ON THE MAXIMAL SUBSEMIGROUPS OF THE SEMIGROUP OF ALL MONOTONE TRANSFORMATIONS

ON THE MAXIMAL SUBSEMIGROUPS OF THE SEMIGROUP OF ALL MONOTONE TRANSFORMATIONS

Iliya Gyudzhenov and Ilinka Dimitrova

South-West University ”Neofit Rilski”

Faculty of Mathematics and Natural Science 2700 Blagoevgrad, Bulgaria

e-mail: iliadg@aix.swu.bg

e-mail: ilinka dimitrova@yahoo.com

Abstract

In this paper we consider the semigroup M

of all monotone trans- formations on the chain X

under the operation of composition of transformations. First we give a presentation of the semigroup M

and some propositions connected with its structure. Also, we give a description and some properties of the class ˜ J

of all monotone transformations with rank n − 1. After that we characterize the maxi- mal subsemigroups of the semigroup M

and the subsemigroups of M

which are maximal in ˜ J

.

Keywords: transformation semigroup, maximal subsemigroups, idempotent, isotone, antitone and monotone transformations, Green’s equivalences.

2000 Mathematics Subject Classification: 20M20.

∗

The author acknowledges support from the ”Center of Excellence of Pure and Applied

Mathematics” to DAAD.

1. Introduction

The maximal subsemigroups of some classes of semigroups have been the object of study by many semigroups theoreticians. In particular, Todorov and Kra˘colova [8] have determined the maximal subsemigroups of transfor- mations of fixed rank of finite chains. Nichols [5] and Reilly [6] considered the maximal inverse subsemigroups of T

. Yang Xiuliang [9] obtained a com- plete classification of the maximal inverse subsemigroups of finite symmetric inverse semigroups, Taijie You and Yang Xiuliang [7] have determined the maximal idempotent-generated subsemigroups of finite singular semigroups.

The semigroup O

of all isotone transformations on the finite chain X

is also of great interest. Yang Xiuliang [10] obtained a complete classification of the maximal subsemigroups of the semigroup O

. Yang Xiuliang and Lu Chunghan [11] received the maximal idempotent-generated and regular subsemigroups of the semigroup O

and etc. Recently Fernandes, Gomes and Jesus [2] gave a presentations for the monoid of all order-preserving and order-reversing transformations on a chain with n elements.

In this paper, we consider the semigroup M

of all monotone trans- formations on X

. This semigroup consists of all isotone transformations together with all antitone transformations. We present two subsets of the class ˜ J

of all monotone transformations with defect 1, which generate the semigroup M

. Here, we characterize all maximal subsemigroups of M

. We also determine all subsemigroups of M

which are maximal in the class ˜ J

. Here, we show that the maximal subsemigroups of the semigroup M

are closely connected with the maximal subsemigroups of the semigroup O

.

2. Preliminary results

For convenience, the following well-known definition and propositions will be used throughout the paper.

Definition 1 . Let A ⊆ X

and let π be an equivalence relation on X

. If

| ¯ x ∩ A | = 1 for all ¯ x ∈ X

/π, then A is called a cross-section of π, denoted by A # π.

Proposition 1 ([8]) . Let α, β ∈ J

. Then there hold 1) αβ ∈ J

⇐⇒ X

α # π

.

2) If X

α # π

and X

β # π

then X

αβ # π

.

Notation 1 . Let U be a subset of T

. Then the set of all idempotents of the semigroup T

, belonging to the set U , is denoted by E(U ).

Proposition 2 ([3]) . Let α and β be two elements of the class J

⊆ T

(1 ≤ k ≤ n − 1). Then the following are equivalent:

1) αβ ∈ J

. 2) αβ ∈ R

∩ L

. 3) L

∩ R

⊆ E(J

).

4) L

R

= J

.

From Proposition 1 and Proposition 2, we have the following two corollaries.

Corollary 1 . If α and β are two transformations of the class J

, then X

α # π

⇐⇒ L

∩ R

⊆ E

.

Corollary 2 . Let α, β

, . . . , β

(2 ≤ p ∈ IN) be transformations of the class J

(1 ≤ k ≤ n − 1). Then α = β

β

· · · β

iff R

= R

, L

= L

and X

β

# π

for all i = 1, . . . , p − 1.

The next proposition determines the number of the idempotents of R

(respectively of L

) for any α ∈ T

.

Proposition 3 ([1]). Let 1 ≤ k ≤ n − 1. Then for all α ∈ J

, there hold 1) |E(R

)| = |¯ x

||¯ x

| · · · |¯ x

| for X

/π

= {¯ x

, ¯ x

, . . . , ¯ x

}.

