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.

  • 3
    $\begingroup$ 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. $\endgroup$ – Sean Eberhard Jul 6 '18 at 8:31
  • $\begingroup$ @SeanEberhard Thanks for your comment. I just wanted to rule out the mentioned case. $\endgroup$ – Meisam Soleimani Malekan Jul 6 '18 at 17:55
  • 1
    $\begingroup$ 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. $\endgroup$ – Steffen Kionke Jul 24 '18 at 9:54
  • $\begingroup$ @SteffenKionke: Thanks for your answer, I think these two sets should have the identity, aren't? $\endgroup$ – Meisam Soleimani Malekan Jul 24 '18 at 14:42
  • $\begingroup$ @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. $\endgroup$ – Steffen Kionke Jul 25 '18 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 $b=(x_1\dots x_{k-1})x_k\in BA^\#$.

  • $\begingroup$ @tomgoowillie Thanks, this answers my question. $\endgroup$ – Meisam Soleimani Malekan Jul 6 '18 at 17:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.