gr.group theory - 〈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.