A finite group G acts freely on a simply connected manifold M

Question

If a finite group G acts freely on a simply connected manifold M,Is the fundamental group $\pi_1(M/G,x_0)$ isomorhphic to G? If it is true,how to prove it?

Answer 1

Note that a free action of finite group on Hausdorff space is a covering space action. Then the quotient map from M to M/G is normal covering map and G is isomorphic to fundamental group of M/G, since M is simply connected.

Answer 2

If the action of $G$ is free, then it is also proper, so $M/G$ is a manifold, moreover it is a covering map and a Serre fibration, apply the exact sequence:

$1\rightarrow \pi_1(M)=1\rightarrow\pi_1(M/G)\rightarrow\pi_0(G)\rightarrow 1$.

Answer 3

This is the property of covering space action.