Permutable (Lie) subgroups 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. 
 A: As desired in the question, the action of $A \times B$ on $G$ is the key point.  This is a transitive action, and it is a general fact in the theory of actions of (separable) Lie groups on manifolds (see Bourbaki, Lie Groups & Lie Algebras, Ch. III, no. 1.7, Proposition 14) that if a Lie group $H$ with at most countably many connected components (equivalently, a countable base for its topology) acts transitively on a smooth manifold $X$ then for any $x_0 \in X$ with stabilizer $H_{x_0} \subset H$ the natural orbit map $H/H_{x_0} \rightarrow X$ is a diffeomorphism. (We need a countability hypothesis on $H$ to rule out the case where $X$ is a positive-dimensional connected Lie group and $H$ is the underlying discrete group acting by left-translation.)  In particular, if the action is simply transitive then the orbit map $H \rightarrow X$ is an $C^{\infty}$-isomorphism.
Now consider a Lie group $G$ with only countably many connected components and $A, B$ closed Lie subgroups of $G$ (so they each have only countably many connected components)  such that the multiplication map $A \times B \rightarrow G$ is surjective. For the Lie subgroup $H := A \cap B$ setting $X=G$ with the indicated transitive action, we get that the natural map $A \times B \rightarrow G$ defined by $(a,b) \mapsto ab^{-1}$ is a $C^{\infty}$-submersion (in fact, a quotient map modulo $A \cap B$), and a $C^{\infty}$-isomorphism when $A\cap B = 1$. In particular, in all such cases ${\rm{Lie}}(A) + {\rm{Lie}}(B) = {\rm{Lie}}(G)$, and the sum is direct if and only if $A \cap B$ is discrete.  (The generalization to Lie subgroups that might not have the subspace topology is left to the interested reader!)
A: Let $i_A : A \to G$ and $i_B : B \to G$ be the embeddings and let $\mu :  G \times G \to G$ be the group multiplication.  Then the composition
$$
A \times B \stackrel{i_A\times i_B}{\longrightarrow} G \times G \stackrel{\mu}{\longrightarrow} G
$$
is a smooth map which is surjective and sends the identity to the identity.  (It is not, however, a group homomorphism.)
In a previous, incorrect, version of this answer, I had claimed that the tangent map at the identity was surjective.  This is of course false.  This was used only to arrive at an inequality:
$$
\dim\mathrm{Lie}(G) \leq \dim\mathrm{Lie}(A) + \dim\mathrm{Lie}(B)~.
$$
In fact, this inequality follows from Sard's theorem and the fact that the map $A \times B \to G$ is surjective, for if $\dim G > \dim A + \dim B$ then the image of the map $A \times B \to G$ would have measure zero.
Having established the above inequality, we now establish the reverse inequality.  To see this simply notice that if $\mathrm{Lie}(A) \cap \mathrm{Lie}(B) = 0$, then the fact that $\mathrm{Lie}(A)$ and $\mathrm{Lie}(B)$ are subspaces of $\mathrm{Lie}(G)$ implies
$$
\dim\mathrm{Lie}(A) + \dim\mathrm{Lie}(B) \leq \dim\mathrm{Lie}(G)~.
$$
So the answers to Q1 and Q2 are both true (as vector spaces not as Lie algebras).
