From: mareg@mimosa.csv.warwick.ac.uk () Subject: Re: A Finite Group Theory Question Date: 29 Dec 1999 17:04:38 GMT Newsgroups: sci.math Keywords: How many simple groups of a bounded order? In article <842csl\$80q\$1@nntp1.atl.mindspring.net>, "Daniel Giaimo" writes: > Let k(n) be the number of simple groups of order <= n. Considering >small values of n (up to, say, 10000) it seems reasonable to conjecture that >lim(n->oo)(k(n)/n) = 0. How would one go about proving this? (Hopefully >without invoking the classification of finite simple groups.) In fact it >seems that even >lim(n->oo)(k(n)/ln(n)) = 0. Is this known to be true? > >--Daniel Giaimo Your final conjecture lim(n->oo)(k(n)/ln(n)) = 0. is false, even assuming that you are restricting your attention to nonabelian simple groups. In fact lim(n->oo)(k(n)/ln(n)) = infinity. For each prime p, the simple group PSL(2,p) has order p(p-1)(p+1)/2, and by the prime number theorem there are certainly asymptotically more than ln(n) numbers of this form less than n. I think your more modest conjecture lim(n->oo)(k(n)/n) = 0. is true, although I don't think you have much hope of proving it without the classification. All enumeration theorems of this kind that I have come across use the classification of finite simple groups at the very least in order to show that there are at most two nonisomorphic simple groups of any given order - without the classification no sensible bound is known. If you are prepared to invoke the classification, then by looking at the orders of the groups in the various families, you should not find it too hard to show that the number of simple groups of order <= n is at most O(pp(n)), where pp(n) is the number of prime powers that are <= n. In fact, you cna probably show that, asymptotically, almost all simple groups are PSL(2,p) for p prime. A good general reference for enumeration theorems on various classes of finite groups is L. Pyber, "An enumeration theorem for finite groups", Ann. Math. 137, 203-220 (1993). Derek Holt.