20080602, 05:52  #1 
Dec 2003
Hopefully Near M48
2·3·293 Posts 
(Z/pZ)* isomoprhic to Z/(p1)Z?
In proving a much larger theorem, my abstract algebra textbook assumes this result without proof. Here p is a prime, (Z/pZ)* is the group of units of Z/pZ, i.e. {1, 2, ... p1} with multiplication modulo p as the operation and Z/(p1)Z is {0, 1, ... p2} with addition modulo p1 as the operation.
This claim is not at all obvious to me. How do I justify it? Thanks 
20080602, 06:57  #2 
"Nancy"
Aug 2002
Alexandria
2,467 Posts 
Both are cyclic, and they have the same order.

20080602, 07:50  #3 
Dec 2003
Hopefully Near M48
2×3×293 Posts 
Thanks. The hint from my teaching assistant was that the group of units of any finite field is cyclic. Unfortunately, I don't know the proof of that and I've been trying to find it. I did find one proof online, but it has a hole in it. Here's my paraphrasing of it, applied to the specific case I'm interested in:
We can regard (Z/pZ)* as a subset of the field Z/pZ. Since (Z/pZ)* = p1, the order of any element in (Z/pZ)* divides p1. Also, for any divisor d of p1, the number of elements such that x^d = 1 is at most d because t^d  1 is a polynomial in Z/pZ and hence has at most d solutions. Let D be the least common multiple of the orders of all the elements in (Z/pZ)*. Then every element is a root of the polynomial t^D  1. Since there are only p1 elements, this forces D >= p1. Since p1 is a multiple of all the orders, D <= p1. Thus D = p1. ... Choose an element with order D... The existence of an element of order p1 shows that (Z/pZ) is cyclic. The problem here is the existence of an element whose order is the least common multiple of all the orders. It's not clear to me why such an element must exist. I do know that in abelian group, order(xy) = order(x)order(y) whenever order(x) and order(y) are relatively prime. But once I drop the condition on relative primality, the result fails completely; I can't even claim that the right hand side is the least common multiple of order(x) and order(y). Last fiddled with by jinydu on 20080602 at 07:52 
20080924, 10:22  #4 
Oct 2007
linköping, sweden
10100_{2} Posts 
I suppose it's too late for your purposes, but then I'm not breaking
the rules. I'm surprised no one has answered your question. It's a standard result in Elementary Number Theory and its proof is given in just about any book on that topic, as well as in several more advanced texts. There are proofs on pp. 70 and 177 in the following document: http://www.mai.liu.se/~pehac/booktot.pdf Last fiddled with by Peter Hackman on 20080924 at 10:23 
20080924, 10:32  #5  
Banned
"Luigi"
Aug 2002
Team Italia
2^{6}·3·5^{2} Posts 
Quote:
Code:
Fel 404: Den sida som du försökte nå är odefinierad, i likhet med uttrycket 0/0. 

20080924, 10:51  #6  
Oct 2007
linköping, sweden
2^{2}×5 Posts 
Quote:
Sorry: http://www.mai.liu.se/~pehac/kurser/TATM54/booktot.pdf 

20080924, 11:34  #7 
Dec 2003
Hopefully Near M48
2×3×293 Posts 

20080924, 13:53  #8 
Einyen
Dec 2003
Denmark
3,037 Posts 

20080924, 14:06  #9 
Banned
"Luigi"
Aug 2002
Team Italia
2^{6}×3×5^{2} Posts 
