〈x,y : x^p = y^p = (xy)^p = 1〉

Question

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 $G$.

Answer 1

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.

Answer 2

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.

Answer 3

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 $p<3$.

• For a nice introduction to 2-dimensional orbifolds, I recommend Peter Scott's article The geometries of 3-manifolds.

Answer 4

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.

Answer 5

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 [2]

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] = [1] b[1]=[-p+2] a[-p+2]=[-p+3] (ab)^{p-1}[-p+2]=[0] b(ab)^{p-1}[-p+2]=b[0]=[1] (ab)^{p}[-p+2]=a[1]=[2].

For you to understand my notation: My notation results in (123)[3]=[1], (34)[5]=[5], ((234)(345)^{2}[2]=[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.