The answer to the question at Does almost every pair of elements in a compact Lie group generates the connected component? says there must be countably many conjugacy classes of closed subgroups of compact connected Lie groups. One of the comments (Does almost every pair of elements in a compact Lie group generates the connected component?) says that this is a direct result of inclusions of closed subgroups being (real) algebraic. However, I don't see how the result follows without using something fancy. Is there an obvious or nice proof I'm missing?
7 years ago I wrote in an unfinished paper a proof of this result; I didn't know then if it was original but it indeed seems to be known to a number of specialists although I'm not aware of a written complete proof. Since I don't want anymore to complete this paper, I reproduce the proof here (not selfcontained, relying on [LW] below) as Proposition 5; possibly Uri has a more direct approach. I include closely related Proposition 1, which was my original goal.
Proposition 1 Let $G$ be a compact group (not necessarily Lie). Consider the action by $G$ by conjugation on $\mathcal{S}(G)$ (the space of subgroups of $G$, endowed with the Chabauty topology). Then the orbits of $G_0$ (connected component of 1) coincide with the connected components of $\mathcal{S}(G)$.
As a corollary, for $G$ a compact group, $\mathcal{S}(G)$ is totally disconnected if and only if $G_0$ is central in $G$ (equivalently, $G/Z(G)$ is totally disconnected), a result obtained in [FG]. More generally, it can be checked that for an arbitrary locally compact group $G$, $\mathcal{S}(G)$ is totally disconnected if and only if either totally disconnected, or $G$ is pointwise elliptic with $G_0$ central. ($G$ pointwise elliptic means that every single element generates a cyclic subgroup with compact closure.)
Lemma 2 Let $G$ be a compact Lie group. Let $F$ be a finite group. Then there are finitely many conjugacy classes of homomorphisms $F\to G$.
Proof. The space $\mathrm{Hom}(F,G)$ can be described as a real algebraic variety, so has finitely many components. By [LW], all homomorphisms in a given component are conjugate under $G_0$. \end{proof}
Lemma 3 Let $G$ be a compact Lie group. Let $S$ be a connected compact semisimple Lie group. Then there are finitely many conjugacy classes of homomorphisms $S\to G$.
Proof. The space $\mathrm{Hom}(S,G)$ is canonically identified with the space of Lie algebra homomorphisms $\mathfrak{s}\to\mathfrak{g}$, which is an algebraic variety as well, so has finitely many connected components. So we can again use [LW].
Lemma 4 Let $G$ be a compact Lie group. Let $D$ be a connected compact abelian Lie group. Then there are countably many conjugacy classes of homomorphisms $D\to G$.
Proof. Any such homomorphism maps into a maximal torus, and all maximal tori are conjugate [Bourbaki, Groupes et Algèbres de Lie, Chap 9, §2.2]. This reduces to the case when $G$ is a torus, but then the result just follows from the fact that $\mathrm{Hom}((\mathbf{R}/\mathbf{Z})^a,(\mathbf{R}/\mathbf{Z})^b)\simeq\mathbf{Z}^{ab}$, which is countable.
Lemma 5 Let $G$ be a compact Lie group. Then there are countably many conjugacy classes of closed connected subgroups in $G$.
Proof. Since there are finitely many isomorphism types of compact semisimple groups in a given dimension, we can fix this type, so by Lemma 3, it is enough to consider the set of closed connected subgroups whose semisimple part is a given subgroup $S$. Therefore, it is enough to count conjugacy classes of closed abelian connected subgroups in $N(S)/S$, where $N(\cdot)$ denotes the normalizer, and Lemma 4 applies.
Proposition 6 Let $G$ be a compact Lie group. Then there are countably many conjugacy classes of closed subgroups in $G$.
Proof. By Lemma 5, we can consider those subgroups for which $H_0=L$ is given. So it is enough to count conjugacy classes of finite subgroups in $N(L)/L$, there are countably many because of Lemma 2.
Proof of Proposition 1 We begin with the case when $G$ is a Lie group. Since $G_0$ has finite index in $G$, Proposition 6 implies that there are countably many $G_0$conjugacy classes of closed subgroups in $G$. Let $C$ be a connected component of $\mathcal{S}(G)$. Clearly, it is invariant under conjugation by $G_0$. We just obtained that the quotient of $C$ by the action of $G_0$ is countable. It is also a compact connected Hausdorff space, so is reduced to a point (it is Hausdorff because the acting group is compact, see [Bourbaki, Topologie Générale, III, §4.1]. This means that $G_0$ acts transitively on each connected component of $\mathcal{S}(G)$, and we are done.
