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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.

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    $\begingroup$ Ops, in the question I meant "every" not "any" $\endgroup$ – Nick Belane Oct 24 '15 at 10:17
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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.

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  • $\begingroup$ Excuse me, but why is the maximal torus of a compact Lie group of positive dimension nontrivial? $\endgroup$ – abenthy Oct 14 '16 at 13:08
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    $\begingroup$ @abenthy: the closure of any one-parameter subgroup is a torus. Actually I guess this is already enough to show that there's a nontrivial torus and there is no need to appeal to the existence of maximal tori. $\endgroup$ – Qiaochu Yuan Oct 15 '16 at 3:47
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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.

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    $\begingroup$ 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. $\endgroup$ – Richard Rast Oct 24 '15 at 11:22
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    $\begingroup$ @RichardRast It's written $a \in \mathbb{R}_{\color{red}+}$, and I assume $\mathbb{R}_+ = \{ x \in \mathbb{R} \mid x > 0 \}$... $\endgroup$ – Najib Idrissi Oct 24 '15 at 17:38
  • $\begingroup$ Ah! I didn't see the plus... I wish I could blame it on my mathjax, but it's probably just my inattention. I could have sworn it just said R, though. $\endgroup$ – Richard Rast Oct 24 '15 at 20:28

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