From: mareg@primrose.csv.warwick.ac.uk (Dr D F Holt) Subject: Re: the number of groups of order n Date: 31 Aug 1999 08:33:22 GMT Newsgroups: sci.math.research Keywords: Pyber's estimate In article , Edwin Clark writes: >----------------------------------------------------------------------------- > This article is being reposted by the moderator because it was > cancelled by someone else on August 14. >----------------------------------------------------------------------------- > > >Let N(n) denote the number of groups of >order n and let SN(n) denote the number of >groups of order <= n. > >Supposedly there is no formula for N(n), >but is there an asympotic formula for SN(n) ? >And does the average SN(n)/n diverge? The best asymptotic upper bound known to date was proved by L. Pyber (Ann. Math. 137, 203-220 (1993)). Let \mu(n) denote the highest exponent e in any prime power p^e that divides n. Then (2/27 + o(1))\mu^2 N(n) <= n as \mu -> infinity For groups of order n = p^e, it was proved a long time ago by G. Higman that (2/27 - c/e)e^2 N(p^e) >= n (c constant) This was proved just by considering special groups of class 2. It is still an open conjecture whether, asymptotically, almost all groups are special 2-groups of class 2, but Pyber's result shows that this is true logarithmically. So yes, SN(n)/n diverges. >and get around 11,000. So I suspect that the average >diverges. But is this due to a relative small number of n for >which there is a large number of groups? Is it possible that >for most n the number of groups of order n >is small ? Yes, the large values of N(n) occur where there are large exponents in prime powers dividing n (so especially for powers of 2). I would guess that N(n) < n for almost all n, but you would need to know more number theory and the relative distribution of numbers without large exponents to investigate that further. Derek Holt.