Transitive action of a lie group in a connected manifold I am trying to see why is it that if we have a transitive action of a lie group $G\rightarrow M$ then if this action is transitive the connected component of the identity $G^0$ also acts transitively on $M$ and that for all $p\in M$, $G/G_0 \cong G_p/(G_p\cap G^0)$.
Since the action is transitive we know that $G/G_p\cong M$ and since the map $G\rightarrow G_p$ is a submersion we get that the map $G\rightarrow M, g\rightarrow g.p$ is open . I don't know how useful this is since we only know that $G^0$ is a closed set I don't think it has to be open since connected components don't necessarily need to be open, and also I am not sure if this map is even closed or if it is how I could try and prove it.
Any enlightment is appreciated. Thanks in advance.
 A: Manifolds (and thus also Lie groups) inherit the property of being locally connected from Euclidean space, and connected components of locally connected spaces are both closed and open. Hence $G_0$ is open in every Lie group.
Further, if we know that for every $q \in M$ the map $\phi_q : G \to M, g \mapsto g \cdot q$ is open, then the orbit $\phi_q(G_0) \subset M$ must be open.  Then, fixing a single $p \in M$, the union $\bigcup_{q \in M \setminus \phi_p(G_0) } \phi_q(G_0)$ is open as a union of open sets. But note that $$M = \bigcup_{q \in M} \phi_q(G_0)=  \phi_p(G_0) \cup \bigcup_{q \in M \setminus \phi_p(G_0) } \phi_q(G_0),$$
because every $m \in M$ is equal to $e \cdot m \in \phi_m(G_0)$ if $e \in G_0$ is the identity.
But then we have written the connected set $M$ as a disjoint union of open sets $\phi_p(G_0) \cup \bigcup_{q \in M \setminus \phi_p(G_0) } \phi_q(G_0)$, and since the first set is nonempty, this must mean that the second one is empty, and hence $\phi_p(G_0) = M$.
(For academic integrity: Pieced this answer together from the comments to the very similar question here.)
