Not very transitive actions 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.
 A: 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.
A: 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 $m\geq 4$.
