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From: ka@hropus.UUCP (Kenneth Almquist)
Newsgroups: net.math
Subject: Re: A simple group theory problem
Message-ID: <245@hropus.UUCP>
Date: Tue, 4-Feb-86 12:59:33 EST
Article-I.D.: hropus.245
Posted: Tue Feb  4 12:59:33 1986
Date-Received: Wed, 5-Feb-86 04:44:56 EST
References: <14907@rochester.UUCP>
Organization: Bell Labs, Holmdel, NJ
Lines: 39

David Sher poses a problem which I will interpret as follows:
	Given an integer n > 1, the set of all positive integers less
	than n whose greatest common denominator with n is 1 form a
	group under the operation of multiplication modulo n.  David
	calls this group the muliplicative group of n, and asks whether
	there exists a finite abelian group which is not isomorphic to
	the multiplicative group of some n.

I contend that the cyclic group of order 3 is such a group.  This group
has 3 elements, and I will prove that no multiplicative group contains
3 elements.

Clearly the multiplicative group of n will contain less than 3 elements
if n <= 3.  If n > 3, then the multiplicative group of n must contain 1
and n-1, which will be distict.  Therefore, a multiplicative group with
3 elements must contain exactly one element in addition to 1 and n-1.
Call this element p.

Now consider n-p.  Clearly, this cannot be 1, or p would be equal to
n-1, which we have assumed it does not.  Similarly, n-p cannot equal
n-1.  Finally, n-p cannot equal p, for if it did, n would equal 2p, and
thus p would not be relatively prime to n.  Therefore, if our group is
to contain exactly 3 elements, n-p cannot be relatively prime to n.

Let g > 1 be the greatest common denominator of n and p-n.  Then for
some integers i and j,
	n-p = ig
	n = jg
It follows that
	p = (j-i)g
which contradicts our assumption that p is relatively prime to n.  Since
our assumption that a multiplicative group with 3 elements leads to
this contradiction, we conclude that there is no such group.

I have just noticed that this proof can be extended to show that no
multiplicative group has an odd number of elements.
				Kenneth Almquist
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