TITLE: Does there exist a non-commutative algebraic structure with the following properties?
QUESTION [9 upvotes]: A magnium is a set M with a binary operation $\cdot$ satisfying:

$|M| \ge 2$
For all $a$, $b$, $c$ $\in M$, $(a \cdot b) \cdot c = a \cdot (b \cdot c)$.
For all $a$, $b$ $\in M$ with $a \ne b$, exactly one of the equations $a \cdot x = b$ and $b \cdot x = a$ has a solution for $x$ in $M$.
For all $a$, $b$ $\in M$, the equation $a \cdot x = b$ has a solution for $x$ in $M$ if and only if the equation $y \cdot a = b$ has a solution for $y$ in $M$.

Examples of magniums are the positive real numbers and the non-negative integers under addition. Another example is the set $\{1, 2, 3, ..., 120\}$ under the operation $x \cdot y = \min\{x + y, 120\}$, which shows that magniums generally do not have the cancellation property.
So the question is, is there a non-commutative magnium? Currently I'm trying to think of some two-valued function $f(x, y)$ on $\Bbb{R}$ satisfying $f(x, y) \ge \max\{x, y\}$ that's associative but not commutative, and I'm not coming up with anything good.

REPLY [4 votes]: 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 $g\in M$.