Let $P$ be a finite set, $\mathscr F(P)$ the free abelian monoid with basis $P$ (which I'll write multiplicatively), $H$ a submonoid of $\mathscr F(P)$, and $\mathcal A(H)$ the set of atoms of $H$ (where an atom is a non-unit element that doesn't factorize in a non-trivial way). We say that $H$ is a $\text{C}_0$-monoid if there exists $\alpha \in \mathbf N^+$ such that $$ H \cap p^{2\alpha} \mathscr F(P) = p^\alpha(H \cap p^\alpha \mathscr F(P)), \quad\text{for all }p \in P. $$ E.g., $\text{C}_0$-monoids are used as canonical models in the study of various phenomena of interest related to the arithmetic of orders in number fields. Given $x \in \mathscr F(P)$, we take $$ {\rm supp}(x) := \{p \in P: {\sf v}_p(x) \ne 0\} \subseteq P $$ to be the support of $x$, where ${\sf v}_p(\cdot)$ denotes the $p$-adic valuation naturally associated with $\mathscr F(P)$. Then, we define $$\Lambda(H) := \{{\rm supp}(x): x \in H\} \setminus \{\emptyset\}.$$ We regard $\Lambda(H)$ as a subposet of the power set of $P$ ordered by $\subseteq$, and we say that $H$ is simple if the minimal elements of $\Lambda(H)$ are singletons.

Question. Suppose that $H$ is a simple $\text{C}_0$-monoid and let $E$ be the set of all $p \in P$ such that ${\rm supp}(x) = \{p\}$ for some $x \in H$. Is it true that, for every $p \in P\setminus E$, there exists $k \in \mathbf N^+$ such that $p^k a \in \mathcal A(H)$ for every $a \in \mathcal A(H)$?

The question has actually an additive flavor, were it only for the fact that it has an equivalent formulation in the monoid $(\mathbf N^{|P|}, +)$.

Note that the restriction, in the question, to the elements of $P \setminus E$ is necessary, since it is not too difficult to show that, if $H$ is a $\text{C}_0$-monoid (either simple or not), then there is a bound $K \in \mathbf N$ such that ${\sf v}_p(a) \le K$ for all $p \in E$ and $a \in \mathcal A(H)$.


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