Now let us deal with a general compact group $G$. (It is no longer true in general that the set of connected components of $\mathcal{S}(G)$ is countable). We write $G$ as a projective limit of compact Lie groups $G/K_i$. Clearly, if two elements $H,H'$ of $\mathcal{S}(G)$ are conjugate under $G_0$, then they lie in the same component. Conversely, suppose they belong to the same connected component of $\mathcal{S}(G)$. Then since $K_i$ is compact, the projection map $\mathcal{S}(G)\to\mathcal{S}(G/K_i)$ is continuous. Because of the case of Lie groups, there exists $g_i\in G_0$ such that $(g_iHg_i^{1}K_i=H'K_i)$ (we use the fact that the inverse image of the unit component of $G/K_i$ is the unit component of $G$). Let $g$ be a limit point of $(g_i)$. If $h\in H$, then $g_ihg_i^{1}$ can be written as $h'_ik_i$ with $h'_i\in H'$ and $k_i\in K_i$. Necessarily $k_i\to 1$, so $h'_i=g_ihg_i^{1}k_i^{1}$ also converges, necessarily to an element $h'$ of $H'$. We have $ghg^{1}=h'$. This proves that $gHg^{1}\subset H'$, and the converse inclusion $g^{1}H'g\subset H'$ is strictly similar.
[FG] S. Fisher, P. Gartside. On the space of subgroups of a compact group II. Topology Appl. 156 (2009) 855861
[LW] D.H. Lee, T.S. Wu. On conjugacy of homomorphisms of topological groups. Illinois J. Math. 13 1969 694699
in [LW] I refer to Theorem 2.6: if $G,H$ are compact Lie groups, then the connected components of $\mathrm{Hom}(G,H)$ are precisely the $H_0$orbits (for action by postconjugation).

2$\begingroup$ That every compact Lie group has at most countably many conjugacy classes of closed subgroups is proved in Corollary 1.7.27 [Palais, Memoirs AMS, The classification of $G$spaces, 1960], see p.39 at vmm.math.uci.edu/PalaisPapers/ClassificationOfGSpaces.pdf. $\endgroup$ – Igor Belegradek May 13 '17 at 12:43
Required Claim: In a compact Lie group there are at most countably many conjugacy classes of closed subgroups.
I am the one who made the comment that the required claim could follow from an algebraicgroupsreasoning. I did not intend to say that this is obvious, rather that there is a conceptual explanation. Roughly, the claim follows from the fact that given compact Lie groups $G$ and $H$, the space of continuous homomorphisms $\text{Hom}(H,G)$ has the structure of an indvariety and the "tangent spaces to $\text{Hom}(H,G)$ modulo $G$conjugation" are trivial. These "tangent spaces" could be identified with first cohomologies of $H$ with coefficients in the Lie algebra of $G$. I will try to make this precise below. In fact, taking into account the density of the algebraic points in $\text{Hom}(H,G)$, I will be able to prove a stronger result which is new to me, see the conceptual claim below.
Convention: Let me denote here by $\bar{\mathbb{Q}}$ the algebraic closure of $\mathbb{Q}$ in $\mathbb{R}$.
Conceptual Claim: Given a compact Lie group $G$ there exists a reductive $\bar{\mathbb{Q}}$algebraic group $\mathbf{G}$ and a Lie groups isomorphism $G\simeq \mathbf{G}(\mathbb{R})$ under which every closed subgroup of $G$ corresponds to a $\mathbf{G}(\mathbb{R})$conjugate of the $\mathbb{R}$points of a $\bar{\mathbb{Q}}$subgroup of $\mathbf{G}$.
Remark: As we'll see below, the $\mathbb{R}$group $\mathbf{G}$ is associated canonically with $G$. It follows from the conceptual claim that $\bar{\mathbb{Q}}$structure is canonical up to conjugation.
Conceptual Claim $\Rightarrow$ Required Claim: Immediate by the fact that $\mathbf{G}$ has at most countably many conjugacy classes of $\bar{\mathbb{Q}}$subgroups and each such subgroup is given by a finitely generated ideal in the countable algebra $\bar{\mathbb{Q}}[\mathbf{G}]$.
