Let $\mathcal{M}$ be a commutative monoid and $x, y \in \mathcal{M}$.
In general, is there something one can say about $x$ and $y$ if for any $n \in \mathbb{N}$ the element $y^n$ divides $x$, i.e., there is a $z_n$ such that $x = y^n \cdot z_n$?
In certain concrete monoids the following possibilities appear.
- $x$ is a zero element, that is, $x \cdot z = x$ for all $z$
- $y$ must be idempotent (e.g., natural numbers, non-negative rationals under addition, non-negative reals under addition ...)
- $y$ is invertible (any group)
What other possibilities are there, if any, and can one say something meaningful about commutative monoids where if $x$ and $y$ satisfy the condition above then either $x$ is a zero element or $y$ is idempotent?
Of course such a monoid cannot have non-idempotent invertible elements.
Recall that an epigroup (also called group-bound semigroup) is a semigroup in which every element has a power that belongs to a subgroup. Note that the identity of the subgroup is an idempotent which is not required to be an identity for the semigroup.
Let $S$ be an epigroup and let $x \in S$. Then some power of $x$, say $x^n$, belongs to group $G$ with identity $e_x$. Note that $e_x$ is entirely determined by $x$. Indeed, suppose that $x^m$ belongs to some group $H$ with identity $e$. Then $x^{nm} \in G \cap H$ and thus $e_x = e$.
Now, I claim that $y^n$ divides $e_y$ for all $n \geqslant 0$. Indeed, since $S$ is an epigroup, there exists $k > 0$ such that $y^k$ belongs to a group $G$ with identity $e_y$. Let us choose $q$ and $r$ such that $n = kq - r$ with $q > 0$ and $r \geqslant 0$. Then $y^ny^r = y^{kq} = (y^k)^q \in G$ and thus $y^n$ divides $y^ny^r$ which divides $e_y$.
It follows that $y^n$ divides $x$ for all $n \geqslant 0$ if and only if $e_y$ divides $x$.