$\DeclareMathOperator\SO{SO}\DeclareMathOperator\SU{SU}\DeclareMathOperator\SL{SL}$A closed subgroup $ \Gamma $ of a Lie group $ G $ is called Lie primitive if it is not contained in any proper positive dimensional closed subgroup.

Let $ G $ be a compact group. Then a Lie primitive subgroup $ \Gamma $ of $ G $ must be finite.

Must a compact simple Lie group $ G $ have only finitely many isomorphism types of Lie primitive subgroups?


This is true for $ \SO_3(\mathbb{R}) $ which has only $ 3 $ Lie primitive subgroups up to conjugacy: the exceptional rotation groups of the platonic solids $ A_4$, $S_4$, $A_5 $.

This is also true for $ \SU_3 $ and $ \SU_4 $ due to work by Blichfeldt. For an explicit list for $ \SU_4 $ for example see Hanany and He - A Monograph on the Classification of the Discrete Subgroups of SU(4).

Since it is true for $ \SU_4 $ then it must be true for $ \SO_6 $ by the double cover $ \SU_4 \to \SO_6 $.

Also this seems relevant: Primitive subgroup of $ SU_n $ contained in maximal finite subgroup?

Some comments about the case when $ G $ is not compact:

A closed subgroup not contained in any positive dimensional subgroup must be dimension $0$ i.e. it must be discrete. Peter McNamara points out that the only proper positive dimensional closed subgroups of $ \SL_2(R) $ are solvable and thus can't contain a free group. Since a free group on any number of letters is a subgroup of $ \SL_2(R) $ that means there are infinitely many non isomorphic Lie primitive subgroups of $ \SL_2(R) $.

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    $\begingroup$ This is very close to the theorem of Jordan-Frobenius-Schur which says that for fixed $GL_n$ there is a bound $F(n)$ such that every finite group in $GL_n$ has a normal abelian subgroup of index at most $F(n)$. For sufficiently large groups, you can enlarge the abelian subgroup to a torus to produce a proper positive dimensional subgroup. Maybe sufficiently large = abelian subgroup is not central in the Lie group? $\endgroup$ Jan 22 at 15:43
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    $\begingroup$ @BenWieland "you can enlarge the abelian subgroup to a torus" -- why? e.g. if we take the subgroup of diagonal matrices in $SO_n$ -- it is a maximal abelian subgroup and it is finite. the problem is that a max torus is a max abelian subgroup but not vice versa. $\endgroup$ Jan 22 at 15:57
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    $\begingroup$ Case of $O(n)$. I've seen this in some treatment of JS. The abelian subgroup preserves some some direct sum decomposition of $\mathbb R^n$. $\Gamma$ is in the Lie group that preserves this decomp into unlabeled summands, the normalizer of the centralizer of the abelian subgroup (which is what preserves the labeled decomp). There are three possibilities for this Lie group. (1) All of $O(n)$: the abelian group is contained in the center and the whole group is at most size $2F(n)$. (2) Positive dimensional but not everything—not primitive. (3) Discrete: $S_n\ltimes O(1)^n$—finitely many cases. $\endgroup$ Jan 23 at 23:29
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    $\begingroup$ I assume that a compactness assumption is missing, as otherwise you can realise a free group on an arbitrarily large finite number of letters as a primitive subgroup of SL_2(R). $\endgroup$ Jan 24 at 1:52
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    $\begingroup$ My argument missed the detail that eigenspaces are not defined over $\mathbb R$ My case (1) should split into (1a) if the abelian group's eigenspace is all of $\mathbb C$, then it is central and $\Gamma$ is small; and (1b) if the abelian group's eigenspace does not split $\mathbb R^n$, but does split $\mathbb C^n$, then it defines a complex structure and is contained in $U(n/2)$. In general, the centralizer of the abelian group is not the group that preserves its real eigenspaces, but the smaller group that preserves its complex eigenspaces, not just a product of $O(m)$ but also $U(m)$. $\endgroup$ Jan 24 at 16:31

1 Answer 1


Yes, it is true. Let us show that, in a compact Lie group $G$ with $G^0$ nonabelian, finite subgroups that are not contained in any proper subgroup of positive dimension have bounded cardinal.

By contradiction, let finite subgroups $G_n$ have cardinal tending to infinity, and not contained in any proper subgroup of positive dimension.

Let $B$ be a neighborhood of 0 in the Lie algebra of $G$ such that $\exp|_B$ is a homeomorphism onto its image, with inverse denoted by $\log$ (we will let $B$ vary, but can assume it is contained in a single such neighborhood $B_0$ once and for all, hence the function $\log$ will not depend on $B$).

Define $T_n^B=\exp\big(\mathrm{span}\big(\log(G_n\cap\exp(B))\big)\big)$.

The first point is that for each $B\subset B_0$ and for $n$ large enough, $G_n\cap\exp(B)$ consists of commuting elements. This is close to Jordan's theorem and is a particular case of Zassenhaus' theorem (see for instance Theorem 1.3 here, with in mind that connected nilpotent subgroups of compact groups are abelian). We can extract once and for all to suppose that this holds for all $n$.

The next point is that $T_n^B$ is a torus. For an arbitrary commutative subset, it might have been dense in a torus. But the point is that elements of finite subgroups are well-behaved. More precisely: for this point, we can suppose that $G$ is embedded in the unitary group $\mathrm{U}(d)$. Every element of $G$ can be diagonalized with eigenvalues $e^{i\pi k_1/m},\dots,e^{i\pi k_d/m}$. We can arrange $B_0$ so that it avoids elements with eigenvalues $-1$ and define the log accordingly, so assume all $k_i$ to be in $]-m,m[$. Hence the log of such an element will be, in the same diagonal basis, the matrix with eigenvalues $i\pi k_1/m,\dots i\pi k_d/m$, and the exp of its span will be the 1-dimensional torus of diagonal matrices with eigenvalues $\theta^{k_1},\dots,\theta^{k_d}$ for $\theta$ in the unit circle. Now when we consider all $G_n\cap \exp(B)$, we just consider the torus generated by several such (commuting) tori.

Now consider $\liminf_n \dim(T_n^B)$. This is positive (because $G_n$ is infinite, so that $G_n\cap\exp(B)$ is never reduced to $\{1\}$. This liming decreases when $B$ decreases, and hence we can find $B_1$ for which its value, say $c>0$, is minimal. After extracting once again, we an suppose that $\lim_n\dim(T_n^{B_1})=c$. Define $T_n=T_n^{B_1}$. Hence, for every neighborhood $B\subset B_1$, there exists $n_0$ (depending on $B$) such that for all $n\ge n_0$ we have $T_n^B=T_n$ (because $T_n^B\subset T_n$ and these are tori of the same dimension).

This observation allows to deduce that $T_n$ is, for large $n$, normalized by $G_n$. Indeed, for $g_n\in G_n$, we have $$g_nT_n^Bg_n^{-1}=\exp(\mathrm{span}\log(G_n\cap \exp(g_nBg_n^{-1})));$$ since $1$ has a basis of conjugation-invariant neighborhoods, we can choose $B$ such that $\bigcup_g gBg^{-1}$ is contained in $B_1$. We deduce that $g_nT_ng_n^{-1}\subseteq T_n$ for large $n$, hence this is an equality.

Then $G_nT_n$ is a positive-dimensional closed subgroup containing $G_n$. By assumption, it equals $G$, and this is a contradiction since $G^0$ is non-abelian.


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