In what follows I will try to explain why the conceptual claim holds true. To get started let me state a standard fact, closely related to "TannakaKrein duality".
Fact 1: There is a fully faithful functor from compact Lie groups to reductive $\mathbb{R}$algebraic groups which is a right inverse to the functor of realpoints (ie, for a compact Lie group $G$ we associate a reductive $\mathbb{R}$algebraic group $\mathbf{G}$ and there is a natural Liegroups isomorphism $G\to \mathbf{G}(\mathbb{R})$).
Idea of proof: Given a compact Lie group $G$, consider the subalgebra of $G$finite functions in the Banach algebra $C(G)$. This is a finitely generated algebra with no nilpotents. Let $\mathbf{G}$ be its spectrum. Give proper attributions to PeterWeyl and StoneWeierstrass in due places.
Fact 2: Every reductive $\mathbb{R}$algebraic group has a $\bar{\mathbb{Q}}$structure.
Idea of proof: This follows by structure theory  every reductive group is a finite extension of a connected reductive group, every connected one is a central toral extension of a semisimple group. (I'd be happy if someone will provide a more conceptual explanation.)
In light of the facts above, in proving the conceptual claim we can abandon compact groups terminology and prove the following purealgebraicgroupsclaim.
AG Claim: For every reductive $\bar{\mathbb{Q}}$algebraic group $\mathbf{H}$ and every $\bar{\mathbb{Q}}$algebraic group $\mathbf{G}$, for every $\mathbb{R}$morphism $\phi:\mathbf{H}\to\mathbf{G}$ there exists $g\in \mathbf{G}(\mathbb{R})$ such that $\text{inn}(g)\circ\phi$ is defined over $\bar{\mathbb{Q}}$, where $\text{inn}(g)$ is the inner automorphism of $\mathbf{G}$ given by $g$.
AG Claim $\Rightarrow$ Conceptual Claim: Given a compact Lie group $G$, let $\mathbf{G}$ be the $\mathbb{R}$group associated with $G$ by the functor given in Fact 1 and use Fact 2 to view $\mathbf{G}$ as a $\bar{\mathbb{Q}}$group. Given $H<G$, let $\mathbf{H}$ be the corresponding $\mathbb{R}$group and $\phi:\mathbf{H}\to \mathbf{G}$ be the $\mathbb{R}$morphism associated with the inclusion $H<G$. Use Fact 2 again to fix a $\bar{\mathbb{Q}}$structure on $\mathbf{H}$ and apply the AG Claim.
Note that, in general, the space of algebraic groups morphisms $\text{Hom}(\mathbf{H},\mathbf{G})$ is not a variety (eg $\text{Hom}(\mathbf{G}_m,\mathbf{G}_m)\simeq\mathbb{Z}$). However it is always an indvariety, that is a compatible ascending union of a sequence of algebraic varieties. Moreover, these varieties could be chosen to be invariant under the $\mathbf{G}$conjugation action.
Edit: When I initially wrote this answer I thought this stuff ought to be well known. I still think so, but in light of the discussion in Why do automorphism groups of algebraic varieties have natural algebraic group structure? and Number of connected components of an Automorphism group I decided to expand a bit this part of our discussion. In particular I added the following lemma and its corollary.
Lemma: Let $k$ be a field and $\mathbf{X}$ and $\mathbf{Y}$ be affine $k$varieties. Let $\mathbf{G}$ and $\mathbf{H}$ be $k$groups acting morphically on $\mathbf{X}$ and $\mathbf{Y}$ correspondingly. Then the space of morphisms $\text{Mor}(\mathbf{Y},\mathbf{X})$ could be represented as an ascending union of a sequence of $\mathbf{H}\times\mathbf{G}$invariant subsets which have compatible invariant structures of $k$varieties.
