Let $p > 3$ be a prime number, and let $G \leq \mathrm{PGL}\_2(\mathbb{F}\_p)$ be a solvable subgroup.


Is it possible that the action of $G$ on $\mathbb{P}^1(\mathbb{F}\_p)$ is transitive?



Can a projective solvable group be transitive?
Yes. Take any generator of the multiplicative group of $ \newcommand{\GF}[1]{\mathbb{F}\_{#1}} \GF{p^2}$ and make it into an element $g$ of $ \DeclareMathOperator{\GL}{GL} \DeclareMathOperator{\PGL}{PGL} \GL\_2( \GF{p} )$. Then $g$ is diagonalizable over $\GF{p^2}$ with two Galois conjugate eigenvalues. Thus when $g^k v = av$ for some $v\in \GF{p}^2$ and $a\in \GF{p}$, then necessarily $g^k = aI$. Thus the image of $\langle g \rangle $ in $\PGL\_2(\GF{p})$ is a regular, cyclic group of order $p+1$. 


**Added later:** As Nick Gill mentions, the $g$ constructed above (or $\langle g \rangle$) is called a **Singer cycle** in the literature. More generally, a Singer cycle $g\in \GL\_n(\GF{q})$, constructed analogously from a generator of $ (\GF{ q^n } )^\*$, induces one long cycle on the elements of $ (\GF{q})^n \setminus \{0\}$. (Thus the name Singer *cycle*.) The corresponding subgroup of $\PGL\_n( \GF{q} )$ permutes transitively the points of $\mathbb{P}^{n-1}(\GF{q})$. 


**Bonus:** Let's just assume that $p$ is a prime power. If the action of a solvable group $G\leq \PGL\_2( \GF{p} )$ on $\mathbb{P}^1(\GF{p} )$ is primitive, then $p=3$ or $p=2^{2^k}$ and $p+1$ is a Fermat prime. 


*Proof*. Take a minimal normal subgroup $N$ of $G$. Then $N$ is elementary abelian and transitive, and thus regular, so $p+1$ must be a prime power, too. Thus either $p$ is a Mersenne prime, or $p = 8$ and $p+1 = 9$ or $p+1$ is a Fermat prime. The last case is of course possible. The Sylow $3$-subgroup of $\PGL\_2( \GF{8} )$ has order $9$, and its normalizer does not act primitively, so the case $p=8$ is ruled out. 


If $p=2^m-1$ is Mersenne, then $N$ is elementary $2$-abelian of order $2^m$. It is not too dificult to see that the Sylow $2$-subgroup of $\PGL\_2( \GF{p} )$ is a dihedral group. It follows that $|N| \leq 2^2$ and thus $p=3$ as claimed.


