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Aug 17, 2019 at 3:37 comment added John Shareshian It is quite reasonable to hope that, with more work than was required in the comment above, one can then show that $[H,G]$ is a Boolean algebra of rank $n-1$, thus rendering further computation unnecessary.
Aug 17, 2019 at 3:34 comment added John Shareshian With respect to the remark: Let's work in $S_n$. It seems that $H$ is a Sylow $2$-subgroup. To prove this, first use the fact that a Sylow $2$-subgroup $H \leq S_{2^n}$ is an $n$-fold iterated wreath product of ${\mathbf Z}_2$ with itself to show by induction that this subgroup stabilizes an equipartition of each possible type and that these equipartitions have the desired refinement property. Then observe that a $2^n$-cycle stabilizes exactly one equipartition of each type. Then show by induction that no element of odd prime order stabilizes all equipartition stabilized by $H$......
Aug 16, 2019 at 19:48 history edited Sebastien Palcoux CC BY-SA 4.0
Minor edit: same shutdown with S_{32}
Aug 14, 2019 at 17:16 history edited Sebastien Palcoux CC BY-SA 4.0
Additional remark on the order of H
Aug 14, 2019 at 16:49 history edited Sebastien Palcoux CC BY-SA 4.0
computation for S_{16}
Aug 14, 2019 at 14:49 history answered Sebastien Palcoux CC BY-SA 4.0