Selçuk J. Appl. Math. Selçuk Journal of Vol. 12. No. 1. pp. 123-126, 2011 Applied Mathematics
Relations between Ranks of Certain Semigroups Fikret Kuyucu
Çukurova University, Faculty of Arts and Science, Department of Mathematics 01330 Adana, Turkiye
e-mail: fkuyucu@ cukurova.edu.tr
Received Date: May 17, 2010 Accepted Date: September 1, 2010
Abstract. Let be a finite semigroup and let be an aquivalence relation on such that is a congruence. Then is a subsemigroup of × , the direct product of semigroups, and = { : ∈ } is a semigroup with ()() = (). In [2,3] Howie and Ribeiro study the small, lower, intermediate, upper and large rank of a finite semigroups . In this paper, we investigate some relation among ranks of semigroups , and .
Key words: Ranks; rank properties; finite semigroups 2000 Mathematics Subject Classification. 20M10. 1.Introduction
The notion of ’rank’ or ’dimension’, belongs primarily to linear algebra. One can define the rank of a (finite-dimensional) vector space either as cardinality of maximal linearly independent subset of , or as the cardinality of a minimal generating set of , and it is an elementary result in linear algebra that these two cardinalities are equal.
In [2,3], Howie and Ribeiro extend these ideas to more general algebraic struc-tures, such as semigroups.
For a set , | | denotes the cardinality of .
If is a semigroup and ⊆ then hi is a subsemigroup of generated by elements of .
Let be a finite semigroup. A subset ⊆ is called independent if, for every ∈ , ∈ h \ {}i. From a study of rank, in the context of universal algebra, by Marczewski [4], we get the following definitions of rank for a semigroup by Howie and Ribeiro [3].
r1(S) =max{ : every subset of of cardinality is independent}
r2(S) =min{ : there exists a subset of of cardinality such that
generates }
r3(S) =max{ : there exists a subset of of cardinality which is
independent and which generates }
r4(S) =max{ : there exists a subset of of cardinality which is
independent}
r5(S) =min{ : every subset of of cardinality generates }.
It is easly seen that 1() ≤ 2() ≤ 3() ≤ 4() ≤ 5(). Also, it can be use
terminology as follows: 1() : 2() : 3() : 4() : 5() :
Here, 2() is the ”classical” study of rank of semigroups and groups, extensively
studied. In [2,3], Howie and Ribeiro were devoted to the study of the ranks of the certain standart semigroups.
Does an arbitrary finite semigoup satisfy 2() = 4() ? The answer is
no. For example, if Z6 = {0 1 2 3 4 5} is the cyclic group of order 6 under
addition, then evidently 2(Z6) = 1, but the set {2 5} is independent, and in
fact 4(Z6) = 2.
Let be a finite semigroup and let be an equivalence relation on . Then is called a congruence if and then for all ∈ . It is clear that if is a congruence on then is a subsemigroup of × , the direct product of semigroups. Let = { : ∈ }, where = { ∈ : }. If is a congruence on then is a semigroup with ()() = (). In this paper, we investigate some relation among ranks of semigroups , and . 2. Main Results
If is a finite semigroup and 2() =| | then is said to be royal. Also if
is a semigroup such that 2= for every ∈ then is called a band.
Theorem 1. Let be a finite semigroup with | | ≥ 2 and let be a congruence on with | | ≥ 2 and | |≥ 2.
(i) If is not a royal semigroup then 1() = 1() = 1().
(ii) If is a royal semigroup then 1() =| | and 1() = | |.
Proof. (i)Let 1() = . Then is equal one of 1, 2 by [3, theorem 2].
Let 1() = 1. Firstly we show that 1() = 1. Since 1() = 1, there exists a
⊆ such that | |= 2 and is not independent Then = {(1 1) (2
2)} ⊆ and (2 2) ∈ h(1 1)i. Therefore there exists an ∈ N such that
(1 1) = (1 1) = (2 2). Then 2∈ h1i. Therefore = {1 2} is not
an independent subset of . Thus 1() = 1.
Secondly we show that 1() = 1. Suppose that 1() 1. Then is
independent for every subset of such that | |= 2 . We take = {1,
2}. Then there exist 1 2 ∈ such that (1 1), (2 2) ∈ . Thus (1
1) ∈ h(2 2)i since 1() = 1. Then 1 = 2 since 1 = 2. Therefore
1 ∈ h 2i and = {1, 2} is not independent. This is a contradiction.
Thus 1() = 1.
