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Assume that $X$ is a Hausdorff compact space such that $\forall n\in \mathbb{N}$, we have a free action of $\mathbb{Z/{2^{n}}\mathbb{Z}}$ on $X$. Must $\mathbb{Z}(2^{\infty})$ act freely on $X$?

Note that we do not assume any compatibility of these actions with the inclusion $\mathbb{Z/{2^{n}}\mathbb{Z}}$ in $\mathbb{Z/{2^{n+1}}\mathbb{Z}}$. That is: the action of $\mathbb{Z/{2^{n}}\mathbb{Z}}$ on $X$ is not necessarily the same action of $\mathbb{Z/{2^{n}}\mathbb{Z}}$ as a subgroup of $\mathbb{Z/{2^{n+1}}\mathbb{Z}}$.

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    $\begingroup$ If you take $X$ to be the product of all groups showing up in your question, you get an example, so the answer is 'yes'. $\endgroup$ Nov 24, 2014 at 21:30
  • $\begingroup$ @FernandoMuro Is this $X$ a counter example? $\endgroup$ Nov 24, 2014 at 21:40
  • $\begingroup$ @FernandoMuro are you meaning that the answer to my question is yes for this particular $X$? Could you please more explain. $\endgroup$ Nov 24, 2014 at 21:45
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    $\begingroup$ Well, you wondered whether something is possible and I've given a self-explanatory example that it is, unless I've made a mistake. $\endgroup$ Nov 24, 2014 at 22:57
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    $\begingroup$ I think the OP wanted to ask "must" rather than "can." $\endgroup$ Nov 25, 2014 at 4:25

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I tried to find a counterexample, but ended up with the following interesting example instead. Put $X=\prod_{n=0}^\infty\{0,1\}$. Define $$ \phi_n : X \to \mathbb{Z}/2^n \times X $$ by $$ \phi_n(a) = \left(\sum_{i=0}^{n-1}2^{n-1-i}a_i,(a_n,a_{n+1},\dotsc) \right). $$ Define $\tau_n:\mathbb{Z}/2^n\to\mathbb{Z}/2^n$ by $\tau_n(i)=i+1$, then define $\sigma_n:X\to X$ by $\sigma_n=\phi_n^{-1}\circ(\tau_n\times 1)\circ\phi_n$. One can check that $\sigma_n^{2^n}=1$ and $\sigma_{n+1}^2=\sigma_n$, so the maps $\sigma_n$ fit together to give an action of $\mathbb{Z}/2^\infty$ on $X$. This is easily seen to be free.

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  • $\begingroup$ thank you for the example: May be for a counter example, the following question could be an starting point : what is an example of two fixed point free homeomorphisms $f$ and $g$ of order $4$ and $2$, res-pectivly , on a compact $X$ such that $f^{2}$ is not conjucate to $f$? $\endgroup$ Nov 24, 2014 at 22:00
  • $\begingroup$ @AliTaghavi, I guess the last $f$ in your comment should be $g$, right? $\endgroup$
    – LSpice
    Nov 26, 2014 at 22:41
  • $\begingroup$ @LSpice Yes thanks. It was my typos. $\endgroup$ Nov 28, 2014 at 17:23

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