TITLE: Group ring and left zero divisor
QUESTION [12 upvotes]: Let $K$ be a finite field and $G$ be a discrete group. 

Is it true that  for every $a,b\in K[G]$ the condition $ab=0$ implies $ba=0$?

It does not seem to be related to zero divisor problem, any ideas if this can be true and for which fields?

REPLY [2 votes]: If $G$ is torsion-free then the question of reversibility of $K[G]$ (that is, does $ab = 0$ imply $ba = 0$) is in fact equivalent to the zero divisor conjecture, for any field $K$.
Connell showed that such a $K[G]$ is prime, so given non-zero $a, b \in K[G]$ there exists $c \in K[G]$ such that $bca \neq 0$. If it were the case that $ab = 0$ then we have $a (bc) = 0$ but $(bc) a \neq 0$ and $K[G]$ is not reversible. This trick is how you show that the unit conjecture implies the zero divisor conjecture.
(In the other direction, if a ring has no zero divisors then reversibility holds trivially.)
Connell, I. G., On the group ring, Can. J. Math. 15, 650-685 (1963). ZBL0121.03502.