# A BF-monoid $H$ s.t. $H^\times$ is not divisor-closed

Let $H$ be a (multiplicative) monoid, and denote by $H^\times$ the set of units of $H$ and by $\mathcal A(H)$ the set of atoms of $H$ (let me recall that an element $a \in H$ is an atom if (i) $a \notin H^\times$ and (ii) $a = xy$ for some $x, y \in H$ implies $x \in H^\times$ or $y \in H^\times$). Given $x \in H \setminus H^\times$, we take $${\sf L}_H(x) := \{k \in \mathbf N^+: x = a_1 \cdots a_k \text{ for some }a_1, \ldots, a_k \in \mathcal A(H)\};$$ moreover, we assume ${\sf L}_H(1_H) := \{0\}$ and ${\sf L}_H(u) := \emptyset$ for all $u \in H^\times \setminus \{1_H\}$. It is said that $H$ is atomic if ${\sf L}_H(x) \ne \emptyset$ for every $x \in H \setminus H^\times$, and a BF-monoid if $H$ is atomic and ${\sf L}_H(x)$ is finite for all $x \in H$. On the other hand, we call $H$ unit-cancellative provided that $xy = x$ or $yx = x$ for some $x, y \in H$ only if $y \in H^\times$.

Cancellative monoids are, of course, unit-cancellative, and it is an easy exercise to check that if $H$ is commutative or unit-cancellative, then $H^\times$ is a divisor-closed subgroup of $H$ (that is, $xy \in H^\times$ for some $x, y \in H$ only if $x, y \in H^\times$). So here is my question:

Q. What about the existence of a (non-commutative, non-unit-cancellative) BF-monoid $H$ for which $H^\times$ is not a divisor-closed subgroup of $H$?

Your monoid cannot be atomic if $H^\times$ is not divisor-closed.
Suppose $xy$ is a unit. Assume $x$ is not a unit. The other case is similar. Then $xyz=1$ for some $z$. Let $a$ be an atom. Then $a=x(yza)$. So by definition of an atom $yza$ must be a unit as $x$ is not. Thus $yzav=1$ for some $v$. Thus $yz$ is both left and right invertible hence invertible. Thus $x=(yz)^{-1}$ is a unit. Contradiction.
• Btw, your proof shows something more, namely, that $H^\times$ is divisor-closed whenever $H=H^\times$ or $\mathcal A(H) \ne \emptyset$. This implies in turn that, instead of assuming (as a convention) that the set of lengths of a unit $u\ne 1_H$ is equal to $\emptyset$, we can derive that from the following definition (alternative to the one in the OP): We take ${\sf L}_H(x) :=\{k\in \mathbf N^+: x=a_1\cdots a_k\text{ for some }a_1,\ldots,a_k\in \mathcal A(H)\}$ for every $x \in H\setminus \{1_H\}$ and ${\sf L}_H(1_H):=\{0\}$. – Salvo Tringali Feb 10 '17 at 18:54