2016
DOI: 10.1007/s10998-016-0134-3
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Restriction semigroups and $$\lambda $$ λ -Zappa-Szép products

Abstract: The aim of this paper is to study λ-semidirect and λ-Zappa-Szép products of restriction semigroups. The former concept was introduced for inverse semigroups by Billhardt, and has been extended to some classes of left restriction semigroups. The latter was introduced, again in the inverse case, by Gilbert and Wazzan. We unify these concepts by considering what we name the scaffold of a Zappa-Szép product S T where S and T are restriction. Under certain conditions this scaffold becomes a category. If one action … Show more

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Cited by 6 publications
(11 citation statements)
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“…Also r = 1 because r(aρ) = S × S. We can therefore assume that r = (w, r, ℓ). If w = v, then (w, r, ℓ) = 1(w, r, ℓ) ν (v, 1 M , v)(w, r, ℓ) = 0, proving (10) in this case.…”
Section: Brandt Semigroupsmentioning
confidence: 77%
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“…Also r = 1 because r(aρ) = S × S. We can therefore assume that r = (w, r, ℓ). If w = v, then (w, r, ℓ) = 1(w, r, ℓ) ν (v, 1 M , v)(w, r, ℓ) = 0, proving (10) in this case.…”
Section: Brandt Semigroupsmentioning
confidence: 77%
“…The next proof, although the calculations are different, follows the exact pattern of the previous, and so we omit it. The monoids in Corollaries 4.10, 4.11 and Remark 4.12 all possess an inverse skeleton [10]; the techniques of [10] would provide an alternative unified approach to these results.…”
Section: The Relation L E Is Defined Dually and We Put Hmentioning
confidence: 99%
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“…This result was mentioned without proof in [10], as an example of a Zappa-Szép product. We include the full argument in this paper for the sake of completeness.…”
Section: Zappa-szép Product Of a Left Restriction Semigroup And A Semmentioning
confidence: 92%
“…They specifically discuss the Baumslag-Solitar groups, the binary adding machine, the semigroup N N ×  , and the ax b + -semigroup Z Z ×  . In [4] they study semigroups possessing E-regular elements, where an element a of a semigroup S is E-regular if a has an inverse a  such that , aa a a   lie in…”
Section: Introductionmentioning
confidence: 99%