A question on almost simple groups 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: I think there is a direct argument. Let $M$ be the unique minimal normal subgroup of $G,$
which is non-Abelian simple. Then $M$ must act faithfully by conjugation
on the $(p+1)$ Sylow $p$-subgroups of $G$- otherwise, $M$
has a normal Sylow $p$-subgroup, which must then be trivial.
But even then, $M$ must normalize, and hence centralize, a Sylow $p$-subgroup $P$
of $G$, as $M$ and $P$ normalize each other and have trivial intersection.
Then $P$ is contained in $C_G(M)=1,$ a contradiction.
Thus $G$ is isomorphic to a subgroup the symmetric
group of degree $p+1$ and a Sylow $p$- subgroup 
of $G,$ say $P,$ has order $p.$ Now $P$ fixes no other Sylow $p$- subgroup
 of $G$ in the conjugation action, so permutes the remaining
$p$ such subgroups in one orbit of length $p.$ Hence $G$
Is doubly transitive. 
Later addition: Let me try to address more precisely Mart's question in the comments-
the argument is less elementary, but still avoids the classification of finite simple 
groups. Let me retain my notation of $M$ for the unique minimal normal subgroup of $G,$ (called $S$ by Derek and Mart) and let $P$ be  a Sylow $p$-subgroup of $G,$ which has order $p,$ as we have seen already.
   The key point I will use is a Theorem of Feit and Thompson (Nagoya J. Math ~1963), which built on an earlier result of Brauer: the combined result asserts that if $X$ is a finite irreducible subgroup of ${\rm GL}(n,\mathbb{C})$ for some $n \leq \frac{p-1}{2},$ where $p$ is a prime, then either $X$ has a normal Sylow $p$-subgroup, or $X/Z(X) \cong {\rm PSL}(2,p).$
Our group $G$ has a transitive faithful permutation action on $p+1$ points, affording a permutation character $\chi,$ say. We are assuming that $M$ has order prime to $p,$ and aiming to derive a contradiction. The orbits of $M$ all have equal length, and are permuted by $G$. If $M$ has two or more orbits, then ${\rm Res}^{G}_{M}(\chi)$ has at least two trivial constituents, and $M$ has a faithful irreducible character of degree at most $\frac{p-1}{2},$ which extends irreducibly to $MP$ (it can't induce irreducibly by degree considerations). Now $P$ is not normal in $MP$ as $[M,P] \neq 1.$ But $MP/Z(MP)$ is not isomorphic to ${\rm PSL}(2,p),$ since ${\rm  PSL}(2,p)$ has no normal $p$-complement, while $MP$ does have a normal $p$-complement (note that we do need $p >3$ here, but $S_{4}$ is solvable and $G$ is not, so we do indeed have $ p >3$). This contradicts the result of Brauer, Feit and Thompson. Hence $M$ is transitive.
    Now $M$ is not doubly transitive, as $p$ does not divide $|M|,$ so that ${\rm Res}^{G}_{M}(\chi)$ is a sum of at least $3$ irreducible characters (allowing multiplicities). However, the trivial character only occurs once, and $M$ has no other linear character. Hence ${\rm Res}^{G}_{M}(\chi)$ has a faithful irreducible constituent $\mu$ of degree at most $\frac{p-1}{2},$ which once more extends irreducibly to $MP,$ and we obtain the same contradiction as above.
Third edit: Actually, there is a simpler argument using less sophisticated representation theory to obtain $p$ divides $|M|$. Suppose otherwise, and retain the notation above. Note that ${\rm Res}^{G}_{MP}(\chi)$ can't have an irreducible constituent of degree $p$ (but does have a trivial constituent): for if $\mu$ were such  constituent, then Clifford's theorem would force $\mu$ to restrict to a sum of non-trivial linear characters of $M$, contrary to the fact that $M$ is perfect. Hence $MP$ has a non-trivial complex irreducible character $\theta$ say, of degree less than $p$ (and $\theta$ is faithful using the simplicity of $M$). Let $r$ be an odd prime divisor of $|M|$, and let $R$ be a $P$-invariant Sylow $r$-subgroup of $M$ (which exists). Then by the theorem of Hall-Higman-Shult, we have $[M,R] \leq {\rm ker} \theta = 1.$ Let $Q$ be a $P$-invariant Sylow $2$-subgroup of $M$. Then as $r$ was arbitrary, we have $M = QC_{M}(P).$ Hence $[M,P] \leq Q.$
But $[M,P] \lhd M$ and $M$ is non-Abelian simple, so $[M,P] = 1$ and $P \leq C_{G}(M) = 1,$ a contradiction.
A: 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.
