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

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 2transitively 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. 


I think there is a direct argument. Let $M$ be the unique minimal normal subgroup of $G,$ which is nonAbelian 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 $n \leq \frac{p1}{2},$ 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 $\frac{p1}{2},$ 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 $\frac{p1}{2},$ 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 nontrivial linear characters of $M$, contrary to the fact that $M$ is perfect. Hence $MP$ has a nontrivial 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 HallHigmanShult, 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 nonAbelian simple, so $[M,P] = 1$ and $P \leq C_{G}(M) = 1,$ a contradiction. 

