# What can we say about the sets X on which a group G acts freely and transitively?

The "philosophy" of (group) actions is such that the most rigid conditions we can think of (abstractly) are freeness and transitivity.

Let me introduce some terminology to make this idea clearer. For $G$ a group, we call a set $X$ a rigid set for $G$ if there exists a group action $. :G \times X \rightarrow X$ that is both free and transitive.

Does any group $G$ admits a rigid set $X$? What can we say about the category of rigid sets for $G$ where morphisms are $G$-equivariant maps?

PS: this question is motivated by the fact that in practice, most natural actions aren't free but only faithful. For example, the natural action of $SO(3)$ on $\mathbb{R}^3$ is faithful and transitive, but certainly not free. Yet, it seems to me that the "good object" on which a group $G$ should act has to be a rigid set for $G$. Does a rigid set for $SO(3)$ is known to exist ?

• I think letting $G$ act on itself by left multiplication should work. – user281392 Oct 14 '16 at 21:45
• Alright, that's true, but I think I'm interested in more "geometric" example :> So I guess the interesting question is the one about the category of rigid sets for $G$ – sure Oct 14 '16 at 21:46
• I am not really sure what you are looking for. Any transitive action is equivalent to $G$ acting on a coset space, and free implies that the coset you are acting on is the trivial group, so $X$ is basically $G$. – user29123 Oct 15 '16 at 0:47

Every free and transitive $G$-set is isomorphic to $G$ acting on itself by left multiplication. Every morphism of $G$-sets between free and transitive $G$-sets is an isomorphism. So the category is equivalent to the category with one object with automorphism group $G$.
A natural example of a space on which $SO(3)$ acts freely and transitively is the space of oriented orthonormal bases of $\mathbb{R}^3$.