Prove that the group ring Z p G is not a domain.

[hsong9]

Prove that the group ring Z_pG is not a domain.

Homework Statement

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

The Attempt at a Solution

Suppose that Z_pG 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 Z_pG is a domain, it follows that g-1 = 0, so that g=1 - a contradiction.

 

[vacuum]

Therefore, ZpG is not a domain.

Explanation:

A domain is a commutative ring in which the product of any two non-zero elements is also non-zero. In this case, we are considering the group ring ZpG, which is the set of all formal sums of the form ∑ r_gg, where r_g in Zp and g in G. Addition and multiplication in this ring are defined in the usual way for polynomials, with the additional rule that the elements of G commute with each other.

Now, let g be an element of G with order p. Since p is a prime and p | |G|, we know that g^p = 1, where 1 is the identity element of G. This means that (g-1)^p = g^p - 1 = 1 - 1 = 0.

If ZpG were a domain, then (g-1)^p = 0 would imply that g-1 = 0, since the product of any two non-zero elements in a domain is also non-zero. However, this would mean that g = 1, which is a contradiction since we assumed that g is an element of G with order p, and g cannot be equal to the identity element 1.

Therefore, we have shown that (g-1)^p = 0 but g-1 ≠ 0, which means that ZpG is not a domain. This completes the proof.