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?
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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 $C_7$.
answered Jun 5, 2015 at 10:23
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$\begingroup$ Very nice. Many thanks. Have you another example in which the orders of $x$ and $y$ are not $3$? One key in your example is that $(1+x+x^2)(1-x)=0$ if $x^3=1$. Anyway, thanks again. $\endgroup$ Commented Jun 5, 2015 at 19:30
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$\begingroup$ Using the above answer let me give an element $\beta$ of $\mathbb{Q}[G]$ such that $(1+x+y)\beta=0$ to record one. Let $G=\langle a,h \;|\; a^3=h^7=1, h^a=h^4 \rangle$. Now take $x:=a$ and $y:=(a^h)^{-1}$. Then $(1+x+y)(x+x^h+\cdots+x^{h^6}-x^{-1}-y-y^{h}-\cdots-y^{h^5})=0$. The support of $\beta$ has size $14$. There is another $\beta$ with support of size $21$. $\endgroup$ Commented Jun 5, 2015 at 19:54
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$\begingroup$ Again to record, for any two numbers $m,n$ (not both zero, of course) we have $(1+x+y)( n(x+x^h+\cdots+x^{h^6})+m(x^{-1}+y+y^h+\cdots+y^{h^5})-(n+m)(1+h+\cdots+h^6) )=0$ $\endgroup$ Commented Jun 6, 2015 at 1:41