11

Suppose $m$ is a positive integer.

I am looking for finite sets with group actions such that the action is transitive on the set of $m$-element subsets, but NOT transitive on the set of $(m+1)$-element subsets.

An example for $m=2$ is a projective space over a finite field.

2 Answers2

11

According to Theorem 4.11 of Peter Cameron's book `Permutation Groups' it follows from the classification of finite simple groups that the only finite 6-transitive groups are (some of the) symmetric and alternating groups in their natural actions, and the only finite 4-transitive groups are symmetric, alternating and the Mathieu groups $M_{11}$, $M_{12}$, $M_{23}$ and $M_{24}$. Thus there will not be very many examples of what you are looking for when $m\geq 4$.

IJL
  • 3,361
  • 1
    Can you make it without CFSG for large $m$? – Anton Petrunin Nov 10 '22 at 20:38
  • 3
    An $m$-transitive group is transitive on ordered $m$-tuples of distinct elements, not $m$-element sets. (The latter property is called $m$-homogeneous.) Thus Theorem 4.11 does not quite give what you want. See, however, https://math.stackexchange.com/questions/4207740. – Richard Stanley Nov 11 '22 at 03:39
  • 1
    @RichardStanley I hadn't noticed this - thank you. – IJL Nov 11 '22 at 10:24
  • 1
    @AntonPetrunin I don't know about removing the reliance on CFSG. – IJL Nov 11 '22 at 10:25
  • 2
    Bill Kantor has a paper that classifies the $k$-homogeneous groups that aren't $k$-transitive. https://pages.uoregon.edu/kantor/PAPERS/k-Homogeneous.pdf I believe that this paper does not use CFSG.... Which means that were one to find a CFSG-free proof of $k$-homogeneous groups for some $k$, then you would have a CFSG-free proof of Jordan's conjecture and that is beyond our current knowledge: https://mathoverflow.net/questions/161280/multiply-transitive-groups-continued – Nick Gill Nov 11 '22 at 14:07
9

Generally any projective group $\mathrm{PGL}(2,\mathbb{F}_q)$ acting on the $q+1$ points of the projective line over $\mathbb{F}_q$ is sharply $3$-transitive. This gives infinitely many $3$-homogeneous but not $4$-homogeneous groups.

A related example I particularly like is the $3$-transitive but $4$-homogeneous group $\mathrm{P}\Gamma\mathrm{L}(2,8)$ obtained by extending $\mathrm{PGL}_2(\mathbb{F}_8)$ by the Frobenius automorphism of order $3$. Such groups are important in the classification of multiplicity-free permutation characters of $S_n$. I have a relevant paper which builds on work of Saxl, and there is also independent work of Godsil and Meagher.

Incidentally, going in the opposite direction, it is not completely obvious that a $k$-homogeneous group on $n$ points is $(k-1)$-homogeneous, provided $k \le n/2$. But this is of course true, and has a neat proof using character theory: the number of orbits of $G \le S_n$ on $k$-subsets is $\langle 1\!\!\uparrow_G^{S_n}, \pi_k \rangle$, where $\pi_k$ is the permutation character of $S_n$ acting on $k$-subsets; now use that $\pi_k - \pi_{k-1}$ is the irreducible character $\chi^{(n-k,k)}$, so the difference has non-negative inner product with any character.

Mark Wildon
  • 10,750
  • 3
  • 44
  • 70