A group 2-transitive with p+1 Sylow subgroups Let $G$ be a simple group of Lie type and $2$-transitive such that the
number of Sylow $p$-subgroups of $G$ is $p+1$. How we can prove that $G$ is
isomorphic to $PSL(2,p)$? Thanks.
 A: Suppose that the group $G$ has $p+1$ Sylow $p$-subgroups for some prime $p$ and let $\Omega = {\rm Syl}_p(G)$ and $P \in \Omega$. Consider the action of $G$ on $\Omega$ by conjugation. Since $P$ cannot normalize any Sylow $p$-subgroup other than $P$ itself, $P$ fixes a unique point in this action, namely $P$ itself. The orbits of $P$ on $\Omega$ have length a power of $p$, so there must be exactly two orbits, of lengths $1$ and $p$. Since $G$ acts transitively (by Sylow's Theorem) and $P$ acts transitively on $\Omega \setminus \{ P \}$, $G$ acts 2-transitively. The stabilizer of $P$ in this action is the normalizer of $P$, so this stabilizer has $P$ as a normal subgroup, and in fact we must have $|P|=p$, so $P$ is a regular normal subgroup of the stabilizer of $P$ in its action on $\Omega \setminus \{ P \}$.
The result of Hering, Kantor and Seitz mentioned earlier is:
Let $G$ be a finite 2-transitive group on a set $\Omega$ and suppose that, for $\alpha \in \Omega$, $G_\alpha$ has a normal subgroup regular on $\Omega \setminus \{ \alpha\}$; then $G$ contains a normal subgroup $M$ and $M$ acts on $\Omega$ as one of the following groups in their usual 2-transitive representation: a sharply 2-transitive group, ${\rm PSL}(2,q)$, ${\rm Sz}(q)$, ${\rm PSU}(3,q)$ or a group of Ree type.
Since your group $G$ is simple, its action on $\Omega$ is faithful, and the hypotheses of this result apply. Simplicity implies that $M=G$. The groups ${\rm PSL}(2,p)$ are the only groups inthe conclusion with Sylow subgroups of order $p$.
