# Does every non-abelian Lie group have a finite subgroup?

I am interested in finite-order elements (different from the identity) of non-abelian Lie group. It seems to me that each non-abelian Lie group has at least one (actually many) finite-order elements or, in other terms, one or more finite subgroups. Is it true?

For example, if we consider the classical lie groups O(n), SO(n), SL(n) etc, it is easy to find finite-order elements considering the diagonal matrices with diagonal elements taken from $\{1,-1\}.$

I know there exist a lot of works on discrete subgroups of simple lie groups and on lattices in lie groups by Margulis, Harish-Chandra etc but I am not able to find existence results about finite subgroups.

• Ops, in the question I meant "every" not "any" – Nick Belane Oct 24 '15 at 10:17

Every group can be made a Lie group by giving it the discrete topology, so any nonabelian group with no nontrivial elements of finite order gives a counterexample. If you want the group to be connected, you can still find counterexamples. For instance, consider the group of affine transformations $x\mapsto ax+b$ for $a\in\mathbb{R}_+$ and $b\in\mathbb{R}$. This is nonabelian and connected, but it is easy to see it has no nontrivial elements of finite order.
• There is a problem with your counterexample. The element $x\to 1-x$ has order two (I believe this is the only finite order you can get). But this is not actually connected; you need to say a is not zero (else it's not a group) but then you have two components (a positive or negative). I think if you restrict to positive a, you're okay, though. – Richard Rast Oct 24 '15 at 11:22
• @RichardRast It's written $a \in \mathbb{R}_{\color{red}+}$, and I assume $\mathbb{R}_+ = \{ x \in \mathbb{R} \mid x > 0 \}$... – Najib Idrissi Oct 24 '15 at 17:38