Every discrete central subgroup of a connected Lie group is finitely generated.

This result was alluded to without comment in a book I was reading (Lie Group Actions in Complex Analysis by D. Akhiezer Proposition on page 38).

Assuming it was a trivial result, I posted to math.stackexchange where YCor was kind enough to give the not so trivial reference: Corollary 8.A.23 here.

That Corollary uses a handful of results + definitions that are not so familiar to me and would take up some time to understand.

So before I begin going through the reference, I wanted to see if someone knew of a more concise/elementary proof of this result.

Note: I am aware that the case where the center of the Lie Group has finitely many components is exercise level.


  • 5
    $\begingroup$ Let $G$ be a connected Lie group, $\Gamma$ a discrete central subgroup. Then $G\to H:=G/\Gamma$ is a connected Galois cover with $\Gamma$ its group of deck transformations, so $\Gamma$ is a quotient of $\pi_1(H)$. Thus, it suffices to show $\pi_1(H)$ is finitely generated for any connected Lie group $H$. But a maximal compact subgroup $K$ of $H$ is a deformation retract (structure of connected Lie groups), so $\pi_1(H)=\pi_1(K)$. Since $\pi_1(K) = {\rm{H}}_1(K,\mathbf{Z})$, and the integral homology of any compact manifold is finitely generated, we are done. $\endgroup$
    – nfdc23
    Jan 1, 2018 at 19:52
  • $\begingroup$ Yes basically I had to rely on simple connectedness of the quotient by a maximal compact subgroup. It's even enough to know the existence of maximal subgroup and their conjugacy. I'll finish writing a post based on this. $\endgroup$
    – YCor
    Jan 1, 2018 at 20:05
  • $\begingroup$ @nfdc23: this was the first argument I thought of but it felt like overkill to me. The special case $G = \mathbb{R}^n$ can be done in a much more elementary way, for example. Is there no hope of an elementary argument in general? If $G$ is a real algebraic group perhaps one could take the Zariski closure of $\Gamma$ or something like that, at least, and argue in that closure. $\endgroup$ Jan 1, 2018 at 22:55
  • $\begingroup$ @QiaochuYuan actually when $G$ is a closed linear group there's a somewhat immediate argument using Zariski closure. $\endgroup$
    – YCor
    Jan 1, 2018 at 23:58
  • $\begingroup$ @QiaochuYuan: Fair point. Those results on maximal compact subgroups (at least for connected Lie groups) are so central to how the structure in general is demystified (at least in my own experience with self-education in the subject) that my instinct is to use them whenever convenient. $\endgroup$
    – nfdc23
    Jan 2, 2018 at 1:00

1 Answer 1


An old result of Iwasawa is that in any connected Lie group $G$, every compact subgroup is contained in a maximal compact subgroup, and all maximal compact subgroups are conjugate.

Let $G$ be a connected Lie group and let $Z$ be a discrete central subgroup. Then $Z$ has an infinite torsion quotient $Z'=Z/B$ (lemma below). Write $H=G/B$. Then $Z'$ is a discrete central subgroup of $G$ and is infinite torsion. Let $K$ be a maximal compact subgroup of $H$. Every finite $F$ subgroup of $Z'$ is contained in a maximal compact subgroup $K'_F$ of $H$; since $F$ is central and $K$ is conjugate to $K'_F$, we deduce that $F\subset K$. Hence $Z'\subset K$, a contradiction.

This would be fine with a proof of Iwasawa's result; the only problem is that I don't know if it relies on the result of finite generation of discrete central subgroups!

Lemma: if $Z$ is an infinitely generated abelian group, then $Z$ has an infinite torsion quotient.

Proof: let $B$ be a maximal free subgroup. Then $Z/B$ is torsion. If $B$ is finitely generated then $Z/B$ is infinite. Otherwise, choose an infinite subset $(e_n)_{n\ge 1}$ in $B$ and replace $e_n$ by $ne_n$ for all $n$, to obtain a smaller free subgroup $B'$, with $B/B'$ infinite torsion; then $Z/B'$ is infinite torsion.

