〈x,y : x^p = y^p = (xy)^p = 1〉

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 $p<3$.

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

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.