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 2008-06-02, 05:52 #1 jinydu     Dec 2003 Hopefully Near M48 175810 Posts (Z/pZ)* isomoprhic to Z/(p-1)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, ... p-1} with multiplication modulo p as the operation and Z/(p-1)Z is {0, 1, ... p-2} with addition modulo p-1 as the operation. This claim is not at all obvious to me. How do I justify it? Thanks
 2008-06-02, 06:57 #2 akruppa     "Nancy" Aug 2002 Alexandria 246710 Posts Both are cyclic, and they have the same order.
 2008-06-02, 07:50 #3 jinydu     Dec 2003 Hopefully Near M48 175810 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)*| = p-1, the order of any element in (Z/pZ)* divides p-1. Also, for any divisor d of p-1, 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 p-1 elements, this forces D >= p-1. Since p-1 is a multiple of all the orders, D <= p-1. Thus D = p-1. ... Choose an element with order D... The existence of an element of order p-1 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 2008-06-02 at 07:52
 2008-09-24, 10:22 #4 Peter Hackman     Oct 2007 linköping, sweden 22·5 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 2008-09-24 at 10:23
2008-09-24, 10:32   #5
ET_
Banned

"Luigi"
Aug 2002
Team Italia

2×2,417 Posts

Quote:
 Originally Posted by Peter Hackman 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
Code:
Fel 404: Den sida som du försökte nå är odefinierad,
i likhet med uttrycket 0/0.
Luigi

2008-09-24, 10:51   #6
Peter Hackman

Oct 2007

101002 Posts

Quote:
 Originally Posted by ET_ Code: Fel 404: Den sida som du försökte nå är odefinierad, i likhet med uttrycket 0/0. Luigi

Sorry:

http://www.mai.liu.se/~pehac/kurser/TATM54/booktot.pdf

2008-09-24, 11:34   #7
jinydu

Dec 2003
Hopefully Near M48

33368 Posts

Quote:
 Originally Posted by Peter Hackman I suppose it's too late for your purposes, but then I'm not breaking the rules.
Far too late. Not only have I finished the course, I've already graduated from university. Anyway, I did eventually finish the problem in time.

2008-09-24, 13:53   #8
ATH
Einyen

Dec 2003
Denmark

2·1,601 Posts

Quote:
 Originally Posted by ET_ Code: Fel 404: Den sida som du försökte nå är odefinierad, i likhet med uttrycket 0/0. Luigi
Translation: "The page that you tried to reach is undefined, like the expression 0/0."

2008-09-24, 14:06   #9
ET_
Banned

"Luigi"
Aug 2002
Team Italia

2·2,417 Posts

Quote:
 Originally Posted by ATH Translation: "The page that you tried to reach is undefined, like the expression 0/0."
Thank you. I had some idea thanks to that 404 and my little knowledge of German

Luigi

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