In this paper, we have studied generating sets of the complete semigroups defined by X-semilattices of the class Σ 2( X, 4).

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

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

α f = ∪ x ∈ X ( { x } × f ( x ) ) for each f. The set of all such binary relations is denoted

by B X ( D ) . It is easy to prove that B X ( D ) is a semigroup with respect to the product operation of binary relations. This semigroup B X ( D ) is called a complete semigroup of binary relations defined by an X-semilattice of unions D. This structure was comprehensively investigated in Diasamidze  and  . We assume that t , y ∈ X , Y ⊆ X , α ∈ B X , T ∈ D and ∅ ≠ D ′ ⊆ D . Then we denote following sets

y α = { x ∈ X | y α x } ,     Y α = ∪ y ∈ Y y α ,

V ( D , α ) = { Y α | Y ∈ D } ,     X ∗ = { Y | ∅ ≠ Y ⊆ X } Y T α = { y ∈ X | y α = T } ,     V ( X ∗ , α ) = { Y α | ∅ ≠ Y ⊆ X } D t = { Z ′ ∈ D | t ∈ Z ′ } ,     B 0 = { α ∈ B X ( D ) | V ( X ∗ , α ) = D }

Let D = { D ⌣ , Z 1 , Z 2 , ⋯ , Z m − 1 } be finite X-semilattice of unions and C ( D ) = { P 0 , P 1 , P 2 , ⋯ , P m − 1 } be the family of pairwise nonintersecting subsets of X. If φ = ( D ⌣ Z 1 ⋯ Z m − 1 P 0 P 1 ⋯ P m − 1 ) is a mapping from D on C ( D ) , then the equalities D ⌣ = P 0 ∪ P 1 ∪ P 2 ∪ ⋯ ∪ P m − 1 and Z i = P 0 ∪ ∪ T ∈ D \ D Z φ ( T ) are valid. These equalities are called formal.

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

of a binary relation α of the form α = ∪ T ∈ V ( X ∗ , α ) ( Y T α × T ) is called quasinormal.

Let P 0 , P 1 , P 2 , ⋯ , P m − 1 be parameters in the formal equalities, β ∈ B X ( D ) , β ¯ 2

be mapping from X \ D ⌣ to D . Then β ¯ = ∪ i = 0 m − 1 ( P i × ∪ t ∈ P i   t β ) ∪ ∪ t ′ ∈ X \ D ⌣ ( { t ′ } × β ¯ 2 ( t ′ ) )

is called subquasinormal represantation of β . It can be easily seen that the following statements are true.

a) β ¯ ∈ B X ( D ) .

b) ∪ i = 0 m − 1 ( P i × ∪ t ∈ P i   t β ) ⊆ β and β = β ¯ for some β ¯ 2 .

c) Subquasinormal represantation of β is quasinormal.

d) β ¯ 1 = ( P 0 P 1 ⋯ P m − 1 P 0 β ¯ P 1 β ¯ ⋯ P m − 1 β ¯ ) is mapping from C ( D ) on D ∪ { ∅ } .

β ¯ 1 and β ¯ 2 are respectively called normal and complement mappings for β .

Let α ∈ B X ( D ) . If α ≠ δ ∘ β for all δ , β ∈ B X ( D ) \ { α } then α is called external element. Every element of the set B 0 = { α ∈ B X ( D ) | V ( X ∗ , α ) = D } is an external element of B X ( D ) .

Theorem 1.  Let X be a finite set and α , β ∈ B X ( D ) . If β ¯ is sub- quasinormal representation of β then α ∘ β = α ∘ β ¯ .

Corollary 1.  Let B ˜ ′ ⊆ B ˜ ⊆ B X ( D ) . If α ≠ δ ∘ β ¯ for α ∈ B ˜ ′ , δ ∈ B ˜ \ { α } , β ¯ ∈ B ˜ \ { 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 X ( D ) in  .

2. Results

In this work by symbol Σ 2.2 ( X ,4 ) we denote all semilattices D = { Z 3 , Z 2 , Z 1 , D ⌣ } of the class Σ 2 ( X ,4 ) which the intersection of minimal elements Z 3 ∩ Z 2 = ∅ . This semilattices graphic is given in Figure 1. By using formal equalities, we have Z 3 ∩ Z 2 = P 0 = ∅ . So, the formal equalities of the semilattice D has a form

D ⌣ = P 1 ∪ P 2 ∪ P 3 Z 1 = P 2 ∪ P 3 Z 2 = P 1 ∪ P 3 Z 3 = P 2 (1)

Let δ , β ¯ ∈ B X ( D ) . If quasinormal representation of binary relation δ has a form δ = ( Y 3 δ × Z 3 ) ∪ ( Y 2 δ × Z 2 ) ∪ ( Y 1 δ × Z 1 ) ∪ ( Y 0 δ × D ⌣ ) then

δ ∘ β ¯ = ( Y 3 δ × Z 3 β ¯ ) ∪ ( Y 2 δ × Z 2 β ¯ ) ∪ ( Y 1 δ × Z 1 β ¯ ) ∪ ( Y 0 δ × D ⌣ β ¯ )

We denote the set

B 32 = { α ∈ B X ( D ) | V ( X ∗ , α ) = { Z 3 , Z 2 , D ⌣ } } B 21 = { α ∈ B X ( D ) | V ( X ∗ , α ) = { Z 2 , Z 1 , D ⌣ } } B 31 = { α ∈ B X ( D ) | V ( X ∗ , α ) = { Z 3 , Z 1 } } B ˜ 32 = { α ∈ B 32 | α = ( Y 3 α × Z 3 ) ∪ ( Y 2 α × Z 2 ) , Y 3 α ∪ Y 2 α = X , Y 3 α ∩ Y 2 α = ∅ } B ˜ 21 = { α ∈ B 21 | α = ( Y 2 α × Z 2 ) ∪ ( Y 1 α × Z 1 ) , Y 2 α ∪ Y 1 α = X , Y 2 α ∩ Y 1 α = ∅ }

It is easy to see that

B 0 ∩ B 32 = B 0 ∩ B 21 = B 0 ∩ B 31 = B 21 ∩ B 32 = B 31 ∩ B 32 = B 21 ∩ B 31 = ∅ .

