〈x,y : x^p = y^p = (xy)^p = 1〉

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 $G$.

Many techniques discussed here: group-pub

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.

• currently the site seems to be offline. – Martin Brandenburg Apr 24 '10 at 23:04
• Hmm, it works for me. I will edit to add the ideas of the answers there. – Steve D Apr 24 '10 at 23:06
• thank you. but $ba$ does not have order $p$. I've checked that many times. is there are typing error? – Martin Brandenburg Apr 25 '10 at 0:02
• I don't know which example you're referring to. If you mean the infinite permutations, then $ab$ does have order $p$, if you let $b$ act first, then $a$. – Steve D Apr 25 '10 at 0:10
• I gave a few example to kbmag -- a program for dealing with automatic groups. Use what's called recursive ordering, it found that the given presentation (with some very minor tweaking) is what's called "confluent". The upshot is that there's a very simple automaton which solves the word problem for this group. Looking at that might help give a simple combinatorial proof. – Victor Miller Apr 25 '10 at 1:51

More generally, let $a,b,c \in \mathbb{Z}^+$ and define the group

$\Delta(a,b,c) = \langle x,y,z \ | \ x^a = y^b = z^c = xyz = 1 \rangle$.

These groups were studied by von Dyck in the late 19th century and are sometimes called the von Dyck groups. The most basic fact about them is that $\Delta(a,b,c)$ is infinite iff $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} \leq 1$. (The groups you ask about are when $p = a = b = c$. Thus $\Delta(2,2,2)$ is finite, and for $p > 2$, $\Delta(p,p,p)$ is infinite.)

Perhaps the nicest way to see this is to realize $\Delta(a,b,c)$ as a discrete group of isometries of a simply connected surface of constant curvature. More precisely, consider a geodesic triangle with angles $\frac{\pi}{a}$, $\frac{\pi}{b}$, $\frac{\pi}{c}$. Then, according to whether $\frac{1}{a} + \frac{1}{b} + \frac{1}{c}$ is greater than, equal to, or less than $1$, these triangles live either in the Riemann sphere, the Euclidean plane or the hyperbolic plane.

Now $\Delta(a,b,c)$ has as a homomorphic image the group generated by three elements $x$,$y$,$z$, each of which is the composition of reflection through two adjacent sides of the triangle. Indeed, an easy calculation shows that $x$, $y$, $z$ satisfy the relations defining $\Delta(a,b,c)$, so that it must be a homomorphic image of it. (In fact the abtract group is isomorphic to the isometry group, but that is a little more delicate to show.) Now there is a corresponding tesselation of the space obtained by repeatedly reflecting copies of one fundamental triangle across each of the sides. If you consider the overgroup $\tilde{\Delta}(a,b,c)$ generated by the reflections themselves and not the rotations -- so that $\Delta(a,b,c)$ is the index $2$ subgroup consisting of orientation-preserving isometries -- then it is immediately clear that $\tilde{\Delta}(a,b,c)$ acts transitively on the triangles in the tesselation. Since the Euclidean and hyperbolic plane each have infinite volume, there are clearly infinitely many triangles in the tesselation, so $\tilde{\Delta}(a,b,c)$ is infinite, and therefore so is its index $2$ subgroup $\Delta(a,b,c)$.

In the case when $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} > 1$, this argument can be modified to show that $\Delta(a,b,c)$ is finite, but in this case a reasonable alternative is brute force, since this is a well-known family of groups: the finite isometry groups of $3$-dimensional Euclidean space (namely $C_n$, $D_n$, $S_4$, $A_4$, $A_5$).

Also either of both of the families of groups $\Delta$ and $\tilde{\Delta}$ are often called triangle groups.

Your group is the fundamental group of a 2-dimensional orbifold with underlying surface the 2-sphere and 3 cone points of order p. It follows that it acts properly discontinuously and cocompactly by isometries on the 2-sphere when p=2, the Euclidean plane when p=3, and the hyperbolic plane when p>3. Hence it's infinite when p>2.

UPDATE:

Pete Clark's answer very eloquently explains the details that I outlined. I'd just like to add a couple of further remarks.

• To determine which sort of geometry (spherical, Euclidean or hyperbolic) an orbifold $O$ admits (ie upon which space your group acts as a discrete group if isometries) you just need to look at the (orbifold) Euler characteristic, defined to be

$\chi(O)=\chi(|O|)+\sum_i (1-1/p_i)$

where $|O|$ is the underlying surface and $p_i$ are the orders of your cone points. So we see that this is positive when $p=2$, zero when $p=3$ and negative when $p<3$.

• I just wanted to add that I don't (yet) speak orbifold, and so this answer -- concerning a construction that I know very well from a more classical perspective -- was very useful for me. +1. – Pete L. Clark Apr 26 '10 at 21:19
• When I first read this, I took "very eloquently" to be modifying how you outlined :). – Steve D Sep 16 '11 at 23:35
• @SteveD - Ah, I edited the offending sentence. – Sam Nead Jul 25 '17 at 19:18

Just want to add that $G$ has a quotient isomorphic to $H:=\mathbb Z_p \ltimes_\varphi \mathbb Z^{p-1}$ where $\varphi(1)\cdot v=Av$ for some $A \in SL_{p-1}(\mathbb Z)$ of order $p$; in particular, $G$ is infinite.

Indeed, such an $A$ exists: take $A$ to be a companion matrix to the polynomial $\frac{x^p-1}{x-1}$. Then any element in $H$ not contained in $\{0\} \times \mathbb Z ^{p-1}$ has order $p$, so mapping $x$ to $(1,0)$ and $y$ to $(1,e_1)$ will give a well-defined homomorphism which can be shown to be surjective.

• For p=3, I believe H = G/γ<sub>∞</sub>(G), the largest residually nilpotent quotient. For larger p, there are a few extra pretty commutators to remove to get exactly H. I prefer to view H by its quotients by its lower central series, where one got very pretty p-groups of maximal class, quite similar to Marty Isaacs's Frobenius group examples. – Jack Schmidt May 7 '10 at 14:06

Please delete this (if you add a comment above answering my question. actually this should be a comment on Steve D.'s answer)

I wanted to comment on the example: 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)... Martin Brandenburg wrote that he doesn't see that (ba)^p is the identity and I don't see it for (ab)^p either. [and surely either both at id or neither is.]

Could someone who believes that Steve D is correct please add a short comment at the appropriate point that makes one see that one gets (ab)^p=id?

I for one get:

(ab)^p applied to [-p+2] goes to 

Here is who I understand the notation to apply. Maybe there is a notational problem: (ab) [-p+2] = [-p+4] (ab) [-p+4] = [-p+6] (ab)^{(p-1)/2} [-p+2] =  b=[-p+2] a[-p+2]=[-p+3] (ab)^{p-1}[-p+2]= b(ab)^{p-1}[-p+2]=b= (ab)^{p}[-p+2]=a=.

For you to understand my notation: My notation results in (123)=, (34)=, ((234)(345)^{2}= I hope that makes it understandable, I guess using the []-brackets i

Please just add a short comment to make the next reader thinking the same thing as I did, where to start the thoughts that he won't get the same result as I did.

• $(ab)^{(p-1)/2}[-p+2]=-p+1$. – Steve D May 8 '10 at 3:51