Generating Sets of the Complete Semigroups of Binary Relations Defined by Semilattices of the Class Σ2 (X,4)

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ISSN Online: 2152-7393 ISSN Print: 2152-7385

DOI: 10.4236/am.2018.91002 Jan. 19, 2018 17 Applied Mathematics

Generating Sets of the Complete Semigroups of

Binary Relations Defined by Semilattices of the

Class

Σ

2

(

X

,4

)

Bariş Albayrak

1

, Omar Givradze

2

, Guladi Partenadze

2

1_{Department of Banking and Finance, School of Applied Sciences, Çanakkale Onsekiz Mart University, Çanakkale, Turkey} 2_{Shota Rustaveli State University, Batumi, Georgia}

Abstract

In this paper, we have studied generating sets of the complete semigroups de-fined by X-semilattices of the class Σ2

(

X, 4

)

.

Keywords

Semigroup, Semilattice, Binary Relations, Idempotent Elements

1. Introduction

Let X be an arbitrary nonempty set and D be a nonempty set of subsets of the set

X. If D is closed under the union, then D is called a complete X-semilattice of unions. The union of all elements of the set D is denoted by the symbol D.

Let BX be the set of all binary relations on X. It is well known that BX is a

semigroup.

Let f be an arbitrary mapping from X into D. Then we denote a binary relation

{ } ( )

(

)

f

x X x f x

α

∈

=

∪

× _{for each}_f_{. The set of all such binary relations is denoted}

by B DX

( )

. It is easy to prove that B DX

( )

is a semigroup with respect to the product operation of binary relations. This semigroup B DX

( )

is called a com-plete semigroup of binary relations defined by an X-semilattice of unions D.

This structure was comprehensively investigated in Diasamidze [1] and [2]. We assume that t y X, ∈ _, Y ⊆X , α ∈BX , T D∈ and ∅ ≠D′⊆D. Then we

denote following sets

{

|

}

, ,

y Y

yα x X y x Yα α yα

∈

= ∈ =

∪

How to cite this paper: Albayrak, B., Gi-vradze, O. and Partenadze,G. (2018) Ge-nerating Sets of the Complete Semigroups of Binary Relations Defined by Semilattices of the Class Σ2

(

X,4

)

. Applied Mathe-matics, 9, 17-27.

https://doi.org/10.4236/am.2018.91002

Received: December 11, 2017 Accepted: January 16, 2018 Published: January 19, 2018

http://creativecommons.org/licenses/by/4.0/

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DOI: 10.4236/am.2018.91002 18 Applied Mathematics

(

) {

}

{

}

{

}

(

)

{

}

{

}

0

{

( )

(

)

}

, | , |

| , , |

| , | ,

T

t X

V D Y Y D X Y Y X

Y y X y T V X Y Y X

D Z D t Z B B D V X D

α

α α

α α α

α α

∗

= ∈ = ∅ ≠ ⊆

= ∈ = = ∅ ≠ ⊆

′ ′

= ∈ ∈ = ∈ =

Let D=

{

D Z Z, , , ,1 2 Zm−1

}

be finite X-semilattice of unions and

( ) {

0, , , ,1 2 m1

}

C D = P P P P− be the family of pairwise nonintersecting subsets of

X. If 1 1

0 1 1

m m

D Z Z

P P P

ϕ −

−

 

=  

 

is a mapping from D on C D

( )

, then the equali-ties D P P P= 0 1 2 Pm−1

∪ ∪ ∪∪ and 0

( )

\ Z

i

T D D

Z P

ϕ

T

∈

= ∪

∪

are valid. These

equalities are called formal.

Let D be a complete X-semilattice of unions α ∈BX . Then a representation

of a binary relation

α

of the form

( )

,

(

)

T T V X

Yα T

α

∗

∈

=

∪

× _{is called quasinormal.}

Let P P P0, , , ,1 2 Pm−1 be parameters in the formal equalities, β∈B DX

( )

, β2 be mapping from X D\ to D. Then 1

(

{ }

2

( )

)

0 i \

m i

i t P t X D

P t t t

β − β β

′

= ∈ ∈

 

′ ′

= _ × _ ×

 ∪

∪

is called subquasinormal represantation of β_{. It can be easily seen that the}

fol-lowing statements are true. a)

β

∈B DX

( )

.

b) 1

0 i

m i i P t Pt

β β

−

= ∈

 

× ⊆

 

 

∪

and β β= for some β2. c) Subquasinormal represantation of β_{is quasinormal.}

d) 0 1 1

1

0 1 1

m m

P P P

β

β β β

−

 

=  

 

is mapping from C D

( )

on D∪

{ }

∅ . 1

β and β2 are respectively called normal and complement mappings for

β_.

