Problem. Suppose that are three prime numbers which are not necessarily distinct. Prove that if is a group of order then is not simple.
Proof. I proved in this post that a group of order where are (not necessarily distinct) primes is not simple. So I may assume that are distinct and, say, . For any prime number let be the number of Sylow -subgroups of Now suppose that is simple. So and, by the Sylow theorem, But we cannot have or because and hence Thus and so, since every Sylow -subgroup of has exactly elements of order the number of elements of order in is We also have and Thus there are more than elements of order Hence the number of elements of order is less than which is impossible because