3 classification

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.

For finite groups, I believe your property

(Thanks to Greg Marks:) The classification of finite group rings over fields where ab=0 implies ba=0 ) holds iff all absolutely simple modules are 1-dimensionalis given in:

Gutan, iff G/OpMarin; Kisielewicz, Andrzej. "Reversible group rings." J. Algebra 279 (G) is abelian2004), no. 1, where p 280–291. MR2078399 DOI:j.jalgebra.2004.02.011.

In particular, K is the characteristic a field of order 22n-1 and G is the field. Actuallyquaternion group of order 8, it looks more like or G itself has to be is abelian. It is necessary that G/OpLi and Parmenter (G) be abelian, but not sufficient2007) extend this to finite group rings over commutative rings with 1 in MR2372321.

2 hrm

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.

For finite groups, I believe it your property (ab=0 implies ba=0) holds iff all absolutely simple modules are 1-dimensional, iff G/Op(G) is abelian, where p is the characteristic of the field. Actually, it looks more like G itself has to be abelian. It is necessary that G/Op(G) be abelian, but not sufficient.

1

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.

For finite groups, I believe it holds iff all absolutely simple modules are 1-dimensional, iff G/Op(G) is abelian, where p is the characteristic of the field.