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# <x,y〈x,y : x^p = y^p = (xy)^p = 1>〉

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Let $p$ be an odd prime and $G := \langle x,y : x^p = y^p = (xy)^p = 1 \rangle$. I want to show that $G$ is infinite and wonder if there is a good way to prove this. I'm familiar with the basics of geometric group theory, but don't know if they can be used here.

Obviously $G^{ab} = \mathbb{Z}/p \times \mathbb{Z}/p$. In particular, $x,y,xy$ have order $p$. The same is true for $yx, x^{p-1} y^{p-1}$ and $y^{p-1} x^{p-1}$. My approach uses the same idea which is used in Serre's book Trees for elements in amalgamated sums. Define $M$ to be the set of formal words of the form $...x^i y^j ...$ (alternating powers of $x$ and $y$), where the exponents are in $[0,p-1]$ and $(xy)^p, (yx)^p, (x^{p-1} y^{p-1})^p, (y^{p-1} x^{p-1})^p$ are no subwords. There is a obvious action from the free group $\langle x,y \rangle$ on $M$. Now it should be obvious that $x^p, y^p$ and $(xy)^p$ act as the identity, but in fact, a proof requires many many cases and would somehow include a solution for the word problem for $G$. Perhaps this is as tedious as making $M$ ad hoc to a group. When this is done, the action extends to $G$. The obvious surjective map $M \to G$ is then injective because the action of $G$ on the empty word yields a inverse map. Now $M$ is infinite, for example it contains all the powers of $x y^2$.

I hope there is a better proof. Perhaps there is a nice action of $G$ on a topological space which makes you see that $x y^2$ has infinite order? By the way, I don't want to use heavy theorems from group theory (Burnside problem etc.)!

Feel free to add other interesting properties of $G$.

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