Is there a Lie group with compact elements?

It is possible that for a (non-abelian) Lie group, the subgroup generated by each element is compact?

• I suppose you want to exclude finite groups? – Eric Wofsey Sep 1 at 13:39
• I don't think this is possible for a connected group, even if it is abelian. All elements would have to be of finite order. – Matt Samuel Sep 1 at 18:31
• Yes, I would like to show that a Lie group with this property is finite. Indeed, I try to show that this property can be inherited locally...We know that Euclidean spaces does not have this property, and lie groups are locally similar to Euclidean spaces. But I could not prove it. – user549766 Sep 7 at 4:52

Proposition. Suppose that $$G$$ is a connected Lie group and $$H< G$$ is a (closed) Lie subgroup such that every element of $$H$$ has finite order (i.e. $$H$$ is a torsion group). Then $$H$$ is finite.
Proof. I consider the case when $$G$$ has finite center (let me know if you are interested in the general case); then the adjoint representation of $$G$$ has finite kernel and sends $$H$$ to a closed Lie subgroup of $$GL(n, {\mathbb R})$$. The image of $$H$$ still is a torsion group. Thus, it suffices to consider the case $$G= GL(n, {\mathbb R})$$.
If $$H$$ has positive dimension then pick a 1-dimensional subspace $$L$$ in its Lie algebra. The restriction of the exponential map to $$L$$ has discrete kernel (trivial or $${\mathbb Z})$$ from which it follows that $$\exp(L)$$ has elements of infinite order.
Thus, $$H$$ has to be zero-dimensional, i.e. is a discrete subgroup of $$G$$; in particular, $$H$$ is countable. Of course, $$S^1$$ contains countable infinite torsion subgroups (say, the group of roots of unity). Hence, we will need to use discreteness.
By Schur's Lemma, every finitely generated torsion subgroup of a matrix group is finite. Since $$H$$ is countable, it contains a sequence of nested finitely generated subgroups $$H_1< H_2 whose union is $$H$$. Each $$H_i$$ is finite, hence, compact. All compact subgroups of $$G=GL(n, {\mathbb R})$$ are conjugate to subgroups of the standard maximal compact subgroup $$O(n)< G$$. Thus, each $$H_i$$ is contained in a conjugate $$K_i$$ of $$O(n)$$. With a bit more work (let me know if you want to see this) one verifies that the subgroups $$K_i$$ can be chosen so that for all sufficiently large $$i, j$$, $$K_i=K_j$$. In other words, $$H$$ is conjugate to a subgroup of $$O(n)$$. But the group $$O(n)$$ is compact, hence, every discrete subgroup of $$O(n)$$ is finite. Thus, $$H$$ is finite. qed
Of course, if you allow disconnected Lie groups $$G$$, the result is false: There even exist infinite finitely generated torsion groups (e.g. the Tarski monster; see here for more examples).