# Distributive nearring

A nearring is a ring-like structure $R$ such that

• $(R, +, 0)$ is a group (possibly non-abelian)
• $(R, \cdot)$ is a semigroup
• $(x+y)\,z=xz+yz$ (right-distributivity)

In other words, three conditions that may fail differ it from an actual ring:

2. There may be no multiplicative identity
3. Left-distributivity may fail

It's easy enough to give examples of nearrings for which arbitrary subsets of these 3 conditions nevertheless hold, except for (2)+(3) and pure (3). It is well-known that existence of identity and two-sided distributivity implies (1), so (2)+(3) without (1) is impossible. What about pure (3)?

Question: are there fully distributive nearrings with non-abelian additive subgroups?

Some thoughts:

Such nearrings are necessarily pretty weird. Adapting the usual proof that (2)+(3) implies (1), we can argue that $$a\,((x+y)-(y+x))=0$$ for all $x,y,z\in R$. This seems a pretty strong condition, even though lack of multiplicative identity does not allow us to conclude immediately $x+y=y+x$.

Definition Let left-distributing element be $a\in R$ such that $a(x+y)=ax+ay$ for all $x, y\in R$

All nearrings are isomprhic to subsets of $M(G)$ - nearring of functions from group $G$ to itself with pointwise addition and composition as multiplication. Left-distributing elements are precisely endomorphisms. Still, set of endomorphisms does contain identity (likely minor obstacle), and for nonabelian $G$ is not a sub-nearring of $R$ (major problem). Product of left-distributing elements is left-distributing, but sum of left-distributing elements may fail to be, so the sub-nearring generated by endomorphisms in general most likely does not satisfy (3). Sub-nearring generated by endomorphisms mapping $G$ to $Z(G)$ would do the trick, but addition would be abelian.

Consider the dihedral group $D_4=\langle a,b :a^4, b^2, (ab)^2\rangle$. Define $x+y=xy$ and $x\cdot y=[x,y]$. Claim: $(D_4,+,\cdot)$ is a distributive near-ring.
Recall commutator identities $[xy,z]=[x,z]^y[y,z]$ and $[z,xy]=[z,y][z,x]^y$. Note that the commutator subgroup coincides with the center. So we have $[xy,z]=[x,z][y,z]$ and $[z,xy]=[z,y][z,x]=[z,x][z,y]$. This translates to $(x+y)\cdot z=x\cdot z+y\cdot z$ and $z\cdot(x+y)=z\cdot x+z\cdot y$.