
> 
> Let $G$ be an algebraic group over an
>  algebraically closed field. If $H$ and
>  $K$ are closed subgroups and one of
>  them is connected, then their
>  commutator $[H,K]$ is also connected.
> 
> 
> 


Is there an easy way to see this fact?


The proof that I see in Springer's book on Linear Algebraic Groups is very long-winded. (Of course, he obtains many results on the way). My question is, is the given statement true for any topological group? If yes, does that proof apply to the algebraic groups' case?


(Remark: The definition of a topological group includes the Hausdorff property, whereas algebraic groups are not so). 



Commutator of algebraic subgroups is connected
The case for $G$ a topological group doesn't look hard; it just comes down to two facts: 


* A union of connected sets which have a point in common is also connected. For example, suppose $K$ is a connected subgroup and $H$ is any set. For each fixed $h \in H$, the set of commutators $[h, K]$ is the image of $K$ under the continuous map $G \to G$ that takes $g$ to $[h, g]$; this image is connected and contains the identity (since $g = e$ belongs to $K$). But then the set of all commutators $[h, k]$ is a union of connected sets $[h, K]$ each containing the identity, so it too is connected.
* If $S \subset G$ is a connected set containing the identity, then so is the group it generates. This is on the same principles as before, that the image of a connected set under a continuous map is connected, and the union of connected sets with an element in common is connected. For example, the set $T = S \cup S^{-1}$ is connected, and then each $T \cdot T \cdot \ldots \cdot T$ is connected, being the image of the connected set $T \times T \times \ldots \times T$ under the multiplication map $G^n \to G$, and finally the union of all these sets gives the group generated by $S$.


Now apply the second fact to the connected set of commutators coming from the first fact, to show that the subgroup $[H, K]$ is connected. 


