Commutator of closed subgroups 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 $G$?
 A: 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 $G$.
A: Maybe I can frame the question further, though I'm not a Lie group specialist.  From the viewpoint of topological groups, just requiring one of the two subgroups to be connected will force the commutator group here to be connected.   But closure is a more delicate issue.   Requiring a simply connected Lie group $G$  at least avoids a standard type of counterexample: the quotient of a simply connected nilpotent Lie group by a discrete subgroup can yield a connected Lie group $G$ for which $[G,G]$ itself fails to be closed.  
For arbitrary linear algebraic groups (initially over an algebraically closed field) the situation is more elementary: here the connectedness of one of the two closed subgroups is enough to make the commutator group both closed and connected.   (Note that one is not dealing with topological groups because the topology is not Hausdorff.   But in the analytic topology, the groups here are actually complex Lie groups if one works over $\mathbb{C}$.)
There are some (but not many) books that treat the structure of Lie groups in general, including older books like The Structure of Lie Groups by Hochschild and Chapter 3 of Bourbaki's treatiste Groupes et algebres de Lie.  Though such books always have some discussion of commutation, I'm unaware of any precise results on simply connected groups that would answer the question here one way or the other.  But I've never delved into the extensive "exercises" for Section 9 of Bourbaki.
Most (but not all) real semisimple Lie groups do turn out eventually to be algebraic groups, where it's more likely that the answer to the question is yes after adapting from $\mathbb{C}$ to $\mathbb{R}$.     
