# Prime numbers dividing the orders of the sporadic groups

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

• For the Monster group the primes are called supersingular and have an interpretation in terms of elliptic curves: en.wikipedia.org/wiki/Supersingular_prime_(moonshine_theory) . I think there are nascent moonshine theories associated to some of the other sporadic groups but nothing concrete. – Qiaochu Yuan Jan 8 '11 at 13:49
• This question is of the same type as "why 3 divides 9 and 2 does not?". Voted to close. – Mark Sapir Jan 8 '11 at 13:55
• There does not seem to be a uniform theory of sporadic groups (yet), so I think the answer to your question is "no". – S. Carnahan Jan 8 '11 at 14:03
• As Qiaochu points out, primes dividing the order of the Monster group were observed several decades ago by Andrew Ogg to have a strange correlation with elliptic modular forms. Similar observations for the entire collection of sporadic simple groups would certainly arouse interest, but in the meantime the answer to your question is "no" as Scott concludes. – Jim Humphreys Jan 8 '11 at 14:21
• Just because the answer to this question is "no" doesn't mean it should be closed. A "yes" answer would be incredibly interesting, and it's not too low-level to be asked. – Peter Shor Jan 8 '11 at 14:53

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. :-).