TITLE: On closed totally disconnected subgroups of connected real Lie groups 
QUESTION [5 upvotes]: So the following statement seems to be obvious but I don't see how to prove it:
Q: How does one prove that a closed totally disconnected subgroup of a connected real Lie group is discrete?
Note that it is essential to take into account the group structure since the Cantor set is a closed totally disconnected subset of $\mathbf{R}$ which is not discrete.

REPLY [2 votes]: To complement what has been said earlier: a Lie group $G$ has no small subgroup, i.e. there exists a neighborhood $U$ of the identity $e\in G$ such that the only subgroup of $G$ contained in $U$, is $\{e\}$. [To see it, let $B$ be a small ball around $0$ in the Lie algebra of $G$, such that the exponential map $\exp$ induces a diffeomorphism between $B$ and its image. Set $U=\exp(B/2)$. Then for every $x\in U\backslash\{e\}$, there exists $p>1$ such that $x^p\notin U$.]
Let then $H$ be a closed subgroup of $G$. If $H$ is totally disconnected, then there is a basis of neighborhoods of $e$ in $H$, consisting of open subgroups of $H$. So $H\cap U=\{e\}$, meaning that $H$ is discrete.