Take the tour ×  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.  share|improve this
question add comment  1 Answer  The short: no. E.g. let X be a
topological group and G be the underlying discrete group of X,
acting on X b