Let's recall that, a group $G$ being given, 
two subgroups $A,B\subset G$ are called 
permutable iff $AB=BA$ for the Minkowski 
law. It is straightforward to see that $(A,B)$ 
are permutable iff $AB$ is a subgroup of $G$. 

> 
> Let now $G$ be a finite dimensional Lie group 
>  (real to begin with) and suppose that if 
>  $A,B\subset G$ are Lie subgroups (then closed 
>  by Cartan's theorem), one can 
>  easily show that, providing $G=AB$ and if it 
>  is a **factorization** (means that the decomposition 
>  is unique or, equivalently, 
>  $A\cap B=\{1\_G\}$), then $Lie(A)\cap Lie(B)=\{0\}$
> 
> 
> 

My questions are the following 

> 
> * Q1) If $G=AB$ is a factorization ($A\cap B=\{1\_G\}$), 
>  do we have 
>  $Lie(A)\oplus Lie(B)=Lie(G)$ ?
>  * Q2) Does the result hold if we just have 
>  $G=AB$ (and still $Lie(A)\cap Lie(B)=\{0\}$) 
>  without supposing that $A\cap B=\{1\_G\}$ ?
>  * Q3) What are the references about these 
>  questions ?
> 
> 
> 

What was (and is still) not clear to me is why the **tangent map** of the multiplication 
$$
A\times B\rightarrow G
$$
must be surjective. More precisely, the fact that it is a factorization provides a section $$s: G\rightarrow A\times B$$ (in fact an inverse bijection) but I am stuck in proving that it is C^1 (continuous would do, I think). Maybe the action 
of $A\times B$ on itself by $(a,b)\*(x,y)=(ax,yb^{-1})$ could be exploited but I do not see how.