Let α∈B DX

( )

. If α δ β≠ for all δ β, ∈B DX

( ) { }

\ α then

α

is called external element. Every element of the set B0=

{

α

∈B D V XX

( )

|

(

∗,

α

)

=D

}

is

an external element of B DX

( )

.

Theorem 1. [1] Let X be a finite set and α β, ∈B DX

( )

. If β is sub

-quasinormal representation of β_thenα β α β = .

Corollary 1. [1] Let B′ ⊆ ⊆B B D X

( )

. If α δ β≠ for α∈B′, δ∈B\

{ }

α ,

{ }

\ B

β∈ α and subquasinormal representation of β∈B\

{ }

α then α δ β≠ .

It is known that the set of all external elements is subset of any generating set of B DX

( )

in [3].

2. Results

In this work by symbol Σ2.2

(

X,4

)

we denote all semilattices D=

{

Z Z Z D3, , ,2 1

}

of the class Σ2

(

X, 4

)

which the intersection of minimal elements Z3∩Z2= ∅.

(3)

[image:3.595.319.432.74.212.2]

Figure 1. Graphic of semilattice D=

{

Z Z Z D3, , ,2 1

}

which

the intersection of minimal elements Z3∩Z2= ∅.

1 2 3

2 1 3

3 2

D P P P

Z P P

Z P

= = = =

∪ ∪ ∪

∪ (1)

Let

δ β

, ∈B DX

( )

. If quasinormal representation of binary relation δ has a

form δ =

(

Y3δ×Z3

) (

Y2δ×Z2

) (

Y1δ×Z1

) (

Y0δ×D

)

∪ ∪ ∪ then

(

Y3δ Z3

) (

Y2δ Z2

) (

Y1δ Z1

) (

Y0δ D

)

δ β = × β ∪ × β ∪ × β ∪ × β

We denote the set

( )

(

) {

}

{

}

( )

(

) {

}

{

}

( )

(

)

{

}

{

}

(

) (

)

{

}

(

) (

)

{

}

32 3 2

21 2 1

31 3 1

32 32 3 3 2 2 3 2 3 2

21 21 2 2 1 1 2 1 2 1

| , , ,

| , ,

X

B B D V X Z Z D

B B D V X Z Z

B B Y Z Y Z Y Y X Y Y

α α α α α α

α α

∗

= ∈ =

= ∈ = × × = = ∅

_∪ _∪ _∩

It is easy to see that

0 32 0 21 0 31 21 32 31 32 21 31

B ∩B =B ∩B =B ∩B =B ∩B =B ∩B =B ∩B = ∅.

Lemma 2. Let D=

{

Z Z Z D3, , ,2 1

}

∈Σ2.2

(

X,4

)

. Then following statements are true for the sets B B B0, 32,32.

a) If α =

(

Y3α×Z3

) (

Y1α×Z1

) (

Y0α×D

)

∪ ∪ for some Y Y Y3α, 1α, 0α∉∅, then

α

is product of some elements of the set B0.

b) If β0=

(

Z Z3× 3

) (

∪

(

X Z\ 3

)

×Z2

)

, then

(

B0

β

0

)

∪B32=B32.

c) If σ1=

(

Z2×Z2

) (

∪

(

X Z\ 2

)

×Z1

)

, then

(

B0σ1

)

∪B21=B21.

d) If σ1=

(

Z2×Z2

) (

∪

(

X Z\ 2

)

×Z1

)

, then B32σ =1 B21.

e) If σ0=

(

Z Z3× 3

) (

∪

(

X Z\ 3

)

×Z1

)

, then B32σ =0 B31.

(4)

{ }

0 32 1

B _∪B _∪ σ _.

