# About the product of finite subsets of a torsion free group

Let $G$ be a torsion free group with identity $e$. For a subset $X$ of $G$, denote by $X^\#$ the set $X\setminus\{e\}$. Let $A$ be a finite subset of $G$ containing $e$. Is there a finite subset $B$ containing $e$ such that $$A\subset B^\#A\quad\text{and}\quad B\subset BA^\#$$ It is obvious there is no such $B$ when $A=\{e,x\}$; this follows from the right inclusion and that $G$ is torsion free, so the $A$ in my question has at least 3 elements.

The above question comes from the zero divisor conjecture; if there exist non zero $\alpha,\beta\in\mathbb C[G]$ such that $\alpha\beta=0$, without loss of generality we can assume $A:=\text{support}(\alpha)$ and $B:=\text{support}(\beta)$ contain $e$, and then $A$ and $B$ should satisfy in above inclusions.

• I'm confused by the quantification in your question. On the face of it it looks like you're asking a question and simultaneously giving a counterexample. Commented Jul 6, 2018 at 8:31
• @SeanEberhard Thanks for your comment. I just wanted to rule out the mentioned case. Commented Jul 6, 2018 at 17:55
• Yes, such sets exist. Let $G$ be a torsion-free group without "unique product property", i.e. there exist finite subsets $A,B \subseteq G$ (containing $e$ if you like) such that every $x \in AB$ can be written in at least 2 ways as $x = ab$ with $a \in A$ and $b\in B$. Then these sets have the property you want. Commented Jul 24, 2018 at 9:54
• @SteffenKionke: Thanks for your answer, I think these two sets should have the identity, aren't? Commented Jul 24, 2018 at 14:42
• @MeisamSoleimaniMalekan Yes, you can arrange that A and B contain the identity element. A simple example can be found in D.S. Promislow's paper: MR0940281. Commented Jul 25, 2018 at 11:38

If such $B$ exists then for some $n>0$ there must exist elements $a_1,\dots ,a_n\in A^\#$ such that $a_1\dots a_n=e$. (The argument is essentially the one you use to rule out the case $\lbrace e,x\rbrace$ when $G$ is torsionfree.)
Conversely, suppose that there exist $a_1,\dots ,a_n\in A^\#$ such that $a_1\dots a_n=e$. Then $B$ can be taken to be $A^n=A\dots A$. In fact,
(1) for any $a\in A$ we have $a=ae\in B^\#A$ if $a\neq e$ and $a=(a_1\dots a_{n-1})a_n\in B^\#A$ if $a=e$,
(2) any $b\in B$ (including $e$) is a product $x_1\dots x_k$ with all $x_j\in A^\#$ and $1\le k\le n$, so that $b=(x_1\dots x_{k-1})x_k\in BA^\#$.