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

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    $\begingroup$ I suppose you want to exclude finite groups? $\endgroup$ – Eric Wofsey Sep 1 at 13:39
  • $\begingroup$ 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. $\endgroup$ – Matt Samuel Sep 1 at 18:31
  • $\begingroup$ 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. $\endgroup$ – 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 $$ H_1< H_2<H_3< ... $$ 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).


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