Proof. It will be enough to show only a, b and g. The rest can be similarly seen. a. Let α =

(

Y3α×Z3

) (

Y1α×Z1

) (

Y0α×D

)

∪ ∪ for some Y Y Y3α, ,1α 0α∉ ∅

{ }

,

0

, B

δ β∈ . Then quasinormal representation of δ has a form

(

Y3δ Z3

) (

Y2δ Z2

) (

Y1δ Z1

) (

Y0δ D

)

δ = × ∪ × ∪ × ∪ ×

where Y Y Y3δ, ,1δ 0δ∉ ∅

{ }

. We suppose that

(

2 3

) (

1 2

) (

3 1

)

(

{ }

2

( )

)

\

t X D

P Z P Z P Z t t

β β

′∈

′ ′

= × × × ×

∪ ∪ ∪

∪

where 1 2 3

1

2 3 1

P P P

Z Z Z

β _{= }∅ _

∅

  is normal mapping for β and β2 is com-

plement mapping of the set X D\ on the set D . So, β∈B0 since

(

,

)

V X∗ β ₌D_{. From the equalities (2.1) and definition of} _β

(

)

(

)

(

)

3 2 3

2 1 3 1 3 2 1 1 2 3 2 3 3 1 1

1 2 3 1 2 3 2 1

Z P Z

Z P P P P Z Z D

Z P P P P Z Z Z

D P P P P P P Z D Z D

β β

β β β β

β β β β β

= =

= = = =

∪ ∪ ∪

∪ ∪ ∪ ∪ ∪ ∪

(

) (

)

(

) (

)

(

) (

(

)

3 3 2 2 1 1 0

3 2 1 0

3 1 2 0 .

Y Z Y Z Y Z Y D

Y D Y D Y D Y D

Y D Y D Y Y D

δ δ δ δ

δ β β β β β

α

= × × × ×

= × × × =

∪ ∪ ∪

b. Let α∈B0β0∪B32. Then α∈B0β0 or α∈B32. If α∈B0β0 then 0

α δ β= for some δ ∈B0. In this case we have

(

Y3δ Z3

) (

Y2δ Z2

) (

Y1δ Z1

) (

Y0δ D

)

δ = × ∪ × ∪ × ∪ ×

{ }

. Also

(

) (

)

(

) (

)

(

) (

(

)

0 3 3 0 2 2 0 1 1 0 0 0

3 3 2 2 1 1 0

3 3 2 2 1 0 32\ 32

Y Z Y Z Y Z Y D

Y Z Y Z Y Y D B B

δ δ δ δ

α δ β= = × β × β × β × β

= × × × ×

= × × × ∈

∪ ∪ ∪

is satisfied. So, we have

(

B0β0

)

∪B32⊆B32 . On the other hand, if

32 32

B B

α∈ ⊆ then

(

B0β0

)

∪B32⊆B32 is satisfied. Conversely, if α ∈B32

then quasinormal representation of

α

has a form

(

Y3α Z3

) (

Y2α Z2

) (

Y0α D

)

α = × ∪ × ∪ ×

where Y Y Y3α, ,2α 0α∉ ∅

{ }

or Y Y3α, 2α∉ ∅

{ }

and Y0α= ∅ . We suppose that

{ }

3 , 2

Y Yα α_{∉ ∅} _{. In this case, we have}

(

) (

)

(

) (

)(

)

0 3 3 0 2 2 0 0 1 0 3 3 2 2 0

Y Z Y Z Y Z

Y Z Y Z Y D

δ δ δ

δ β β β β

α

= × × ×

= × × × =

∪ ∪

∪

for δ =

(

Y3α×Z3

) (

∪ Y2α×Z2

) (

∪ Y0α×Z1

)

∈B0. So, we have B32⊆

(

B0β0

)

∪B32.

(5)

(

)

32 0 0 32

B B B

α∈ ⊆ β ∪ . So,

(

B0β0

)

∪B32=B32.

g. From the statement c, we have that

(

B0β0

)

∪B32=B32 where β0∈B32 by definition of β0. Thus, every element of the set B32 is product of elements

of the set B0∪B32.

Lemma 3. Let D=

{

Z Z Z D3, , ,2 1

}

∈Σ2.2

(

X,4

)

. If X D\ ≥1 _{then the}

following statements are true.

a) If α = ×X D then

α

is product of elements of the set B0.

b) If α = ×X Z1 then

α

is product of elements of the set B0.

c) If α =

(

Y3α×Z3

) (

∪ Y1α×Z1

)

for some Y Y3α, 1α∉∅, then

α

is product of elements of the B0.

d) If α=

(

Y3α×Z3

) (

Y0α×D

)

∪ for some Y Y3α, 0α∉∅, then

α

e) If α =

(

Y2α×Z2

) (

Y0α×D

)

α

f) If α =

(

Y1α×Z1

) (

Y0α×D

)

α

Proof. c. Let quasinormal representation of

α

has a form

(

Y3α Z3

) (

Y1α Z1

)

α= × ∪ × where Y Y3δ, 1δ∉ ∅

{ }

. By definition of the semilattice

D, X ≥3. We suppose that Y3α ≥1 and Y1α ≥2. In this case, we suppose that

(

2 3

) (

(

1 3

)

