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

Question

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.

Answer 1

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.

Answer 2

$G$ 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.)

Answer 3

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.

Answer 4

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.

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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.

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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