Proving that a finite simple group (order < 100) is either abelian or has order 60

Question

As the title suggests, I'm asked to

Prove that a finite simple group $G$ of order less than $100$ is either abelian or has order $60.$

I approached the problem by saying that $G$ could either have a prime order or a non-prime order and have already proven that, if $G$ has order prime, it has to be abelian.

However, I'm stuck on what to do for the second case. I've seen examples online where they prove that the order of a finite simple nonabelian group $G$ is less than $60$, but how do I prove that there is no other order that $G$ can be if it is nonabelian that is between $61$ and $100?$

Answer 1

This is a classic problem which is an exercise in casework. I won't do the casework for you, but here are some key observations that will make your casework a lot easier. Let $G$ be a simple non-abelian group.

1. $G$ can't have a prime power order (consider its center)
2. $G$ is isomorphic to a subgroup of the alternating group on the left coset space of $G/H$ for any proper subgroup $H$.
3. The Sylow number equals the index of the Sylow normalizer for any $p$ prime.
4. For any $H\subseteq G$, the order of $G$ divides $\frac{[G:H]!}2$
5. $G$'s order can't be $p\cdot q\cdot r$, where $p,q,r$ are distinct primes.
6. $A_5$ is a simple non-abelian group.