Let $H$ be a monoid (written multiplicatively) with the property that $H = H^\times A H^\times$ for some finite $A \subseteq H$ (shortly, an f.g.u. monoid), where $H^\times$ is the group of units of $H$.



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> **Question.** Is it true that, if $H$ is duo (i.e., $aH = Ha$ for every $a \in H$), then it's also unit-duo (i.e., $aH^\times = H^\times a$ for every $a \in H$)?
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It's pretty obvious that the answer is yes provided that $H^\times$ is trivial, or $H$ is acyclic (i.e., $uxv \ne x$ for all $u, v, x \in H$ such that either of $u$ or $v$ is a non-unit) or commutative.



An f.g.u. duo monoid is unit-duo: True or false?
This is false. Let $G$ be a group with a non-normal subgroup $A$ with finitely many double cosets (eg $G=S\_3$ and $A$ generated by a transposition). Consider $M=A'\cup G$ where $A'$ is a group isomorphic to $A$ via $a'\mapsto a$. Here $G$ is a two-sided ideal of $M$, $A'$ is the group of units, $G$ and $A'$ multiply as before and $a'g=ag$, $ga'=ga$ for $a\in A$, $g\in G$.  

This monoid is duo since it is inverse Clifford. It is not unit duo because $A$ is not normal.


The finiteness of two-sided orbits of the group of units follows from finitely many double cosets of $A$.


