This question is a follow-up to [Classification of (not necessarily connected) compact Lie groups](https://mathoverflow.net/questions/377981). 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](https://mathoverflow.net/questions/377981). A concise rephrasing of the proof is given in my answer below.




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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.



Improved classification of compact Lie groups
@LSpice has already proven my revised conjecture in the updated answer to [Classification of (not necessarily connected) compact Lie groups](https://mathoverflow.net/questions/377981/classification-of-not-necessarily-connected-compact-lie-groups?noredirect=1#comment959636_377981), 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?](https://mathoverflow.net/questions/378218/), 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?](https://mathoverflow.net/questions/150949), $K$ 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.




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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.


