Let $G$ be a simple group such that
1) $|G|\mid|\mathrm{Alt}_{p}|$
2) $p\mid | G|$, and $p>13$ is prime.
3) $G$ hasn't any elements of order $rp$ for every prime number $r$.
My question: (without classification theorem) How we can prove that $G$ isn't isomophic to a simple group of Lie type.
I edit my question by adding the following assumption:
4) For every prime $t$ with $\frac{p+1}{2}<t<p$, we have $t\in \pi (G)$.
Note: This answers the original question, not the revised version: I'm afraid we can't: not because CFSG is necessary, but because ${\rm PSL}(2,p)$ satisfies those conditions for every prime $p > 13$, and is a simple group of Lie type in characteristic $p$ by any reasonable definition. The only thing that needs any checking is that $|{\rm PSL}(2,p)|$ divides $|A_{p}|$ for all primes $p > 13.$ But $A_{p}$ will always contain an element of order $\frac{p+1}{2}$ ( either a single cycle, or a product of a transposition and a $\frac{p+1}{2}$ cycle disjoint from it), and an element of order $\frac{p-1}{2}$, so it is just a question of checking the size of Sylow $2$-subgroups, and $|A_{p}|$ has plenty of powers of $2$ to spare. For an explicit example, ${\rm PSL}(2,17)$ has order $16 \times 9 \times 17$ and $A_{17}$ has a Sylow $17$-subgroup of order $17$, a Sylow $3$-subgroup of order $729$ and a Sylow $2$-subgroup of order $2^{14}.$
What IS true ( and was already known to Galois, so certainly does not use CFSG) is that ${\rm PSL}(2,p)$ is not isomorphic to a subgroup of $A_{p}$ for any prime $p > 11.$