Let 1() = 2. Then, for every , ∈ , {( ) ( )} ⊆ is independent.
Thus ( ) ∈ h( )i. Therefore ( ) 6= ( ) for every ∈ N. Then
∈ hi. Similary ∈ hi. Thus { } is an independent subset of . Then 1() ≥ 2. Since 1() = 2, there exist (1 1), (2 2), (3 3) ∈ such
that {(1 1) (2 2) (3 3)} is not independent subset of . Assume that
(1 1) ∈ h{(2 2), (3 3)}i. Then
(1 1) = (1 1)( ) = (1 1), where ∈ {2 3}
Therefore 1∈ h{2 3}i since 1= 1.Thus {1 2 3} is not
indepen-dent subset of . Then 1() ≤ 2. Therefore 1() = 2.
Secondly we show that 1() = 2. Since 1() = 2 and is not a royal
semigroup, is a band by [3, theorem 2]. Let ∈ for ∈ . Since (2
2) = ( ) for ( ) ∈ , 2= . Therefore = 2. Thus is a band.
Then 1() = 2 since is not royal semigroup.
(ii)Let be a royal semigroup. Then 1() =| | by [3, theorem 2].
Firstly we show that 1() =| |. Suppose that 1() 6=| |. Then there exists
an 0 ∈ such that 0 ∈ h \ {0}i. Therefore there exist ∈ \ {0} for
= 1 2 such that 0= 12. Then (1 1)(2 2)( ) = (0
0) ∈ . Therefore (0 0) ∈ h \ {(0 0)}i. Then 1() | |. This is a
contradiction. Thus 1() =| |.
Secondly we show that 1() =| |. Suppose that 1() 6=| |. Then
there exists an 0 ∈ such that 0 ∈ h \ {0}i. Therefore there
exist 1 2 ∈ \ {0} such that 0 = 1 = 1. Since
0 1 ∈ , then there exist 0 1 ∈ such that (0 0),
(1 1), , ( ) ∈ . Let = 12, = 12. Then (1 1)(2
2)( ) = ( ). Since is compactible, then ( ) ∈ . Therefore
{( ) (1 1) ( )} is not independent subset of . Then 1() | |.
This is a contradiction. Thus 1() =| |.
Theorem 2. Let be a finite semigoup with | | ≥ 2 and let be a congruence on with | | ≥ 2 and | |≥ 2. If = or then ( ) ≤ () for
= 2 3 4 5 In general there is not equal.
Proof. Assume that = and = 2. If 2() = then there exist
a subset = {(1 1) (2 2) ( )} of such that hi = . Let
= {1 2 } and ∈ for a ∈ . Since ( ) ∈ and hi = ,
there exists a {(1 1) ( )} ⊆ such that
( ) = (1 1)( ) = (1 1).
Then = (1) = 1. Therefore hi = . But, even if 6=
it can be = . Then | |≤ . Thus 2() ≤ 2().
Similary, it can be shown that ( ) ≤ () for = 3 4 5
In general, there is not equal. For example, we consider = {1 2 3 4} which is the chain of order 4, and the congruence generated by {(2 3)} as a congruence. Then = {(1 1) , (2 2) , (3 3) , (4 4) , (2 3) , (3 2)} and = {1, 2, 4}. Then 2() = 5 and 2() = 3. Therefore is a band yet not royal, but
is royal. Then 1() = 2 and 1() = 3. Also it is easly seen that
3() = 4() = 5() = 3 and 3() = 4() = 5 and 2() = 6.
An element of a semigroup is said to irreducible if there do not exist elements in \ {} such that = .
Corollary 1. Let be a finite semigroup and a congruence on . If 5() =
| | then 5() =| |.
Proof. If 5() =| | then contains an irreducible element by [3, Corollary
4]. If ∈ is an irreducible element then clearly ( ) ∈ is an irreducible element too. Thus 5() =| | by [3, Corollary 4].
References
1. Howie, J.M. Foundations of Semigroup Theory, Clarendon Press, Oxford, 1995. 2. Howie, J.M. and Ribeiro,M.I.M. Rank Properties in Finite Semigroups, Communi-cation in Algebra., 27(11), 1999, 5333- 5347.
3. Howie, J.M. and Ribeiro, M.I.M. Rank Properties in Finite Semigroups II: The Small Rank and the Large Rank, Southeast Asian Bulletin of Math., 24, 2000, 231-237.
4. Marczewski, E. Independence in abstract algebras: Results and problems, Colloq. Math. 14(1966), 169-188.
