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

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).