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Rings, finite groups, and domains

  • Thread starter hsong9
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  • #1
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Homework Statement



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.

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.
 

Answers and Replies

  • #2
matt grime
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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.
That does not follow.

there exists an element (a m ) p such that ( a m )p* (am)q = 0
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.
 
  • #3
80
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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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