A non-reduced, commutative BF-monoid s.t. $au = u$ for all $a \in \mathcal A(H)$ and $u \in H^\times$

Question

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 $x \in H \setminus H^\times$.

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 $\kappa \ne 0$?

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 $H$.

Answer 1

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.

Answer 2

Benjamin Steinberg's example is neat and brilliant. Here is a different construction, which, though blatantly complicate, could help with related questions (I had had it in mind for days, but for some reason I couldn't make it work until minutes ago...).

To start with, let $M$ be an additively written monoid (either commutative or not). I'll denote by $\mathcal P(M)$ the extended power monoid of $M$, that is, the monoid obtained by endowing the set of all subsets of $M$ with the binary operation $$ (X, Y) \mapsto X+Y := \{x+y: x \in X \text{ and }y \in Y\}, $$ and by $\mathcal P_{\rm fin}(M)$ the submonoid of $\mathcal P(M)$ consisting of all non-empty finite subsets of $M$ (namely, the power monoid of $M$).

Accordingly, let $\mathbb N = (\mathbf N, +)$ be the additive monoid of non-negative integers and $G$ an additively written abelian group of cardinality $\kappa$, and take $H$ to be the smallest submonoid of $\mathcal P(G \times \mathbb N)$ containing all $1$-element subsets of $G \times \{0\}$, as well as all sets of the form $G \times X$ for which $X$ is a finite subset of $\mathbf N$ with $0 \in X$ and $|X| \ge 2$. Then $H$ is a commutative monoid with $$H^\times = \bigl\{\{(g,0)\bigr\}: g \in G\} \simeq_{\sf Grp} G \quad\text{and}\quad \mathcal A(H) = \{G \times X: 0 \in X \in \mathcal A(\mathcal P_{\rm fin}(\mathbb N))\},$$ and actually a (commutative) BF-monoid, as it follows from considering that $H$ is unit-cancellative and the function $H \to \mathbf N: (A, B) \mapsto |B| - 1$ is a length function (here we use a very basic result from additive number theory, i.e., that $|X+Y| \ge |X| + |Y| - 1$ for all non-empty $X, Y \subseteq \mathbf N$). Moreover, $A + U = A$ for every $U \in H^\times$ and $A \in \mathcal A(H)$, since $\{(g,0)\} + G \times X = G \times X$ for every $X \in \mathcal P_{\rm fin}(\mathbb N)$.

Notes. A monoid $M$ is called unit-cancellative provided that $xy = x$ or $yx = x$ for some $x, y \in M$ only if $y \in M^\times$, and a function $\lambda: M \to \mathbf N$ is a length function if $\lambda(x) < \lambda(y)$ for all $x, y \in M$ for which $y = uxv$ for some $u, v \in M$ with $u \notin M^\times$ or $v \notin M^\times$. If $M$ is unit-cancellative, then it's possible to prove that $M$ is BF iff $M$ has a length function (if requested, I can give a reference).