TITLE: Example of a non-commutative monoid
QUESTION [8 upvotes]: Is there an explicit example of a non-commutative monoid $M$ such that for all elements $m,n \in M$ and $p \in \mathbb{N}$ we have $(m \cdot n)^p=m^p \cdot n^p$?
It suffices to construct a semigroup $H$ with an absorbing element $0$ such that $a^2=0$ for all $a$, because then $M := H \cup \{e\}$ will do the job. This sounds easy at first glance, but I cannot find an example.

REPLY [4 votes]: If this question is still interested for you:
Let $P(\le)$ be a poset. Define a multiplication on pairs $\{(a,b)|a\le b\}\subset P\times P$ with an extra zero by the rule:
$(a,b)(c,d)=(a,d)$ if $b=c$, otherwise $(a,b)(c,d)=0$.