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If $G$ acts freely and cellularly 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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Let

If $H$ be G$acts freely and cellularly on a CW-complex, then the stabilizer of$\sigma$. By a cell must be finite (and therefore trivial, as pointed out in the question). This can be shown by induction on the dimension, the case of cells0-cells being trivial. If$\sigma$is an$n$-cell, with$n\geq 1$, you can assume let$H$freely be the stabilizer of$\sigma$. Then$H$permutes the cells set of smaller cells with dimension involved less than$n$in the boundary closure of$\sigma$. But there are only finitely many such cells, so and inductively each has finite (indeed, trivial) stabilizer. Thus$H$must be is finite. 2 added 21 characters in body Let$H$be the stabilizer of$\sigma$. By induction on dimension of cells, you can assume$H$freely permutes the cells of smaller dimension involved in the boundary of$\sigma$. But there are only finitely many such cells, so$H\$ must be finite.

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