Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
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
add comment

1 Answer

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 $0$.

share|improve this answer
thanks for the fast reply, it is correct. –  Kate Juschenko May 1 '11 at 17:42
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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