0
$\begingroup$

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 $$sign(\sigma) \prod_i a_i =1 $$

$\endgroup$
  • $\begingroup$ 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 $n-1$. 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

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.