# Finite maximal closed subgroups of Lie groups

$$\newcommand{\G}{\mathcal{G}} \newcommand{\K}{\mathcal{K}} \DeclareMathOperator\SU{SU}\DeclareMathOperator\PSU{PSU}\DeclareMathOperator\SO{SO}$$Let $$\G$$ be a Lie group.

I am interested in finite maximal closed subgroups of $$G$$.

I'm guessing that $$\G$$ has a finite maximal closed subgroup if and only if $$\G$$ is simple and compact. Does anyone have other examples of finite maximal closed subgroups?

• @LSpice I think I posted this one first on the 9th, didn't get any answers, cross posted to MO a few days later on the 11th, got interesting comments from Ycor on the MO post about non. connected. counterexamples, sort of wrote up his comments 4 days later on the 15th as an answer to my own MO question, but my original self answer on MO was really bad and also quite short. I also went back to the MSE question and put in a connectedness assumption to avoid Ycor's counterexamples. Sep 26, 2022 at 12:46
• 8 months later on June 16th I came back after learning 8 months worth of stuff and edited my self answer on MO, basically writing a whole new answer. Then 5 days after that took the MO answer and pasted it back into an answer to the original MSE question. So the answer to your question is that the MO version is just a cross-post but overall is slightly better/more comprehensive than the original MSE version Sep 26, 2022 at 12:49

WARNING: This answer has lots of true things in it but is also riddled with false claims, I keep meaning to fix it but don't get around to it. $$\newcommand{\G}{\mathcal{G}} \newcommand{\K}{\mathcal{K}} \DeclareMathOperator\SU{SU}$$There exist Lie groups $$\G$$ which are not compact or simple but have finite maximal closed subgroups. For example the five element cyclic subgroup $$C_5$$ of $$\mathbb{R}^2 \rtimes C_5$$ described at my answer to Finite maximal closed subgroups of Lie groups.

These sorts of examples are interesting in their own right but are not really what I was thinking about when I asked the question. So from now on I will confine myself to the case that $$\G$$ is connected.

In other words I will consider the statement "A connected Lie group $$\G$$ has a finite maximal closed subgroup $$G$$ if and only if $$\G$$ is compact and simple."

The first implication is true.

Claim 1: Let $$G$$ be a connected Lie group. Suppose that $$G$$ has a finite subgroup which is maximal among the closed subgroups of $$G$$. Then $$G$$ must be compact and simple.

Proof:

Call such a finite maximal closed subgroup $$\Gamma$$. $$\Gamma$$ is finite thus compact. So $$\Gamma$$ must be contained in some maximal compact subgroup $$K$$ of $$G$$. Since $$\Gamma$$ is maximal then we have that either $$K=\Gamma$$ or $$K=G$$. Suppose for the sake of contradiction that $$K=\Gamma$$. But according to

[https://mathoverflow.net/questions/140622/are-maximal-compact-subgroups-of-connected-groups-connected]

the maximal compact subgroup $$K$$ of a connected group $$G$$ is always itself connected. So if $$\Gamma=K$$ then $$K$$ is connected and finite thus trivial. But the trivial subgroup can never be a maximal closed subgroup in a connected Lie group $$G$$. The trivial subgroup of a connected Lie group will always be contained in a closed 1-parameter subgroup, which is either a circle or line. So by maximality of $$\Gamma$$ that implies the whole group $$G$$ is either a circle or a line. But the trivial subgroup is not a maximal closed subgroup of $$S^1$$ or $$\mathbb{R}$$. Thus we have a contradiction. So we can conclude that $$K \neq \Gamma$$ and thus $$K=G$$ that is, $$G$$ is compact. Now we will show that $$G$$ must be simple. Suppose for the sake of contradiction that $$G$$ is not simple. Then $$G$$ must have some normal subgroup $$N$$ such that $$0 < dim(N) < dim(G)$$ . So there is a surjective map $$\pi: G \to G/N$$ Then $$\pi(\Gamma)$$ is a maximal closed subgroup of $$G/N$$ ( $$\pi(\Gamma) \neq G/N$$ since $$dim(\pi(\Gamma))=0< dim(G/N))$$ ). Note that $$\pi^{-1}[\pi(\Gamma)]$$ is a closed subgroup of $$G$$ containing $$\Gamma$$. But $$\pi^{-1}[\pi(\Gamma)]$$ contains $$N$$ so $$0 thus $$\Gamma \neq \pi^{-1}[\pi(\Gamma)]$$ so by maximality $$\pi^{-1}[\pi(\Gamma)]=G$$ But that implies that the image of $$\pi$$ is contained in $$\pi (\Gamma)$$. However that is impossible because $$\pi$$ is surjective and $$\pi(\Gamma)$$ is zero dimensional while $$Im(\pi)=G/N$$ is positive dimensional. Thus if a connected Lie group $$G$$ has a finite maximal closed subgroup then we can conclude that $$G$$ is both compact and simple.

