Are there a finite group $G$ and a field $\mathbb{F}$ such that $\gcd(3,|G|)=1$ and the group algebra $\mathbb{F}[G]$ contains a zero divisor whose support is of size $3$?


Recall that the support of an element $\alpha$ of $\mathbb{F}[G]$ is the set $\{x\in G \;|\; \alpha(x)\neq 0\}$.


This question is related to the following one:


[Zero divisors of the form $1+x+y$ in the rational group algebra](https://mathoverflow.net/questions/208506/zero-divisors-of-the-form-1xy-in-the-rational-group-algebra)


One motivation to propose the question is the following well-known observation: if $a$ is an element of order $3$ in a group $G$ then $1+a+a^2$ is a zero divisor over any group algebra of $G$ (whose support is of size $3$). So the hypothesis $\gcd(3,|G|)=1$.



Zero divisors with support of size 3 in group algebras of finite groups
Multiplication by $\sum\_{g\in G}g$ annihilates any element whose sum of coefficients is zero. It remains to choose such element with support of size 3, which is possible whenever $\mathbb F\neq GF(2)$.