1

)

(

{ }

2

( )

)

\

t X D

P Z P P Z t t

β β

′∈

′ ′

= × ∪ ∪ × ∪

∪

×

where 1 2 3

1

1 3 1

P P P

Z Z Z

β _{= }∅ _

∅

  is normal mapping for β and β2 is comple-

ment mapping of the set X D× _{on the set}D Z Z\

{

₃, ₁

} { }

= Z₂ (by suppose

\ 1

X D ≥ ). So, β∈B0 since V X

(

∗,β

)

=D . Also, Y3δ =Y3α and 2 1 0 1

Yδ _∪Yδ _∪Yδ ₌Yδ _since

3 1, 2 1, 1 1, 0 0

Yδ _≥ Yδ _≥ Yδ _≥ Yδ _≥ _{. From the equalities}

(2.1) and definition of β we obtain that

(

)

(

)

(

)

3 2 3

2 1 3 1 3 1 1 1 1 2 3 2 3 3 1 1

1 2 3 1 2 3 1 3 1 1

Z P Z

Z P P P P Z Z Z

D P P P P P P Z Z Z Z

β β

β β β β

β β β β β

= =

= = = =

∪ ∪ ∪

∪ ∪ ∪ ∪ ∪ ∪

(

) (

)

(

) (

)

(

) (

(

)

3 3 2 2 1 1 0

3 3 2 1 1 1 0 1

3 3 2 1 0 1

Y Z Y Z Y Z Y D

Y Z Y Z Y Z Y Z

Y Z Y Y Y Z

δ δ δ δ

δ β β β β β

α

= × × × ×

= × × =

∪ ∪ ∪

Now, we suppose that Y3α ≥2 and Y1α ≥1. In this case, we suppose that

(

)

(

2 3 3

)

(

1 1

)

(

{ }

2

( )

)

\ t X D

P P Z P Z t t

β β

′∈

′ ′

= ∪ × × ×

∪ ∪

∪

where 1 2 3

1

1 3 3

P P P

Z Z Z

β _{= }∅ _

∅

(6)

\ 1

X D ≥ ). So, β∈B0 since V X

(

∗,β

)

=D . Also, Y3δ ∪Y1δ =Y3α and 2 0 1

Yδ _∪Yδ ₌Yα_since

3 1, 2 1, 1 1, 0 0

Yδ _≥ Yδ _≥ Yδ _≥ Yδ _≥ _{. From the equalities (2.1)}

and definition of β we obtain that

(

)

(

)

(

)

3 2 3

2 1 3 1 3 1 3 1 1 2 3 2 3 3 3 3

1 2 3 1 2 3 1 3 3 1

Z P Z

Z P P P P Z Z Z

D P P P P P P Z Z Z Z

β β

β β β β

β β β β β

= =

= = = =

∪ ∪ ∪

∪ ∪ ∪ ∪ ∪ ∪

(

) (

)

(

) (

)

(

)

(

)

(

)

3 3 2 2 1 1 0

3 3 2 1 1 3 0 1

3 1 3 2 0 1

Y Z Y Z Y Z Y D

Y Z Y Z Y Z Y Z

Y Y Z Y Y Z

δ δ δ δ

δ β β β β β

α

= × × × ×

= × × =

∪ ∪ ∪

Lemma 4. Let D=

{

Z Z Z D3, , ,2 1

}

∈Σ2.2

(

X,4

)

,

(

) (

(

)

0 Z Z3 3 X Z\ 3 Z1

σ = × ∪ × and σ1=

(

Z2×Z2

) (

∪

(

X Z\ 2

)

×Z1

)

. If X D=

then the following statements are true

a) If α =

(

Y3α×Z3

) (

Y0α×D

)

∪ for some Y Y3α, 0α∉ ∅

{ }

, then

α

is product

of elements of the B0∪B32.

b) If α=

(

Y2α×Z2

) (

Y0α×D

)

{ }

, then

α

is product

of elements of the B32∪

{ }

σ1 . c) If α=

(

Y1α×Z1

) (

Y0α×D

)

{ }

, then

α

is product of

elements of the B32∪

{

σ σ0, 1

}

.

