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Rings, finite groups, and domains

  1. May 5, 2009 #1
    1. The problem statement, all variables and given/known data

    Let G be a finite group and let p >= 3 be a prime such that p | |G|.
    Prove that the group ring ZpG is not a domain.
    Hint: Think about the value of (g − 1)p in ZpG where g in G and where
    1 = e in G is the identity element of G.

    3. The attempt at a solution

    G is a finite ring and p is a prime number such that p divides the order of G for k times.
    since p is a prime less than the order of G , there exists an element a in G such that (ap) k = n where n is the order of G.
    there exists an element (a m ) p such that ( a m )p* (am)q = 0
    so, (a m )p and (am)q are the zero divisors in Zp G.
    ∴ Zp(G) is not an integral domain.
     
  2. jcsd
  3. May 5, 2009 #2

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    Re: RIng

    That does not follow.

    That is the product of two elements in the group, i.e. a third element in the group, so it cannot be zero in the group ring.

    Note that you have ignored the hint given to you, and in particular at no point have you used the fact that the field has characteristic p.
     
  4. May 6, 2009 #3
    Re: RIng

    How about this??

    Suppose that Zp[G] is a domain.

    Find some g in G with order p. Note that g is not 1.

    (g-1)^p = g^p - 1 = 1 - 1 = 0
    However, since we assumed that Zp[G] is a domain, it follows that g-1 = 0, so that g=1 - a contradiction.
     
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