Lemma 2. Let D = { Z 3 , Z 2 , Z 1 , D ⌣ } ∈ Σ 2.2 ( X , 4 ) . Then following statements are true for the sets B 0 , B 32 , B ˜ 32 .

a) If α = ( Y 3 α × Z 3 ) ∪ ( Y 1 α × Z 1 ) ∪ ( Y 0 α × D ⌣ ) for some Y 3 α , Y 1 α , Y 0 α ∉ ∅ , then α is product of some elements of the set B 0 .

b) If β 0 = ( Z 3 × Z 3 ) ∪ ( ( X \ Z 3 ) × Z 2 ) , then ( B 0 ∘ β 0 ) ∪ B ˜ 32 = B 32 .

c) If σ 1 = ( Z 2 × Z 2 ) ∪ ( ( X \ Z 2 ) × Z 1 ) , then ( B 0 ∘ σ 1 ) ∪ B ˜ 21 = B 21 .

d) If σ 1 = ( Z 2 × Z 2 ) ∪ ( ( X \ Z 2 ) × Z 1 ) , then B 32 ∘ σ 1 = B 21 .

e) If σ 0 = ( Z 3 × Z 3 ) ∪ ( ( X \ Z 3 ) × Z 1 ) , then B 32 ∘ σ 0 = B 31 .

f) Every element of the set B 32 is product of elements of the set B 0 ∪ B ˜ 32 .

g) Every element of the set B 21 is product of elements of the set B 0 ∪ B ˜ 32 ∪ { σ 1 } .

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

a. Let α = ( Y 3 α × Z 3 ) ∪ ( Y 1 α × Z 1 ) ∪ ( Y 0 α × D ⌣ ) for some Y 3 α , Y 1 α , Y 0 α ∉ { ∅ } , δ , β ¯ ∈ B 0 . Then quasinormal representation of δ has a form

δ = ( Y 3 δ × Z 3 ) ∪ ( Y 2 δ × Z 2 ) ∪ ( Y 1 δ × Z 1 ) ∪ ( Y 0 δ × D ⌣ )

where Y 3 δ , Y 1 δ , Y 0 δ ∉ { ∅ } . We suppose that

β ¯ = ( P 2 × Z 3 ) ∪ ( P 1 × Z 2 ) ∪ ( P 3 × Z 1 ) ∪ ∪ t ′ ∈ X \ D ⌣ ( { t ′ } × β ¯ 2 ( t ′ ) )

where β ¯ 1 = ( ∅ P 1 P 2 P 3 ∅ Z 2 Z 3 Z 1 ) is normal mapping for β ¯ and β ¯ 2 is com-

plement mapping of the set X \ D ⌣ on the set D ˜ . So, β ¯ ∈ B 0 since V ( X ∗ , β ¯ ) = D . From the equalities (2.1) and definition of β ¯

Z 3 β ¯ = P 2 β ¯ = Z 3 Z 2 β ¯ = ( P 1 ∪ P 3 ) β ¯ = P 1 β ¯ ∪ P 3 β ¯ = Z 2 ∪ Z 1 = D ⌣ Z 1 β ¯ = ( P 2 ∪ P 3 ) β ¯ = P 2 β ¯ ∪ P 3 β ¯ = Z 3 ∪ Z 1 = Z 1 D ⌣ β ¯ = ( P 1 ∪ P 2 ∪ P 3 ) β ¯ = P 1 β ¯ ∪ P 2 β ¯ ∪ P 3 β ¯ = Z 2 ∪ D ⌣ ∪ Z 1 = D ⌣

δ ∘ β ¯ = ( Y 3 δ × Z 3 β ¯ ) ∪ ( Y 2 δ × Z 2 β ¯ ) ∪ ( Y 1 δ × Z 1 β ¯ ) ∪ ( Y 0 δ × D ⌣ β ¯ ) = ( Y 3 δ × D ⌣ ) ∪ ( Y 2 δ × D ⌣ ) ∪ ( Y 1 δ × D ⌣ ) ∪ ( Y 0 δ × D ⌣ ) = ( Y 3 δ × D ⌣ ) ∪ ( Y 1 δ × D ⌣ ) ∪ ( ( Y 2 δ ∪ Y 0 δ ) × D ⌣ ) = α .

b. Let α ∈ B 0 ∘ β 0 ∪ B ˜ 32 . Then α ∈ B 0 ∘ β 0 or α ∈ B ˜ 32 . If α ∈ B 0 ∘ β 0 then α = δ ∘ β 0 for some δ ∈ B 0 . In this case we have

