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About the order of finite simple groups there exists a very interesting result which stated as follows:

Let $G$ be an non-solvable simple group of order $g$. If $p\mid g$, where $p>g^{1\over 3}$ is a prime, then $p>3$ and either $G\cong L_2(p)$ or $p$ is a Fermat prime and $G\cong L_2(p-1)$.

This result proved by Brauer in 1958 and surely they proved it directly. Now using the classification of finite simple groups, we may prove this result too.

As a similar result we need all finite simple groups $S$ such that $S$ is of order $g$ and $p\mid g$, also $g$ is a divisor of ${p(p-1)^2(p+1)\over 4}<{p^4\over 4}$. I guess that if $p>5$, then $L_2(p)$ is the only answer. Any references or hints would be highly appreciated. Thanks in advance.

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up vote 11 down vote accepted

I think you will just have to work through the orders of the finite simple groups and show that only $L_2(p)$ is possible.

Let's try $L_3(q)$. This has order $q^3(q^2+q+1)(q+1)(q-1)^2/(q-1,3)$. Your prime $p$ must divide one of these factors, so the largest possible value of $p$ is $q^2+q+1$, and you will find that $|L_3(q)| > (q^2+q+1)^4/4$ for $q>13$, and smaller values can be ruled out easily.

$U_3(q)$ of order $q^3(q^2-q+1)(q-1)(q+1)^2/(q+1,3)$ is similar.

For larger rank groups of Lie type, the number of factors increases and so the argument is easier. For example, for $L_4(q)$, the largest possible prime is $q^3+q^2+q+1$ (which in fact can't be prime), but $|L_4(q)|$ is about $q^{15}$.

The other low rank groups, $^2B_2(q)$ and $^2G_2(q)$ might need a bit of thought. We have $|^2B_2(q)| = |{\rm Sz}(q)| = q^2(q^2+1)(q-1)$ but, since $q$ is an odd power of $q$, we have the factorization $q^2+1 = (q + \sqrt{2q}+1)(q-\sqrt{2q}+1)$, so the largest possible $p$ is $q + \sqrt{2q}+1$. There is a similar factorization of $|^2G_2(q)|$ with $q$ an odd power of $3$.

But I am afraid that you will need to the hard work yourself.

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