An example of a certain kind of monoid which is not pure Given a monoid $M$, let $U$ be its set of invertible elements. It can be easily proven that the set $\{aU | a \in M\}$ is a partition of $M$. Call this partition $P$. Suppose that $P$ is closed under set multiplication. Then, $P$ forms a monoid with identity element $U$. My question is, must $P$ be a pure monoid, that is, $U$ is the only invertible element? I have tried to prove that it must be pure, but to no avail. I am suspecting that there is a case where $P$ is closed under set multiplication, but is not a pure monoid. Note, however, that if such a non-pure monoid exists, then $aU$ is not always equal to $Ua$, and a fortiori, $M$ cannot be a commutative monoid.
 A: If I used the right definitions, then I think $P$ is actually pure. So I'll start by clarifying my understanding of your definitions.
An element $m\in M$ is invertible if and only if there exists an element $m^{-1}\in M$ such that $mm^{-1} = 1 = m^{-1}m$ where $1$ is the unit of $M$.
Let $a,b\in M$. Then the multiplication of $aU$ and $bU$ is the set
$$aUbU = \{aubu'|u,u'\in U\}\,.$$
With that out of the way, let's prove that $P$ is pure, i.e. $1U = U$ is the only invertible element. Suppose $a,b\in M$ such that
$$aUbU = U = bUaU\,.$$
We need to show that $a,b\in U$ holds. Since $aUbU = U$ we know that $aubu' \in U$ for all $u,u'\in U$, in particular for $u, u' = 1$. Thus we get that $ab\in U$ and therefore we can find a $x\in U$ with $(ab)x = 1$. Apply associativity to see that
$$a(bx) = (ab)x = 1\,,$$
so we find that $a$ has a right inverse, let's call it $\tilde{x}$. But now from $bUaU = U$ we also know that $ba\in U$ and therefore find that there exists a $y\in U$ with $y(ba) = 1$. This gives us a left inverse
$$(yb)a = y(ba) = 1\,,$$
let's call it $\tilde{y}$.
Now it's a general fact from monoid theory that if an element has a left and a right inverse that those agree and the element is invertible. In this particular instance  we have
$$\tilde{x} = 1\tilde{x} = (\tilde{y}a)\tilde{x} = \tilde{y}(a\tilde{x}) = \tilde{y}1 = \tilde{y}\,.$$
Therefore $a$ is invertible and thus $a\in U$ and $aU = U$.
Now either use the same way to show that $b\in U$ or notice that if $ab\in U$ and $a\in U$ you need to have $b\in U$ as well since $U$ is a group. Therefore $bU = U$ as well and we see that $U$ is the only invertible element in $P$.