δ = ( Y 3 δ × Z 3 ) ∪ ( Y 2 δ × Z 2 ) ∪ ( Y 1 δ × Z 1 ) ∪ ( Y 0 δ × D ⌣ )

where Y 3 δ , Y 1 δ , Y 0 δ ∉ { ∅ } . Also

α = δ ∘ β 0 = ( Y 3 δ × Z 3 β 0 ) ∪ ( Y 2 δ × Z 2 β 0 ) ∪ ( Y 1 δ × Z 1 β 0 ) ∪ ( Y 0 δ × D ⌣ β 0 ) = ( Y 3 δ × Z 3 ) ∪ ( Y 2 δ × Z 2 ) ∪ ( Y 1 δ × Z 1 ) ∪ ( Y 0 δ × D ⌣ ) = ( Y 3 δ × Z 3 ) ∪ ( Y 2 δ × Z 2 ) ∪ ( ( Y 1 δ ∪ Y 0 δ ) × D ⌣ ) ∈ B 32 \ B ˜ 32

is satisfied. So, we have ( B 0 ∘ β 0 ) ∪ B ˜ 32 ⊆ B 32 . On the other hand, if α ∈ B ˜ 32 ⊆ B 32 then ( B 0 ∘ β 0 ) ∪ B ˜ 32 ⊆ B 32 is satisfied. Conversely, if α ∈ B 32 then quasinormal representation of α has a form

α = ( Y 3 α × Z 3 ) ∪ ( Y 2 α × Z 2 ) ∪ ( Y 0 α × D ⌣ )

where Y 3 α , Y 2 α , Y 0 α ∉ { ∅ } or Y 3 α , Y 2 α ∉ { ∅ } and Y 0 α = ∅ . We suppose that Y 3 α , Y 2 α ∉ { ∅ } . In this case, we have

δ ∘ β 0 = ( Y 3 δ × Z 3 β 0 ) ∪ ( Y 2 δ × Z 2 β 0 ) ∪ ( Y 0 δ × Z 1 β 0 ) = ( Y 3 δ × Z 3 ) ∪ ( Y 2 δ × Z 2 ) ( Y 0 δ × D ⌣ ) = α

for δ = ( Y 3 α × Z 3 ) ∪ ( Y 2 α × Z 2 ) ∪ ( Y 0 α × Z 1 ) ∈ B 0 . So, we have B 32 ⊆ ( B 0 ∘ β 0 ) ∪ B ˜ 32 . Now suppose that Y 3 α , Y 2 α ∉ { ∅ } and Y 0 α = ∅ . In this case, we have α ∈ B ˜ 32 ⊆ ( B 0 ∘ β 0 ) ∪ B ˜ 32 . So, ( B 0 ∘ β 0 ) ∪ B ˜ 32 = B 32 .

g. From the statement c, we have that ( B 0 ∘ β 0 ) ∪ B ˜ 32 = B 32 where β 0 ∈ B ˜ 32 by definition of β 0 . Thus, every element of the set B 32 is product of elements of the set B 0 ∪ B ˜ 32 .

Lemma 3. Let D = { Z 3 , Z 2 , Z 1 , D ⌣ } ∈ Σ 2.2 ( 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 B 0 .

b) If α = X × Z 1 then α is product of elements of the set B 0 .

c) If α = ( Y 3 α × Z 3 ) ∪ ( Y 1 α × Z 1 ) for some Y 3 α , Y 1 α ∉ ∅ , then α is product of elements of the B 0 .

d) If α = ( Y 3 α × Z 3 ) ∪ ( Y 0 α × D ⌣ ) for some Y 3 α , Y 0 α ∉ ∅ , then α is product of elements of the B 0 .

e) If α = ( Y 2 α × Z 2 ) ∪ ( Y 0 α × D ⌣ ) for some Y 2 α , Y 0 α ∉ ∅ , then α is product of elements of the B 0 .

f) If α = ( Y 1 α × Z 1 ) ∪ ( Y 0 α × D ⌣ ) for some Y 1 α , Y 0 α ∉ ∅ , then α is product of elements of the B 0 .

Proof. c. Let quasinormal representation of α has a form α = ( Y 3 α × Z 3 ) ∪ ( Y 1 α × Z 1 ) where Y 3 δ , Y 1 δ ∉ { ∅ } . By definition of the semilattice D, | X | ≥ 3 . We suppose that | Y 3 α | ≥ 1 and | Y 1 α | ≥ 2 . In this case, we suppose that

β ¯ = ( P 2 × Z 3 ) ∪ ( ( P 1 ∪ P 3 ) × Z 1 ) ∪ ∪ t ′ ∈ X \ D ⌣ ( { t ′ } × β ¯ 2 ( t ′ ) )

where β ¯ 1 = ( ∅ P 1 P 2 P 3 ∅ Z 1 Z 3 Z 1 ) is normal mapping for β ¯ and β ¯ 2 is comple-

ment mapping of the set X × D ⌣ on the set D ˜ \ { Z 3 , Z 1 } = { Z 2 } (by suppose | X \ D ⌣ | ≥ 1 ). So, β ¯ ∈ B 0 since V ( X ∗ , β ¯ ) = D . Also, Y 3 δ = Y 3 α and Y 2 δ ∪ Y 1 δ ∪ Y 0 δ = Y 1 δ since | Y 3 δ | ≥ 1 , | Y 2 δ | ≥ 1 , | Y 1 δ | ≥ 1 , | Y 0 δ | ≥ 0 . From the equalities (2.1) and definition of β ¯ we obtain that