Actually, instead of all the power of Iwasawa's result, it's enough to prove the result using the weaker result: ($*$) for any connected Lie group $G$ and any increasing sequence $(K_n)$ of compact subgroups of $G$, we have $\overline{\bigcup K_n}$ compact.

Possibly this can done by hand (say, without using such things as Levi factors, just more basic Lie theory), I'll think twice.

Edit: here's a proof of the OP's question relying on little (only on the semisimple case, where the center is discrete and finitely generated):

Let $G$ be a counterexample of minimal dimension. By the previous lemma, we can suppose (up to mod out by a discrete normal subgroup) that $G$ has an infinite, discrete torsion central subgroup $Z$. Taking the semisimple case for granted, $G$ is not semisimple. Let $V$ be the closure of the last nontrivial term of the derived series of its radical. Let $p$ be the projection $G\to G/V$. Since $\dim(G/V)<\dim(G)$, we have $\overline{p(Z)}$ compact. Hence its unit component has finite index, and hence some finite index subgroup of $Z$ is contained in the inverse image $H$ in $G$ of $\overline{p(Z)}^\circ$. Hence $H$ is a counterexample; by minimality, we deduce $\dim(H)=\dim(G)$ and hence $H=G$. That is, $\overline{VZ}$ is dense. We have $[V,Z]\subset V\cap Z$, which is a torsion discrete subgroup of the connected abelian Lie group $V$, and hence is finite. Hence $[G,G]$ is contained in $[V,Z]$, which is finite; by connectedness of $G$, we deduce that $G$ is abelian, and in turn this implies $Z$ finitely generated, and a contradiction.

As regards the semisimple case, if by contradiction $G$ is semisimple and $Z$ is an infinitely generated central subgroup, then the quotient $H$ of $G$ by its center is semisimple and not compactly presented. Then one way to get a contradiction is to use that $G$ is quasi-isometric to its symmetric space, which is non-positively curved and hence large-scale simply connected, and for $G$ this means compactly presented, a contradiction. Of course this latter proof relies on some Riemannian and metric material.

  • $\begingroup$ Thank you! This reduces my difficulty to understanding Iwasawa's result; which I'm very happy to do! $\endgroup$ Jan 1, 2018 at 20:54
  • 2
    $\begingroup$ The proof that compact subgroups of connected Lie groups lie in maximal ones, that all maximal ones are conjugate, and that the ambient group is diffeomorphic to a direct product of a maximal compact subgroup against a Euclidean space does not require any knowledge about finiteness properties of discrete subgroups of the center. These matters are all treated in Chapter XV of Hochschild's book Structure of Lie Groups, assuming the component group is merely finite rather than trivial (very useful for working with the group of $\mathbf{R}$-points of a linear algebraic $\mathbf{R}$-group). $\endgroup$
    – nfdc23
    Jan 1, 2018 at 22:32
  • $\begingroup$ @nfdc23 thanks! the second result on increasing unions of maximal compact subgroups trivially passes from connected to virtually connected Lie groups. Unlike the fact of conjugacy of maximal compact subgroups, and the fact that $G=G^0K$ for virtually connected Lie group; this latter fact is due to Mostow a few years after Iwasawa. $\endgroup$
    – YCor
    Jan 1, 2018 at 23:56
  • $\begingroup$ @YCor: Thanks for the additional details on the arguments. I should have mentioned that the proofs in Hochschild's Chapter XV avoid using general theorems in Riemannian geometry. Some parts of his proofs are inspired by general theorems in Riemannian geometry, but carried out directly in the situations that arise. $\endgroup$
    – nfdc23
    Jan 2, 2018 at 0:53
  • $\begingroup$ @nfdc23 thanks, I'll look up in Hochschild's book when I have a chance. If I can write a fairly simple argument in the semisimple case (to which I can reduce) I'll do. $\endgroup$
    – YCor
    Jan 2, 2018 at 1:42

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