What can the order of the generators of a non abelian group tell us about the group?

Question

What can the order of the generators of a non abelian group tell us about the group?

For example, if we have $G$ as the non abelian group of order $8$, representing the quaternions; $$ G=\{s,t\mid s^4=t^4=1, s^2=t^2, st=ts^{-1}\} $$

and we consider $N$, the subgroup of ${\rm Perm}(G)$ generated by $\lambda(s),\rho(t)$ where $\lambda,\rho\in {\rm Perm}(G)$ define the left and right regular representations respectively.

I know that $\lambda(s)$ and $\rho(t)$ both have order $4$ but it's not immediately obvious to me what $N$ would look like. By lagrange's theorem, $N$ must be a multiple of $4$ and since $\lambda(s), \rho(t)$ are distinct $N$ must be at least $8$ (?)

How would we know the exact structure of $N$ without explicitly calculating each element?

Answer 1

In any group $G$, $\lambda(G)$ and $\rho(G)$ centralize each other - in fact they are equal to each others centralizers in ${\rm Sym}(G)$.

In your example $s^2 = t^2 \in Z(G)$, so $\rho(s^2) = \lambda(t^2)$ and hence $\langle \lambda(s),\rho(t) \rangle$ is abelian of order $8$ (it is isomorphic to $C_4 \times C_2$).

Answer 2

Relatively little. The fact that one has finite-order generators does not imply that the group itself is finite; this is the (in)famous Burnside problem solved by Golod & Shafarevich. Golod & Shafarevich's example requires the generators to have unbounded order; classifying the groups with bounded exponent is a major area of open research. If we know that the finitely-presented group is in fact finite, then one can at least control the group, but this required a Fields medal to show.

I don't know enough about Zelmanov's Fields-medal–winning work to try to summarize it here; hopefully this is enough keywords to start Googling.

Answer 3

If you have a two generated group $G=\langle s, t\rangle$, then the two orders $p=o(s), q=o(t)$ are not enough to conclude very much about the group. In particular, the free product $\mathbb{Z}_p\ast\mathbb{Z}_q$ is infinite with generators of the specified orders, but there are in general many non-abelian finite groups $G=\langle s, t\rangle$ whose generators have the prescribed orders (e.g. assuming $1<p\leq q$, take $s=(1, 2, \ldots, p)$ and $t=(2, ..., q+1)$ in the symmetric group $S_{q+1}$).

The exception here is when $o(s)=2=o(t)$, then $G=\langle s, t\rangle$ is dihedral: here $\mathbb{Z}_2\ast\mathbb{Z}_2$ is the infinite dihedral group, and all its non-abelian quotients are dihedral groups.

However, the three orders $o(s), s(t), o(st)$ can be enough to conclude that your group is finite: if $1/o(s)+1/s(t)+1/o(st)>1$ then $G=\langle s, t\rangle$ is a finite group. (To see this, combine Sections 2.2 and 3 of Wikipedia's article on triangle groups.)