Let $G$ be a connected Lie group with Lie algebra $\mathfrak{g}$ of dimension $d$. Choose a basis $B$ for $\mathfrak{g}$. From the $d$ elements of $B$, we obtain $d$ one-parameter subgroups of $G$. Let $H$ be the smallest subgroup of $G$ that contains all of these $d$ one-parameter subgroup.
Is the closure of $H$ equal to $G$ necessarily? Is it true under some conditions such as "$G$ is compact/semisimple"?
Best Answer
Yes, $\overline H=G$. That is so because, by Cartan's theorem, $\overline H$ is a Lie subgroup of $G$ and, by your assumptions, if $\mathfrak g$ and $\mathfrak h$ are the Lie algebras of $G$ and $H$ respectively, then $\mathfrak g=\mathfrak h$. So, the connected component of $e$ in $G$ is $H$. But, since we are assuming that $G$ is connected, what this means is that $G$ is $H$.