Idea of proof: Fix a finite dimensional generating vector space $U<k[\mathbf{X}]$, let $S(U)$ be the free commutative algebra on $U$ and consider the obvious map $\pi:S(U)\to k[\mathbf{X}]$. Let $W<S(U)$ be a finite dimensional vector space containing $U$ and a generating set of $\text{Ker}(\pi)$. Let $U_0$ be a finite dimensional $\mathbf{G}$invariant vector spaces containing $\pi(W)$. Note that for every $k$algebra $A$ and a $k$linear map $\phi:U_0\to A$, $\phi$ extends to a $k$algebra morphism $k[\mathbf{X}]\to A$ iff its restriction to $U_0$ is multiplicative in the following sense: for every $u,u'\in U_0$, if $uu'\in U_0$ then $\phi(uu')=\phi(u)\phi(u')$. In case $\phi$ extends, the extension is unique. Fix $V_n<k[\mathbf{Y}]$, an ascending sequence of finite dimensional $\mathbf{H}$invariant vector spaces such that $\cup V_n=k[\mathbf{Y}]$. The required ascending union of algebraic varieties is then given by the collections of multiplicative maps $U_0\to V_n$.
It is obvious to check that the indvariety structure on $\text{Mor}(\mathbf{Y},\mathbf{X})$ has the expected functorial properties. Also the map $\text{Mor}(\mathbf{Y},\mathbf{X})\to \text{Mor}(\mathbf{Y}\times \mathbf{Y},\mathbf{X}\times \mathbf{X})$, $\phi\mapsto\phi\times \phi$ is a morphism. In case $\mathbf{Y}=\mathbf{H}$ and $\mathbf{X}=\mathbf{G}$ are algebraic group, the space of algebraic group morphisms $\text{Hom}(\mathbf{H},\mathbf{G})$ is the equalizer of the two maps $\text{Mor}(\mathbf{H},\mathbf{G})\to \text{Mor}(\mathbf{H}\times\mathbf{H},\mathbf{G})$ given by $$ \phi\mapsto \phi\circ m_{\mathbf{H}} \quad \text{and} \quad \phi\mapsto \phi\times\phi \mapsto m_{\mathbf{G}}\circ(\phi\times\phi)$$ where $m_{\mathbf{H}}$ and $m_{\mathbf{G}}$ are correspondingly the multiplication maps $\mathbf{H}\times\mathbf{H}\to \mathbf{H}$ and $\mathbf{G}\times\mathbf{G}\to \mathbf{G}$.
Corollary: For $k$algebraic groups $\mathbf{G}$ and $\mathbf{H}$, the the space of algebraic group homomorphisms $\text{Hom}(\mathbf{H},\mathbf{G})$ could be represented as an ascending union of a sequence of left and right $\mathbf{H}$ and left and right $\mathbf{G}$ invariant subsets which have compatible invariant structures of $k$varieties.
We go back to proving the AG Claim now. We fix an $\mathbb{R}$morphism between $\bar{\mathbb{Q}}$groups $\phi:\mathbf{H}\to\mathbf{G}$ with $\mathbf{H}$ reductive, and argue to show that up to a conjugation by an element of $\mathbf{G}(\mathbb{R})$, $\phi$ is defined over $\bar{\mathbb{Q}}$. Using the Corollary we put $\phi$ in a subset $\mathbf{R}\subset \text{Hom}(\mathbf{H},\mathbf{G})$ which is $\mathbf{G}$conjugation invariant and has the structure of a $\bar{\mathbb{Q}}$algebraic variety. We view $\phi$ as an element of $\mathbf{R}(\mathbb{R})$. We need to show the following.
AG Calim, equivalent formulation: $\mathbf{G}(\mathbb{R})\phi\cap\mathbf{R}(\bar{\mathbb{Q}})\neq \emptyset$.
This follows by showing that there exists a (real topological) neighborhood of $\phi$ in $\mathbf{R}(\mathbb{R})$ which consists of $\mathbf{G}(\mathbb{R})$conjugations of $\phi$, and that this neighborhood contains also a $\bar{\mathbb{Q}}$point. For the second statement, note the following.
Fact 3: For every $\bar{\mathbb{Q}}$variety, the set of $\bar{\mathbb{Q}}$points is dense in the set of $\mathbb{R}$point (with respect to the real topology).
An analogue of this is true for every local field of characteristic 0, but this MSE post by Eric Wofsey gives a proof for the reals ending with a cymbal sound.
For the first statement we use
Fact 4: Given a reductive $\mathbb{R}$group $\mathbf{H}$, for every $\mathbb{R}$representation $W$ of $\mathbf{H}$, $H^1(\mathbf{H}(\mathbb{R}),W(\mathbb{R}))=0$.
The argument in this MO post of Brian Conrad should convince you of that.
