A compact Lie group has descending chain condition on closed subgroups.

Question

Proposition: Let $G$ be a compact Lie group and let $$G\supset G_1\supset G_2\supset\ldots$$ be a chain of closed subgroups $G_i$ of $G$. Then this chain must eventually stabilize.

Question: The hint for the proof is that $G_i$ are closed manifolds and if $G_{i+1}\neq G_i$, either $\dim G_{i+1}<\dim G_i$ or $G_{i+1}$ has fewer connected components than $G_i$. I understand first part since closed subgroup of compact Lie group is subgroup that is also (embedded) submanifold, but I don't understand second part of the hint, ie. if $G_{i+1}\neq G_i$, either $\dim G_{i+1}<\dim G_i$ or $G_{i+1}$ has fewer connected components than $G_i$. Can you please help me with that? And I believe the rest of the proof is (please correct me if i'm wrong or if I missed something): Since dimensions are finite they must stabilize, ie. there exist $n_0$ such that $\dim G_{n_0}=\dim G_{i}=\dim G_{i+1}=\ldots=constant$ But then (from the hint) the number of components of $G_{i+1}$ $<$ number of components of $G_i$ $<$ number of components of $G_{n_0}$. Since $G_{n_0}$ is closed in $G$ it is compact and therefore has a finite number of connected components (since compact and locally-connected topological space has at most finitely many connected components). So it follows that eventually these number must also stabilize and proposition is true. Thank you!

Answer 1

Since your argument is perfectly fine, I'll only explain the point you ask about. In a nutshell, this follows from the fact that a Lie subgroup with $\dim{G_{i+1}} = \dim{G_{i}}$ is an open subgroup.

Good examples to have in mind are the inclusion of $SO(n)$ in $O(n)$ or subgroups of $SO(n) \times M$ of the form $SO(n) \times N$, where $N$ runs through the subgroups of a finite group $M$.

A few generalities: Notice that the connected component $G^0$ of a locally connected topological group $G$ is an open subgroup (take a connected neighborhood $U$ of the identity and look at the subgroup $H = \bigcup_{n \in \mathbb{Z}} U^n$ generated by it: it is connected, open and thus it is also closed, hence $G^0 = H$ by connectedness, see here for more details). Moreover, $G^0$ is normal since $gG^0 g^{-1}$ is a connected subgroup containing the identity. The quotient group $\pi_0 G = G/G^0$ of $G^0$-cosets consists of the connected components of $G$ and it has the discrete topology since the quotient map $G \to G/G^0$ is open and since $G^0$ is open.

Coming back to your question, the dimension requirement $\dim{G_{i+1}} = \dim{G_i}$ together with the fact that the Lie algebra $\mathfrak{h}$ of a closed subgroup $H$ embeds as a Lie subalgebra of $\mathfrak{g}= \operatorname{Lie}{G}$, imply that $\mathfrak{g}_{i+1} = \mathfrak{g}_i$. The exponential map is a local diffeomorphism from $\mathfrak{g}_{i+1} = \mathfrak{g}_i$ to the connected component $G_{i+1}^0$ of $G_{i+1}$ and thus $G_{i+1}^0$ is an open subgroup of $G_{i}^0$, so $G_{i+1}^0 = G_{i}^0$. It follows that $\pi_{0}G_{i+1} = G_{i+1}/G_{i+1}^0$ is a subgroup of $\pi_{0}G_i = G_{i}/G_{i}^0$ and both groups are finite since they are compact and discrete. But as I explained in he second paragraph, $\# \pi_0 G_{i+1} = \# (G_{i+1}/G_{i+1}^0)$ is the number of connected components of $G_{i+1}$ and thus it divides the number of connected components of $G_{i}$.