When are finite-index subgroups of a Lie group closed?

Question

Let $G$ be a Lie group (or, if necessary, a reductive Lie group) and $H$ a subgroup of $G$. If $\lbrack G:H\rbrack < \infty$, is it true that $H$ is closed? If not, are there any broad assumptions on $G$ which make this true?

Answer 1

Yes, this is true. It suffices to show that any homomorphism $\pi : G \to F$ from a Lie group $G$ to a finite group $F$ (with the discrete topology) is continuous, or equivalently has the property that its kernel contains the connected component $G_0$ of the identity in $G$. To prove this it suffices to show that some neighborhood $U$ of the identity in $G$ is contained in the kernel. Pick a neighborhood $U$ which is contained in the image of the exponential map $\exp : \mathfrak{g} \to G$. Then for any $g \in U$ there exists some one-parameter subgroup $\varphi : \mathbb{R} \to G_0$ such that $\varphi(1) = g$. I claim that the composition $\pi \circ \varphi : \mathbb{R} \to F$ is constant, from which the conclusion follows.

But this is clear. The image of any such homomorphism is necessarily a divisible group, and the trivial group is the only divisible finite group.

Answer 2

If $H\subseteq G$ is finite index, then $H$ is made out of components of $G$. In particular, $H$ is closed (and open) in $G$.

(This proof is heavily borrowed from The Lie theory of connected Pro-Lie groups, by Hoffman and Morris, though I honestly couldn't tell you what a Pro-Lie group is!)

Lemma 0: We may assume $G$ is connected.

Proof: If $G^0$ denotes the identity component of $G$, then $H\cap G^0 \subseteq G^0$ is finite index. This follows because the inclusion map $i:G^0\rightarrow G$ induces an injective map $i:G^0/(H\cap G^0)\rightarrow G/H$. The map $i$ is injective, because if $i([g_1]) = i([g_2])$, then $g_1 = g_2 h$ for some $h\in H$. Then $g_2^{-1} g_1 = h\in H$. Since $G^0$ is a subgroup of $G$, this implies $h\in G^0$, so $g_1 \cong g_2 \in G^0/(H\cap G^0).$ Since $i$ is injective, we have $|G^0/(H\cap G^0)|\leq |G/H| = [G:H]$.

Lemma 1: $G$ is generated by its divisible subgroups.

Proof: There is an open set $U\subseteq G$ for which the exponential map $\exp:\mathfrak{g}\rightarrow G$ is onto. Since every connected Lie group is generated by an open set around $e$, it's enough to show that every element in $U$ lies in a divisible subgroup of $G$.

But, if $u = \exp(X)\in U$, then $u$ lies in the divisible subgroup $\{exp(tX):t\in \mathbb{R}\}$.

Lemma 2: We may assume wlog that $H$ is normal.

Proof:

Suppose $G = g_1 H \cup ... \cup g_n H$. Then $N = \bigcap_{i=1}^n g_i H g_i^{-1}$ is finite index and normal.

Lemma 3: $H = G$.

Proof: Assuming $H$ is normal, we have a group homomorphism $\pi:G\rightarrow G/H$ onto a finite group. But since $G$ is generated by its divisible subgroups, so is $\pi(G) = G/H$. But since $G/H$ is finite, it has no nontrivial divisible groups. Thus, $G/H$ is trivial, i.e., $H = G$.