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Suppose $S$ is a semigroup such that for all $x,y \in S$ and all natural $n$ we have $$(xy)^n = x^n y^n.$$

If $S$ group, then it is Abelian; indeed a stronger statement holds, see here.

Does there exist a semigroup with this property that fails to be Abelian? If so, I am interested in whether there is a finite axiomatization of this property for semigroups.

Motivation. Let $D$ denote a (not-necessarily Abelian) semigroup, written additively. Then we have an action of $\mathbb{N}' = \{1,2,3,\cdots\}$ on $A$, defined by $nd = \underbrace{d + \cdots + d}_n$.

Now clearly:

  1. $1d = d$
  2. $(n+n')d = nd + n'd$
  3. $(nn')d = n(n'd)$

Thus, ignoring the fact that $\mathbb{N}'$ isn't a ring, we have that $D$ is very nearly a module over $D$. The only axiom that's missing is:

  • $n(d+d') = nd+nd'$

which is just the additive analogue of $(xy)^n = x^n y^n$.

Community
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goblin GONE
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2 Answers2

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Consider a set $S$ with at least two elements and the operation $x\cdot y =y$. Then $(xy)^n=y=x^ny^n$ for all $n\ge 1$. But of course $xy\ne yx$ if $x\ne y$.

Hagen von Eitzen
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Hagen von Eitzen gave the simplest example. These semigroups are known as left-zero bands in the literature. Here is another type of example. Take an infinite square-free word $t$ and let $F$ be the set of factors of $t$. Then $S = F \cup \{0\}$ is a semigroup under the operation $$ uv = \begin{cases} uv &\text{if $u, v, uv \in F$} \\ 0 &\text{otherwise} \end{cases} $$ Since $t$ is square-free, this semigroup satisfies the identity $x^2 = 0$ for all $x \in S$ and hence $(xy)^n = x^ny^n$ for all $n \geqslant 1$.

J.-E. Pin
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