When I am reading one paper, I have met the following statement:


It is impossible to define a $Z\_{2}\times Z\_{2}$ action on a connected closed curve on a compact Riemann surface.


The claim is equivalent to say that the existence of two fixed point free involutions on a circle is not true.


The author just took this statement as an obvious fact and have used it to prove a lemmer about the antiholomorphic involutions on the compact Riemann surface. It seems the claim is pretty simple, but I cannot find an elegant way to prove this is indeed true. 


I have tried to use the the following fact to prove it:


the one-dimensional manifold can be considered as a circle under the homeomorphism, and it is diffeomorphic to the one dimensional real projective plane, which inherits the automorphism group of $PGL(2, \mathbb{R})$ from $\mathbb{R}^{2}.$ And in $PGL(2, \mathbb{R}),$ the element corresponding to the antipodal map is unique, so the above statement is true.


Is this a correct proof?



The antipodal action on a connected one dimensional manifold 
Assuming you want *fixed-point free* actions...


Since $\pi\_1(S^1)$ is cyclic, all connected coverings of $S^1$ have cyclic group of covering transformations. Now, if $\mathbb Z\_2^2$ acted without fixed points on $S^1$, the corresponding quotient map would be a covering $S^1\to S^1$ of group $\mathbb Z\_2^2$, which is impossible.


**Later:** a bit less technological: suppose $G=\mathbb Z\_2^2$ acts freely on $S^1$ and pick a point $x$. The orbit of $x$ cuts $S^1$ in $4$ segments, and one of them, call it $I$, has $x$ and $\sigma(x)$ as endpoints for some $\sigma\in G\setminus\{1\}$. But the map $\sigma$ maps $I$ to itself, so by continuity, $\sigma$ fixes a point in $I$, which it didn't.


