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 $G$.



〈x,y : x^p = y^p = (xy)^p = 1〉
Many techniques discussed here: [group-pub](http://lists.maths.bath.ac.uk/sympa/arc/group-pub-forum/2010-01/msg00005.html)


EDIT: Some of the ideas, in the above thread, I like the most:


If $q$ is a prime congruent to $1$ mod $p$, then consider the Frobenius group $H\rtimes K$ with $H$ cyclic of order $q$ and $K$ cyclic of order $p$. Then if $a$ generates $H$ and $b$ generates $K$, you can show $b$ and $ab$ are such that $b$, $ab$ and $ab^2$ have order $p$, so this group is a quotient of your group. Choosing $q$ arbitrarily large suffices to show your group is infinite.


You can also show that the two infinite permutations on $\mathbb{Z}$ $a=...(-p+1,-p+2,...,-1,0)(1,2,...,p)(p+1,...,2p)... $ $b=...(-p+2,-p+3,...,0,1)(2,...,p+1)(p+2,...,2p+1)...$  

are such that $a$, $b$, and $ab$ have order $p$, and yet generate an infinite group (it acts transitively on $\mathbb{Z}$).


Fox calculus can be used to show $\lt\lt xy^{-1}\gt\gt$ has infinite abelianization.


