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



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> Is it true that for every $a,b\in K[G]$ the condition $ab=0$ implies $ba=0$?
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It does not seem to be related to zero divisor problem, any ideas if this can be true and for which fields?



Group ring and left zero divisor
Let G be non-abelian of order 6, with x of order 2 and y of order 3. In such a group yxy = x, since both x and xy have order 2. Let K be a field with 2 elements. Then (x+y)⋅(1+xy) = x+y + y+yxy = x+y + y+x = 0, but (1+xy)⋅(x+y) = x+y + xyx+xyy = x+y + yy + xyy ≠ 0.


You may be thinking of the property: if a⋅b = 0 then there is some non-zero c such that c⋅a = 0. This holds in all (two-sided) Artinian rings (because elements are either units or zero-divisors). I believe this is true for two-sided self-injective rings as well. I don't know if it is possessed by group rings of infinite groups over finite fields.


(Thanks to Greg Marks:) The classification of finite group rings over fields where ab=0 implies ba=0 is given in:



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> Gutan, Marin; Kisielewicz, Andrzej. "Reversible group rings."
>  J. Algebra **279** (2004), no. 1, 280–291.
>  MR[2078399](http://www.ams.org/mathscinet-getitem?mr=2078399) DOI:[j.jalgebra.2004.02.011](http://dx.doi.org/10.1016/j.jalgebra.2004.02.011).
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In particular, K is a field of order 22n-1 and G is the quaternion group of order 8, or G is abelian. Li and Parmenter (2007) extend this to finite group rings over commutative rings with 1 in MR[2372321](http://www.ams.org/mathscinet-getitem?mr=2372321).


