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](https://mathoverflow.net/a/259790/16537)* provided that $xy = x$ or $yx = x$ for some $x, y \in H$ only if $y \in H^\times$. 


[Cancellative monoids](https://en.wikipedia.org/wiki/Cancellative_semigroup) 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 $H$?
> 
> 
> 



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