This significantly narrows down the the possibilities for $$G$$ to the well known classification of compact simple Lie groups.

However the reverse implication does not hold: $$\operatorname{SU}_{15}$$ is an example of a compact connected simple Lie group with no finite maximal closed subgroups.

To see why this is the case it is important to note that

Claim 2: For a compact connected simple Lie group $$\G$$, $$G$$ is a finite maximal closed subgroup of $$\G$$ if and only if $$G$$ is Ad-irreducible and $$G$$ is a maximal finite subgroup of $$\G$$.

The implication Ad-irred+ max finite implies finite and max closed follows from Corollary 3.5 of Sawicki and Karnas - Universality of single qudit gates. The reverse implication, that finite and max closed implies Ad-irred+ max finite follows from the following argument (no it doesn't the reverse implication is false see the EDIT): Let $$\Gamma$$ be finite and maximal closed. SSOC that $$\Gamma$$ is Ad-reducible. Then $$\Gamma$$ acting by conjugation preserves some proper nonzero Lie subalgebra $$\mathfrak{h}$$ (EDIT: THIS IS FALSE, $$\mathfrak{h}$$ is a sub vector space but not necessarily a subalgebra, so the entire reverse implication falls apart, in particular I believe that the $$SL(2,8)$$ subgroup of the exceptional Lie group $$G_2$$ is an example of a finite maximal closed subgroup which is not Ad-irreducible) . So $$\Gamma$$ acting by conjugation preserves some connected nontrivial proper subgroup $$H$$ of $$G$$ with Lie algebra $$\mathfrak{h}$$. That implies that $$\Gamma$$ normalizes $$H$$. That is, $$\Gamma$$ is contained in $$N(H)$$. But $$\Gamma$$ is finite while $$N(H)$$ contains the positive dimensional subgroup $$H$$. Thus the closed subgroup $$N(H)$$ properly contains $$\Gamma$$. But $$\Gamma$$ is maximal by assumption so that implies $$N(H)=G$$. However $$0 < dim(H) < dim(G)$$ and $$H$$ is normal, so that contradicts the fact that $$G$$ is simple.

Since a finite subgroup of $$\SU_n$$ is Ad-irreducible if and only if it is a unitary 2-design we have

Claim 3: $$G$$ is a finite maximal closed subgroup of $$\SU_n$$ if and only if $$G$$ is a maximal unitary 2-group in $$\SU_n$$.

By inspecting Theorem 3 of Bannai, Navarro, Rizo, and Pham Huu Tiep - Unitary $$t$$-groups one immediately determines that $$\SU_{15}$$ has no finite maximal closed subgroups.

Some of the main examples of finite maximal closed subgroups of $$\SU_n$$ include the normalizer in $$\SU(p^n)$$ of an extra-special group $$p^{2n+1}$$. Here $$p$$ is an odd prime. There is also a similar construction $$p=2$$. Then there are infinite families of examples relating to the Weil module for $$\operatorname{PSp}_{2n}(3)$$ and another family related to $$\operatorname{PSU}_n(2)$$. Plus many exceptional cases.

• I may be misreading, but it seems to me that the first paragraph in the proof of claim establishes that $\mathcal{G}$ is compact OR $G$ is trivial. To rule out the second possibility, note the closure of any 1 parameter subgroup is closed and contains $G$, so is everything. In particular, the big group is abelian and simple, so is one dimensional. Thus, it's $\mathbb{R}$ or a circle. And in both cases, the trivial subgroup is not maximal among closed subgroups. Oct 13, 2022 at 2:10