# Is there a Lie group with compact elements?

by user549766   Last Updated September 11, 2019 18:20 PM - source

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

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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).

Moishe Kohan
September 11, 2019 18:03 PM