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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 groups 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 and or the hyperbolic plane.

Now $\Delta(a,b,c)$ has as a homomorphic image the groups group generated by three elements $x$,$y$,$z$, each of which is the composition of reflection through two adjacent sides of the triangle. The point is that it is Indeed, an immediate easy calculation to see 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 obained 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.

2 added 168 characters in body

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 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 $a = b = c$. 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 discrete groups 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 and the hyperbolic plane.

Now $\Delta(a,b,c)$ has as a homomorphic image the groups generated by three elements $x$,$y$,$z$, each of which is the composition of reflection through two adjacent sides of the triangle. The point is that it is an immediate calculation to see 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 obained by 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.

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

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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 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 $a = b = c$. 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 discrete groups 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 and the hyperbolic plane.

Now $\Delta(a,b,c)$ has as a homomorphic image the groups generated by three elements $x$,$y$,$z$, each of which is the composition of reflection through two adjacent sides of the triangle. The point is that it is an immediate calculation to see 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 obained by 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.