MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $G$ be a finite abelian group. Is it true that the following element of the group ring ${\mathbb Z}[G]$: $$ \prod_{g\ne 1}(1-g) $$ is non-zero?

share|cite|improve this question
up vote 17 down vote accepted

Counterexample : $G=(\mathbb{Z}/2)^2$. The product is $1-a-b+ab -ab +a^2b+ab^2 -a^2b^2$, with $a^2=b^2=1$, $ab=ba$.

PS: on the other hand, this is true if (and only if?) $G$ cyclic, since then you have an injective character $\chi : G\to \mathbb{C}^\times$, whose linear extension to $\mathbb{Z}[G]$ has nonzero value on you element.

PPS: your expression is indeed $0$ in $\mathbb{Z}[G]$ whenever $G$ isn't cyclic. It is enough to shows it's zero in $\mathbb{C}[G]$, which (by elementary representation theory) is a product of fields $\mathbb{C}_\chi$ over the characters $\chi\in \hat{G}$. The result follows, since no character is injective when $G$ isn't cyclic.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.