Relay-Version: version B 2.10 5/3/83; site utzoo.UUCP
Posting-Version: version B 2.10.2 9/5/84; site sjuvax.UUCP
Path: utzoo!watmath!clyde!burl!ulysses!allegra!princeton!astrovax!sjuvax!bhuber
From: bhuber@sjuvax.UUCP (B. Huber)
Newsgroups: net.math
Subject: Re: Re: Finding groups of order N
Message-ID: <2563@sjuvax.UUCP>
Date: Mon, 25-Nov-85 10:07:31 EST
Article-I.D.: sjuvax.2563
Posted: Mon Nov 25 10:07:31 1985
Date-Received: Thu, 28-Nov-85 03:42:25 EST
References: <1208@bbncc5.UUCP> <9600026@uiucdcsp>
Organization: St. Joseph's University, Phila. PA.
Lines: 34

> 
> 
> 
> This is a very hard question for most n.  It is easy to show, however, that
> the number of nonisomorphic groups of order n is at most n^(n^2).

The method amounts to counting the number of distinct ways of filling in the
entries of a multiplication table for the group; it is an n X n table, with
n possibilities for each entry.  One can vastly improve upon this estimate with
the following elementary observations:

	1. Because a group always has an identity element, only an (n-1) X (n-1)	   subtable needs to be determined;

	2. Because every element in a group has an inverse, every row and every
	   column of the table has no repeated entries.

A crude way of utilizing these results, then, is to overcount the number of
multiplication tables like this:

	There are (n-1)! ways of filling in the first row of the subtable;
	there are (n-1)! ways at most of filling each of the next rows; but:
	the last row is completely determined by those preceding it.

This gives at most ((n-1)!) ** (n-2) different multiplication tables.

Now, if we permute the rows or permute the columns, we get a different table,
for the same group.  Thus we can divide our count by (n-1)! at least (but not
by the square of (n-1)!, because certain combinations of row and column
permutations will fix the entries in the table).

That nets  ((n-1!))  **  (n-3)  different, nonisomorphic tables of order n.
This is much smaller than n ** (n**2), but still is (probably) much larger
than the actual number, since we haven't considered the consequences of the
associative property.