2) |E(L

)| = (a

−a

)(a

−a

) · · · (a

−a

) for X

α = {a

, a

, . . . , a

}.

Here, we consider the elements of the set X

α as cardinals.

3. Structure of the semigroup M

We call a transformation α on X

isotone if x ≤ y =⇒ xα ≤ yα and

antitone if x ≤ y =⇒ xα ≥ yα. A transformation α is called monotone if

it is isotone or antitone. The product of two isotone transformations or two

antitone transformations is an isotone transformation and the product of an

isotone transformation with an antitone transformation, or vice-versa, is an

antitone transformation.

We denote by O

the semigroup of all isotone transformations with defect ≥ 1 under the operation of composition of transformations and by Q

the set of all antitone transformations on X

with defect ≥ 1. The set M

:= O

∪ Q

forms a semigroup under the operation of composition of transformations, which is called the semigroup of all monotone transformations.

Since Q

is a subset of the semigroup T

, each L - class L

of Q

is uniquely determined by the image X

α and each R - class R

of Q

is uniquely determined by the kernel π

. From the definition of the antitone transformation, it follows that each H - class H

of Q

consists of exactly one transformation, namely α (i.e. the H - classes of Q

are trivial). The H - classes of the semigroup O

are also trivial and since M

= O

∪ Q

, it follows that each H - class of M

consists of exactly two transformations with the same image and the same kernel - one isotone and one antitone

Definition 2 . Let S be a subsemigroup of the semigroup M

. We call an H - class H of the semigroup S full if |S ∩ H| = 2.

The semigroup S is called H - full if all H - classes of S are full.

Definition 3 . For 1 ≤ k ≤ n − 1 we put 1) J ˆ

:= {α ∈ O

: |X

α| = k}.

2) ˆ I

:= ∪{ ˆ J

: 1 ≤ i ≤ k}.

3) J ˇ

:= {α ∈ Q

: |X

α| = k}.

4) ˇ I

:= ∪{ ˇ J

: 1 ≤ i ≤ k}.

5) J ˜

:= {α ∈ M

: |X

α| = k}.

6) ˜ I

:= ∪{ ˜ J

: 1 ≤ i ≤ k}.

Clearly, ˜ I

forms an ideal of M

. It is easy to verify that ˜ J

= ˆ J

∪ ˇ J

and I ˜

= ˆ I

∪ ˇ I

.

In this paper, we pay particular attention to the J - class ˜ J

at the

top of the semigroup M

.

Remark 1 . The R -, L - and H - classes in the class ˜ J

have the following form:

R

:= {α ∈ ˜ J

⊆ M

: iα = (i + 1)α} for 1 ≤ i ≤ n − 1, L

:= {α ∈ ˜ J

⊆ M

: X

α = X

\ {k}} for 1 ≤ k ≤ n,

H

:= R

∩ L

.

Each H - class H

of the semigroup M

contains exactly two transforma- tions, one isotone and one antitone, which are denoted by α

and γ

, respec- tively. We write ε

for the isotone transformation, if it is an idempotent.

Only the classes H

and H

(1 ≤ i ≤ n − 1) each contains an idempotent.

Each R - class R

for some 1 ≤ i ≤ n − 1, and each L - class L

for some 2 ≤ k ≤ n − 1 contains only two idempotents. The L - classes L

and L

each contains only one idempotent.

Notation 2 . We denote by E

:= E(M

) ∩ ˜ J

the set of all idempotents of the class ˜ J

for all 1 ≤ k ≤ n − 1.

Corollary 3 . For all α, β ∈ ˜ J

there hold:

1) Let α ∈ L

for some 1 ≤ k ≤ n. Then X

α # π

iff β ∈ {R

, R

}.

2) Let α ∈ R

for some 1 ≤ i ≤ n−1. Then X

β # π

iff β ∈ {L

, L

}.

Remark 2 . Let α, β ∈ M

and X

α # π

. Then X

γ # π

for all γ ∈ H

and all δ ∈ H

, since the transformations of one H - class have the same image and the same kernel.

In the next two lemmas, we will show the connection between the class of all monotone transformations ˜ J

and its subclasses ˆ J

and ˇ J

. Lemma 1 . Let 1 ≤ k ≤ n − 1 and let γ be an antitone transformation with rank k. Then the set ˆ J

∪ {γ} generates ˜ J

.

