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.