# Free $\mathbb{Z}_2$-actions match at some point

I have in front of me a proof of this lemma:

If $f$ and $g$ are free $\mathbb{Z}_2$-actions on $S^1$, then $f(x)=g(x)$ for some $x \in S^1$.

A $\mathbb{Z}_2$-action on the unit circle $S^1$ is a homeomorphism $f \;:\; S^1 \rightarrow S^1$ such that $f(f(x))=x$ for all $x \in S^1$; and $f$ is free if $f(x) \neq x$ for all $x \in S^1$.

The proof (in a paper I'm refereeing) is clear but somewhat laborious. It would be nice to either have a succinct proof, or a reference, rather than a detailed proof from first principles. Has anyone seen this before? Thanks!

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I wonder why this false notation, $\mathbb{Z}_n$ instead of $\mathbb{Z}/n$, is so common. – Martin Brandenburg Nov 25 '11 at 17:00
What do you mean by "false notation"? For many places it is the one true notation, and saying "zee-p" is a lot easier than "zee-mod-p-zee", which has only become popular in the last twenty years, probably due to bourbakist influence... – Igor Rivin Nov 25 '11 at 20:09
@Martin: I agree with Igor, what's a "false notation"? A notation is a notation, once it's generally agreed upon it's not false. If anything the notation ℤ/n is a bad one IMO. If you want to be precise then you should write something like $\mathbb Z/\langle n\rangle$ or ℤ/nℤ. The latter is fairly common but I like $\mathbb Z_n$ just fine. – Vitali Kapovitch Nov 25 '11 at 21:28
How could ${\mathbf Z}/n$ be considered bad while ${\mathbf Z}_n$ is fine? They both have the same amount of content. (I know what you mean about ${\mathbf Z}/n{\mathbf Z}$ being more precise, but if you can handle the idea of writing ${\mathbf Z}_n$ then ${\mathbf Z}/n$ should be okay too.) – KConrad Nov 26 '11 at 1:31
It seemed to me at some point, probably when I was in grad school (late fifties and early sixties) that the topologists used the notation ${\mathbb{Z}}_n$ while we number-theorists used the notations ${\mathbb{Z}}/n$ or ${\mathbb{Z}}/(n)$ or ${\mathbb{Z}}/n{\mathbb{Z}}$. – Lubin Nov 26 '11 at 6:05

we can clearly assume that $f(z)=-z$ in the standard metric on $S^1\subset \mathbb C$ (as we can assume that $f$ is isometric with respect to some Riemannnian metric on $S^1$). Then $g(z)=z\cdot e^{i\alpha(z)}$ with $0<\alpha(z)<2\pi$. If $\alpha(z_0)<\pi$ then $\alpha(g(z_0))>\pi$ and there is a point $z_1$ with $\alpha(z_1)=\pi$ by the intermediate value theorem.
@Vitali: Thanks!!! And apologies re $\mathbb{Z}^2$ vs. $\mathbb{Z}_2$. – Joseph O'Rourke Nov 25 '11 at 16:14
Is it completely clear that we can choose $\alpha$ to be continuous? – Martin Brandenburg Nov 25 '11 at 17:03
@ Martin: yes, it's completely clear. $i\alpha(z)=log(\frac{g(z)}{z})$ and its branch is well defined since $g$ is is fixed point free. – Vitali Kapovitch Nov 25 '11 at 17:13