Converse to a theorem regarding monoids and their set of invertibles

Question

In this paper, there is a theorem regarding monoids and their set of invertible elements. Let $(M,*,1)$ be a monoid, and let $U$ be the set of invertible elements of $M$. The theorem states that if $aU=Ua$, for every $a$ in $M$, then the collection of sets $\{aU | a \in M\}$ forms a partition of $M$. What about the converse? Is it true that if the collection of sets $\{aU | a \in M\}$ forms a partition of $M$, then every element $a$ of $M$ commutes with $U$? I would be very surprised if there was a counterexample.

Answer 1

Let $M=\{f:X\to X\ $functions$\}$ where, say, let $X=\{1,2\}$ for simplicity.
Then $|M|=4$ and $\ U=\{{\rm id}_X,\ (1\mapsto 2,\,2\mapsto 1)\}$.
The other two elements of $M$ are the constant maps ${\bf 1}:x\mapsto 1$ and ${\bf 2}:x\mapsto 2$.

If $a={\bf 1}$ or $\bf 2$, then $aU=\{a\}$, otherwise $aU=U$.
So these do form a partition of $M$.

However, $Ua=\{{\bf 1},\,{\bf 2}\}\ \ne aU$ for $a={\bf 1},\,{\bf 2}$.