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?

  • 1
    $\begingroup$ For a start: if $H$ is closed, then every coset of $H$ is closed, so the complement of $H$ is closed, so $H$ is open and $G$ is disconnected. $\endgroup$ Jul 29, 2013 at 0:39

2 Answers 2


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.

  • $\begingroup$ Hm, wish I knew you were typing such a similar (and more succinct!) answer while I was. +1 from me. I'll delete my momentarily. $\endgroup$ Jul 29, 2013 at 1:39
  • $\begingroup$ @Jason: I think you should keep your answer around. The details are helpful and I omitted a lot of them. $\endgroup$ Jul 29, 2013 at 1:44
  • 1
    $\begingroup$ Fair enough. I suppose details never hurt anyone $\endgroup$ Jul 29, 2013 at 1:54

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.


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., $H = G$.

  • 1
    $\begingroup$ Looking back on this 6 years later, in the proof of Lemma 1, I mean there is an open set $U\subseteq G$ for which $U\subseteq \exp(\mathfrak{g})$. That edit doesn't seem to be worth bumping this question to the top... $\endgroup$ Oct 12, 2019 at 22:04
  • $\begingroup$ I think this answer is very much detailed, but actually I cannot see why $N$ is finite index and normal.... $\endgroup$
    – Kei
    Nov 12, 2020 at 0:07
  • $\begingroup$ I've got some information from here. $\endgroup$
    – Kei
    Nov 12, 2020 at 0:33
  • 1
    $\begingroup$ @Kei: The intersection is finite index by math.stackexchange.com/questions/128538/… and induction. Normality follows from math.stackexchange.com/questions/369629/… together with the fact that for any $g\in G$, $gH = g_iH$ for some $i$. $\endgroup$ Nov 12, 2020 at 0:35
  • $\begingroup$ Thank you so much! Finite index subgroup intersection lemma you have introduced to me is more general and useful I think. $\endgroup$
    – Kei
    Nov 12, 2020 at 0:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.