I am looking for a classification of compact (not necessarily connected) Lie groups. Clearly, all such groups are extensions of a finite "component group" $\pi\_0(G)$ by a compact connected Lie group $G\_0$:
$\require{AMScd}$
\begin{CD}
0 @>>> G\_0 @>>> G @>p>> \pi\_0(G) @>>> 0
\end{CD}
The classification of compact connected Lie groups is familiar to me, so my question is how to classify such extensions.




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UPDATE: I suspect the following is true (due to @LSpice, with my added requirement that $H$, $P$ are finite):


**Hypothesis**: $G$ can always be written as
$$
G= \frac{G\_0 \rtimes H}{P}
$$
for finite groups $H,P$, where $P \subseteq Z(G\_0 \rtimes H)$.


UPDATE 2: @LSpice has **proven this below** for the weaker requirement that $P$ intersects $G\_0$ within $Z(G\_0)$, and provided a counterexample where $P$ cannot be taken to be central.


UPDATE 3: See [Improved classification of compact Lie groups](https://mathoverflow.net/questions/378160/) for a follow-up question (which I won't write here to avoid excessive clutter.)




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A less useful claim from my original question: any such $G$ can be constructed from $G\_0$ in three steps:


1. Take the direct product of $G\_0$ with a finite group.
2. Quotient the result by a finite subgroup of its center.
3. Extend a finite subgroup of $\mathrm{Out}(G\_0)$ by the result.


(Step 3 ~~may always be~~ is not a semidirect product in general.)



Classification of (not necessarily connected) compact Lie groups
$\DeclareMathOperator\U{U}$Consider the matrices $u = \begin{pmatrix}
0 & 1 \\
-1 & 0 \\
&& 0 & 1 \\
&& 1 & 0
\end{pmatrix}$ and $v = \begin{pmatrix}
0 && 1 \\
& 0 && 1 \\
-1 && 0 \\
& -1 && 0
\end{pmatrix}$. These belong to the finite group of signed permutation matrices, so the group that they generate is finite and we call it $H$. Put $G\_0 = \left\{d(z, w) \mathrel{:=} \begin{pmatrix} z \\ & z^{-1} \\ && w \\ &&& w^{-1} \end{pmatrix} \mathrel: z, w \in \U(1)\right\}$. Since $u d(z, w)u^{-1} = d(z^{-1}, w^{-1})$ and $v d(z, w)v^{-1} = d(w, z)$, the group $G$ generated by $G\_0$, $u$, and $v$ has $G\_0$ as its identity component. Now let $G\_0 \rtimes H \to G$ be any cover restricting to the inclusion $G\_0 \to G$, and let $\tilde u$ be an element of $H$ whose image lies in $u G\_0$; say the image is $u d(z, w)$. Then $\tilde u^2$ maps to $(u d(z, w))^2 = u^2 = d(-1, 1)$, so $d(-1, 1) \rtimes \tilde u^2$ lies in $\ker(G\_0 \rtimes H \to G)$. If $\tilde v$ is an element of $H$ whose image lies in $v G\_0$, then $\tilde v(d(-1, 1) \rtimes \tilde u^2)\tilde v^{-1}$ lies in $d(1, -1) \rtimes H$, hence does not equal $d(-1, 1) \rtimes H$. That is, $\ker(G\_0 \rtimes H \to G)$ is not central in $G\_0 \rtimes H$.




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What we can do is find (in general, not just for the specific example above) a finite subgroup $H$ of $G$ such that the multiplication map $G^\circ \times H \to G$ is surjective, and its kernel centralises $G^\circ$. (In the specific example above, we could take $H = \langle u, v\rangle$.)


$\DeclareMathOperator\Ad{Ad}\DeclareMathOperator\Gal{Gal}\DeclareMathOperator\Norm{Norm}\DeclareMathOperator\Weyl{W}\DeclareMathOperator\Zent{Z}\newcommand\C{{\mathbb C}}\newcommand\R{\mathbb R}\newcommand\adform{\_\text{ad}}\newcommand\scform{\_\text{sc}}\newcommand\X{\mathcal X}$To prove this, I'll use a few pieces of structure theory:


1. All maximal tori in $G$ are $G^\circ$-conjugate.
2. All Borel subgroups of $G\_\C$ are $G^\circ\_\C$-conjugate.
3. For every maximal torus $T$ in $G$, the map $\Weyl(G^\circ, T) \to \Weyl(G^\circ\_\C, T\_\C)$ is an isomorphism.
4. If $G\scform$ and $(G\_\C)\scform$ are the simply connected covers of the derived groups of $G^\circ$ and $G^\circ\_\C$, then $(G\scform)\_\C$ equals $(G\_\C)\scform$.
5. [Every compact Lie group has a finite subgroup that meets every component](https://mathoverflow.net/questions/150949/in-any-lie-group-with-finitely-many-connected-components-does-there-exist-a-fin).


