# 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?

• Since $\langle x, y \mid x^2, y^3 \rangle\cong\Bbb Z_2\ast\Bbb Z_3$ is a free product, perhaps Subsection 4.3: "Subgroup Theorems for Free and Amalgamated Products" of "Combinatorial Group Theory" by Magnus et al. might help. – Shaun May 27 '18 at 14:54
• Indeed, there is such a geometric approach, via the Ping Pong Lemma. This answer shows the group at hand is $\text{PSL}_2 (\mathbb{Z})$, which comes equipped with an action by Mobius transformations on the space of irrational numbers. Can you use the Ping Pong Lemma with this action to prove the subgroup is free? – Sameer Kailasa May 27 '18 at 16:02
• I think this example really is straightforward. If you let $t=yxy$ and $u=xyxyx$, then $u^k = x(yxy)^kx$, and apart from that there is no cancellation when you form reduced words in $t^{\pm 1}$ and $u^{\pm 1}$, so no such reduced word can represent the identity and hence $\langle t,u \rangle$ is free. – Derek Holt May 27 '18 at 16:30
• It can also be done computationally (in GAP for example), because the subgroup has index $6$ in $G$, so you could use Reidemeister-Schreier to compute a subgroup presentation. – Derek Holt May 27 '18 at 16:32
• By Bass-Serre theory, this group acts on a bipartite tree where one set of vertices has valence 3 and is stabilized by the conjugates of $y$, and the other set of vertices has valence 2 and is stabilized by the conjuates of $x$. Check that $yxy$ and $xyxyx$ are translations along axes whose intersection, if nonempty, is finite. It follows that they generate a free group. – Lee Mosher May 28 '18 at 3:50

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).