Extracting connectedness of a topological group G from its action on a connected space X

Question

Let $G$ be a topologival group that acts on a topological space $X$. Assume the following conditions:

$(1)$ There is $x_0 \in X$ such that the orbit map $ G \longrightarrow X$ sending $g$ in $G$ to $gx_0$ in $X$ is a quotient map.

$(2)$ The action of $G$ on $X$ is transitive

$(3)$ The stabilizer $H$ in $G$ of $x_0$ is connected

$(4)$ $X$ is connected

Then $G$ is connected.

Here is how I tried to show it:

If $G$ were not connected then there would be a non-constant continuous map $f: G \longrightarrow D$ from $G$ to a discrete topological space $D$. This map would naturally induce a continuous map $G/H \longrightarrow D$ ($G/H$ endowed with the quotient topology) which would be constant on each coset of $G$ by $H$ since $H$ is connected by hypothesis.

Also, because of $(2)$ we can form a well defined map $ \gamma: X \longrightarrow D$ taking $gx_0$ to $f(g)$. Furthermore, since we have $(1)$, it follows from the universal property of the quotient maps that such map is continuous. Hence, by $(4)$ $\gamma$ must be constant and thus so must be $f$. This implies $G$ must be equal to $H$ and therefore connected.

Is this all right? Is my last assertion that $G=H$ correct? I think that way because since $f$ is constant and it is consant on each coset by $H$ it must follow that $G=H$. And why does $H$ have to be the stabilizer of some point of $X$? Wouldn't it suffice to be a connected subgroup of $H$ containing the identity?

Answer 1

Is this all right? Hence, by $(4)$ $\gamma$ must be constant and thus so must be $f$.

You arguments are right up to this sentence and you can finish your proof here, because you obtained a contradiction with the non-constantness of the map $f$

This implies $G$ must be equal to $H$ and therefore connected.

In fact, a contradiction implies any claim, :-) but in general even a connected group need not be equal to a (connected) stabilizer of a point. For instance, let $G=\Bbb R^2$ and $X=\Bbb R$ endowed with the standard topology and $G$ acts on $X$ as $(x,y)z=x+z$ for each $x,y,z\in\Bbb R$. For any point $x_0\in X$ its stabilizer $H$ equals $\{0\}\times \Bbb R\ne G$.

I think that way because since $f$ is constant and it is consant on each coset by $H$ it must follow that $G=H$.

If the map $f$ is constant then it is obviously constant on each coset by $H$, so this proves nothing. The reasoning in the opposite way is also trivial: if $G=H$ and the map $f$ is constant on each coset by $H$ then the map $f$ is constant. Also, in general, if the map $f$ is constant on each coset by $H$ then it is not necessarily constant on $G$. For instance, in the above example you may put $f(x,y)=x$ for each $(x,y)\in G$.

And why does $H$ have to be the stabilizer of some point of $X$? Wouldn't it suffice to be a connected subgroup of $H$ containing the identity?

I don’t quite understand this question. You started to deal with $H$, defining it as the stabilizer. If we redefine $H$ as an arbitrary connected subgroup of the stabilizer, containing the identity, we may gain nothing (for instance, when $H=\{e\}$) and probably we shall not be able to prove that $G=H$. Also if the stabilizer of the point $x_0$ is not connected then the map $\gamma$ may be not well-defined anymore.