Proof. First, remark that Z3 0σ =Z3, Z2 0σ =Dσ0 =Z1

, Z3 1σ =Z1, Z2 1σ =Z2,

1

Dσ =D.

a. Let α=

(

Y3α×Z3

) (

Y0α×D

)

∪ for some Y Y3α, 0α∉∅. In this case, we suppose that

(

Y3δ Z3

) (

Y2δ Z2

) (

Y0δ D

)

δ = × ∪ × ∪ ×

and

(

) (

(

)

(

)

1 Z Z3 3 Z Z1\ 3 Z1 X Z\ 1 D

β = × ∪ × ∪ ×

where Y Y3δ, 2δ∉ ∅

{ }

. It is easy to see that δ ∈B32 and β1 is generating by

elements of the B0 by statement b of Lemma 2. Also, Y3δ =Y3α and 2 0 0

Yδ _∪Yδ ₌Yα _since

3 3

Z β =Z , Z2β =Dβ =D

and Y3δ ≥1,Y2δ ≥1,Y0δ ≥0. So,

α

is product of elements of the B0∪B32. □

Lemma 5. Let

{

3, , ,2 1

}

2.2

(

,4

)

D= Z Z Z D ∈Σ X

and

(

) (

(

)

1 Z2 Z2 X Z\ 2 Z1

σ = × ∪ × .

If X D\ ≥1 then S1=B0∪B32∪

{ }

σ

1 is an irreducible generating set for the

semigroup B DX

( )

.

(7)

ele-DOI: 10.4236/am.2018.91002 23 Applied Mathematics

ments of S1. Let α∈B DX

( )

and

(

Y3α Z3

) (

Y2α Z2

) (

Y1α Z1

) (

Y0α D

)

α= × ∪ × ∪ × ∪ ×

where Y3α∪Y2α∪Y1α∪Y0α =X and Y3α∩Y2α = ∅,

(

0≤ ≠ ≤i j 3

)

. We suppose that V X

(

∗,

α

)

₌1_{. Then we have}

(

)

{

{ } { } { }

{ }

}

3 2 1

, , , ,

V X∗

α

_∈ Z Z Z D _{. If}

(

,

)

{

{ } { } { }

3 , 2 , 1

}

V X∗ α _∈ Z Z Z _then

3

X Z

α = × or α = ×X Z2 or α = ×X Z1.

Quasinormal representations of δ β β, ,1 2 and β3 has form

(

) (

)

(

) (

(

)

(

) (

(

)

(

) (

(

)

3 3 2 2 1 1 0

1 3 2

2 2 1

3 1 2

\ \

\

Y Z Y Z Y Z Y D

D Z X D Z

δ δ δ δ

δ β β

β

= × × × ×

= × ×

∪ ∪ ∪

∪

{ }

. So, δ ∈B0 , β1∈B32 and β β ∈2, 3 B21 since

\ 1

X D ≥ . From the definition of δ β β, ,1 2 and β3 we obtain that

(

) (

)

(

) (

)

(

)

1 3 3 1 2 2 1 1 1 1 0 1 3 3 2 3 1 3 0 3 3 2 1 0 3 3

Y Z Y Z Y Z Y D

Y Z Y Z Y Z Y Z

Y Y Y Y Z X Z

δ δ δ δ

δ β = × β × β × β × β

= × × × ×

= × = ×

∪ ∪ ∪

(

) (

)

(

) (

)

(

)

2 3 3 2 2 2 2 1 1 2 0 2 3 2 2 2 1 2 0 2

3 2 1 0 2 2

Y Z Y Z Y Z Y D

Y Z Y Z Y Z Y Z

Y Y Y Y Z X Z

δ δ δ δ

δ β = × β × β × β × β

= × × × ×

= × = ×

∪ ∪ ∪

(

) (

)

(

) (

)

(

)

3 3 3 3 2 2 3 1 1 3 0 3 3 1 2 1 1 1 0 1

3 2 1 0 1 1

Y Z Y Z Y Z Y D

Y Z Y Z Y Z Y Z

Y Y Y Y Z X Z

δ δ δ δ

δ β = × β × β × β × β

= × × × ×

= × = ×

∪ ∪ ∪

That means, X Z X Z× 1, × 2 and X Z× 3 are generated by B0∪B32,

0 21

B ∪B and B0∪B21 respectively. By using statement g and h of Lemma 3,

we have X Z X Z× 1, × 2 and X Z× 3 are generated by B0∪B32∪

{ }

σ1 . On the

other hand, if V X

(

∗,α

) { }

₌ D _then _{α = ×}_{X D} _{By using statement a of}

Lemma 3, we have

α

is product of some elemets of B0.

So, S1 is generating set for the semigroup B DX

( )

. Now, we must prove that

{ }

1 0 32 1

S =B ∪B ∪ σ is irreducible. Let α ∈S1.

