Relay-Version: version B 2.10 5/3/83; site utzoo.UUCP
Posting-Version: version B 2.10.1 6/24/83; site alice.UUCP
Path: utzoo!watmath!clyde!burl!ulysses!allegra!alice!amo
From: amo@alice.UucP (n)
Newsgroups: net.math
Subject: Re: Finding groups of order N
Message-ID: <4591@alice.UUCP>
Date: Wed, 20-Nov-85 07:37:29 EST
Article-I.D.: alice.4591
Posted: Wed Nov 20 07:37:29 1985
Date-Received: Thu, 21-Nov-85 07:29:40 EST
Organization: Bell Labs, Murray Hill
Lines: 21

In general, the problem of determining the number of non-isomorphic groups
of a given order is very hard, and it seems hardest when the order is
a prime power.  (The latest result on groups of order 2**k is due to
E. Rodemich, but I don't have the reference at hand.)  On the other hand,
when the order is square-free, there is a remarkable formula for this
number, due to O. Hoelder (Nachr. Koenigl. Ges. Wiss. Goettingen Math.-Phys.
Klasse 1 (1895), pp. 211-229).  Define 

  f( n, m ) = product( gcd( n, q-1 ), q runs over prime divisors of m ).

Then, for n square-free, the number of non-isomorphic groups of order n is

     sum( g( n, d ), sum over d dividing n ),

where

   g( n, d ) = product( ( f(p,n/d) - 1 ) / ( p - 1 ), product over primes 

                                                      p dividing d ).
References: <1208@bbncc5.UUCP>