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- All that I have found on near-rings features one-sided distributivity. Why not two-sided? I post some results that I have derived for the two-sided case, like lack of a multiplicative identity for nontrivial multiplication.
As an introduction, an abstract-algebra ring generalizes integers with addition and multiplication. The multiplication operation is distributive over the addition operation, this multiplication forms a semigroup over set R of its elements, and this addition forms an abelian group over R. But can addition be made noncommutative? Near-ring - Wikipedia describes what one finds when one does so and makes the distributive property one-sided. But what happens if one makes it two-sided? I have derived some results.
From two-sided distributivity,
Additive identity 0: a*0 = 0*a = 0 for all a in R -- multiplicative zero / annihilator / absorber
Additive inverse -(): (-a)*b = a*(-b) = -(a*b) for all a,b in R
(a+b)*(c+d) - take left, then right distributivity and subtract right, then left distributivity. One finds:
a*c + b*d = b*d + a*c for all a,b,c,d in R
In short, R*R is commutative. It includes the additive identity and it is closed under additive inversion. What happens when multiplication has an identity? Set a = b = 1. Then c + d = d + c and the additive group is commutative. So for that group to be noncommutative, multiplication must not have an identity.
Consider fixing one multiplicand and varying the other one over R. Thus,
a*R = (some abelian subgroup of the additive group)
R*a = (some abelian subgroup of the additive group, not necessarily the same one)
These subgroups may differ for different a's; a = 0 gives a*R = R*a = {0}.
Both multiplications are group homomorphisms, and for nonabelian +, the only trivial one is to {0}. Thus, for nonzero products, the additive group must have a nontrivial composition series with an abelian quotient group in it. Thus, the additive group must not be simple, though it need not be solvable.
The quaternionic icosahedral group is not possible, because its center is {I,-I} with quotient group (icosahedral group), a simple group.
However, the reflection icosahedral group is possible, at least if one ignores multiplicative associativity, because it is {I,-I} * (icosahedral group) and that can have quotient group Z2, from the {I,-I} part.
From two-sided distributivity,
Additive identity 0: a*0 = 0*a = 0 for all a in R -- multiplicative zero / annihilator / absorber
Additive inverse -(): (-a)*b = a*(-b) = -(a*b) for all a,b in R
(a+b)*(c+d) - take left, then right distributivity and subtract right, then left distributivity. One finds:
a*c + b*d = b*d + a*c for all a,b,c,d in R
In short, R*R is commutative. It includes the additive identity and it is closed under additive inversion. What happens when multiplication has an identity? Set a = b = 1. Then c + d = d + c and the additive group is commutative. So for that group to be noncommutative, multiplication must not have an identity.
Consider fixing one multiplicand and varying the other one over R. Thus,
a*R = (some abelian subgroup of the additive group)
R*a = (some abelian subgroup of the additive group, not necessarily the same one)
These subgroups may differ for different a's; a = 0 gives a*R = R*a = {0}.
Both multiplications are group homomorphisms, and for nonabelian +, the only trivial one is to {0}. Thus, for nonzero products, the additive group must have a nontrivial composition series with an abelian quotient group in it. Thus, the additive group must not be simple, though it need not be solvable.
The quaternionic icosahedral group is not possible, because its center is {I,-I} with quotient group (icosahedral group), a simple group.
However, the reflection icosahedral group is possible, at least if one ignores multiplicative associativity, because it is {I,-I} * (icosahedral group) and that can have quotient group Z2, from the {I,-I} part.