# Proving $S^{4}/G$ is simply connected where $G$ is not a free group action

Consider the sphere $S^{4}$ as a subset of $\mathbb{R}^{5}$ and consider the action of the group $G$ of homeomorphisms generated by $(x_1, x_2, x_3, x_4, x_5) \rightarrow (-x_2, x_1, -x_4, x_3, x_5)$

The problem asked to show that the action is free on $M=S^{4}\setminus \{(0, 0, 0, 0, \pm 1\}$ (easy), and then to compute the fundamental group of $M/G$. So, $M$ is simply connected, the action by $G$ on each fiber is $\mathbb{Z}_4$, and the action is proper, free, and continuous, so by the quotient manifold theorem and the automorphism structure theorem, the fundamental group of $M/G$ is $\mathbb{Z}_4$. (It would be nice if someone could double check this).

The last part of the problem that I am stuck on however is showing that $S^{4}/G$ is simply connected. Since the action is no longer free in that case, I do not have any structure theorems (that I know of) to attack this problem. I can't imagine what this space might look like, so no Van Kampen's theorem. And now I am out of ideas.

Thanks for any help!

• This is a special case of Armstrong's (1968) theorem: fundamental group of the quotient is the quotient of G by subgroup generated by elements which do not act freely. – Moishe Kohan Dec 31 '14 at 22:03
• @studiosus Would you happen to have a reference somewhere? Or how I could prove that? I tried google, but did not find it :P – TheManWhoNeverSleeps Dec 31 '14 at 22:20
• math.stackexchange.com/questions/656858/…, see my answer with references there. – Moishe Kohan Dec 31 '14 at 22:24
• Your space is $\Sigma(S^3/G)$ — so it's simply connected. – Grigory M Dec 31 '14 at 22:34
• @GrigoryM I think I can see that, thanks :) – TheManWhoNeverSleeps Dec 31 '14 at 22:53