Z 3 β ¯ = P 2 β ¯ = Z 3 Z 2 β ¯ = ( P 1 ∪ P 3 ) β ¯ = P 1 β ¯ ∪ P 3 β ¯ = Z 1 ∪ Z 1 = Z 1 Z 1 β ¯ = ( P 2 ∪ P 3 ) β ¯ = P 2 β ¯ ∪ P 3 β ¯ = Z 3 ∪ Z 1 = Z 1 D ⌣ β ¯ = ( P 1 ∪ P 2 ∪ P 3 ) β ¯ = P 1 β ¯ ∪ P 2 β ¯ ∪ P 3 β ¯ = Z 1 ∪ Z 3 ∪ Z 1 = Z 1

δ ∘ β ¯ = ( Y 3 δ × Z 3 β ¯ ) ∪ ( Y 2 δ × Z 2 β ¯ ) ∪ ( Y 1 δ × Z 1 β ¯ ) ∪ ( Y 0 δ × D ⌣ β ¯ ) = ( Y 3 δ × Z 3 ) ∪ ( Y 2 δ × Z 1 ) ∪ ( Y 1 δ × Z 1 ) ∪ ( Y 0 δ × Z 1 ) = ( Y 3 δ × Z 3 ) ∪ ( ( Y 2 δ ∪ Y 1 δ ∪ Y 0 δ ) × Z 1 ) = α

Now, we suppose that | Y 3 α | ≥ 2 and | Y 1 α | ≥ 1 . In this case, we suppose that

β ¯ = ( ( P 2 ∪ P 3 ) × Z 3 ) ∪ ( P 1 × Z 1 ) ∪ ∪ t ′ ∈ X \ D ⌣ ( { t ′ } × β ¯ 2 ( t ′ ) )

where β ¯ 1 = ( ∅ P 1 P 2 P 3 ∅ Z 1 Z 3 Z 3 ) is normal mapping for β ¯ and β ¯ 2 is com-

plement mapping of the set X × D ⌣ on the set D ˜ \ { Z 3 , Z 1 } = { Z 2 } (by suppose | X \ D ⌣ | ≥ 1 ). So, β ¯ ∈ B 0 since V ( X ∗ , β ¯ ) = D . Also, Y 3 δ ∪ Y 1 δ = Y 3 α and Y 2 δ ∪ Y 0 δ = Y 1 α since | Y 3 δ | ≥ 1, | Y 2 δ | ≥ 1, | Y 1 δ | ≥ 1, | Y 0 δ | ≥ 0 . From the equalities (2.1) and definition of β ¯ we obtain that

Z 3 β ¯ = P 2 β ¯ = Z 3 Z 2 β ¯ = ( P 1 ∪ P 3 ) β ¯ = P 1 β ¯ ∪ P 3 β ¯ = Z 1 ∪ Z 3 = Z 1 Z 1 β ¯ = ( P 2 ∪ P 3 ) β ¯ = P 2 β ¯ ∪ P 3 β ¯ = Z 3 ∪ Z 3 = Z 3 D ⌣ β ¯ = ( P 1 ∪ P 2 ∪ P 3 ) β ¯ = P 1 β ¯ ∪ P 2 β ¯ ∪ P 3 β ¯ = Z 1 ∪ Z 3 ∪ Z 3 = Z 1

δ ∘ β ¯ = ( Y 3 δ × Z 3 β ¯ ) ∪ ( Y 2 δ × Z 2 β ¯ ) ∪ ( Y 1 δ × Z 1 β ¯ ) ∪ ( Y 0 δ × D ⌣ β ¯ ) = ( Y 3 δ × Z 3 ) ∪ ( Y 2 δ × Z 1 ) ∪ ( Y 1 δ × Z 3 ) ∪ ( Y 0 δ × Z 1 ) = ( ( Y 3 δ ∪ Y 1 δ ) × Z 3 ) ∪ ( ( Y 2 δ ∪ Y 0 δ ) × Z 1 ) = α

Lemma 4. Let D = { Z 3 , Z 2 , Z 1 , D ⌣ } ∈ Σ 2.2 ( X , 4 ) , σ 0 = ( Z 3 × Z 3 ) ∪ ( ( X \ Z 3 ) × Z 1 ) and σ 1 = ( Z 2 × Z 2 ) ∪ ( ( X \ Z 2 ) × Z 1 ) . If X = D ⌣ then the following statements are true

a) If α = ( Y 3 α × Z 3 ) ∪ ( Y 0 α × D ⌣ ) for some Y 3 α , Y 0 α ∉ { ∅ } , then α is product of elements of the B 0 ∪ B 32 .

b) If α = ( Y 2 α × Z 2 ) ∪ ( Y 0 α × D ⌣ ) for some Y 2 α , Y 0 α ∉ { ∅ } , then α is product of elements of the B 32 ∪ { σ 1 } .

c) If α = ( Y 1 α × Z 1 ) ∪ ( Y 0 α × D ⌣ ) for some Y 1 α , Y 0 α ∉ { ∅ } , then α is product of elements of the B 32 ∪ { σ 0 , σ 1 } .

