1975
DOI: 10.1007/bf01895947
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Free inverse semigroups are not finitely presentable

Abstract: In the memory of Professor A. Kert&zFree inverse semigroups became a subject of intense studies in the last few years. Their existence was proved long ago: as algebras with two operations (binary multiplication and unary involution) inverse semigroups may be characterized by a finite system of identities, i.e. they form a variety of algebras [10]. Therefore, free inverse semigroups do exist.A construction of a free algebra in a variety of algebras (as a quotient algebra of an absolutely free word algebra) is w… Show more

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Cited by 28 publications
(30 citation statements)
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“…A question then arises as to the relationship between monoid presentations and inverse monoid presentations for a given inverse monoid S. It is easy to see that every monoid presentation for S is also an inverse monoid presentation for S. By way of contrast, Schein [22] proved that the free monogenic inverse monoid is not finitely presented as a monoid.…”
Section: N Ruškucmentioning
confidence: 98%
See 1 more Smart Citation
“…A question then arises as to the relationship between monoid presentations and inverse monoid presentations for a given inverse monoid S. It is easy to see that every monoid presentation for S is also an inverse monoid presentation for S. By way of contrast, Schein [22] proved that the free monogenic inverse monoid is not finitely presented as a monoid.…”
Section: N Ruškucmentioning
confidence: 98%
“…Finally, the relation (26) can be eliminated by using (22) Proof. The "if" part follows from Corollary 4.7 and Theorem 4.1, and the "only if" part follows from the fact that every monoid presentation for S is also an inverse monoid presentation for S.…”
Section: −1 Iamentioning
confidence: 99%
“…Indeed, in [40] it was shown that the free inverse semigroup (of any rank) is not finitely presented as a semigroup. Throughout this paper we shall work with inverse semigroups, inverse semigroup presentations, and by finitely presented we shall always mean finitely presented as an inverse semigroup.…”
Section: Inverse Semigroup Presentationsmentioning
confidence: 99%
“…Indeed, if S is inverse and 5 = Mon{X\R) then, by adding the standard inverse monoid relations (see (T2) below), S = Im{X\R). The converse does not hold in general: in [10] it is shown that the free inverse monoid is not finitely presented as a monoid (this result also appeared as Theorem IX.4.7 of [7]). …”
Section: Preliminariesmentioning
confidence: 99%