sher@rochester.UUCP (02/11/86)
From: David Sher <sher> The most complete and pleasant reply to my original simple group theory problem was this letter by Bill Huber which I include herewith. Date: Mon, 3 Feb 86 17:12:16 est >From: allegra!sjuvax!bhuber (B. Huber) Subject: Re: A simple group theory problem Newsgroups: net.math In-Reply-To: <14907@rochester.UUCP> Organization: Saint Josephs Univ. Phila., Pa. Cc: Status: RO In article <14907@rochester.UUCP> you write: >I have been bothered for some time (so long that I don't remember when >I thought of it) by a relatively simple group theory problem. > Here is the problem: Is >every abelian (commutative) group the multiplicative group of some >number n? Random facts that seem relevant are that the multiplicative >group of every prime p might be (I'm not sure I remember this right) >isomorphic to the cyclic group in p - 1 elements. Also the >multiplicative group of 8 is not cyclic but generated by two >generators of size (I know I got this word wrong but you understand me >if not work out the group yourself) 2. I accept mail or you can post >if you think your answer is of general interest. You have an excellent memory: your assertions are correct. An infinite abelian group of course cannot be the multiplicative group modulo an integer n. The cyclic group of three elements does not arise as such a group. The reason is that the order of the multiplicative group modulo n is given by Euler's phi-function. There is a formula: one writes n as a product of positive powers of distinct primes. If p^e is one such factor, then it introduces a factor of p^e - p^(e-1) (which is exactly the order of the multiplicative group modulo p^e) in phi(n). Thus, for example, phi(3) = 2, phi(4) = 2, phi(6) = 2. Now these terms factor; p^e - p^(e-1) = (p-1)*(p^(e-1)). Since 3 cannot be so written, no matter what the prime p is, there is no integer n whose multiplicative group has order 3. You can produce other counterexamples in a similar fashion. Natural questions, such as given m, which n have phi(n) = m, generally are only partially answered or completely unanswered to date. You could go further and ask yourself (and the Net, if you like) whether every finite abelian group whose order is phi(n) for some integer n is actually isomorphic to some multiplicative group. Again, the answer is no. For counterexample, observe that the only n with phi(n) = 8 is n= 2^4 = 16, n = 5 * 2^2 = 20, n = 5 * 6 = 30, n = 3 * 5 = 15, or n= 3 * 2^3 = 24. No element has order 8 in any of these groups (exercise; I hope I'm right!). But the cyclic group of eight elements has seven elements of order 8, and so cannot be isomorphic to one of these. Bill Huber. -- -David Sher sher@rochester seismo!rochester!sher