The Wikipedia page for monoid says
Any commutative monoid is endowed with its algebraic preordering $\leq$, defined by $x\leq y$ if there exists $z$ such that $x + z = y$.
I was wondering why we need commutativity here. It seems to me that for non-commutative monoids, we also have this algebraic preordering. To be precise, suppose $M$ is a generally non-commutative monoid and we define $x\leq y$ if there exists $z$ such that $xz= y$. We have
- (Reflexivity) For any $x$, $x\leq x$ since $xe=x$, where $e$ is the identity.
- (Transitivity) Assume $x\leq y$ and $y\leq z$. Then there exists $a$ and $b$ such that $xa=y$ and $yb=z$. Thus $x(ab)=z$. Therefore we deduce that $x\leq z$.
I guess we do not need commutativity here. However, in almost all related books that I read, when they talk about this natural pre-ordering, they assume commutative monoids. Thank you for your help :-)
The relation you define is the opposite of the Green's preorder $\leqslant_\mathcal{R}$ on a (not necessarily commutative) monoid $M$, defined as follows: $$ x \mathrel{\leqslant_\mathcal{R}} y \quad \text{if and only if} \quad xM \subseteq yM \quad \text{if and only if} \quad \text{there exists $z \in M$ such that $x= yz$} $$ This definition is quite standard, see for instance [1, p. 288] or [2, Chap. 2]. See also this related answer.
[1] S. Eilenberg, Automata, languages, and machines. Vol. B. With two chapters by Bret Tilson. Pure and Applied Mathematics, Vol. 59. Academic Press [Harcourt Brace Jovanovich, Publishers], New York-London, 1976.
[2] P. A. Grillet, Semigroups, An Introduction to the Structure Theory, Marcel Dekker, Inc., New York, 1995.