We know that if $G$ is a simple group with $p+1$ Sylow $p$-subgroups, then $G$ is 2-transitive. Now let $G$ be almost simple group with $p+1$ Sylow $p$-subgroups. Is $G$ 2-transitive group?



A question on almost simple groups
I am sure that the answer is yes, but you might have to do a bit of work to write down a completely rigorous proof.


Let $S \unlhd G$ with $S$ simple and $G \le {\rm Aut}(S)$, and suppose that $G$ has $p+1$ Sylow $p$-subgroups. If $p$ divides $|S|$, then $S$ has at most $p+1$ and hence exactly $p+1$ Sylow $p$-subgroups, and so $S$ and hence $G$ act 2-transitively by conjugation on the set of Sylow $p$-subgroups of $S$ (or of $G$).


Using the classification, we find that the only finite simple groups $S$ that have an automorphism of prime order $p$ that does not divide $|S|$ are groups $S=L(r^{p^k})$ of Lie type that have a field automorphism of order $p^k$. So a Sylow $p$-subgroup $P$ of $G$ is cyclic of order $p^l$ for some $l \le k$. Then the centralizer of $P$ in $S$ is isomorphic to $L(r^{p^{k/l}})$, which has index greater than $p+1$ in $S$, so it is not possible for $G$ to have $p+1$ Sylow $p$-subgroups in this situation.


