1986
DOI: 10.1016/0021-8693(86)90200-0
|View full text |Cite
|
Sign up to set email alerts
|

Completely 0-simple semigroups of quotients

Help me understand this report

Search citation statements

Order By: Relevance

Paper Sections

Select...
4
1

Citation Types

0
87
1

Year Published

1990
1990
2007
2007

Publication Types

Select...
8
1

Relationship

1
8

Authors

Journals

citations
Cited by 49 publications
(88 citation statements)
references
References 3 publications
0
87
1
Order By: Relevance
“…Now if 5 is a weak order in Q and q e Q, then q = a*b for some a,b e 5. But then q = (a 2 )*ab and ab^^a 2 Consider the three element semilattice Q = {a, /3, y} where a < /3 and a ^ y. Clearly Q is an order in Q\ if &{Q) were a weak order in 8P(Q) then {/3, -y} could be written as {0, y} = U*V for some U,V e 3>(Q) with V< X U.…”
Section: P(g) Is Idempotent If and Only If E Is A Finite Subgroup Momentioning
confidence: 99%
“…Now if 5 is a weak order in Q and q e Q, then q = a*b for some a,b e 5. But then q = (a 2 )*ab and ab^^a 2 Consider the three element semilattice Q = {a, /3, y} where a < /3 and a ^ y. Clearly Q is an order in Q\ if &{Q) were a weak order in 8P(Q) then {/3, -y} could be written as {0, y} = U*V for some U,V e 3>(Q) with V< X U.…”
Section: P(g) Is Idempotent If and Only If E Is A Finite Subgroup Momentioning
confidence: 99%
“…Inspired by the ring theory concepts of orders and classical rings of quotients, Fountain and Petrich introduced the notion of a completely 0-simple semigroup of quotients in [19]. This was generalised to a much wider class of semigroups by Gould in [20].…”
Section: Introductionmentioning
confidence: 99%
“…We call such an isomorphism an isomorphism over S. Our notion of a semigroup of left quotients is that introduced by Fountain and Petrich in [6] and developed in [7]; in Section 2 we give a precise definition, restricting ourselves here to remarking that the approach is via consideration of group inverses of elements. If Q is a semigroup of left quotients of S then we also say that S is a left order in Q.…”
Section: Introductionmentioning
confidence: 99%
“…Unfortunately, the corresponding result is not true for semigroups. In the first paper of this area, Fountain and Petrich [6] give an example of a semigroup having two nonisomorphic semigroups of left quotients. However, by restricting the class of semigroups of left quotients under consideration, one can obtain some positive results.…”
Section: Introductionmentioning
confidence: 99%