This question is a follow-up to Classification of (not necessarily connected) compact Lie groups. In the answer to that question, @LSpice proved that any compact, not necessarily connected Lie group $G$ takes the form $$ G = \frac{G_0 \rtimes R}{P} $$ where $G_0$ is the identity component of $G$, $R$ is a finite group, and $P$ is a finite, common subgroup of $G_0$ and $R$ that is central within $G_0$ (but need not be central within $R$).

Nonetheless, there are many possibilities for the semi-direct product. To narrow the list, it would be convenient to separate out those elements of $R$ that act by non-trivial outer automorphisms on $G_0$ and modify the rest so that they commute with $G_0$.

UPDATE: my original hypothesis (below) is false. A weaker, possibly correct version is:

Hypothesis: $R$ and $P$ can be chosen above such that every element of $R$ either (1) acts by a non-trivial outer automorphism on $G_0$ or (2) acts trivially on $G_0$.

UPDATE 2: @LSpice proved this in the updated answer to Classification of (not necessarily connected) compact Lie groups. A concise rephrasing of the proof is given in my answer below.

By comparison, this is false:

Hypothesis: Any compact Lie group $G$ can be written in the form $$ G = \frac{(G_0 \times H) \rtimes R}{P} $$ where $H, R, P$ are finite groups and non-trivial elements of $R$ act by non-trivial outer automorphisms on $G_0$.

Counterexample: consider $G = U(1) \rtimes \mathbb{Z}_4$, where the generator $r$ of $\mathbb{Z}_4$ acts by the ``charge conjugation'' outer automorphism $r^{-1} e^{i \theta} r = e^{-i \theta}$ on $U(1)$. In any finite extension $G'$ of this group, elements of $\pi_0(G)$ that act by charge conjugation will never square to the identity in $G'$, so $G'$ never takes the required $(G\times H) \rtimes \mathbb{Z}_2$ form with $\mathbb{Z}_2$ acting on $U(1)$ by charge conjugation.

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    $\begingroup$ I have deleted my wrong answer after you produced a counterexample. $\endgroup$
    – LSpice
    Commented Dec 5, 2020 at 19:53
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    $\begingroup$ Do you agree that my answer to your other question also answers the revised version of this question? $\endgroup$
    – LSpice
    Commented Dec 6, 2020 at 0:21
  • $\begingroup$ @LSpice Yes. I'm also writing a short alternative answer to this question to explain the proof method I had in mind (very similar to your proof but not identical). $\endgroup$ Commented Dec 6, 2020 at 6:06

1 Answer 1


@LSpice has already proven my revised conjecture in the updated answer to Classification of (not necessarily connected) compact Lie groups, but let me give another, closely related proof.

Since $1\to \mathrm{Inn}(G_0) \to \mathrm{Aut}(G_0) \to \mathrm{Out}(G_0) \to 1$ always splits, see Does Aut(G) → Out(G) always split for a compact, connected Lie group G?, we can choose a subgroup $R_0 \subseteq \mathrm{Aut}(G_0)$ for which the restriction of $\mathrm{Aut}(G_0) \to \mathrm{Out}(G_0)$ is an isomorphism. The inverse image of $R_0$ under the map $f:G \to \mathrm{Aut}(G_0)$ induced by conjugation is a subgroup $K \subseteq G$ whose intersection with $G_0$ is $Z(G_0)$.

Multiplying any $g\in G$ by arbitrary $h \in G_0$ multiplies the associated $f(g) \in \mathrm{Aut}(G_0)$ by an arbitrary inner automorphism $f(h) \in \mathrm{Inn}(G_0)$, without changing $g$'s connected component. Thus, $K$ meets every connected component of $G$.

Using the result of In any Lie group with finitely many connected components, does there exist a finite subgroup which meets every component?, $K$ has a finite subgroup $R$ that meets every component of $K$, hence it meets every component of $G$ as well, and intersects $G_0$ within $Z(G_0)$. By design, the elements of $R$ either act by non-trivial outer automorphisms on $G_0$ or they act trivially on $G_0$. This proves my (revised) conjecture.

COMMENT ADDED: An interesting, yet false, generalization is stated and disproven below.

It is well known that any compact, connected Lie group $G_0$ takes the form $$G_0 = \frac{T^k \times G_1 \times \ldots \times G_\ell}{P}$$ where $T^k$ denotes a $k$-torus, $G_1, \ldots, G_\ell$ are compact, simply connected, simple Lie groups, and $P$ is central. One might think that the quotients in the expressions for $G$ and $G_0$ could be combined, so that any compact Lie group $G$ would take the form: $$ G = \frac{(T^k \times G_1 \times \ldots \times G_\ell) \rtimes R}{P} $$ where as before each element of $R$ acts by a non-trivial outer or acts trivially on $T^k \times G_1 \times \ldots \times G_\ell$. However, this is false.

Counterexample: Consider $G=(\mathrm{SO}(2k) \rtimes \mathbb{Z}_4) / \mathbb{Z}_2$, where the generator $r \in \mathbb{Z}_4$ acts by parity on $\mathrm{SO}(2k)$ and $r^2 = -1 \in SO(2k)$. Now let $G’=(\mathrm{Spin}(2k) \rtimes R)/P$ be a cover of $G$ whose connected component is $G_0'=\mathrm{Spin}(2k)$. There is some element $r'$ of $R$ that projects to $r$, hence $r’$ acts on $\mathrm{Spin}(2k)$ by parity. If $k$ is odd, then $Z(G_0') = \mathbb{Z}_4$, and $(r’)^2$ must be one of the two elements of order 4 in $Z(G_0')$ to project to $(r)^2 = -1$. However, parity exchanges these two elements, so we find $(r’)^{-1} (r’)^2 r’ \ne (r’)^2$, which is a contradiction. The case of even $k$ is very similar.

  • $\begingroup$ If you use the subgroup $R_0$ from the answer to mathoverflow.net/questions/378218, then your $K$ is $\ker(p)$ from the answer to mathoverflow.net/questions/150949. $\endgroup$
    – LSpice
    Commented Dec 6, 2020 at 12:31
  • $\begingroup$ @LSpice Great, so this approach really is very closely related to your original proof, as I suspected. $\endgroup$ Commented Dec 6, 2020 at 15:37
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    $\begingroup$ Since @LSpice is not going to duplicate the updated proof from mathoverflow.net/questions/378218 here, I've accepted my own answer to indicate that it is correct (and essentially summarizes LSpice's proof in alternate form.) $\endgroup$ Commented Dec 7, 2020 at 16:07

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