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Let $K$ be a finite field and $G$ be a discrete group.

Is it true that for every $a=e+a_1+\ldots+a_n,b=e+b_1+\ldots+b_m\in K[G]$ with $b_i\neq e,a_j\neq e$ the condition $ab=0$ implies $ba=0$?

It is related to this question. The examples proposed there do not violate the condition above.

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Why did you put the tag "sofic groups"? – Andreas Thom May 1 '11 at 17:12
This comes as a subquestion of the condition ab=1 implies ba=1. Maybe this question is true for some simple reasons, which will restrict the class of groups that satisfy ab=1. – Kate Juschenko May 1 '11 at 17:24
up vote 5 down vote accepted

I hope I understand the question right: the $a_i$ are to be distinct elements of $G$, as are the $b_j$?

If so, then the answer is no. Let $K$ be $\mathbb Z/2$ and suppose that $x$ and $y$ are elements of $G$ such that $x^2=e$ and $xy$ is not equal to $yx$. Let $a=e+x$ and let $b=(e+x)(e+y)=e+x+y+xy$. Then $ab=0$ but $ba=y+xy+yx+xyx$ is not $0$.

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thanks for the fast reply, it is correct. – Kate Juschenko May 1 '11 at 17:42

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