Conjecture. There exists a function $f:\mathbb{N} \rightarrow \mathbb{N}$ such that if $\alpha$ and $\beta$ are non-zero elements of the complex group algebra $\mathbb{C}[G]$ of a finite group $G$ such that $1\in \text{supp}(\alpha) \cap \text{supp}(\beta)$, $|\text{supp}(\alpha)|\leq |\text{supp}(\beta)|$ and $\alpha \cdot \beta =0$, then $\text{exp}\big (\langle \text{supp}(\alpha) \rangle \big) \leq f(|\text{supp}(\beta)|)$, where $\text{exp}(H)$ denotes the exponent of a finite group $H$ and $\text{supp}(\gamma)$ for an element $\gamma=\sum\_{g\in G} \gamma\_g \; g \in \mathbb{C}[G]$ denotes the set $\{ g\in G \;|\; \gamma\_g \not=0\}$.


Motivation. If the conjecture is true, then the support of any zero divisor of the complex group algebra of any residually finite group generates a finite subgroup.


Proof. Let $\alpha$ be a non-zero element of $\mathbb{C}[G]$ for some residually finite group $G$ such that $\alpha \cdot \beta =0$ for 
some non-zero $\beta \in \mathbb{C}[G]$. Let $H=\langle \text{supp}(\alpha) \rangle$. By Zelmanov's celebrated result on restricted Burnside problem, it is enough to show that the exponent of $H$ is finite. Since $H$ is finitely generated residually finite, there exists a descending series $H=N\_1\geq N\_2 \geq \cdots$ of normal subgroups $H$ of finite index such that $\cap\_{i\in\mathbb{N}} N\_i=1$. Note that there exists $k\in\mathbb{N}$ such that $\bar{\alpha} \cdot \bar{\beta}=0$ in $\mathbb{C}[H/N\_i]$ for all $i\geq k$ and $|\text{supp}(\bar{\beta})|=|\text{supp}(\beta)|=:t$, where $\bar{}$ is the natural ring epimorphism from $\mathbb{C}[H]$ onto $\mathbb{C}[H/N\_i]$. Now if the above conjecture is true then $\text{exp}(H/N\_i)\leq f(t)$ and so $\text{exp}(H)$ is finite. This completes the proof. 


If the above proof and conjecture are true then the complex group algebras of torsion-free residually finite groups have no zero divisor.



Zero divisors in complex group algebras of residually finite groups
The conjecture does not hold. 


Let $m \geq 1$ be an arbitrary integer.
Let $G = \langle x,y | x^m, y^5 , [x,y]\rangle$ be the finite abelian group isomorphic to a product of two cyclic groups $C\_m \times C\_5$.
Let $\alpha = xy +y -x -1$ and let $\beta = y^4 +y^3 +y^2 +y +1$ in $\mathbb{Z}[G]$.
Then $\alpha \beta = (x+1)(y-1)\beta= 0$.
The support of $\beta$ has $5$ elements and the support of $\alpha$ has $4$ elements. Moreover, the support of $\alpha$ generates the group $G$, which has exponent $\mathrm{exp}(G) \geq m$.


