Non-abelian simple groups of odd order less than $10000$. I am trying to solve problem 6.2.16 from Dummit and Foote, namely

Prove there are no non-abelian simple groups of odd order $< 10000$.

I did something similar for order $<100$, where I showed the only non-abelian simple group of order $< 100$ is $A_5$, but this was done by looking at many different forms of order in terms of prime arrangements and picking a few special cases aside to rule everything out but order $60$. The order is too big to do this here. The only thought I have is that $10000=100^2$ and so taking $G$ to be a minimal counterexample with $|G|>100$ we can write $|G|=ab$ with $a<100, b\geq 100$. Apart from that not much that I feel could be constructive. Any help would be appreciated for getting to an elegant proof that doesn't grind through every prime arrangement.
 A: If $n$ is the order of a nonabelian finite simple group then $n$ has at least 3 prime factors (Burnside's $p^a q^b$ theorem) and Sylow's theorem implies that the number $n_p$ of Sylow $p$-subgroups is a nontrivial divisor of $n / p^{v_p(n)}$ congruent to $1$ mod $p$.
Moreover, Burnside's transfer theorem implies

*

*if $v_p(n) = 1$ then $n / n_p$ shares a factor with $p-1$,

*if $v_p(n) = 2$ then $n / n_p$ shares a factor with $(p-1)(p+1)$.

Proof: Let $P \leq G$ be a Sylow $p$-subgroup. Assuming $v_p(n) \leq 2$, $P$ is abelian. By Burnside's transfer theorem (see the answer at https://math.stackexchange.com/a/484777/23805 for discussion), $N_G(P)$ acts nontrivially on $P$. Since $P$ is abelian, $P \leq C_G(P)$, so $N_G(P)/C_G(P)$ is isomorphic to a nontrivial subgroup of $\operatorname{Aut}(P)$ of $p'$-order. Hence $|N_G(P)| = n / n_p$ and $|\operatorname{Aut}(P)|$ share a prime factor other than $p$. Now consider the possibilities for $P$ and $\operatorname{Aut}(P)$. $\square$
Trawling through all odd $n \leq 10000$ (I recommend using technology) shows that we have reduced to three possibilities:

*

*$3^3 \cdot 5^2 \cdot 7$.

*$3^{4} \cdot 7 \cdot 13$.

*$3^{3} \cdot 5^{2} \cdot 11$.

Probably @DerekHolt can rule these out using Sylow's theorem in clever ways.
