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Let $p$ be a prime and $\mathbb{Z}_p$ denote as usual the field of order $p$. Let $AL(p)$ be the affine linear group $\{x\mapsto ax+b \;|\; a\in \mathbb{Z}_p\setminus \{0\}, b\in\mathbb{Z}_p\}$. For a given integer $1\leq k \leq p$, it is known that each orbit of the natural action of $AL(p)$ on $k$-subsets of $\mathbb{Z}_p$ is a $2$-design $D_k(p)$.

Are all such $2$-designs $D_k(p)$ known?

Any information on a generic $D_k(p)$ for an arbitrary $p$ and $k$ is apreciated.

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For fixed $k \ge 3$, the generic such design is a full orbit with "index" $\lambda = k(k-1)$ and is probably not of much (combinatorial) interest. Short orbits are, of course, very interesting. As an example, developing $\{1,2,4\}$ under the field of order $7$ gives the Fano plane. Replace $7$ by a larger prime and this is just a rather boring (setting aside intricate structural questions) design with index six.

I believe most researchers in combinatorics and group theory "know" about these designs. But if by "known" you are asking about the question of classification, then difference sets and their multipliers can yield challenging sub-problems.

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  • $\begingroup$ As far as I know for the orbit of a $k$-subset $S$ of $\mathbb{Z}_p$, $\lambda=\frac{k(k-1)}{|Stab_{AL(p)}(S)|}$, where $Stab_{AL(p)}(S)=\{\alpha\in AL(p) \;|\; S^\alpha=S\}$. So you are claiming that $|Stab_{AL(p)}(S)|=1$. Could you please give a hint for this? $\endgroup$
    – Alireza Abdollahi
    Nov 8, 2014 at 6:30
  • $\begingroup$ I just meant that in most cases the stabilizer is trivial. That is what I thought you meant by "generic". $\endgroup$
    – Peter Dukes
    Nov 9, 2014 at 9:11
  • $\begingroup$ so I would like to know how you arrived to the point that in the generic case The stabilizer is trivial? $\endgroup$
    – Alireza Abdollahi
    Nov 9, 2014 at 14:00

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