I would like to know which is the commutator subgroup of the group of rotational isometries of the $n$dimensional cube. The group i am talking about is the subgroup of $\{ \pm 1 \} \wr S_n$ consisting of the pairs $(a, \sigma)$ such that $$sign(\sigma) \prod_i a_i =1 $$
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$\begingroup$ For $n \ge 3$, the commutator subgroup is the subgroup of index $2$ consisting of those elements in which $\sigma$ is even. This subgroup is perfect for $n \ge 5$. $\endgroup$ – Derek Holt Aug 9 '15 at 15:02

$\begingroup$ @DerekHolt This is obvious for $n$ odd, but is it also obvious for $n$ even? $\endgroup$ – Igor Rivin Aug 9 '15 at 15:43

$\begingroup$ @IgorRivin I wasn't claiming it was obvious! The permutation module for $S_n$ over any field has exactly two nonzero proper submodules, of dimensions $1$ and $n1$. That is a known result. My claim above follows from this (together with $[S_n,S_n]=A_n$), but I wouldn't describe it as obvious. If the characteristic of the field divides $n$ then the smaller submodule is contained in the larger submodule  otherwise they are disjoint. $\endgroup$ – Derek Holt Aug 9 '15 at 19:45