When we consider the list of the prime numbers that divide the order of the 26 (or 27 if you include Tits group T) sporadic groups, we find that they all are among the 20 smallest prime numbers. In fact, the order of the monster sporadic group M is divided by the 15 following primes: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 41, 47, 59 and 71. Moreover, the order of the sporadic group J4 is divided by: 2, 3, 5, 7, 11, 23, 29, 31, 37 and 43, and the order of the sporadic group Ly is divided by 2, 3, 5, 7, 11, 31, 37 and 67. So that, because no other prime is dividing the order of another sporadic group, we find that considering the list of the 20 smallest prime numbers, only the 16th prime, 53, and the 19th prime, 61, are omitted of the list of the divisors of the orders of the sporadic groups. Do we know an explanation for this curious fact ? Gérard Lang
Take the 2minute tour
×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

A trivial remark: Since the number of these primes is less than the number of sporadic groups, the multiplicative subgroup of ${\mathbb Q}^{\times}$ generated by the orders of sporadic groups is not free on 26 (or 27) generators. :). 

