Timeline for For which $n$ can $S_n$ act transitively on $n+k$ elements?
Current License: CC BY-SA 4.0
15 events
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| Sep 5, 2022 at 11:47 | comment | added | The Amplitwist |
The bit.ly link in a comment above points to the following Wikipedia page: O'Nan–Scott theorem. Just posting this in case the URL shortener ends up breaking in the future.
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| Jan 19, 2021 at 7:50 | comment | added | Derek Holt | Yes, sorry for the mistakes, I wrote my comment late last night! | |
| Jan 19, 2021 at 3:48 | answer | added | Mikko Korhonen | timeline score: 9 | |
| Jan 19, 2021 at 2:03 | comment | added | YCor | @GeoffRobinson I should have read more carefully :) this seems to the only exception anyway, and finiteness is still OK. | |
| Jan 18, 2021 at 23:43 | comment | added | Geoff Robinson | @YCor: : I don't think that is correct in general, for example because of the transitive action of $S_{n}$ on $2n$ points (with point stabilizer $A_{n-1}).$ | |
| Jan 18, 2021 at 21:52 | comment | added | YCor | Note that the answers so far already answer positively the finiteness question, with an explicit bound: if $n>5$, $k>0$ and $n>3/2+(2k+9/4)^{1/2}$ then $S_n$ doesn't act transitively on $n+k$ elements. | |
| Jan 18, 2021 at 21:33 | comment | added | Geoff Robinson | @DerekHolt : The bound you give works for primitive actions, but is not right for transitive actions in general. | |
| Jan 18, 2021 at 21:31 | comment | added | Geoff Robinson | @Wojowu: Yes it does, as I said above . I think Derek Holt was probably thinking of maximal subgroups and primitive actions, and $A_{n-1}$ is not maximal. | |
| Jan 18, 2021 at 21:23 | comment | added | Wojowu | @DerekHolt Doesn't $S_n$ act transitively on $S_n/A_{n-1}$ which has size $2n$? | |
| Jan 18, 2021 at 21:21 | comment | added | Derek Holt | In fact for $n>5$, the smallest set of size greater than $n$ on which $S_n$ acts transitively has size $n(n-1)/2$. The corresponding result for $A_n$, with a lot more detail, is proved in Theorem 5.2A of Dixon & Mortimer - the proof is indeed based on the O'Nan-Scott Theorem. Almost simple maximal subgroups are not problematic, because they are generally much smaller than the intransitive and imprimitive maximal subgroups. | |
| Jan 18, 2021 at 21:14 | comment | added | Ian Agol | Just here to point out that this is the same as determining the number of subgroups of index $n+k$, corresponding to the conjugacy class of a fixed point of the transitive action. In particular, by Lagrange's theorem, one needs $(n+k)|n!$. But that's not very restrictive unless $n+k$ happens to be prime. For sufficiently large $n$, if one can prove that all maximal subgroups of $S_n$ are of index $>n+k$ (other than $S_{n-1}$), then this would suffice. One might be able to prove this using the O'Nan-Scott Theorem. But analyzing the almost simple case might be intricate. bit.ly/3nWXOZc | |
| Jan 18, 2021 at 20:57 | comment | added | Geoff Robinson | Of course I forgot that $A_{n-1}$ is a subgroup of index $2n$ in $S_{n}.$ | |
| Jan 18, 2021 at 19:51 | comment | added | abx | I think you'll find a good overview of what is known in Dixon-Mortimer's Permutation groups. | |
| Jan 18, 2021 at 19:38 | comment | added | Geoff Robinson | I think that for $n >7$, there is a gap between $n-1$ and $\frac{(n-1)(n-2)}{2}$ for irreducible character degrees of $S_{n}$ . On the other hand $S_{n}$ does have a transtive permutation representation of degree $\frac{n(n-1)}{2}$ (on unordered pairs of distinct elements). Yoou may find more information in the books of Fulton and Harris, or James. | |
| Jan 18, 2021 at 19:25 | history | asked | M. Winter | CC BY-SA 4.0 |