Proof. First, remark that Z 3 σ 0 = Z 3 , Z 2 σ 0 = D ⌣ σ 0 = Z 1 , Z 3 σ 1 = Z 1 , Z 2 σ 1 = Z 2 , D ⌣ σ 1 = D ⌣ .

a. Let α = ( Y 3 α × Z 3 ) ∪ ( Y 0 α × D ⌣ ) for some Y 3 α , Y 0 α ∉ ∅ . In this case, we suppose that

δ = ( Y 3 δ × Z 3 ) ∪ ( Y 2 δ × Z 2 ) ∪ ( Y 0 δ × D ⌣ )

and

β 1 = ( Z 3 × Z 3 ) ∪ ( ( Z 1 \ Z 3 ) × Z 1 ) ∪ ( ( X \ Z 1 ) × D ⌣ )

where Y 3 δ , Y 2 δ ∉ { ∅ } . It is easy to see that δ ∈ B 32 and β 1 is generating by elements of the B 0 by statement b of Lemma 2. Also, Y 3 δ = Y 3 α and Y 2 δ ∪ Y 0 δ = Y 0 α since Z 3 β ¯ = Z 3 , Z 2 β ¯ = D ⌣ β ¯ = D ⌣ and | Y 3 δ | ≥ 1, | Y 2 δ | ≥ 1, | Y 0 δ | ≥ 0 . So, α is product of elements of the B 0 ∪ B 32 . □

Lemma 5. Let

D = { Z 3 , Z 2 , Z 1 , D ⌣ } ∈ Σ 2.2 ( X , 4 )

and

σ 1 = ( Z 2 × Z 2 ) ∪ ( ( X \ Z 2 ) × Z 1 ) .

If | X \ D ⌣ | ≥ 1 then S 1 = B 0 ∪ B ˜ 32 ∪ { σ 1 } is an irreducible generating set for the semigroup B X ( D ) .

Proof. First, we must prove that every element of B X ( D ) is product of elements of S 1 . Let α ∈ B X ( D ) and

α = ( Y 3 α × Z 3 ) ∪ ( Y 2 α × Z 2 ) ∪ ( Y 1 α × Z 1 ) ∪ ( Y 0 α × D ⌣ )

where Y 3 α ∪ Y 2 α ∪ Y 1 α ∪ Y 0 α = X and Y 3 α ∩ Y 2 α = ∅ , ( 0 ≤ i ≠ j ≤ 3 ) . We suppose

that | V ( X ∗ , α ) | = 1 . Then we have V ( X ∗ , α ) ∈ { { Z 3 } , { Z 2 } , { Z 1 } , { D ⌣ } } . If

V ( X ∗ , α ) ∈ { { Z 3 } , { Z 2 } , { Z 1 } } then α = X × Z 3 or α = X × Z 2 or α = X × Z 1 . Quasinormal representations of δ , β 1 , β 2 and β 3 has form

δ = ( Y 3 δ × Z 3 ) ∪ ( Y 2 δ × Z 2 ) ∪ ( Y 1 δ × Z 1 ) ∪ ( Y 0 δ × D ⌣ ) β 1 = ( D ⌣ × Z 3 ) ∪ ( ( X \ D ⌣ ) × Z 2 ) β 2 = ( D ⌣ × Z 2 ) ∪ ( ( X \ D ⌣ ) × Z 1 ) β 3 = ( D ⌣ × Z 1 ) ∪ ( ( X \ D ⌣ ) × Z 2 )

where Y 3 δ , Y 2 δ , Y 1 δ ∉ { ∅ } . So, δ ∈ B 0 , β 1 ∈ B ˜ 32 and β 2 , β 3 ∈ B 21 since | X \ D ⌣ | ≥ 1 . From the definition of δ , β 1 , β 2 and β 3 we obtain that

δ ∘ β 1 = ( Y 3 δ × Z 3 β 1 ) ∪ ( Y 2 δ × Z 2 β 1 ) ∪ ( Y 1 δ × Z 1 β 1 ) ∪ ( Y 0 δ × D ⌣ β 1 ) = ( Y 3 δ × Z 3 ) ∪ ( Y 2 δ × Z 3 ) ∪ ( Y 1 δ × Z 3 ) ∪ ( Y 0 δ × Z 3 ) = ( Y 3 δ ∪ Y 2 δ ∪ Y 1 δ ∪ Y 0 δ ) × Z 3 = X × Z 3

δ ∘ β 2 = ( Y 3 δ × Z 3 β 2 ) ∪ ( Y 2 δ × Z 2 β 2 ) ∪ ( Y 1 δ × Z 1 β 2 ) ∪ ( Y 0 δ × D ⌣ β 2 ) = ( Y 3 δ × Z 2 ) ∪ ( Y 2 δ × Z 2 ) ∪ ( Y 1 δ × Z 2 ) ∪ ( Y 0 δ × Z 2 ) = ( Y 3 δ ∪ Y 2 δ ∪ Y 1 δ ∪ Y 0 δ ) × Z 2 = X × Z 2

δ ∘ β 3 = ( Y 3 δ × Z 3 β 3 ) ∪ ( Y 2 δ × Z 2 β 3 ) ∪ ( Y 1 δ × Z 1 β 3 ) ∪ ( Y 0 δ × D ⌣ β 3 ) = ( Y 3 δ × Z 1 ) ∪ ( Y 2 δ × Z 1 ) ∪ ( Y 1 δ × Z 1 ) ∪ ( Y 0 δ × Z 1 ) = ( Y 3 δ ∪ Y 2 δ ∪ Y 1 δ ∪ Y 0 δ ) × Z 1 = X × Z 1

That means, X × Z 1 , X × Z 2 and X × Z 3 are generated by B 0 ∪ B ˜ 32 , B 0 ∪ B 21 and B 0 ∪ B 21 respectively. By using statement g and h of Lemma 3, we have X × Z 1 , X × Z 2 and X × Z 3 are generated by B 0 ∪ B ˜ 32 ∪ { σ 1 } . 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 B 0 .

