# How to approach proofs similar to "Show a group, $G$, is infinite if $G = \langle r, s, t\mid rst = 1\rangle$"

How to approach proofs similar to "Show a group, $$G$$, is infinite if $$G = \langle r, s, t\mid rst = 1\rangle$$"

I have not worked much with relations and tend to get lost in notation. I am practicing solving problems like the one in the title but am having a hard time as I am not sure the tricks to try or areas to investigate first in trying to make a proof. What are some hints for starting a proof about some quality of a group defined by a relation?

So far the only relations I know about are the dihedral groups of order $$2n$$, the quaternions, and cyclically generated groups so comparisons to how we show properties of those might be illuminating.

$$G$$ is the set of words on $$r,s,t$$ subject to the relation $$rst=1$$.

The relation $$rst=1$$ means that you can replace every occurrence of $$t$$ by $$(rs)^{-1}=s^{-1}r^{-1}$$.

Therefore, $$G$$ is the set of words on $$r,s$$, that is, the free group in two letters.

Alternatively, the set $$\{1,r,r^2, r^3, \dots \}$$ is an infinite subset of $$G$$ because these words do not contain $$s$$ or $$t$$ and so cannot be further reduced or to one another.

(By words on $$S$$, I mean words on the elements of $$S$$ and their inverses.)

One thing I often find clarifying is to try adding relations. If you still get an infinite group after you added a relation then you must have started with an infinite group.

Here, for instance, set $$r=e$$. Then the new group is generated by $$s,t$$ with $$s=t^{-1}$$. Hence it is generated by $$t$$ with no relations, so the new group is $$\mathbb Z$$. As that is infinite, so must $$G$$ have been.

• Thank you! This is a very help tool for addressing these cardinality questions with relations. I will likely select this as the answer but will wait a couple more hours as to not discourage others from answering. (+1) Commented Jan 7, 2020 at 0:55

Consider $$f:\{r,s,t\}\rightarrow\mathbb{Z}$$ defined by $$f(r)=1, f(s)=-1, f(t)=0$$, $$f(r)+f(s)+f(t)=0$$ implies that $$f$$ extends to a morphism of groups $$g:G\rightarrow\mathbb{Z}$$. The fact $$g(r^n)=n$$ implies that$$g$$ is surjective and $$G$$ infinite.

• This is, in essence, what @lulu says in the answer with that username. The notation is helpful here to give clarity though (+1).
– Shaun
Commented Jan 7, 2020 at 0:17
• Thank you, as @Shaun said the notation is certainly illuminating and I had almost forgotten that the most obvious way to show infinite cardinatlity is with a map onto the integers or natural numbers! (+1) Commented Jan 7, 2020 at 0:57

I want to stick to generalities with this answer, but the underlying point is: you can see that your group is infinite, and indeed is "large", simply by looking at the presentation. No calculations are needed.

One way to try and prove that a group is infinite is to compute the abelianization of the group (that is, force the generators to pairwise commute) and see if the resulting group is infinite (this is a special case of @lulu's answer). The abelianisation of the group you have here is $$\mathbb{Z}\times\mathbb{Z}$$. For general methods to compute abelinisations, you might find this question useful.

Now, by considering the abelinisation it can be proven that a presentation with more generators than relators defines an infinite group. In particular, every group with at least two generators and a single defining relation is infinite (these are called "one relator groups", and there is a rich theory of these groups). Equipped with this result, you can see that your group is infinite without having to do any calculations.

A group is large if it has a finite index subgroup which maps onto a non-abelian free group. Clearly, large groups are infinite. In a pleasingly short paper, Benjamin Baumslag and Stephen J. Pride* proved that a presentation with two more generators than relators defines a large group. Hence, your group is large. Gromov then proved that a presentation with more generators than relators such that one relator is a proper power (so of the form $$w^n$$, $$n>1$$) defines a large group. Equipped with the Baumslag-Pride result, you can see that your group is large without having to do any calculations (this observation is weaker than @lhf's answer).

*"Groups with two more generators than relators." Journal of the London Mathematical Society 2.3 (1978): 425-426

• Thank you! I wish I could select two answers as this one does an excellent job of addressing how to solve "similar proofs" as in a whole class of this type of proof on relations! What an interesting result, I will have to put this reading aside and get back to it after my qualifying exam =) Commented Jan 7, 2020 at 17:50