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?



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> Does $G = \PSL(2, q)$ for $ q \equiv 1 \pmod 4$ acting on $\projline$ have a regular subset?
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The answer is no for $q = 5, 9, 13, 17$, by direct computation.



Regular subsets of $\text{PSL}(2, q)$
This is not a complete solution of the problem, but only an outline of a possible strategy, along with some numerical evidence:
Let $S$ be a subset of $G=\operatorname{PSL}(2,q)$. For $g\in G$ let $P\_g$ be the permutation matrix of $g$, and let $J$ be the all-$1$-matrix of the same size. Then regularity of $S$ is equivalent to $\sum\_{g\in S}P\_g=J$. Thus the existence of a regular subset of $G$ is equivalent to the existence of a solution of
\begin{equation}
\sum\_{g\in G}x\_gP\_g=J
\end{equation}
for $x\_g\in\{0,1\}$.


If there is a regular subset, then we may assume that it contains $1$, so the remaining elements are fixed-point-free. Let $G^\star$ be the set of fixed-point-free elements from $G$ together with $1$. Then the existence of a regular subset is equivalent to the $0$-$1$-solvability of
\begin{equation}
\sum\_{g\in G^\star}x\_gP\_g=J.
\end{equation}
Now it seems to be the case that this system of linear equations in the $x\_g$ isn't even solvable over $\mathbb F\_2$. I have verified that for all $q\le 89$.


How could one prove that in general? Let $V$ be the space of $(q+1)\times(q+1)$ matrices over $\mathbb F\_2$. Then $V\times V\to\mathbb F\_2$, $(U,V)\mapsto\operatorname{trace}(UV)$ is a nondegenerate bilinear form on $V$.


From that we get the following: If the system is not solvable, then there is a matrix $M$ which is orthogonal to the space of matrices generated by the left hand side, but is not orthogonal to $J$. The converse holds of course too.


Thus we are left to find $M$ such that $\operatorname{trace}(P\_gM)=0$ for all $g\in G^\star$, and $\operatorname{trace}(JM)=1$.


One may even make some additional assumptions on $M$. For $h\in G$ set $M^h=P\_h^{-1}MP\_h$. If $H$ is a subgroup of $G$ of odd order, and $M$ fulfills the assumptions from above, then $\sum\_{h\in H}M^h$ fulfills this assumption too.


Thus in the search for $M$ we may restrict to those matrices for which $M^h=M$ for all $h\in H$. If $q$ is a power of the prime $p$, and $H$ is a $p$-Sylow subgroup of $G$, then it seems to be that there are exactly $8$ choices for such an $M$, no matter what $q$ is.


A working family of matrices $M$ seems to be as follows: Use $\mathbb F\_q\cup\{\infty\}$ to number the rows and columns of $M$. Then $M[i,j]=1$ if and only if either $j=\infty$, $i\in\mathbb F\_q$, or $i,j\in\mathbb F\_q$ are distinct and $i-j$ is a square.


To prove that this $M$ works reduces to a concrete question about $\mathbb F\_q$ which shouldn't be hard to prove.


I would be curious to learn the background or the motivation of this question.


*Remark*: It would be interesting to see other methods for proving non-existence of regular sets of permutations. Some years ago I worked with Theo Grundhöfer and Gabor Nagy on such questions. In most cases, the non-solvability of the first displayed equation modulo a prime was the most successful technique. In those cases, one could even choose $M$ as a rank $1$ matrix, which allowed for a direct and easy to use argument (we called it *contradicting subsets lemma*, see Lemma 2 in <https://arxiv.org/abs/1005.1598>). The special features of your question seem to be: (a) One really has to work with the second displayed equation, the first one does not give a contradiction, and (b) For $q\ge9$ it seems that there is no matrix $M$ as above (with no assumption on $H$-symmetry) of rank $1$, so the contradicting subsets lemma does not work here. 