So, S 1 is generating set for the semigroup B X ( D ) . Now, we must prove that S 1 = B 0 ∪ B ˜ 32 ∪ { σ 1 } is irreducible. Let α ∈ S 1 .

If α ∈ B 0 then α ≠ σ ∘ τ for all σ , τ ∈ B X ( D ) \ { α } from Lemma 2. So, α ≠ σ ∘ τ for all σ , τ ∈ S 1 \ { α } . That means, α ∉ B 0 .

If α ∈ B ˜ 32 then the quasinormal representation of α has form α = ( Y 3 α × Z 3 ) ∪ ( Y 2 α × Z 2 ) for some Y 3 α , Y 2 α ∉ ∅ . Let α = δ ∘ β for some δ , β ∈ S 1 \ { α } .

We suppose that δ ∈ B 0 \ { α } and β ∈ S 1 \ { α } . By definition of B 0 , quasinormal representation of δ has form

δ = ( Y 3 δ × Z 3 ) ∪ ( Y 2 δ × Z 2 ) ∪ ( Y 1 δ × Z 1 ) ∪ ( Y 0 δ × D ⌣ )

where Y 3 δ , Y 2 δ , Y 1 δ ∉ { ∅ } . By using Z 3 ⊂ Z 1 ⊂ D ⌣ and Z 2 ⊂ D ⌣ we have Z 3 β and Z 2 β are minimal elements of the semilattice { Z 3 β , Z 2 β , Z 1 β , D ⌣ β } . Also, we have

( Y 3 α × Z 3 ) ∪ ( Y 2 α × Z 2 ) = α = δ ∘ β = ( Y 3 δ × Z 3 β ) ∪ ( Y 2 δ × Z 2 β ) ∪ ( Y 1 δ × Z 1 β ) ∪ ( Y 0 δ × D ⌣ β )

Since Z 3 and Z 2 are minimal elements of the semilattice { Z 3 , Z 2 , D ⌣ } , this equality is possible only if Z 3 = Z 3 β , Z 2 = Z 2 β or Z 3 = Z 2 β , Z 2 = Z 3 β . By using formal equalities and P 3 β , P 2 β , P 1 β ∈ D , we obtain

Z 3 = Z 3 β = P 2 β and Z 2 = Z 2 β = P 1 β = P 3 β Z 2 = Z 3 β = P 2 β and Z 3 = Z 2 β = P 1 β = P 3 β

respectively. Let Z 3 = P 2 β and Z 2 = P 1 β = P 3 β . If β ¯ is sub-quasinormal representation of β then δ ∘ β = δ ∘ β ¯ and

β ¯ = ( ( P 1 ∪ P 3 ) × Z 2 ) ∪ ( P 2 × Z 3 ) ∪ ∪ t ′ ∈ X \ D ⌣ ( { t ′ } × β ¯ 2 ( t ′ ) )

where β ¯ 1 = ( ∅ P 1 P 2 P 3 ∅ Z 2 Z 3 Z 2 ) is normal mapping for β ¯ and β ¯ 2 is com-

plement mapping of the set X × D ⌣ on the set D ˜ = { Z 3 , Z 2 , Z 1 } . From formal equalities, we obtain

β ¯ = ( Z 2 × Z 2 ) ∪ ( Z 3 × Z 3 ) ∪ ∪ t ′ ∈ X \ D ⌣ ( { t ′ } × β ¯ 2 ( t ′ ) ) ∈ S 1 \ { α }

and by using Z 1 ∩ Z 2 ≠ ∅ , Z 3 ∪ Z 2 = D and | Y 1 δ ∪ Y 0 δ | ≥ 1 , we have

δ ∘ β ¯ = ( Y 3 δ × Z 3 β ¯ ) ∪ ( Y 2 δ × Z 2 β ¯ ) ∪ ( Y 1 δ × Z 1 β ¯ ) ∪ ( Y 0 δ × D ⌣ β ¯ ) = ( Y 3 δ × Z 3 ) ∪ ( Y 2 δ × Z 2 ) ∪ ( Y 1 δ × D ⌣ ) ∪ ( Y 0 δ × D ⌣ ) = ( Y 3 δ × Z 3 ) ∪ ( Y 2 δ × Z 2 ) ∪ ( ( Y 1 δ ∪ Y 0 δ ) × D ⌣ ) ≠ α

This contradicts with α = δ ∘ β . So, δ ∉ B 0 \ { α } .

Now, we suppose that δ ∈ B ˜ 32 \ { α } and β ∈ S 1 \ { α } . Similar operations are applied as above, we obtain δ ∉ B ˜ 32 \ { α } .

Now, we suppose that δ = σ 1 and β ∈ S 1 \ { α } . Similar operations are applied as above, we obtain δ ≠ σ 1 .

That means α ≠ δ ∘ β for any α ∈ B ˜ 32 and δ , β ∈ S 1 \ { α } .

If α = σ 1 , then by the definition of σ 1 , quasinormal representation of α has a form α = ( Z 2 × Z 2 ) ∪ ( ( X \ Z 2 ) × Z 1 ) . Let α = δ ∘ β for some δ , β ∈ S 1 \ { σ 1 } .