It follows in particular that $H^1(\mathbf{H}(\mathbb{R}),{\frak g})=0$, where ${\frak g}$ is the Lie algebra of $\mathbf{G}(\mathbb{R})$ viewed as a an $\mathbf{H}(\mathbb{R})$module via $\phi$. Informally, we think of $H^1(\mathbf{H}(\mathbb{R}),{\frak g})$ as the tangent spaces at $\phi$ to $\text{Hom}(H,G)$ modulo $G$conjugation. Formally, we use the same standard reasoning one uses when showing that the outer automorphism group of a semisimple Lie group is finite. This argument is written in a way that fits our needs here in Theorem 6.7 and Lemma 6.8 of Raghunathan's book.

$\begingroup$ I'd expect that there is a similar sketch, but possibly slightly simpler, based on (1) Every complex reductive group has countably many conjugacy classes of (complex) reductive subgroups. It is easy to deduce the result since the $\mathbf{R}$points of a transitive $G$set has finitely many $G(\mathbf{R})$orbits (BorelSerre). As there are countably many reductive groups, (1) follows from (2): for all $H,G$ complex reductive, $Hom(H,G)$ has countably many classes (up to $G$conjugacy). (2) seems to follow from your approach, but the whole avoids passing to $\bar{\mathbf{Q}}$points. $\endgroup$ – YCor Apr 11 '17 at 15:42

$\begingroup$ Thanks, Yves. Actually, I also thought of using BorelSerre and the analogous complex result. I would, if I have found a simpler proof going this way, but I have not. So, if anyway giving the localrigidity proof, why not providing a stronger result (one should be able to replace $\mathbb{R}$ with any local field)? $\endgroup$ – Uri Bader Apr 11 '17 at 16:09
The countability of the number of conjugacy classes of closed subgroups of any compact Lie group is, in fact, a simple corollary of ideas that predate the end of World War II. Unfortunately, the earliest quotable reference that I could find was given already in Igor Belegradek's comment, as follows.
Originally, Richard Palais observed this fact [1960, 1.7.27] and his proof uses the Peter–Weyl theorem [1927b] (harmonic analysis), as well as ChungTao Yang's theorem [1957] (differential geometry) on the local finiteness of orbit types within smooth $K$manifolds, when $K$ is a compact Lie group.
Alternatively, I show how to achieve the fact using those earlier methods internal to the topology of $K$.
0 Selection Theorem [1927a, Satz 28:VI]
Let $X$ be a metric space.
Equip the set $\mathrm{Cpt}(X)$ of nonempty compact subsets of $X$ with the Pompeiu–Hausdorff metric $d$.
If $X$ is compact then $\mathrm{Cpt}(X)$ is compact.
1 Closed Subgroup Theorem [1930, Section 27]
Any closed subgroup of a Lie group is a realanalytic submanifold and furthermore a Lie subgroup.
2 Neighboring Subgroups Theorem [1942]
Any compact subgroup $H$ of an arbitrary Lie group $G$ admits a neighborhood $O$ in $G$ such that any subgroup contained in $O$ is $G$conjugate to a subgroup of $H$.
Equivalently, there exists $\varepsilon>0$ such that any compact subgroup $H'$ of $G$ is conjugate to a subgroup of $H$ if $d(H',H)<\varepsilon$.
3 Corollary [1960, Corollary 1.7.27]
Any compact Lie group $K$ has only countably many conjugacy classes of closed subgroups.
3 Proof [2019, Proof 3.8]
Write $\mathrm{CptSgp}(K) \subset \mathrm{Cpt}(K)$ for the set of closed (hence compact) subgroups of $K$.
By continuity of the group operations of $K$ and the definition of the Pompeiu–Hausdorff metric, it follows that any sequence in $\mathrm{CptSgp}(K)$ that converges in $\mathrm{Cpt}(K)$ has its limit point already in $\mathrm{CptSgp}(K)$.
Since $\mathrm{Cpt}(K)$ is compact (0), we have $\mathrm{CptSgp}(K)$ is also compact.
Observe that any proper closed subgroup of $K$, as a point of the compact metric space $\mathrm{CptSgp}(K)$, lies in the complement of the open ball $B(K,1/n)$ for some integer $n > 0$.
