How do you prove that a group specified by a presentation is infinite?

Solution 1:

$\langle x,y \; | \; x^2=y^3=1 \rangle \cong \operatorname{PSL}_2(\mathbb Z)$ and this isomorphism identifies G with $\operatorname{PSL}_2/T^7=1$ (where $T:z\mapsto z+1$). Result is the symmetry group of the tiling of the hyperbolic plane. From this description one can see that G is infinite (e.g. because there are infinitely many triangles in the tiling and G acts on them transitively).

Solution 2:

Grigory has already answered your particular question. However, I wanted to point out that your question "How do you prove that a group specified by a presentation is infinite?" has no good answer in general. Indeed, in general the question of whether a group presentation defines the trivial group is undecidable.

Solution 3:

The group:

$$G = \left\langle x, y \mid x^l = y^m = (xy)^n = 1 \right\rangle$$ is triangular group so, if $\frac{1}{l} + \frac{1}{m} + \frac{1}{n} \lt 1$, then this group is infinite...

Solution 4:

One general way to do this, (which is not guaranteed to work, as Noah points out), is to exhibit infinitely many different homomorphic images of the group you start with . In this case, any group generated by an element of order $2$ and an element of order $3$ whose product has order $7$ is a homomorphic image of the group $G$. Such a group is a Hurwitz group, and these are well studied, for example in the work of M. Conder and of G. Higman, among others. Infinitely many finite simple groups are known to be Hurwitz groups, I believe.

Solution 5:

Your group is the alternating subgroup of a Coxeter group

$$\langle x, y, z | x^2 = y^2 = z^2 = (xy)^2 = (yz)^3 = (zx)^7 = 1 \rangle$$

which is not on the (well-understood) list of finite Coxeter groups, and the alternating subgroup always has index $2$. (The connection to the hyperbolic plane is that Coxeter groups of rank $3$ always act as symmetries of a tiling of either the sphere, the Euclidean plane, or the hyperbolic plane by triangles.)