We suppose that δ ∈ B 0 \ { σ 1 } and β ∈ S 1 \ { σ 1 } . By definition of B 0 , quasinormal representation of δ has form

δ = ( Y 3 δ × Z 3 ) ∪ ( Y 2 δ × Z 2 ) ∪ ( Y 1 δ × Z 1 ) ∪ ( Y 0 δ × D ⌣ )

where Y 3 δ , Y 2 δ , Y 1 δ ∉ { ∅ } . By using Z 3 ⊂ Z 1 ⊂ D ⌣ and Z 2 ⊂ D ⌣ we have Z 3 β and Z 2 β are minimal elements of the semilattice { Z 3 β , Z 2 β , Z 1 β , D ⌣ β } . Also, we have

( Z 2 × Z 2 ) ∪ ( ( X \ Z 2 ) × Z 1 ) = α = δ ∘ β = ( Y 3 δ × Z 3 β ) ∪ ( Y 2 δ × Z 2 β ) ∪ ( Y 1 δ × Z 1 β ) ∪ ( Y 0 δ × D ⌣ β )

From Z 2 and Z 1 are minimal elements of the semilattice { Z 2 , Z 1 , D ⌣ } , this equality is possible only if Z 2 = Z 3 β , Z 1 = Z 2 β or Z 2 = Z 2 β , Z 1 = Z 3 β . By using formal equalities, we obtain

Z 2 = Z 3 β = P 2 β and Z 1 = Z 2 β = P 1 β ∪ P 3 β Z 1 = Z 3 β = P 2 β and Z 2 = Z 2 β = P 1 β = P 3 β

respectively. Let Z 2 = P 2 β and Z 1 = P 1 β ∪ P 3 β where P 1 β , P 3 β ∈ { Z 3 , Z 1 } . Then subquasinormal representation of β has one of the form

β ¯ 1 = ( P 1 × Z 3 ) ∪ ( P 2 × Z 2 ) ∪ ( P 3 × Z 1 ) ∪ ∪ t ′ ∈ X \ D ⌣ ( { t ′ } × β ¯ 2 ( t ′ ) )

β ¯ 2 = ( P 3 × Z 3 ) ∪ ( P 2 × Z 2 ) ∪ ( P 1 × Z 1 ) ∪ ∪ t ′ ∈ X \ D ⌣ ( { t ′ } × β ¯ 2 ( t ′ ) )

β ¯ 3 = ( P 2 × Z 2 ) ∪ ( ( P 1 ∪ P 3 ) × Z 1 ) ∪ ∪ t ′ ∈ X \ D ⌣ ( { t ′ } × β ¯ 2 ( t ′ ) )

where

β ¯ 1 1 = ( ∅ P 1 P 2 P 3 ∅ Z 3 Z 2 Z 1 ) , β ¯ 1 2 = ( ∅ P 1 P 2 P 3 ∅ Z 1 Z 2 Z 3 ) , β ¯ 1 3 = ( ∅ P 1 P 2 P 3 ∅ Z 1 Z 2 Z 1 )

are normal mapping for β ¯ , β ¯ 2 is complement mapping of the set X × D ⌣ on the set D ˜ = { Z 3 , Z 2 , Z 1 } and δ ∘ β = δ ∘ β ¯ i . From formal equalities, we obtain

β ¯ 1 = ( ( Z 2 \ Z 1 ) × Z 3 ) ∪ ( ( Z 1 \ Z 2 ) × Z 2 ) ∪ ( ( Z 2 \ Z 1 ) × Z 1 ) ∪ ∪ t ′ ∈ X \ D ⌣ ( { t ′ } × β ¯ 2 ( t ′ ) )

β ¯ 2 = ( ( Z 2 ∩ Z 1 ) × Z 3 ) ∪ ( ( Z 1 \ Z 2 ) × Z 2 ) ∪ ( ( Z 2 \ Z 1 ) × Z 1 ) ∪ ∪ t ′ ∈ X \ D ⌣ ( { t ′ } × β ¯ 2 ( t ′ ) )

β ¯ 3 = ( ( Z 1 \ Z 2 ) × Z 2 ) ∪ ( Z 2 × Z 1 ) ∪ ∪ t ′ ∈ X \ D ⌣ ( { t ′ } × β ¯ 2 ( t ′ ) )

and by using | Y 1 δ ∪ Y 0 δ | ≥ 1 , we have

δ ∘ β ¯ 1 = δ ∘ β ¯ 2 = δ ∘ β ¯ 3 = ( Y 3 δ × Z 3 β ¯ 1 ) ∪ ( Y 2 δ × Z 2 β ¯ 1 ) ∪ ( Y 1 δ × Z 1 β ¯ 1 ) ∪ ( Y 0 δ × D ⌣ β ¯ 1 ) = ( Y 3 δ × Z 2 ) ∪ ( Y 2 δ × Z 1 ) ∪ ( Y 1 δ × D ⌣ ) ∪ ( Y 0 δ × D ⌣ ) = ( Y 3 δ × Z 2 ) ∪ ( Y 2 δ × Z 1 ) ∪ ( ( Y 1 δ ∪ Y 0 δ ) × D ⌣ ) ≠ α

This contradicts with α = δ ∘ β . So, δ ∉ B 0 \ { σ 1 } .

Now, we suppose that δ ∈ B ˜ 32 \ { σ 1 } and β ∈ S 1 \ { σ 1 } . Similar operations are applied as above, we obtain δ ∉ B ˜ 32 \ { σ 1 } .