Since $X_n := \mathrm{CptSgp}(K)  B(K,1/n)$ is compact, there exists a finite cover $\mathcal{U}_n = \{ B(H_{n,i}, \varepsilon_{n,i}) ~~ 1 \leqslant i \leqslant r_n \}$ for some integer $r_n>0$ such that each $\varepsilon_{n,i}>0$ occurs from (2) for $H_{n,i} \in X_n = \bigcup \mathcal{U}_n$.
Since each proper closed subgroup $H_{n,i}$ of $K$ is Lie (1), inductively each $\mathrm{CptSgp}(H_{n,i})$ is partitioned into countably many $H_{n,i}$conjugacy classes. Indeed, for the double induction on dimension and number of components of a compact Lie group, the basic case is the trivial group.
Thus $X_n$ has only countably many $K$conjugacy classes of closed subgroups. Therefore $\mathrm{CptSgp}(K) = \{K\} \cup \bigcup_{n=1}^\infty X_n$ consists of only countably many $K$conjugacy classes of closed subgroups. $\square$
Furthermore, there is a more general statement that applies to noncompact Lie groups $G$, which must have only countably many connected components.
4 Theorem [2019, Corollary 3.9]
Any Lie group $G$ has only countably many conjugacy classes of compact subgroups.
This would be immediate if we already knew whether the ancillary space $\mathrm{Cpt}(G)$ is Lindelöf, so as instead to use a countable union in the above proof. However, I don't know this (equivalently, if the metric space is separable).
Instead, I shall use the following tool, which is a more elaborate corollary of (2) in terms of Moore–Smith nets and the Axiom of Choice, since secondcountability of $\mathrm{Cpt}(G)$ is not assumed.
5 Proposition [2019, Proposition 3.6]
Let $C$ be a compact set in an arbitrary Lie group $G$.
Any compact subgroup of $G$ contained in $C$ is contained in a conjugacymaximal compact subgroup within $C$.
Furthermore, there are only finitely many conjugacymaximal compact subgroups within $C$.
4 Proof
The hemicompact space $G$ is the countable ascending union of compact sets in such a way that any compact set is contained in a member of that exhaustion. The rest follows from (5), (1), and (3). $\square$
References and Historical Remarks
[1916]
W Blaschke, Kreis und Kugel, Veit
[1927a]
F Hausdorff, Mengenlehre, Walter de Gruyter
Satz 28:VI formally generalizes to metric spaces from the compact convex sets of [1916, §18].
[1927b]
F Peter and H Weyl, Die Vollständigkeit der primitiven Darstellungen einer geschlossenen kontinuierlichen Gruppe, Math Ann 97:737–755
[1930]
E Cartan, La théorie des groupes finis et continus et l'Analysis situs, Mémorial des Sciences Mathématiques 42
[1942]
D Montgomery and L Zippin, A theorem on Lie groups, Bull Amer Math Soc 48:448–452
For a proof that is purely topological and avoids differential geometry, see [1961, 4.2].
[1957]
CT Yang, On a problem of Montgomery, Proc Amer Math Soc 8:255–257
[1960]
R Palais, The Classification of $G$spaces, Mem Amer Math Soc 36
[1961]
R Palais, On the existence of slices for actions of noncompact Lie groups, Ann Math(2) 73:295–323
[2019]
Q Khan, Countable approximation of topological $G$manifolds, III: arbitrary Lie groups $G$, preprint submitted for publication.

$\begingroup$ The statement "There exists $n$ and proper closed subgroups $H_1,\dots,H_n$ of $G$ such that $CptSgp(K)\subset U_{H_1}\cup\dots\cup U_{H_n}$ is clearly false (since $K$ is missing), but even if you mean $CptSgp(K)\subset U_{H_1}\cup\dots\cup U_{H_n}\cup\{K\}$, it's false when $K$ is the circle group. (I'm also confused you define $U(H)$ and then denote $U_H$.) $\endgroup$ – YCor Nov 19 '19 at 10:19

$\begingroup$ Probably you can prove with such an argument that there exists a sequence of closed proper subgroups $(U_n)$ such that every proper subgroup is conjugate into one of the $U_n$, and then the argument works as soon as you know that there is no properly decreasing sequence of closed subgroups. $\endgroup$ – YCor Nov 19 '19 at 10:21

$\begingroup$ @YCor : I've added an edit to address your two comments. I hope that my amended answer is satisfying from a purely classical point of view. $\endgroup$ – Qayum Khan Nov 20 '19 at 5:58