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Theorem: :$\mathbb{Z}_2$ is the only nontrivial group that can act freely on $S^n$ if $n$ is even.

Proof: Since the degree of a homeomorphism must be $\pm 1$, an action of a group on $S^n$ determines a degree function $d:G\rightarrow \{\pm 1\}$. This is a homomorphism since $\deg fg=\deg f\deg g$. If the action is free then $d$ send every nontrivial element of $G$ to $(-1)^{n+1}$ by property (g) above.Thus, when $n$ is even $d$ has trivial kerneel, so $G\subset \mathbb{Z}^2$.

Question: So what I don't get is how does a group action determine a degree function in this case. I thought that an action of $G$ on $S^n$ in this case is a homomorphism $\phi:G\rightarrow \text{Aut}(S^n)$ where $\text{Aut}(S^n)$ is the group of homeomorphisms of $S^n$. How does such a map induce a degree function?

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    $\begingroup$ Hint: $G \to Aut(S^n) \to \{1,-1\}$.Compose! $\endgroup$ – Arpit Kansal Jan 30 '16 at 4:44
  • $\begingroup$ Woops. That was dumb of me. Thanks. $\endgroup$ – Enigma Jan 30 '16 at 4:45
  • $\begingroup$ Dear @Enigma You're welcome.(I've copied my comment as an answer) $\endgroup$ – Arpit Kansal Jan 30 '16 at 4:49
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Hint: $G \to Aut(S^n) \to \{1,-1\}$.Compose!

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