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 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 at 15:30
  • Yes, thanks and sorry. I didn't pay attention... – Steffen Kionke Oct 24 at 16:10
up vote 6 down vote accepted

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 $\mathrm{exp}(G) \geq m$.

  • 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 at 15:34

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