I only need (4) to prove that, for every maximal torus $T$ in $G$, the map from $T$ to the set of conjugation-fixed elements of $T/\Zent(G^\circ)$ is surjective. This is probably a well known fact in its own right for real-group theorists.


Now consider triples $(T, B\_\C, \X)$ as follows: $T$ is a maximal torus in $G$; $B\_\C$ is a Borel subgroup of $G^\circ\_\C$ containing $T\_\C$, with a resulting set of simple roots $\Delta(B\_\C, T\_\C)$; and $\X$ is a set consisting of a real ray in each complex simple root space (i.e., the set of positive real multiples of some fixed non-$0$ vector). (Sorry about the pair of modifiers "complex simple".) I will call these 'pinnings', although it doesn't agree with the usual terminology (where we pick individual root vectors, not rays). I claim that $G^\circ/\Zent(G^\circ)$ acts simply transitively on the set of pinnings.


Once we have transitivity, freeness is clear: if $g \in G^\circ$ stabilises some pair $(T, B\_\C)$, then it lies in $T$, and so stabilises every complex root space; but then, for it to stabilise some choice of rays $\X$, it has to have the property that $\alpha(g)$ is positive and real for each simple root $\alpha$; but also $\alpha(g)$ is a norm-$1$ complex number, hence trivial, for each simple root $\alpha$, hence for each root $\alpha$, so that $g$ is central.


For transitivity, since (1) all maximal tori in $G$ are $G^\circ$-conjugate, so (2) for every maximal torus $T$ in $G$, the Weyl group $\Weyl(G^\circ\_\C, T\_\C)$ acts transitively on the Borel subgroups of $G^\circ\_\C$ containing $T\_\C$, and (3) $\Weyl(G^\circ, T) \to \Weyl(G^\circ\_\C, T\_\C)$ is an isomorphism, it suffices to show that all possible sets $\X$ are conjugate. Here's the argument that I came up with to show that they are even $T$-conjugate; I think it can probably be made much less awkward. Fix a simple root $\alpha$, and two non-$0$ elements $X\_\alpha$ and $X'\_\alpha$ of the corresponding root space. Then there are a positive real number $r$ and a norm-$1$ complex number $z$ such that $X'\_\alpha = r z X\_\alpha$. Choose a norm-$1$ complex number $w$ such that $w^2 = z$. There is then a unique element $s\adform$ of $T\_\C/\Zent(G^\circ\_\C)$ such that $\alpha(s\adform) = w$, and $\beta(s\adform) = 1$ for all simple roots $\beta \ne \alpha$. By (4), we can choose a lift $s\scform$ of $s\adform$ to $(G\scform)\_\C = (G\_\C)\scform$, which necessarily lies in the preimage $(T\_\C)\scform$ of (the intersection with the derived subgroup of) $T$, and put $t\scform = s\scform\cdot\overline{s\scform}$. Then
$$
\alpha(t\scform) = \alpha(s\scform)\overline{\overline\alpha(s\scform)} = \alpha(s\scform)\overline{\alpha(s\scform)^{-1}} = w\cdot\overline{w^{-1}} = z,
$$
and, similarly, $\beta(t\scform) = 1$ for all simple roots $\beta \ne \alpha$. Now the image $t$ of $t\scform$ in $G^\circ\_\C$ lies in $T\_\C$ and is fixed by conjugation, hence lies in $T$; and $\Ad(t)X\_\alpha = z X\_\alpha$ lies on the ray through $X'\_\alpha$.


Since $G$ also acts on the set of pinnings, we have a well defined map $p : G \to G^\circ/\Zent(G^\circ)$ that restricts to the natural projection on $G^\circ$. Now $\ker(p)$ meets every component, but it contains $\Zent(G^\circ)$, so it need not be finite. Applying (5) to the Lie group $\ker(p)$ yields the desired subgroup $H$. Note that, as requested in your [improved classification](https://mathoverflow.net/questions/378160/improved-classification-of-compact-lie-groups), conjugation by any element of $H$ fixes a pinning, hence, if inner, must be trivial.


