# Example of "ring" without the distributive property?

Can anyone give an example of an "non-artificial" algebraic structure that fails to be a ring only because of a lack of one- and two-sided distributive property?

• There are a bunch of good examples here: en.wikipedia.org/wiki/Near-ring Commented Oct 3, 2013 at 8:54
• @Casteels: This is an answer, not just a comment. Commented Oct 3, 2013 at 9:22
• This is an example of structure that satisfies all axioms of a ring except distributivity (and it is commutative): mathoverflow.net/questions/424372/… Commented Jun 23, 2022 at 0:56

Consider a multiplicatively written abelian group $G$ (more later), and the set $N$ of maps $G \to G$, written as exponents.
Define operations $+$ and $\cdot$ on $N$ by $$g^{m + n} = g^{m} g^{n},$$ and $\cdot$ is composition $$g^{m n} = (g^{m})^{n}.$$ Since $G$ is abelian, addition is commutative.
As to the distributive properties, $m (n + k) = m n + m k$ holds by the definition of $+$, but $$(m + n) k = m k + n k\tag{distr}$$ will fail if $k$ is not an endomorphism. To see this, take any two elements $a, b \in G$, consider any two maps $m, n$ such that $m : 1 \mapsto a$ and $n : 1 \mapsto b$, then if (distr) holds you have $$(a b)^{k} =(1^{m} 1^{n})^{k} = (1^{m + n})^{k}= 1^{(m + n) k} = 1^{m k + n k}= 1^{m k} 1^{n k} = a^{k} b^{k}.$$
So if $G$ has at least two elements, (distr) fails.