If α ∈B0 then α σ τ≠ for all σ τ, ∈B DX

( ) { }

\ α from Lemma 2. So,

α σ τ≠ for all σ τ, ∈S1\

{ }

α . That means, α ∉B0.

If α∈B32 then the quasinormal representation of

α

has form

(

Y3α Z3

) (

Y2α Z2

)

α= × ∪ × for some Y Y3α, 2α∉∅ . Let α δ β= for some

{ }

1

, S \

δ β∈ α .

We suppose that δ∈B0\

{ }

α and β∈S1\

{ }

α . By definition of B0,

quasi-normal representation of δ _{has form}

(

Y3δ Z3

) (

Y2δ Z2

) (

Y1δ Z1

) (

Y0δ D

)

(8)

{ }

. By using Z3⊂Z1⊂D

and Z2⊂D

we have Z3β

and Z2β are minimal elements of the semilattice

{

Z3β,Z2β,Z1β,Dβ

}

. Also, we have

(

) (

)

(

) (

)

3 3 2 2

3 3 2 2 1 1 0

Y Z Y Z

Y Z Y Z Y Z Y D

α α

δ δ δ δ

α δ β

β β β β

× × = =

= × × × ×

∪

∪ ∪ ∪

Since Z3 and Z2 are minimal elements of the semilattice

{

Z Z D3, ,2

}

, this equality is possible only if Z3=Z3β, Z2 =Z2β or Z3 =Z2β , Z2=Z3β . By

using formal equalities and P P P3β, 2β β ∈, 1 D, we obtain

3 3 2 2 2 1 3

2 3 2 3 2 1 3

and and

Z Z P Z Z P P

β β β β β

= = = = =

respectively. Let Z3 =P2β and Z2=P1β =P3β. If β is sub-quasinormal

re-presentation of β_thenδ β δ β = and

(

)

(

1 3 2

)

(

2 3

)

(

{ }

2

( )

)

\ t X D

P P Z P Z t t

β β

′∈

′ ′

= × × ×

∪ ∪ ∪

∪

where 1 2 3

1

2 3 2

P P P

Z Z Z

β _{= }∅ _

∅

plement mapping of the set X D× _{on the set}D=

{

Z Z Z₃, ,₂ ₁

}

. From formal equalities, we obtain

(

2 2

) (

3 3

)

(

{ }

2

( )

)

1

{ }

\ \

t X D

Z Z Z Z t t S

β β α

′∈

′ ′

= × × × ∈

∪ ∪

∪

and by using Z1∩Z2≠ ∅,Z3∪Z2=D and Y1δ∪Y0δ ≥1, we have

(

) (

)

(

) (

)

(

) (

(

)

3 3 2 2 1 1 0

3 3 2 2 1 0

Y Z Y Z Y Z Y D

Y Z Y Z Y D Y D

Y Z Y Z Y Y D

δ δ δ δ

δ β β β β β

α

= × × × ×

= × × × ≠

∪ ∪ ∪

This contradicts with α δ β= _{. So,} δ∉B₀\

{ }

α .

Now, we suppose that

δ

∈B32\

{ }

α

and β∈S1\

{ }

α . Similar operations are applied as above, we obtain

δ

∉B32\

{ }

α

.

Now, we suppose that δ σ= 1 and β∈S1\

{ }

α . Similar operations are ap-plied as above, we obtain δ σ≠ 1.

That means α δ β≠ for any α∈B32 and δ β, ∈S1\

{ }

α .

If α σ= 1, then by the definition of σ1, quasinormal representation of

α

has

a form α=

(

Z2×Z2

) (

∪

(

X Z\ 2

)

×Z1

)

. Let α δ β= for some δ β, ∈S1\

{ }

σ1 . We suppose that δ∈B0\

{ }

σ1 and β∈S1\

{ }

σ1 . By definition of B0,

qua-sinormal representation of δ _{has form}

(

Y3δ Z3

) (

Y2δ Z2

) (

Y1δ Z1

) (

Y0δ D

)

δ = × ∪ × ∪ × ∪ ×

{ }

. By using Z3⊂Z1⊂D

and Z2⊂D

we have Z3β

and Z2β are minimal elements of the semilattice

{

Z3β,Z2β,Z1β,Dβ

}

(9)

we have

(

) (

(

)

(

23 2 3

) (

2 2 2 1

) (

1 1

) (

0

)

