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. 



Permutable (Lie) subgroups
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!)


