In Hatcher's book on Algebraic Topology, in the first sentence of example 1.43 he states: $$ \text{The Antipodal map of $S^n$, $x\to -x$, generates an action of $\mathbb{Z}_2$ on $S^n$ with orbit space $\mathbb{R}P^n$.} $$How do we know that the action generated is $\mathbb{Z}_2$?
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2Applying the action twice fixes the point. – Future Feb 28 '16 at 00:33
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I'm going to repeat what Prospect mentioned. Since applying multiplication by $-1$ twice gives you the identity, then the group involved must be $\langle a | a^2=e \rangle$ which is just $\mathbb{Z}_2$.
If you need to check this is actually a $\mathbb{Z}_2$-action is now easy (checking the axioms etc.).
Jack Davies
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1Why must the group involved be $\mathbb{Z}/2$? I could define a $\mathbb{Z}/4$-action on $S^n$ in much the same way, no? – Alex Provost Feb 28 '16 at 01:17
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You could, and depending on what you do you may end up with a Lens Space in that case. In this case though, the Hatcher defines a $\mathbb{Z}_2$-action. – Jack Davies Feb 28 '16 at 01:21