# Semisimple Lie group topologically generated by two finite order elements

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

This is partially inspired by

https://mathoverflow.net/questions/59213/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 https://mathoverflow.net/questions/254164/is-every-finite-quasi-simple-group-generated-by-2-elements

Note that $$G=U_1$$ is not a counterexample because $$U_1$$ is not semisimple. The universal cover of $$SL_2(\mathbb{R})$$ is not a counterexample since it is not algebraic.

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 $$SU_2$$.

Update: A lot of simple compact Lie groups have Lie primitive subgroups (not contained in any proper positive dimensional closed subgroups) which are finite simple groups (or at least quasisimple). Since quasi simple groups are 2-generated you can take $$a,b$$ generators for the Lie primitive (quasi)simple group and then randomly pick a 3rd finite order element $$c$$ and almost surely $$a,b,c$$ will generate a dense subgroup of $$G$$. Indeed if you pick a maximal Lie primitive subgroup $$\Gamma$$, that is also (quasi)simple, then any choice of a finite order element $$c \not \in \Gamma$$ is guaranteed to topologically generate $$G$$. Papers like

https://arxiv.org/abs/math/0502080

list (quasi)simple finite subgroups of many simple Lie groups.

Here's a weaker result. It's apparently known that every finite-dimensional semisimple Lie algebra over $$\mathbb{R}$$ can be generated by two elements $$X, Y$$. It follows that a compact connected semisimple Lie group $$G$$ can be densely generated by two one-parameter subgroups $$e^{tX}, e^{tY}$$ (in the sense that $$G$$ is the closure of the subgroup generated by the images of the one-parameter subgroups). Each of these are $$S^1$$ so contain a dense subgroup isomorphic to $$\mathbb{Z}$$. Edit: As Jason DeVito points out in the comments, this is incorrect but it can be salvaged: each one-parameter subgroup has closure a torus $$T^n$$ and we can find a copy of $$\mathbb{Z}$$ inside each one-parameter subgroup which remains dense in $$T^n$$. So:
Proposition: $$G$$ can be densely generated by two elements of infinite order.
We know that the free group $$F_2$$ embeds into, say, the free product $$C_2 \ast C_3$$, so given a dense copy of $$F_2$$ we may be able to embed it into a dense copy of some free product $$C_n \ast C_m$$, but I don't see how at the moment.
• @IanGershonTeixeira: yes, although that's a stronger statement because it doesn't assume compactness. In this argument I need compactness so I can assume that one-parameter subgroups are isomorphic to $S^1$. Sep 29, 2022 at 16:57
• I don't think compactness gets you the fact that 1-parameter subgroups are circles. On $T^2$, there are dense windings which are 1-parameter subgroups, but $T^2$ is a subgroup of any compact Lie group of rank $2$ or more. That said, I think the overall argument is ok. The closure of a $1$-parameter subgroup is a torus (of some rank), which has a dense copy of $\mathbb{Z}$, and I think that's enough to keep using your argument. Sep 29, 2022 at 21:46