P roof. The class R

certainly contains at least one idempotent ε (see

Proposition 3). Hence for each transformation β ∈ L

we have R

∩ L

=

ε ⊆ E

. By Proposition 2, we have that the product βγ belongs to ˜ J

and

βγ ∈ R

∩ L

. Since β is an isotone transformation and γ is an antitone

transformation, it follows that βγ is antitone. Therefore, L

γ = L

⊆ ˜ J

.

Moreover, we certainly have that the class L

contains at least one idempo- tent δ (see Proposition 3). Thus from Proposition 2, we obtain L

R

= ˜ J

, and so h ˆ J

, γi ⊇ ˜ J

.

Lemma 2 . Let 1 ≤ k ≤ n − 1. Then the class ˇ J

generates the class ˜ J

, i.e. ˜ J

⊆ h ˇ J

i.

P roof. Let α ∈ ˆ J

be an isotone transformation with rank k. For each transformation α of this type we can find antitone transformations β

, β

∈ J ˇ

, such that π

= π

and X

β

= X

α. We choose the image of the trans- formation β

and the kernel of β

, such that the condition X

β

# π

will hold. Then by Corollary 2 we have α = β

β

, i.e. each isotone transforma- tion α ∈ ˆ J

can be represented as a product of two antitone transformations β

, β

∈ ˇ J

. Hence h ˇ J

i ⊇ ˆ J

∪ ˇ J

= ˜ J

.

Theorem 1 . M

= hE

, γi = h ˇ J

i for any γ ∈ ˇ J

.

P roof. The set of all idempotents of the class ˆ J

generates the semigroup O

, i.e. hE

i = O

(see [4]). By Lemma 1 we have that ˆ J

∪ {γ}

generates ˜ J

, for any γ ∈ ˇ J

. From Corollary 1, we have that X

α # π

iff L

∩ R

⊆ E

for all α, β ∈ ˜ J

. We also know that only the classes H

and H

(1 ≤ i ≤ n − 1) of the class ˜ J

each contains an idempotent.

Therefore, there are an isotone transformation α ∈ ˜ J

and an antitone transformation β ∈ ˜ J

such that X

α is not a cross-section of π

, and so the product αβ = γ

belongs to the class ˇ J

(see Proposition 1). Hence from Lemma 1, we have that h ˆ J

, γ

i = ˜ J

. Continuing in this way, we find h ˆ J

, γ

i = ˜ J

for all 2 ≤ k ≤ n − 1 and γ

∈ ˇ J

, i.e.

we obtain the ideal ˜ I

. Since ˜ I

∪ ˜ J

= M

, we deduce that the set E

∪ {γ} generates the semigroup M

.

From Lemma 2 we have that the set ˇ J

of all antitone transformations with defect 1 generates the class ˜ J

. Since the set E

∪ {γ}, where γ ∈ ˇ J

, is a subset of the class ˜ J

we obtain that ˇ J

generates the semigroup M

.

4. The maximal subsemigroups of the semigroup M

In this section, we give a complete classification of the maximal subsemi-

groups of the semigroup M

of all monotone transformations. Here, we can

use the classification of the maximal subsemigroups of O

given by Yang

Xiuliang ([10]).

Any maximal subsemigroup S of O

has the form S = ˆ I

∪ U for some U ⊆ ˆ J

. Now, we will study the elements of S \ ˜ I

for any maximal subsemigroup S of M

.

Lemma 3 . Let S be a maximal subsemigroup of M

. Then S ∩ E

6= ∅.

P roof. Let U := S \ ˜ I

and let us assume that U ∩ E

= ∅, i.e. no one of the idempotents

ε ∈ 

H

∪ H



∩ O

for i = 1, . . . , n − 1

belongs to the semigroup S. On the other hand, the product of any two antitone transformations is an isotone transformation. Hence the semigroup S certainly contains isotone transformations. If α

∈ S ∩ H

is such a transformation, then by Corollary 2 and Corollary 3 we get the following equations:

(1) ε

= α

α

= α

α

, ε

= α

α

= α

α

.

Therefore, from the assumption that α

∈ U and ε

, ε

∈ U , it follows that /

(2) n

α

, α

, α

, α

o

⊆ ˆ J

\ U.

We have that S is a maximal subsemigroup of the semigroup M

, and so hS, αi = M

for each transformation α ∈ M

\ S. In particular, we have hS, ε

i = M

. The transformation α

∈ M

\ S is expressible as a product of transformations in the following way:

(3) α

= βε

γ ∈ ˆ J

for some β, γ ∈ S. From Corollary 2 we have (4) R

= R

, L

= L

, X

β # π

i

, X

ε

# π

.

From the last requirements and Corollary 3, it follows that

(5) β ∈ H

∪ H

⊆ ˜ J

and γ ∈ H

∪ H

⊆ ˜ J

.

