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Recall that a CW-complex $X$ with an action of a group $G$ which permutes the cells (i.e., for any $g \in G$ and any cell $\sigma \subseteq X$, $g\sigma$ is a cell) is called a $G$-complex. If the action permutes the cells freely ($g\sigma = \sigma$ implies $g=1$), $X$ is a free G-complex.

Clearly, if $X$ is a free $G$-complex, then the $G$-action on $X$ is free (i.e., for any $g \in G$ and any $x \in X$, $gx = x$ implies $g=1$). A question that pops to my mind every once in a while is the following: is a $G$-complex with a free $G$-action a free $G$-complex? I see that if $g\sigma = \sigma$ for some nontrivial $g \in G$ and a cell $\sigma$, then $g$ has infinite order (for a finite group cannot act freely on a contractible space), but this doesn't seem to get me anywhere.

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  • $\begingroup$ The usual definition of a $G$-CW-complex also requires that a group element fixing a cell setwise also fixes it pointwise. For example the unit interval with the $\mathbb{Z}/2$-action given by reflection at $1/2$ and the usual G-CW structure is not a G-CW-complex. I never came across the notion of a $G$-complex. $\endgroup$
    – HenrikRüping
    Commented Dec 11, 2012 at 13:33
  • $\begingroup$ I have a question, why can we use finite group cannot act freely on a contractible space? A CW-complex is not necessary to be contractible. $\endgroup$
    – 6666
    Commented Apr 7, 2017 at 22:53
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    $\begingroup$ @6666 That fact is being applied to the action of the cyclic subgroup generated by $g$ on the single cell $\sigma$, which is contractible. $\endgroup$
    – Jeremy Rickard
    Commented Apr 8, 2017 at 8:28

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If $G$ acts freely on a CW-complex, permuting the cells, then the stabilizer of a cell must be finite (and therefore trivial, as pointed out in the question).

This can be shown by induction on the dimension, the case of 0-cells being trivial. If $\sigma$ is an $n$-cell, with $n\geq 1$, let $H$ be the stabilizer of $\sigma$. Then $H$ permutes the set of cells with dimension less than $n$ in the closure of $\sigma$. But there are only finitely many such cells, and inductively each has finite (indeed, trivial) stabilizer. Thus $H$ is finite.

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    $\begingroup$ This argument applies when the cell has positive dimension, because then there is at least one lower-dimensional cell involved in the boundary. Fortunately a separate argument is available to show that an infinite group cannot act freely on a $0$-cell. $\endgroup$
    – Tom Goodwillie
    Commented Dec 11, 2012 at 14:04
  • $\begingroup$ A separate argument is also needed if the complex $X$ is not locally finite (in which case the number of lower-dimensional cells whose images intersect the boundary of the $n$-cell, could be infinite). $\endgroup$
    – Misha
    Commented Dec 11, 2012 at 16:46
  • $\begingroup$ @Misha: Doesn't the boundary of an $n$-cell $\sigma$ always involve only finitely many cells in the $(n-1)$-skeleton? Local finiteness means there are only finitely many higher dimensional cells whose boundaries involve $\sigma$, and I don't think my argument uses that. $\endgroup$
    – Jeremy Rickard
    Commented Dec 11, 2012 at 18:38
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    $\begingroup$ @Tom: True. I've slightly edited my answer so that I don't lie about 0-cells. :-) $\endgroup$
    – Jeremy Rickard
    Commented Dec 11, 2012 at 20:18

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