\

Z Z X Z Z

Yδ Z Yδ Z Yδ Z Yδ D

α δ β

β β β β

× × = =

= × × × ×

∪

∪ ∪ ∪

From Z2 and Z1 are minimal elements of the semilattice

{

Z Z D2, ,1

}

, this equality is possible only if Z2 =Z3β, Z1=Z2β or Z2=Z2β, Z1=Z3β. By

using formal equalities, we obtain

2 3 2 1 2 1 3

1 3 2 2 2 1 3

and and

Z Z P Z Z P P

β β β β β

= = = =

= = = = =

∪

respectively. Let Z2=P2β and Z1=P1β∪P3β where P P1β, 3β∈

{

Z Z3, 1

}

. Then subquasinormal representation of β _{has one of the form}

(

) (

)

(

{ }

( )

)

1

1 3 2 2 3 1 2 \

t X D

P Z P Z P Z t t

β β ′∈ ′ ′ = × × × × ∪ ∪ ∪

∪

(

) (

)

(

{ }

( )

)

2

3 3 2 2 1 1 2 \

t X D

P Z P Z P Z t t

β β ′∈ ′ ′ = × × × × ∪ ∪ ∪

∪

(

) (

(

)

(

{ }

( )

)

3

2 2 1 3 1 2 \

t X D

P Z P P Z t t

β β ′∈ ′ ′ = × × × ∪ ∪ ∪

∪

where

1 2 3

1 1

3 2 1

P P P

Z Z Z

β = ∅_∅ 

 ,

1 2 3

2 1

1 2 3

P P P

Z Z Z

β = ∅_∅ 

 ,

1 2 3

3 1

1 2 1

P P P

Z Z Z

β = ∅_∅ 

 

are normal mapping for β, β2 is complement mapping of the set X D×

on the set D=

{

Z Z Z3, ,2 1

}

and δ β δ β = i. From formal equalities, we obtain

(

)

(

)

(

)

(

)

(

{ }

( )

)

1

2 1 3 1 2 2 2 1 1 2 \

\ \ \

t X D

Z Z Z Z Z Z Z Z Z t t

β β ′∈ ′ ′ = × ∪ × ∪ × ∪

∪

×

(

)

(

)

(

)

(

)

(

{ }

( )

)

2

2 1 3 1 2 2 2 1 1 2 \

\ \

t X D

Z Z Z Z Z Z Z Z Z t t

β β ′∈ ′ ′ = × × × × ∩ ∪ ∪ ∪

∪

(

)

(

)

(

)

(

{ }

( )

)

3

1 2 2 2 1 2 \

\

t X D

Z Z Z Z Z t t

β β ′∈ ′ ′ = × × × ∪ ∪

∪

and by using Y1δ∪Y0δ ≥1, we have

(

) (

)

(

) (

)

(

) (

(

)

1 2 3

1 1 1 1

3 3 2 2 1 1 0 3 2 2 1 1 0 3 2 2 1 1 0

Y Z Y Z Y Z Y D

Y Z Y Z Y D Y D

Y Z Y Z Y Y D

δ δ δ δ

δ β δ β δ β

β β β β

α = = = × × × × = × × × × = × × × ≠ ∪ ∪ ∪ ∪ ∪ ∪ ∪ ∪ ∪

This contradicts with α δ β= . So, δ∉B0\

{ }

σ1 .

Now, we suppose that δ∈B32\

{ }

σ1 and β∈S1\

{ }

σ1 . Similar operations are applied as above, we obtain δ∉B32\

{ }

σ1 .

That means α δ β≠ _{for any}α∈B₃₂_and _{δ β}_, ∈_S₁_\

{ }

_α . □

Lemma 6. Let D=

{

Z Z Z D3, , ,2 1

}

∈Σ2.2

(

X,4

)

, σ0=

(

Z Z3× 3

) (

∪

(

X Z\ 3

)

×Z1

)

and σ1=

(

Z2×Z2

) (

∪

(

X Z\ 2

)

×Z1

)

. If X D=

then S2 =B0∪B32∪

{

σ σ

0, 1

}

(10)

Theorem 7. Let D=

{

Z Z Z D3, , ,2 1

}

∈Σ2.2

(

X,4

)

,

(

) (

(

)

0 Z Z3 3 X Z\ 3 Z1

σ = × ∪ × and σ1=

(

Z2×Z2

) (

∪

(

X Z\ 2

)

×Z1

)

