TITLE: A followup on non-homogeneous spaces.
QUESTION [5 upvotes]: This question asks for an example of a manifold which is not a homogeneous space of any Lie Group, and many examples are given in the answers. However: is there a an example known with a metric of positive sectional curvature (this, apparently, is a question asked by Marcel Berger many years ago, so was open then, but I have no idea of its current status...)?

REPLY [4 votes]: Yes, some quotients of the 3-sphere are examples. Namely, view $S^3$ as the group of quaternions of unit one. Consider two finite subgroups $F_1,F_2$ of $S^3$, with $F_1$ not abelian, and $F_2$ cyclic of odd prime order $p$ not dividing the order of $F_1$. So $F_1\times F_2$ acts freely on the sphere by $(g,h)\cdot z=gzh^{-1}$. I claim that the quotient $X=F_1\backslash S^3/F_2$ is not homeomorphic to a homogeneous space of any connected Lie group $G$.
Assume the contrary. First, by a result of Montgomery (Proc AMS 1950), any maximal compact subgroup of $G$ acts transitively (it's a general fact, using only that $X$ has finite fundamental group). So we can assume that $G$ is compact. Then a result of Montgomery-Samelson (Annals 1943) tells us that some simple $S$ subgroup acts transitively. (The result concerns actions on spheres, and is applied by considering the action of a finite covering of $G$ on $S^3$.) The isometry group of a 3-dimensional Riemannian manifold is at most 6-dimensional, and the only simple compact Lie group of dimension at most 6 is $SO(3)$ (and its universal covering). Thus $X=S^3/F$ for some finite subgroup $F$ of $S^3$, isomorphic to $F_1\times F_2$. But in $S^3$, the centralizer of any nontrivial element of odd order is abelian. This is a contradiction.
These examples are not simply connected, of course. It seems that the other answers deal with the much more delicate simply connected case.