Does there exist a non-commutative algebraic structure with the following properties?

Let $M=(\mathbb{Q}_+\times\{0\})\cup(\mathbb{Q}\times\{1\})\cup\{\infty\}$ and consider the binary operation on $M$ defined as follows:

  • $(q,0)\cdot(r,0)=(q+r,0)$ for all $q,r\in \mathbb{Q}_+$
  • $(q,0)\cdot(r,1)=(q+r,1)$ for all $q\in\mathbb{Q}_+,r\in\mathbb{Q}$
  • $(r,1)\cdot(q,0)=(2q+r,1)$ for all $q\in\mathbb{Q}_+,r\in\mathbb{Q}$
  • $(q,1)\cdot(r,1)=\infty$ for all $q,r\in\mathbb{Q}$
  • $\infty\cdot x=x\cdot\infty=\infty$ for all $x\in M$

A bit of casework shows that this is associative. It also has the property that $a\cdot x=b$ and $x\cdot a=b$ each have a solution (for $a\neq b$) iff $a<b$, where $<$ is the total order on $M$ defined by ordering each of $\mathbb{Q}_+\times\{0\}$ and $\mathbb{Q}\times\{1\}$ according to their first coordinate and saying that every element of $\mathbb{Q}_+\times\{0\}$ is less than every element of $\mathbb{Q}\times\{1\}$ and that $\infty$ is the greatest element. It follows that your properties (3) and (4) hold, so $M$ is a magnium. However, it is not commutative.

As another way to get counterexamples, let $G$ be any totally ordered nonabelian group, and let $M$ be the monoid of nonnegative elements of $G$. Properties (3) and (4) follow from the fact that $a^{-1}b$ and $ba^{-1}$ are each nonnegative iff $a\leq b$. An explicit example of such a $G$ is the group of affine maps $K\to K$ of positive slope for any ordered field $K$. The subset $M$ can then be explicitly described as the set of maps of the form $x\mapsto ax+b$ where $a\geq 1$ and if $a=1$ then $b\geq 0$. (When $K=\mathbb{Q}$, this is closely related to the first example above, identifying $(q,0)$ with $x\mapsto x+q$ and $(r,1)$ with $x\mapsto 2x+r$.)


This is not an answer, but some commenters were searching for finite noncommutative magniums (magnia?), and the proof is too long for a comment, so I put it here.

Theorem. Every finite magnium is commutative.

Suppose $M$ is a noncommutative finite magnium. Let $\lt$ be the linear order on $M$ such that $a\lt b$ if $a\ne b$ and the equation $ax=b$ has a solution.

Call a pair $\{a,b\}\subseteq M$ special if it satisfies the conditions:
(1) $uv=vu$ whenever $u\le a$ and $v\le b$;
(2) $ab\notin\{a,b\}$;
(3) there is no element $g\in M$ such that $\{a,b\}\subseteq\langle g\rangle=\{g^n:n=1,2,3,\cdots\}$.

Let $a$ be the least element of $M$ which is in a noncommuting pair, and let $b$ be the least element which does not commute with $a$; so $ab\ne ba$ and $a\lt b$. Since $a\lt b$, there are elements $x,y\in M$ such that $ax=ya=b$. Now $x$ and $y$ can't both equal $b$; without loss of generality we assume that $x\ne b$, so $x\lt b$. Now it is easy to verify that $\{a,x\}$ is a special pair. We have proved that, in a finite magnium, the existence of a noncommuting pair implies the existence of a special pair.

Now let $c$ be the least element of $M$ which is in a special pair, and let $d$ be the least element such that $\{c,d\}$ is a special pair. Since $c\lt d$, there is an element $e$ such that $ce=d$. Then $e\lt d$, and it is easy to verify that $\{c,e\}$ is a special pair, contradicting the minimality of $d$.

Remark. We have actually shown that, if $M$ is a finite magnium, then for any $a,b\in M$ either $ab=ba=\max\{a,b\}$, or else $a$ and $b$ are both powers of some element $g\in M$.

Tags:

Abstract Algebra

Semigroups

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