If $H$ is a closed subgroup of Lie group $G$, then show that $\mathfrak{h}=0$ if and only if $H$ is discrete, where $\mathfrak{h}$ is the Lie algebra of $H$.
We know that $\mathfrak{h}=\{X\in \mathfrak{g}: \mathrm{exp}(tX)\in H\ \forall t\in \mathbb{R}\}$. By closed subgroup theorem, we know that $H$ is an embedded Lie subgroup of $G$. Now, how to use the fact that $H$ is discrete?
Thanks!
Suppose that $\dim\mathfrak h>0$. There is then some $X\in\mathfrak h\setminus\{0\}$. For such a $X$, $(\forall t\in\mathbb R):\exp(tX)\in H$. So, consider the map$$\begin{array}{rccc}\gamma\colon&\mathbb R&\longrightarrow&H\\&t&\mapsto&\exp(tX).\end{array}$$Then $\gamma'(0)=X\neq0$ and so $\gamma$ is not constant near $0$. But then every neighbourhood of $e$ contains elements of the form $\gamma(t)\in H$ which are distinct from $e$. This is impossible, since $H$ is discrete.