Moreover, since α

and ε

are isotone transformations, it follows that the transformations β and γ are both isotone or both antitone transformations.

Let us now consider separately each of these two cases.

Let β and γ be isotone transformations. We know that each H - class of the class ˜ J

contains exactly one isotone and one antitone transformation.

In this case from (2) and (5) we have:

β ∈ H

∪ H

= n α

o

∪ n α

o

⊆ ˆ J

\ S

γ ∈ H

∪ H

= n α

o

∪ n α

o

⊆ ˆ J

\ S, i.e. β and γ do not belong to the semigroup S.

Now let β and γ be antitone transformations. Then from γ ∈ H

∪H

, it follows that γ

is an idempotent and from the assumption it does not belong to S. Therefore, the transformation γ does not belong to S.

Thus we deduce that the transformations β and γ do not belong to the semigroup S. This contradicts the assumption that β, γ ∈ S.

The argument above hold for any representation of the transformation α

, since the first and the last element in that product have to satisfy the conditions (3) and (4). Therefore, U ∩ E

6= ∅.

For the proof of the main result, which characterizes the maximal subsemigroups of the semigroup M

, we need the following lemmas.

Lemma 4 . If the maximal subsemigroup S of M

contains an H - class which is not full then S contains an idempotent ε such that H

is not full.

P roof. Let H

= {α

, γ

} be one H - class of the semigroup S which is not full and let α

∈ S and γ

∈ S. From Lemma 3, we have that the / semigroup S contains an idempotent ε. We will prove that H

⊆ S is not full. From Corollary 2 and Corollary 3 we obtain:

γ

= α

γ

= α

γ

= γ

α

= γ

α

∈ S. /

This shows that the antitone transformations γ

, γ

, γ

, γ

do not

belong to S, since α

∈ S and γ

∈ S. The respective H - classes of /

these transformations contain the idempotents ε

, ε

, ε

, ε

. If at least

one of these idempotents belongs to S then the semigroup S contains an

idempotent whose H - class is not full.

Assume that all these idempotents do not belong to S. Then from Corollary 2 and Corollary 3 we have:

ε

= α

α

= α

α

=⇒ α

, α

∈ ˆ J

\ S.

Since S is a maximal subsemigroup and ε

∈ S, it follows that hS, ε /

i = M

. On the other hand, again by Corollary 2 and Corollary 3 we have:

α

= ε

α

,

where α

, ε

and α

do not belong to S (as we mentioned above). This shows that the transformation α

∈ hS, ε /

i, i.e. hS, ε

i 6= M

. Hence the assumption that all idempotents ε

, ε

, ε

, ε

do not belong to S contradicts the condition that S is a maximal subsemigroup of M

.

Analogously, one can obtain the same result for the converse case, when the antitone transformation γ

of the class H

belongs to S and the isotone transformation α

does not belong to S.

Lemma 5 . Let S be a maximal subsemigroup of the semigroup M

. Then either S \ ˜ I

= ˆ J

or S is H - full.

P roof. Let U = S \ ˜ I

and let S be a maximal subsemigroup of the semigroup M

. Then from Lemma 3, it follows that S contains at least one idempotent. Let us assume that S is not an H - full semigroup, i.e. there is at least one element of U , whose H - class is not full. From Lemma 4, we have that U contains an idempotent whose H - class is not full.

Let the idempotent ε

∈ H

belong to S and let the antitone transforma- tion γ

not belong to S. We will show that all H - classes of the semigroup S, which contain idempotents, are not full.

From the condition γ

∈ S and the equations (see Corollary 2, / Corollary 3)

γ

= ε

γ

= γ

ε

,

it follows that the antitone transformations γ

, γ

∈ ˇ J

do not belong to S. For this two transformations, we can construct the following product:

(6) γ

= αγ

β,

and by Corollary 2, it follows that

R

= R

, L

= L

, X

α # π

i

, X

γ

# π

, i.e. α ∈ H

∪ H

and β ∈ H

∪ H

(see Corollary 3).

Obviously, the last conditions completely define the R - and L - classes of the transformations α and β. Moreover, they are isotone or antitone at the same time.

Since S is maximal in M

and γ

, γ

∈ S then hS, γ /

i = M

and thus in the equation (6) the transformations α and β must belong to S.

Now we will consider the cases, when they are isotone and when they are antitone transformations.

If α and β are antitone transformations then we have:

α = γ

∈ H

and β = γ

∈ H

or β = γ

∈ H

.

