Is there a finite non-ablelian group $G$ generated by $x$ and $y$ such that $1+x+y$ is a zero divisor in the rational group algbera $\mathbb{Q}[G]$ and also $x^2$, $y^2$ and $(x^{-1} y)^2$ are all non-trivial?



Zero divisors of the form $1+x+y$ in the rational group algebra
If $G$ has a cyclic quotient $G/N$ of order $3$ in which $x$ and $y$ map to the two non-identity elements, then $1+x+y$ acts as zero on the non-trivial linear complex representations of $G/N$. So $1+x+y$ does not have full rank in the regular representation of $G$, and is therefore a zero divisor.


For example, take $G=C\_7\rtimes C\_3$, and $x$ and $y$ generators of two different Sylow $3$-subgroups in different cosets of $C\_7$.


