Let $H$ be a monoid, and denote by $H^\times$ and $\mathcal A(H)$, respectively, the *set of units* (or *invertible elements*) and the *set of atoms* (or *irreducible elements*) of $H$ (an element $a \in H$ is an atom if $a \notin H^\times$ and $a = xy$ for some $x, y \in H$ implies $x \in H^\times$ or $y \in H^\times$). 


Given $x \in H$, we set $\mathsf L\_H(x) := \{k \in \mathbf N^+: x = a\_1 \cdots a\_k \text{ for some }a\_1, \ldots, a\_k \in \mathcal A(H)\}$ if $x \ne 1\_H$ and $\mathsf L\_H(x) := \{0\} \subseteq \mathbf N$ otherwise (in factorization theory, this is referred to as the *set of lengths* of $x$ (relative to the atoms of $H$)). We say that $H$ is a *BF-monoid* if $\mathsf L\_H(x)$ is non-empty and finite for every $x \in H \setminus H^\times$.



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> **Q.** Does there exist a commutative BF-monoid $H$ such that $H \ne H^\times$ and $au = a$ for all $a \in \mathcal A(H)$ and $u \in H^\times$? If so, can we make $|H^\times| = \kappa$ for every fixed (small) cardinal $\kappa \ne 0$?
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My guess is that the answer to both questions is positive, but so far I haven't been able to construct an example to prove it. And though the question is not so important, a positive answer would shed light on the relation (and the contrast) between two different "philosophies" beyond the definition of what is called the *factorization monoid* of $H$.



A non-reduced, commutative BF-monoid s.t. $au = u$ for all $a \in \mathcal A(H)$ and $u \in H^\times$
The answer seems yes if I understood the question. Take $G$ to be any commutative group and $N$ be the free semigroup on one-generator $x$ (so isomorphic to the positive natural numbers under $+$, but I want to use multiplicative notation). Let $H=G\cup N$ where the operation on $G$ and $N$ are their original operations and the product $gn=n=ng$ for any $g\in G$ and $n\in N$. Then $H^\times=G$ and your property is satisfied. The element $x$ is the only atom of $H$ and each non-unit is uniquely a product of atoms so it is a BF-monoid. Of course $G$ can have any cardinality.


