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

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.

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What have you tried? –  Jason DeVito May 3 '12 at 13:59
Well $(g,x)\mapsto g x$ is an open map simply because for fixed $g$ the map $x\mapsto gx$ is open. But what i really need is, that for fixed $x$ the map $g\mapsto g x$ is open. This would suffice to prove the upper statement, but I think it is not true. –  fk44 May 3 '12 at 15:29
Are you sure there aren't any other hypothesis on $X$? I think I have a stupid counter example. Take $H$ to be your favorite connected topological group and let $G = H\times \mathbb{Z}/2\mathbb{Z}$. So the identity component of $G$ is $H$. Let $X=G$ as a set and a group, but with the indiscrete topology. Then $X$ is connected, $G$ acts transitively on $X$ by the usual group multiplication, but $H$ does not. –  Jason DeVito May 3 '12 at 18:02
Dear fk44, Isometry groups of Riemannian manifolds are pretty far from the counterexamples that you and @Jason are considering. In the context of Dave Witte Morris's book, all the actions $G\times X \to X$ will be very well-behaved, and in particular, the openness property that you want will be true. Regards, –  Matt E May 4 '12 at 11:11
It's entirely possible that the author has, in a preface or something, a statement along the lines of "Every topological space considered will be Hausdorff...". But barring that, I'm in agreement that the counterexample works. Then again, I'm biased and am often wrong... ;-). –  Jason DeVito May 4 '12 at 13:47
Once you assume the space $X$ is a connected homogeneous space with the group acting on it as $G$, the quotient map $\pi:G\to G/G_x$, where $G_x$ is a stabilizer, is an open map and $G/G_x\to X$ is a homeomorphism. This tells us that $g\mapsto gx$ is an open map. Now it follows that (in particular) $G_0$, the identity component, acts transitively on $X$ (since X is connected). This is what Morris assumed in (the exercise) his book I think.