# Zero divisors in complex group algebras of residually finite groups

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.

• There are finitely generated, infinite groups with finite exponent. So I think the argument is not quite complete, but for applications to torsion-free groups, this should not be a problem. – Steffen Kionke Oct 24 '18 at 11:36
• @SteffenKionke The condition residually finite" on a finitely generated group of finite exponent garantees that the group is finite. – Alireza Abdollahi Oct 24 '18 at 15:30
• Yes, thanks and sorry. I didn't pay attention... – Steffen Kionke Oct 24 '18 at 16:10

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 $$\mathrm{exp}(G) \geq m$$.
• Thanks! Actually I undertand from your example that I need a notion of irreducibilty" on a zero divisor something like the following: the element should not be written as the product of two group algebra element with support size less than the element and maybe this property that (one of them is a zero divisor). I will pose my question for the first unsettled case of existence of zero divisors with respect to the support size. – Alireza Abdollahi Oct 24 '18 at 15:34