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.
 A: Everything is much better if you additionally require that your Lie groups are compact. Every compact connected Lie group of positive dimension has a nontrivial maximal torus which has many elements of finite order. 
In the noncompact connected case we're still fine if the maximal compact subgroup is interesting. Unfortunately there are some connected Lie groups, such as nilpotent Lie groups, which have trivial maximal compact subgroups, and since finite subgroups are in particular compact subgroups, these have no nontrivial finite subgroups either. 
A: 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.
