Conjecture. There exists a function $f:\mathbb{N} \rightarrow \mathbb{N}$ such that if $\beta$ is a non-zero element of the complex group algebra $\mathbb{C}[G]$ of a finite group $G$ such that $1\in \text{supp}(\beta)$, and $(1+x+y) \cdot \beta =0$ or $(1+x-y) \cdot \beta =0$ for some distinct and nontrivial $x,y\in G$, then $\text{exp}\big (\langle x,y \rangle \big) \leq f(|\text{supp}(\beta)|)$, 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 form $1+x+y$ or $1+x-y$ of the complex group algebra of any residually finite group generates a finite subgroup.

Proof. Let $\alpha=1+x+y$ or $1+x-y$ 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 integral group algebras of torsion-free residually finite groups have no zero divisor with support size $3$.


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