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Let $G$ be an arbitrary torsion-free group. For $x,y\in G$, which of these elements‌ can be decided immediately not to be zero divisors in $\mathbb ZG$ (or in $\mathbb CG$)? $$1+x+y,\quad 4+x+x^{-1}+y+y^{-1},\quad\text{or}\quad 3+x+x^{-1}+y+y^{-1}$$

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    $\begingroup$ The second one cannot be a zero-divisor in $\mathbb CG$. This is a standard argument looking at the action of $\mathbb CG$ on $\ell^2(G)$. If some non-zero vector $\xi \in \ell^2(G)$ is annihilated by $4+x+x^{-1}+y+y^{-1}$ one can show using computations with scalar products that $-x\xi = \xi$. But that is impossible if $x$ has infinite order. For the other two, I do not think that it is known if they can be zero-divisors or not. $\endgroup$ Oct 10, 2018 at 14:50
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    $\begingroup$ For any group $G$ and any fixed nonzero $w\in ZG$, an easy argument shows that $w$ is zero divisor in $ZG$ iff it's zero divisor in $CG$ (iff it's in $KG$ for $K$ of char 0). And this implies $w$ being zero divisor in $(Z/pZ)G$ for each prime $p$. $\endgroup$
    – YCor
    Oct 10, 2018 at 15:10
  • $\begingroup$ @YCor: I do not understand why if $w$ is zero divisor in $\mathbb CG$, it is also a zero divisor in $\mathbb ZG$? $\endgroup$ Oct 10, 2018 at 15:48
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    $\begingroup$ Let me prove it for $1+x+y$, the general case is similar. If it's zero divisor in $KG$ ($K$ field of char 0 or more generally ring that's a torsion-free $Z$-module), it means that there's a nonzero finitely supported family $(r_g)_{g\in G}$ such that $r_g+r_{x^{-1}g}+r_{y^{-1}g}=0$ for all $g$. Let $A$ be the abelian subgroup generated by these elements, and choose a homomorphism $f:A\to Z$ not vanishing on all $r_g$. Then $f$ maps $(f(r_g))_{g\in G}$ is a family with the same property in $Z$ instead of $K$ (note that the product of $K$ does not appear!). So $1+x+y$ is zero divisor in $ZG$. $\endgroup$
    – YCor
    Oct 10, 2018 at 15:52
  • $\begingroup$ @AndreasThom: can you please explain why $-x\xi=\xi$ is impossible? $\endgroup$ Oct 10, 2018 at 15:53

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