assume $G \times X \to X$ is a continuous group action of a compact Hausdorff topological group $G$ such that the action is transitive, i.e. $X$ is a homogeneous space. Let $G_x$ be the stabilizer subgroup of $x \in X$ and assume $G_x$ is closed. Is it then true that $X$ is a Hausdorff space?
My ideas:
We can define the map $\varphi: G/H \to X$, $[g] \mapsto gx$, which is a bijective continuous function. Then it is well-known that $G/H$ is Hausdorff since $H$ is closed. However, I'm not sure if we can show that $\varphi$ is a homeomorphism. To deduce this directly would already require the Hausdorff property for $X$.
Best Answer
Let $G=S_{3}$, the symmetric group over $3$ elements, equipped with the discrete topology (so it's a compact Hausdorff group). There is a natural action $\varphi\colon G\curvearrowright X=\{1,2,3\}$ given by $\sigma\cdot x=\sigma(x)$. The action is transitive, and for $x=1$, $G_{x}=\{e,(23)\}$ is closed. Nevertheless, if we give $X$ the trivial topology, $X$ cannot be Hausdorff (and the action is still continuous because $X$ is trivial).