TITLE: Regular subsets of $\text{PSL}(2, q)$
QUESTION [10 upvotes]: Suppose a group $G$ acts on a set $\Omega$. Call a subset $A \subset G$ regular (or sharply transitive or simply transitive or...) if for every two points $\omega_1, \omega_2 \in \Omega$ there is a unique $a \in A$ such that $\omega_1^a = \omega_2$. This generalizes the concept of a regular subgroup to subsets.
Consider the action of $G = \def\PSL{{\text{PSL}}}\def\F{{\mathbf{F}}}\def\projline{{\mathbf{P}^1 \F_q}}\PSL(2, q)$ on the projective line $\Omega = \projline$. It is not hard to prove that $G$ has a regular subgroup if and only if $q \not\equiv 1 \pmod 4$. Indeed if $q$ is even then there is a cyclic regular subgroup, and if $q \equiv 3 \pmod 4$ then there is a dihedral regular subgroup, while if $q \equiv 1 \pmod 4$ then any subgroup of order $q+1$ must contain an element of order $2$, but every element of order $2$ has eigenvalues $\pm i$ and hence two fixed points. But could $G$ yet have a regular subset?

Does $G = \PSL(2, q)$ for $ q \equiv 1 \pmod 4$ acting on $\projline$ have a regular subset?

The answer is no for $q = 5, 9, 13, 17$, by direct computation.

REPLY [4 votes]: I have come across some references for this problem, which amount to a solution very different from the (beautiful) one given by Peter Mueller, and which also goes further. It turns out that it is possible to show directly that a sharply transitive subset of $\mathrm{PGL}_2(q)$ acting on $\mathbf{P}^1$ must be a coset of a subgroup. Since the subgroups of $\mathrm{PGL}_2(q)$ are classified, so are regular subsets.
This result is due to Bader, Lunardon, and Thas, who classified flocks of the hyperbolic quadric $\mathcal{Q} \subset \mathbf{P}^3$.
The solution was given in the following two papers.
Thas, J. A., Flocks, maximal exterior sets, and inversive planes, Finite geometries and combinatorial designs, Proc. AMS Spec. Sess., Lincoln/NE (USA) 1987, Contemp. Math. 111, 187-218 (1990). ZBL0728.51010.
Bader, Laura; Lunardon, Guglielmo, On the flocks of (Q^+(3,q)), Geom. Dedicata 29, No. 2, 177-183 (1989). ZBL0673.51010.
On the face of it the problem these authors study is different, but it is equivalent.
A flock is a partition of a quadric into conics.
The quadric $\mathcal{Q}$ can be taken without loss of generality to be the subset $\det A = 0$ of the space $M_2(K) / Z$ of $2\times2$ matrices up to scalars. Each conic can be identified with the plane it spans and hence by duality with a point of $\mathbf{P}^3 = M_2(K)/Z$. The condition that the planes intersect $\mathcal{Q}$ in disjoint subsets is equivalent to saying that the the corresponding points define a regular subset (or maximal exterior set, as they call it).
The equivalence is explicit in the introduction here:
Bonisoli, Arrigo; Korchmáros, Gábor, Flocks of hyperbolic quadrics and linear groups containing homologies, Geom. Dedicata 42, No. 3, 295-309 (1992). ZBL0756.51009.
I took some time to boil off all the geometry, and I wrote up what was left here: https://arxiv.org/abs/2105.02760