On a subgroup of the deck transformation of a covering space

Question

I'm stuck with an exercise. Suppose you have a covering space $M \rightarrow X$, and you define $G:=\{\tau \in Deck(M)|\tau(S)=S\}$, for some $2$-sphere $S$ in $M$, and $G$ acts freely by isometries in $M$. How do I prove that $G$ must be trivial or isomorphic to $\mathbb{Z_2}$? I've tried but I can't prove that there cannot be two non trivial elements in $G$ without fixed points. Here, $M$ and $X$ are both Riemannian manifolds.

Answer 1

$S$ is a topological surface with Euler Characteristic $\chi(S) = 2$. Now $G$ acts on $S$, then $S/G$ is covered by $S$ and so $\chi(S) $ is divisible by $\chi(S/G)$. Thus $|G| = 1$ or $2$.

Remark That $G$ is a subgroup of the group of deck transformations implies that $G$ acts properly discontinuously on $S$: For all $s\in S$, $s\in M$ and so there is an open set $U \subset M$ containing $s$ so that $gU\cap U = \emptyset$ for all elements $g$ in $\text{Deck}(M)$. Pick $V = U\cap S$, then $gV \cap V = \emptyset $ for all $g\in G$. That the action is proper is obvious as $S$ is compact.

Note also that $\pi |_S : S\to \pi(S)$ is a covering, so $S/G$ is also a topological surface.