Suppose we have a simply-connected Lie group $G$. Let $G\_1$ and $G\_2$ be two closed and connected subgroups of $G$. Is it true that the commutator $[G\_1,G\_2]$ is a closed subgroup of $G$?



Commutator of closed subgroups
No. Let's make an example in which both $G\_1$ and $G\_2$ are one-dimensional. 


Start by choosing $x$ and $y$ in $\frak{sl}\_2(\mathbb R)$ such that the subgroup determined by $[x,y]$ is a circle (the square of this matrix has negative trace) but the subgroups generated by $x$ and by $y$ are isomorphic to $\mathbb R$ (their squares have positive trace).


Now in $\frak{sl}\_2(\mathbb R)\times \frak{sl}\_2(\mathbb R)$ consider the elements $(x,x)$ and $(y,cy)$, where $c$ is some irrational number. These give closed noncompact one-dimensional subgroups $G\_1$ and $G\_2$, but their commutator gives a dense line in a torus.


$SL\_2(\mathbb R)\times SL\_2(\mathbb R)$ is not simply connected, so embed it in $SL\_4(\mathbb R)$ and take that, or rather its double cover, to be $G$.


