On semidirect product of semigroups

Abstract

Let X be a subset of a semigroup S. We denote by E(X) the set of idempotent elements Of X.An element a of a semigroup S is called E-inverse if there exists $x ∈ E(S)$ such that $ax ∈ E(S)$. We note that the definition is not one-sided. Indeed, an a element of a semigroup S is E-inversive if there exists $y ∈ S$ such that $ay, ya ∈ E(S)$ (see [7], [l] p. 98). A semigroup S is called E-inversive if all its elements are E-inversive. This class of semigroups is extensive. All semigroups with a zero and all eventually regular semigroups [2] are E-inversive semigroups.Recently E-inversive semigroups reappeared in a paper by Hall and Munn [3] and in a paper by Mitsch [5]. The special case of E-inversive semigroups with pairwise commuting idempotents, called E-dense, was considered by Margolis and Pin [4]. Let S and T be semigroups, and let $? : S → End(T)$ be a homomorphism of S into the endomorphism semigroup of T. If $s ∈ S$ and $t ∈ T$, denote $t(sa) by t^s$. Thus, if $s,s' ∈ S$ and $t ∈ T$ then ${t^s}^s = t^ss'$. The semidirect product of S and T ,in that order, with strutture map (Y, consists of the set S x T equipped with the product $(s,t)(s',t') = (ss', {t^s'}{t'})$. This product will be denoted by S ?_{?}T. In this note we determine which semidirect products of semigroups are E-inversive semigroups and E-dense semigroups, respectively. It turns out that the case in which S induces only automorphism on T allows a particularly simple description. In [6], Preston has answered the analous question for regular semigroups and for inverse semigroups. For the terminology and for the definitions of the algebraic theory of semigroups, we refer to [1].