$G$ acts transitively on connected space, then so does identity component

Question

Suppose $G$ is a topological group that acts on a connected topological space $X$. Show that if this action is transitive (and continuous), then so is the action of the identity component of the group.

Answer 1

I was also working on this exercise and think I have a solution. I can't figure out what's wrong with Jason De Vito's counter example, but definitely if we assume $X$ to be Hausdorff, these examples are excluded. Here are details that fill in Poovu's answer.

Suppose $G$ acts transitively on $X$ by homeomorphisms and $G_x$ is the stabilizer of any point $x \in X$. Define a map $$ \alpha_x:G \longrightarrow X \quad \quad \alpha_x(g) = gx$$ Then $\alpha_x$ is continuous because the action is, and surjective because the action is transitive. If $gG_x = g'G_x$, then $g = g'h$ for some $h \in G_x$, and thus $g(x) = g'h(x) = g'(x)$. This means $\alpha_x$ is constant on cosets of $G_x$ and by the universal property of qoutients, we get a continuous map $\bar{\alpha}_x:G/G_x \longrightarrow X$. That $\bar{\alpha}_x$ is also surjective follows from $\alpha_x$ being surjective, and by modding out cosets of $G_x$ we have made $\bar{\alpha}_x$ injective. Thus $\bar{\alpha}_x$ is a homeomorphism.

Let $\pi:G \longrightarrow G/G_x$ be the projection. It's clear that $\pi$ is open, and $G^0$ is clopen because it's a connected component, so $$\bar{\alpha}_x\pi(G^0) = \alpha_x(G^0) = G^0x $$

is open. What we really have shown is that the orbit of any point under an open subgroup of $G$ is open. As for any action, we can write $X$ as the disjoint union of $G^0$ orbits, and each of these is open. This means the complement of an orbit is a union of open sets (all the other orbits) which means each orbit is also closed. Thus $G^0x$ is clopen. If $X$ is connected this forces $X = G^0x$ and $G^0$ acts transivitely.

I think we might be able to push further and show these orbits are exactly the connected components of $X$ in general.