of $\langle x, y \mid x^2, y^3 \rangle$. I am trying to prove that the subgroup of $\langle x, y \mid x^2, y^3 \rangle$ generated by $yxy$ and $xyxyx$ is free. I know that I must show every reduced word can by described uniquely by these generators, but I still lack a general approach for this type of exercise. 
Even the brute force approach of comparing $w = w'$ and showing that the generators used must be the same does not work for me, but I also wonder if there is some more general approach using geometric group theory for these problems?
 A: This follows from the normal form theorem for free products (you should think of this solution as being the "brute force" solution*). First, a definition: A reduced sequence is a sequence $g_1, g_2, \ldots, g_n$, $n\geq0$, of elements of $A\ast B$ such that each $g_i\neq1$, each $g_i$ is in one of the factors, $A$ or $B$, and successive $g_i, g_{i+1}$ are not in the same factor. (We allow $n=0$ for the empty sequence.)

Theorem. (Lyndon and Schupp, Combinatorial group theory, Theorem IV.1.2, p175)
  If $w=g_1\cdots g_n$ where $g_1, \ldots, g_n$ is a reduced sequence, then $w\neq 1$ in $A\ast B$.

For your example we can use the idea pointed out in Derek Holt's comment: the elements you are looking at are words over $yxy$ and $x(yxy)x^{-1}$ (note that $x^{-1}=x$). Then each non-empty word $w$ can be thought of as freely-reduced "power-free" word $u$ over $\{(yxy)^k, x(yxy)^kx\mid k\in\mathbb{Z}\setminus\{0\}\}$ (by "power-free" I mean $a^2, a^3, \ldots\not\leq u$, etc.). The word $u$ can be chosen so that no free reduction occurs when forming the word $u$. Hence, $u$ corresponds to some reduced sequence, and so $u\neq1$.
*In practice I would use the action of $A\ast B$ on a tree, as in Lee Mosher's comment. But this is harder to explain without pictures! If you are interested, then you could look up Section 3.5 of the book "Groups, graphs and trees" by John Meier (Section 3.1.3 also contains a Ping-Pong argument, c.f. Sameer Kailasa's comment).
