We know all 2transitive simple groups by Dixon's book (Permutation groups). Now let $G$ be finite simple group $2$transitive and $p(p^{2}1)/2$ divides order $G$ and also $\pi (G)\subseteq \pi (p(p^{2}1))$. Is it true $G$ isomorphic to $L_{2}(p)$?

$\begingroup$ I have been assuming that $p$ is prime. Is that right? $\endgroup$ – Derek Holt Apr 22 '12 at 11:09

$\begingroup$ Thank you for you answer. Yes $p$ is prime. Also let the number of Sylow $p$subgroup $G$ equal to the number of Sylow $p$subgroup $PSL(2,p)$. Now: whether $G$ isomorphic to $PSL(2,p)$? $\endgroup$ – R K Apr 22 '12 at 13:00

2$\begingroup$ If $G$ has $p+1$ Sylow $p$subgroups, then its action by conjugation on the set of Sylow $p$subgroups is 2transitive, and the point stabilizer has order $p(p1)/2$ with a normal subgroup of order $p$. A group with those properties is isomorphic to ${\rm PSL}(2,p)$  you don't even need the classification of fintie simple groups for that  it follows from the result proved in: Hering, Christoph; Kantor, William M.; Seitz, Gary M. Finite groups with a split BNpair of rank 1. I. J. Algebra 20 (1972), 435–475. $\endgroup$ – Derek Holt Apr 23 '12 at 9:14
If I have understood the question correctly, then there seem to be lots of small counterexamples, such as $G=A_6, p=5$; $G=L_2(8)$ or $U_3(3)$, $p=7$; $G=M_{11}$ or $M_{12}$, $p=11$.
Added later: I thought of two more examples: $G=L_2(27), p=13$ and $G=L_3(5), p=31$. The interesting question is whether there are only finitely many examples. I would guess yes, but it could be hard to prove it.