. If X is a

finite set and X =n then the following statements are true

a) If X D\ ≥1 then

{ }

1 2

0 32 1 4n 3n 2n 2

B _∪B _∪ σ ₌ ₋ + ₊ + ₋

b) If X D= _then

{

}

1 2

0 32 0, 1 4n 3n 2n 1

B _∪B _∪ σ σ ₌ ₋ + ₊ + ₋

Proof. Let

{

}

{

}

{

| : 1,2, , 1,2, , ,one to one mapping

}

n i i

S = ϕ ϕ M= n →M= n

be a group, ϕ ϕi1, , ,i2 ϕim∈Sn

(

m n≤

)

and Y Yϕ1, ϕ2, , Yϕm be partitioning of

X. It is well known that

{

₁ ₂

}

(

( )

) (

)

1

1 , , ,

1 ! !

m

m i m

m n

i

k Y Y Y

i m i

ϕ ϕ ϕ

+

=

−

= =

− −

∑

. If m=2,3, 4

then we have

2 1

3 1 1

4 1 1 1

2 1

1 ₃ ₂ 1

2 2

1 ₄ 1 ₃ 1 ₂ 1

6 2 2 6

n n

n

n n n

n

k

−

− −

− − −

= −

= ⋅ − +

= ⋅ − ⋅ + ⋅ −

If Y Yϕ1, ϕ2 are any two elements of partitioning of X and

(

Y Tϕ1 1

) (

Yϕ2 T2

)

β

= × ∪ × where T T D1, 2∈ and T T1≠ 2, then the number of

different binary relations β of semigroup B DX

( )

is equal to 2

2 2n 2

n

k

⋅ = − (2) If Y Y Yϕ1, ϕ2, ϕ3 are any three elements of partitioning of X and

(

Y Tϕ1 1

) (

Yϕ2 T2

)

(

Yϕ3 T3

)

β

= × ∪ × ∪ × where T T T1, ,2 3 are pairwise different

ele-ments of D, then the number of different binary relations β of semigroup

( )

X

B D is equal to

3

6 3n 3 2n 3

n

k

⋅ = − ⋅ + (3) If Y Y Y Yϕ1, ϕ2, ϕ3, ϕ4 are any four elements of partitioning of X and

(

Y Tϕ1 1

) (

Yϕ2 T2

)

(

Yϕ3 T3

) (

Yϕ4 T4

)

β

= × ∪ × ∪ × ∪ × where T T T T1, , ,2 3 4 are pairwise

different elements of D, then the number of different binary relations β of se-migroup B DX

( )

is equal to

4

24 4n 4 3n 3 2n 4

n

k

⋅ = − ⋅ + ⋅ − (4) Let α ∈B0. Quasinormal represantation of

α

has form

(

Y3α Z3

) (

Y2α Z2

) (

Y1α Z1

) (

Y0α D

)

α= × ∪ × ∪ × ∪ ×

where Y Y Y3α, ,2α 1α∉ ∅

{ }

. Also, Y Y Y3α, 2α, 1α or Y Y Y Y3α, 2α, 1α, 0α are partitioning of X for X ≥4. By using Equations (2.3) and (2.4) we obtain

1

0 4 3n n 3 2 1n

B ₌ ₋ + _{+ ⋅} ₋

Let α∈B32. Quasinormal represantation of

α

has form

(

Y3α Z3

) (

Y2α Z2

)

(11)

32 2n 2

B = −

So, we have

{ }

{

}

1 2 0 32 1

1 2 0 32 0 1

4 3 2 2

, 4 3 2 1

n n n

B B B B

σ σ σ

+ +

= − + −

∪ ∪

since B0∩B32=B0∩

{

σ σ0, 1

}

=B32∩

{

σ σ0, 1

}

= ∅. □

Acknowledgements

Sincere thanks to Prof. Dr. Neşet AYDIN for his valuable suggestions.

References

[1] Diasamidze, Y. and Makharadze, S. (2013) Complete Semigroups of Binary Rela-tions. Kriter Yay Nevi, Istanbul.

[2] Diasamidze, Y., Ayd, N.N. and Erdoğan, A. (2016) Generating Set of the Complete Semigroups of Binary Relations. Applied Mathematics, 7, 98-107.

https://doi.org/10.4236/am.2016.71009

[3] Bakuridze, A., Diasamidze, Y. and Givradze, O. Generated Sets of the Complete Semigroups of Binary Relations Defined by Semilattices of the Class Σ1

(

X,2 .

)

(In

http://www.scirp.org/journal/am

: 10.4236/am.2018.91002

http://creativecommons.org/licenses/by/4.0/

https://doi.org/10.4236/am.2016.71009