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

The group:

$$G = \left\langle x, y \; \left| \; x^2 = y^3 = (xy)^7 = 1\right. \right\rangle$$

is infinite, or so I've been told. How would I go about proving this? (To prove finiteness of a finitely presented group, I could do a coset enumeration, but I don't see how this helps if I want to prove that it's infinite.)

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It feels like the element $xy^2$ should be of infinite order here, but I'm not very sure of how to prove this. –  Akhil Mathew Jul 22 '10 at 22:02
Is the operation simply multiplication here? –  Noldorin Jul 23 '10 at 8:02
@Noldorin: This is an abstract group specified by a presentation, and, the operation is the group operation. This is not supposed to be a subset of C. –  Simon Nickerson Jul 23 '10 at 12:04
@Simon: I misunderstood clearly. However, surely by the definition x can only have at most 2 values and y at most 3 values, even in the complex plane? It can't be infinite that way. –  Noldorin Jul 23 '10 at 12:09
@Noldorin: The group is the "freest" group that is generated by the elements x and y such that x^2 = y^3 = (xy)^7 = 1. As another example <x : x^n = 1> would be the cyclic group of order n. –  Eric O. Korman Jul 23 '10 at 13:15

## 6 Answers

<x,y|x^2=y^3=1>≅PSL_2(Z) and this isomorphism identifies G with $PSL_2/T^7=1$ (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).

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What does PSL_2 mean? –  Casebash Jul 23 '10 at 23:38
SL_2 is 2x2 matrices with determinant 1. PSL_2 is the quotient of this group module its center, namely the scalar matrices. –  Noah Snyder Jul 24 '10 at 0:14

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.

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And an increadibly nice where one can read about this is John Stillwell's Classical topology and combinatorial group theory –  Mariano Suárez-Alvarez Jul 30 '10 at 16:23
It's certainly true that there is no universal algorithm for solving such problems. But quite universal method of showing that a group is infinite (or non-trivial) is constructing transitive action on infinite (resp. non-trivial) set. –  Grigory M Jul 31 '10 at 8:12

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.

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Another approach, which can't work in general (see Noah's answer), but will surely work in this case, is to find a normal form for each element of the group, and then see whether there are finitely or infinitely many.

In practice, that means imagining a word in x and y, and then applying the relations as much as possible to simplify it, and then trying to figure out (and prove!) what the possible different "irreducible words" (i.e. words that can no longer be simplified) are.

In your case, the first thing one would note is that x can only appear to the 1st power (since any higher power can be simplified using x^2 = 1), while y can only appear to the powers +1 or -1 (for the same reason). Also, we can't have too many expressions of the form xy or yx in a row, because of the third relation.

One can keep going like this. I didn't, but what I imagine is that one can have expressions of the form x y x y^{-1} x y x y^{-1} ... that are arbitrarily long, and inequivalent, explaining the infinite order of the group.

It shouldn't be so hard to settle the question from this point of view by sitting down with pencil and paper and just playing around with different words to get a feel for what kinds of reductions can take place. (In geometric arguments like Grigory's, one uses a geometric context, such as an action on a hyperbolic tiling, as a more conceptual way of understanding the relations and distinguishing inequivalent words. But in this case I'm sure it won't be hard to see everything directly from the presentation.)

Added after rereading the question: what I am suggesting is precisely that even when the answer might be infinite, you can still hope to find a coset enumeration, at least in a case like this with relatively simple relations.

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Usually it's easy to show that some relations hold but quite hard to show that some relations doesn't (i.e. that constructed "canonical form" is indeed canonical)... –  Grigory M Jul 31 '10 at 8:14
Of course that is true in general, since the word problem is not solvable in general, but not in this case. –  Matt E Jul 31 '10 at 12:25

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...

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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.)

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