Action of a quotient of a compact Lie Group on M Let $M$ be a smooth manifold and $G$ a compact Lie Group acting transitively on $M$. Let $H$ be the subgroup of $G$ defined by $\{ g \in G | g \cdot m = m, \forall m \in M \} = \operatorname{Stab}_g(M)$, i.e $H$ is the "global-stabilizer" subgroup of $G$. 
Firstly, is $H$ a normal subgroup ( of $G$ ? I think so because $H =  \bigcap_{m \in M} \operatorname{Stab}_g(m)$. 
If so, then does the quotient  $M/ (G/H)$ have the structure of a smooth manifold?  Again, I imagine this would be the case since $G/H$ would act freely on $M$. 
Is something amiss in my reasoning ? 
 A: There seems to be some confusion here.
First, yes, $H$ is a normal subgroup of $G$.  But why does it follow from writing $H$ as an intersection of Stabilizer subgroups?  Here's an altnerate way of thinking.  By definition, a smooth group action is a homomorphism $G\rightarrow Diff(M)$.  This homomorphism has a kernel and kernels are always normal.  Prove that this kernel is nothing but $H$.
Second, yes $M/(G/H)$ has the structure of a smooth manifold, but it's not very interesting.  The point is that the action by $G/H$ has the same orbits as the action by $G$.  You've assumed your action is transitive, so $M/(G/H)$ consists of a single point - the unique connected 0-dimensional manifold.
Third, there is no reason why the action of $G/H$ on $M$ must be free.  Here's a really simple example to keep in mind.  Let $M = \{x,y,z\}$ consist of two points.  Let $G = S_3\times \mathbb{Z}_2$ act on $M$ with the symmetric group $S_3$ acting in the usual way and the $Z_2$ factor doing nothing.
Then $H = \{e\}\times \mathbb{Z}_2$, but the action by $G/H \cong S_3$ is not free - the transpositions all fix points.
If you'd like a less dumb example, consider the action of $G = Sp(1) = \{p\in \mathbb{H}: |p| = 1\}$ on $S^2 = \{v=ai+bj+ck\in \operatorname{\mathbb{H}}| a^2 + b^2 + c^2 = 1\}$ given by $p\ast v = pvp^{-1} = pv\overline{p}$.
This action is transitive, $H = \{\pm 1\}$, but the action by $G/H$ on $S^2$ is not free.  In fact, every element of $G/H$ fixes some points.  For if $[p]\in G/H$ is not the identity, then $\operatorname{Im}(p)\neq 0$.  Let $v = \frac{\operatorname{Im(p)}}{|\operatorname{Im(p)}|}\in S^2$.  Then $p\ast v = pvp^{-1} = v$.
