# Semisimple compact Lie group topologically generated by two finite order elements

Edit: I'm specializing this question to the compact case I'll ask about the noncompact case as a new question.

Let $$G$$ be a compact connected semisimple Lie group. Do there always exist two finite order elements of $$G$$ which generate a dense subgroup?

Example: $$\frac{i}{\sqrt{2}}\begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}$$ and $$\begin{bmatrix} \overline{\zeta_{16}} & 0 \\ 0 & \zeta_{16} \end{bmatrix}$$ generate a dense subgroup of $$SU_2$$.

This question is partially inspired by

Generating finite simple groups with $2$ elements

which shows that every finite simple group is 2-generated. Indeed even every finite quasisimple group is 2-generated Is every finite quasi-simple group generated by 2 elements? (however this fails for infinite simple groups: $$\mathrm{PSL}_2(\mathbb{Q})$$ is not 2-generated indeed not even finitely generated).

This is a cross-post of https://math.stackexchange.com/questions/4537024/dense-subgroups-generated-by-two-finite-order-elements an answer posted there points out that every compact semisimple Lie group can be topologically generated by 2 infinite order elements.

• I think that there's a paper that proves this by some mechanism like showing the set of pairs that generate has positive Haar measure. I'll see if I can dig it up. Commented Sep 29, 2022 at 15:25
• Theorem 2.1 in this J. Algebra paper of Breuillard-Gelander should help: indeed it is then enough to show that every (connected, finite center) semisimple Lie group is (topologically) generated by two circle subgroups.
– YCor
Commented Sep 29, 2022 at 16:10
• [Finite center should be assumed. Or, at least one has to exclude the case when $G$ surjects onto the universal cover of $\mathrm{SL}_2(\mathbf{R})$, which is torsion-free.]
– YCor
Commented Sep 29, 2022 at 16:11
• Good point I've added the assumption of finite center Commented Apr 10, 2023 at 14:46
• You should add the assumption that $G$ is semisimple, otherwise the claim is clearly false since $G$ can be torsion-free, e.g. the subgroup $B< SL(2,{\mathbb R})$ of non-strictly upper triangular matrices with positive diagonal entries. Commented Apr 10, 2023 at 14:57

There's also a 1999 Proc AMS paper by M. Field, (freely accessible at AMS site — MR number for subscribers: MR1618662), which shows that the set of pairs generating a dense open subset of $$G$$ is non-empty Zariski open in $$G \times G$$ when $$G$$ is compact connected semisimple. This should lead to a proof that the elements can be taken to have finite order in this case, as the set of pairs of elements of finite order is dense in $$G\times G$$. Of course, in the non-compact case some extra condition on $$G$$ is required.