That means α ≠ δ ∘ β for any α ∈ B ˜ 32 and δ , β ∈ S 1 \ { α } . □

Lemma 6. Let D = { Z 3 , Z 2 , Z 1 , D ⌣ } ∈ Σ 2.2 ( X , 4 ) , σ 0 = ( Z 3 × Z 3 ) ∪ ( ( X \ Z 3 ) × Z 1 ) and σ 1 = ( Z 2 × Z 2 ) ∪ ( ( X \ Z 2 ) × Z 1 ) . If X = D ⌣ then S 2 = B 0 ∪ B ˜ 32 ∪ { σ 0 , σ 1 } is irreducible generating set for the semigroup B X ( D ) .

Theorem 7. Let D = { Z 3 , Z 2 , Z 1 , D ⌣ } ∈ Σ 2.2 ( X , 4 ) , σ 0 = ( Z 3 × Z 3 ) ∪ ( ( X \ Z 3 ) × Z 1 ) and σ 1 = ( Z 2 × Z 2 ) ∪ ( ( X \ Z 2 ) × Z 1 ) . If X is a finite set and | X | = n then the following statements are true

a) If | X \ D ⌣ | ≥ 1 then | B 0 ∪ B ˜ 32 ∪ { σ 1 } | = 4 n − 3 n + 1 + 2 n + 2 − 2

b) If X = D ⌣ then | B 0 ∪ B ˜ 32 ∪ { σ 0 , σ 1 } | = 4 n − 3 n + 1 + 2 n + 2 − 1

Proof. Let

S n = { φ i | φ i : M = { 1,2, ⋯ , n } → M = { 1,2, ⋯ , n } , onetoonemapping }

be a group, φ i 1 , φ i 2 , ⋯ , φ i m ∈ S n ( m ≤ n ) and Y φ 1 , Y φ 2 , ⋯ , Y φ m be partitioning of

X. It is well known that k n m = | { Y φ 1 , Y φ 2 , ⋯ , Y φ m } | = ∑ i = 1 m ( − 1 ) m + i ( i − 1 ) ! ( m − i ) ! . If m = 2 , 3 , 4

then we have

k n 2 = 2 n − 1 − 1 k n 3 = 1 2 ⋅ 3 n − 1 − 2 n − 1 + 1 2 k n 4 = 1 6 ⋅ 4 n − 1 − 1 2 ⋅ 3 n − 1 + 1 2 ⋅ 2 n − 1 − 1 6

If Y φ 1 , Y φ 2 are any two elements of partitioning of X and β ¯ = ( Y φ 1 × T 1 ) ∪ ( Y φ 2 × T 2 ) where T 1 , T 2 ∈ D and T 1 ≠ T 2 , then the number of different binary relations β ¯ of semigroup B X ( D ) is equal to

2 ⋅ k n 2 = 2 n − 2 (2)

If Y φ 1 , Y φ 2 , Y φ 3 are any three elements of partitioning of X and β ¯ = ( Y φ 1 × T 1 ) ∪ ( Y φ 2 × T 2 ) ∪ ( Y φ 3 × T 3 ) where T 1 , T 2 , T 3 are pairwise different elements of D, then the number of different binary relations β ¯ of semigroup B X ( D ) is equal to

6 ⋅ k n 3 = 3 n − 3 ⋅ 2 n + 3 (3)

If Y φ 1 , Y φ 2 , Y φ 3 , Y φ 4 are any four elements of partitioning of X and β ¯ = ( Y φ 1 × T 1 ) ∪ ( Y φ 2 × T 2 ) ∪ ( Y φ 3 × T 3 ) ∪ ( Y φ 4 × T 4 ) where T 1 , T 2 , T 3 , T 4 are pairwise different elements of D, then the number of different binary relations β ¯ of semigroup B X ( D ) is equal to

24 ⋅ k n 4 = 4 n − 4 ⋅ 3 n + 3 ⋅ 2 n − 4 (4)

Let α ∈ B 0 . Quasinormal represantation of α has form

α = ( Y 3 α × Z 3 ) ∪ ( Y 2 α × Z 2 ) ∪ ( Y 1 α × Z 1 ) ∪ ( Y 0 α × D ⌣ )

where Y 3 α , Y 2 α , Y 1 α ∉ { ∅ } . Also, Y 3 α , Y 2 α , Y 1 α or Y 3 α , Y 2 α , Y 1 α , Y 0 α are partitioning of X for | X | ≥ 4 . By using Equations (2.3) and (2.4) we obtain

| B 0 | = 4 n − 3 n + 1 + 3 ⋅ 2 n − 1

Let α ∈ B ˜ 32 . Quasinormal represantation of α has form α = ( Y 3 α × Z 3 ) ∪ ( Y 2 α × Z 2 ) where Y 3 α , Y 2 α ∉ { ∅ } . Also, Y 3 α , Y 2 α are partitioning of X. By using (2.2) we obtain

| B ˜ 32 | = 2 n − 2

So, we have

| B 0 ∪ B ˜ 32 ∪ { σ 1 } | = 4 n − 3 n + 1 + 2 n + 2 − 2 | B 0 ∪ B ˜ 32 ∪ { σ 0 , σ 1 } | = 4 n − 3 n + 1 + 2 n + 2 − 1

since B 0 ∩ B ˜ 32 = B 0 ∩ { σ 0 , σ 1 } = B ˜ 32 ∩ { σ 0 , σ 1 } = ∅ . □

Acknowledgements

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

Cite this paper

Albayrak, B., Givradze, O. and Partenadze, G. (2018) Generating Sets of the Complete Semigroups of Binary Relations Defined by Semilattices of the Class . Applied Mathematics, 9, 17-27. https://doi.org/10.4236/am.2018.91002

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