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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.

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2 Answers 2

up vote 7 down vote accepted

Let $M:=\{1,a,b\}$ with the following multiplication: $$\begin{align} a\cdot b:=a & \quad b\cdot a:=b \\ a\cdot a:= a & \quad b\cdot b:= b \end{align} $$ (In other words, $xy=x$ holds for all $x\ne 1$, so associativity comes easily.)

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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 $(a,b)(c,d)=0$.

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