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Every finite group whose order is divisible by at most two distinct primes is a solvable group. A nonabelian simple group must therefore have at least three distinct prime divisors.

For every $\epsilon >0$ there is $\lambda (\epsilon )$ such that every non-abelian finite simple group $G$ and every finite symmetric generating set $S$ containing the identity satisfyThe proof applies the one-scale virtual nilpotence theorem below the diameter and then uses simplicity to eliminate both the bounded-index subgroup and the finite normal subgroup.

Every finite group of odd order is a solvable group. In particular, a nonabelian simple group has even order and therefore has an involution by Cauchy's theorem for finite groups. This is a deep theorem, not an elementary consequence of the Sylow theorems.

A finite nonabelian simple group is a finite group that is a nonabelian group and a simple group. Every such group is a perfect group because its commutator subgroup is a nontrivial normal subgroup.

For distinct primes $p,q$, the Sylow theorems force a proper nontrivial normal subgroup in any group of order $p^{2}q$. If $p>q$, the Sylow $p$-subgroup is unique. If $p<q$ and both Sylow counts were nontrivial, the count congruences force $(p,q)=(2,3)$. In order twelve, four Sylow $3$-subgroups exhaust eight nonidentity elements; only three remain for a Sylow $2$-subgroup, making that subgroup unique. Thus the group is not a simple group.

An infinite residually finite group is not a simple group. For a nonidentity element, residual finiteness supplies a map to a finite group in which it survives. Its kernel is proper and cannot be trivial, since an infinite group cannot inject into a finite group.