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:
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> This article is being reposted by the moderator because it was
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>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.