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 :-)