**sher@rochester.UUCP (David Sher)** (01/30/86)

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. I will try to define it unambiguously. Consider the operation multiply on Z sub n (integers mod n). Consider the set of integers less than n relatively prime to n including 1. This set when combined with the operation of multiplication mod n is a group with identity 1. Call this group the multiplicative group of n. 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. -- -David Sher sher@rochester seismo!rochester!sher

**weemba@brahms.BERKELEY.EDU (Matthew P. Wiener)** (02/01/86)

In article <14907@rochester.UUCP> sher@rochester.UUCP (David Sher) writes: > Here is the problem: Is >every abelian (commutative) group the multiplicative group of some >number n? For n>2, -1 != 1 mod n. But (-1)^2 = 1, so every such group for n>2 contains an element of order 2. Thus the cyclic group of order 3, for example, cannot be such a group. A much harder approach (ie the one I first tried) is to find out what every such group looks like. The random facts you quoted are part of that method. See T Apostol, _Introduction to Analytic Number Theory_ for a nice low-key discussion. ucbvax!brahms!weemba Matthew P Wiener/UCB Math Dept/Berkeley CA 94720

**ka@hropus.UUCP (Kenneth Almquist)** (02/04/86)

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 ihnp4!houxm!hropus!ka (official name) ihnp4!opus!ka (shorter path)