Indecomposable weirdos (cnt.)

Question

This post is a continuation of 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.

Answer 1

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.)