Question

Let X be a contractible metric space and G a topological group acting transitively on X (i.e. given any two points x,y \in X, there exists g \in G such that gx=y).

My question is the following: is it true that given any x \in X its stabilizer Stab(x)={ g \in G : gx=x } and the whole group G have the same homotopy type?

If the answer is "no", I'd like to know some "mild" hypothesis that could be add to have an affirmative response.

For instance, I know that whenever G is a Lie group and H < G is a closed subgroup such that G/H is contractible, then G and H are homotopically equivalent (in this case H can be seen as the stabilizer of the coset H under the natural G-action on G/H). However, to assume that G is a Lie group seems to be too restrictive. In fact, I'd like to apply this "result" to some groups which are not locally compact.

Answer 1

The short: no. E.g. let X be a topological group and G be the underlying discrete group of X, acting on X by left translation.

One standard hypothesis is the existence of slices. For some (hence any) point x in X, if there is an open neighborhood U of x and a section s:U -> G such that s(u)*x = u for all u in U, then the map p:G -> X given by p(g) = g*x is a fiber bundle with fiber Stab(x).

This hypothesis automatically implies that Stab(x) is weakly equivalent to G by the long exact sequence of the fibration. If G and Stab(x) have the homotopy type of CW-complexes then they are homotopy equivalent by the Whitehead theorem.