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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.

2 added 221 characters in body

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.

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