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Every finite abelian group is the direct product of its cyclic groups of prime order, and every commutative monoid divides a product of its cyclic submonoids. Could these results generalized to locally commutative semigroups? Here a semigroup is called locally commutative if for every idempotent $e \in S$ the semigroup $eSe$ is commutative.

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  • $\begingroup$ What do you have in mind? If you take a locally trivial semigroup ($eSe = e$ for each idempotent $e$), how do you want to decompose it? $\endgroup$
    – J.-E. Pin
    Jan 14, 2014 at 12:05
  • $\begingroup$ I do not believe that there is any nice decomposition result for locally commutative semigroups. Even the natural guess for a semigroup product decomposition, $\mathbf {Com}\ast \mathbf D$, doesn't hold by a result of Therien and Weiss. $\endgroup$ Jan 14, 2014 at 14:45
  • $\begingroup$ I had in mind some decomposition in easy semigroups whose recongizable languages are easy to describe for a variety correspondence result, specifically I am looking for a variety of $\omega$-semigroups recognizing languages of the form $\bigcup B_n(\xi) = \{ \eta \in X^{\omega} : \xi[1\ldots n] = \eta[1\ldots n] \land F_n(\xi) = F_n(\eta) \}$ where $F_n(\xi)$ denotes the factors of length $n$ of $\xi$. $\endgroup$
    – StefanH
    Jan 14, 2014 at 15:01
  • $\begingroup$ For finite words it seems like you would be looking at local semilattices, which do have a semidirect product decomposition. I don't really know the infinite word version but perhaps @J.-E.Pin can answer that. $\endgroup$ Jan 14, 2014 at 15:31

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