Two almost disjoint submonoids whose union is the whole monoid

Question

Does there exist a monoid $(M;*,1)$ which has two submonoids $M'$ and $M''$, such that neither $M'$ nor $M''$ is equal to $M$, and neither $M'$ nor $M''$ is the trivial monoid $\{1\}$, the intersection of $M'$ and $M''$ is $\{1\}$ and the union of $M'$ and $M''$ is $M$?

Answer 1

Yes. A nice example is eg $M = \Bbb Z$ under multiplication, $M' = \{0, 1\}$ and $M'' = \Bbb Z \setminus \{0\}$.

In fact "monoid with a zero element" is an idea we can use in more generality - if $M$ is any monoid, form a monoid $M_0 = M \cup \{0\}$ with a new element called $0$, and define $x \cdot 0 = 0 \cdot x = 0$ for all $x \in M$. Then $M_0$ is a monoid too, and we have a decomposition of the sort you're interested in - $M_0 = M \cup \{0, 1\}$. Therefore there are also a few examples of order $3$ (which of course is the smallest possible order).