This post is a continuation of [Weirdos but algebraic](https://mathoverflow.net/questions/360746/weirdos-but-algebraic).


Logically, the quoted post could follow the present one rather than precede it.


**Question** Does there exist an indecomposable weirdo which is neither an Abelian group nor is based on a module over certain left and right rings, with a distinguished invertible element for each of these rings?


These weirdos were described in my previous post. Also, *indecomposable* means indecomposable into a direct (Cartesian) product of two nontrivial weirdos.


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**Examples** of indecomposable weirdos:


* every weirdo such that the number of its elements is a prime number is indecomposable;
* every cyclic Abelian group of finite order $\ p^n,\ $ where $\ p\ $ is a prime, is indecomposable;
* the additive Abelian group $\ \Bbb Z\ $ is indecomposable;
* the set of ring $\ X:=\Bbb Z[\frac 12]\ $ is an indecomposable weirdo with respect to:


$$ \forall\_{x\ y\in X}\quad \sigma(x\ y)\ :=\ \frac{x+y}2. $$


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**EDIT**


Please. check my "EDIT" from my previous post.


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**PS.**   Within a day or two, I will leave MO, this time for good.


MO had great potential, I saw it for all these years. It was not fulfilled at all (just the opposite). So be it.



Indecomposable weirdos (cnt.)
Not sure if the following example counts as being different
from those listed in the problem, but consider the *weirdo*
with universe $\mathbb N = \{0,1,2,\ldots\}$ and
$\sigma(x,y)=x+y$  

$\lambda(x,y)=\rho(y,x) = x\ominus y$,   

where $x\ominus y = x-y$ if $x\geq y$
while $x\ominus y = 0$ if $x<y$.
**Comments.**


Everybody knows that a **quasigroup** is a structure $\langle Q; \*, /,\backslash \rangle$
satisfying the laws


(Axioms 1) $(x\* y)/y = x$, $x\backslash(x\* y) = y$, and   

(Axioms 2) $(x/y)\* y = x$, $x\*(x\backslash y) = y$.


These say that the operation table of $x\* y$ is a Latin square
on $Q$, or equivalently that the maps $x\mapsto a\* x$
and $x\mapsto x\* b$ are permutations of $Q$ with inverses
$x\mapsto a\backslash x$ and $x\mapsto x/b$. 


If we take $\sigma(x,y)=x\* y$, $\lambda(x,y) = x/y$,
and $\rho(x,y)=x\backslash y$ and only impose (Axioms 1), then
we get what the OP calls an ``eccentric''.
(Axioms 1) guarantee that multiplication is is cancellative
in each variable, but do not seem to force multiplication
to be a permutation in each variable.




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The **medial law** for multiplication
is $(x\* y)\* (u\* v) = (x\* u)\* (y\* v)$.


The OP's *weirdos* are medial groupoids satisfying (Axioms 1).
Weirdos satisfying (Axioms 2) are medial quasigroups.


The following theorem from the 1940's is relevant to the question here:
**Theorem.**
(Bruck (1944), Murdoch (1941), Toyoda (1941))
The following are equivalent for a quasigroup $Q$.  



(1) $Q$ satisfies the medial law for multiplication.  

(2) There is an abelian group structure on $Q$
and two commuting abelian group automorphisms $\alpha, \beta: Q\to Q$
such that $x\* y = \alpha(x)+\beta(y) + c$ for some $c\in Q$.
Thus, one does get an affine representation for
*weirdos* satisfying (Axioms 2).




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If we delete (Axioms 2) we are only guaranteed that
multiplication is cancellative.
But the following is a consequence of Corollary 1.2 of
my paper 
A quasi-affine representation.  

Internat. J. Algebra Comput. 5 (1995), 673--702.
**Consequence.**
If $\langle Q; \* \rangle$ is a medial, cancellative groupoid, then
$Q$ has a quasi-affine representation iff
$Q$ is abelian in the sense of commutator theory.

Thus abelian *weirdos* will have quasi-affine representations.
(Saying that $Q$ has a quasi-affine representation means
that there is an $R$-module $M$ containing
$Q$ as a subset and the multiplication is representable
as $x\*y = \alpha(x)+\beta(y) + c$ for some
not-necessarily-invertible
ring elements $\alpha, \beta\in R$ and some $c\in Q$.
You will see from the first example above, involving $\mathbb N$,
why we need to talk about subsets of modules rather than full modules.)




