Since the transformation γ

does not belong to the semigroup S from the equation γ

= γ

ε

∈ S it follows that the transformation γ /

= α does not belong to S. This contradicts the condition α ∈ S. Therefore, the equation (6) if α and β are antitone transformations contradicts the maximality of S.

If α and β are isotone transformations then we have:

α = ε

∈ H

or α = α

∈ H

and

β = ε

∈ H

or β = α

∈ H

. From equation (6) we obtain:

γ

= ε

γ

ε

= ε

γ

α

= α

γ

ε

= α

γ

α

and since ε

∈ S, so in order for at least one of these equations to hold it is enough that at least one of the transformations either ε

or α

belongs to S. If α

∈ S, then ε

= α

ε

∈ S. Hence in both cases the idempotent ε

belong to S. Its H - class is not full in S, since the antitone transformation γ

does not belong to S.

Since the transformation γ

does not belong to the semigroup S, from

the equation γ

= γ

ε

∈ S, it follows that the antitone transforma- /

tion γ

does not belong to S.

Further, from the conditions γ

, γ

∈ S and hS, γ /

i = M

for the transformation γ

we have

γ

= αγ

β, and by Corollary 2 we have:

R

= R

, L

= L

, X

α # π

i−1

, X

γ

# π

, i.e. α ∈ H

∪ H

and β ∈ H

∪ H

(see Corollary 3).

Since S is maximal in M

, the transformations α and β belong to S.

They can not be antitone, since the antitone transformations γ

and γ

, of the respective H - classes, do not belong to S.

Therefore, α and β are isotone transformations and thus α = ε

∈ H

, α = ε

∈ H

and

β = ε

∈ H

, β = α

∈ H

.

Hence we obtain that the idempotent ε

belongs to S. Its H - class is not full, since the antitone transformation γ

does not belong to S.

Thus the assumption that the H - class of the idempotent ε

is not full, implies the same deduction for the H - class of the idempotent ε

. Continuing by induction on the indices k ≤ i of the idempotents ε

, we obtain that all idempotents

ε

, ε

, ε

, ε

, . . . , ε

, ε

, ε

belong to S and moreover, their H - classes are not full.

Analogously, starting from the idempotent ε

and the conditions hS, γ

i = M

, and γ

∈ S first we obtain that the idempotent ε /

belongs to S and its H - class is not full and then we obtain the same result for the idempotent ε

.

Continuing by induction on the indices k ≥ i of the idempotents ε

we deduce that all idempotents

ε

, ε

, ε

, . . . , ε

, ε

, ε

belong to S and their H - classes are not full.

Hence if S contains an H - class which is not full then all idempotents of the class ˜ J

belong to S and their H - classes are not full.

By Howie ([4]), hE

i = ˆ J

and thus ˆ J

⊆ S. Let us assume that S contains at least one antitone transformation γ ∈ ˇ J

. From Theorem 1 we have h ˆ J

, γi = M

, and so M

= S. Therefore, S ∩ ˇ J

= ∅, since S is a maximal subsemigroup of M

and S = ˜ I

∪ ˆ J

, i.e. U = ˆ J

.

We have shown that if S contains an H - class which is not full then S contains all idempotents of the class ˜ J

and thus the set U coincides with the class ˆ J

. If S contains an antitone transformation γ ∈ ˇ J

then S does not contain all idempotents of the class ˜ J

, since h ˆ J

, γi = M

. Thus from the argument above, it follows that S is an H - full semigroup.

Lemma 6 . Let S

be a subsemigroup of O

and γ ∈ ˇ J

with γ

∈ S ∩ E

. Then S = hS

, γi is an H - full subsemigroup of M

and S ∩ O

= S

.

P roof. Let U

= S

\ ˆ I

and let L

(U

) (respectively R

(U

)) be the set of all elements of the class L

(respectively R

), that belong to the semigroup S

, i.e. L

(U

) = L

∩ S

for all 1 ≤ k ≤ n and R

(U

) = R

∩ S

for all 1 ≤ i ≤ n − 1. From γ

= ε we have that H

= H

, i.e. L

= L

and R

= R

. Let γ

= ε ∈ L

(U

) ∩ R

(U

). Then for each α ∈ L

(U

) we have αγ = δ ∈ R

∩ L

⊆ ˇ J

, since ε ∈ L

∩ R

(see Proposition 2). The transformation δ is antitone, since α is isotone and γ is antitone. We also have that L

= L

= L

. This shows that H

= H

. Therefore, for each α ∈ L

(U

) we have αγ = δ ∈ H

∩ Q

.

If we denote by L

(U ) the L - class of all elements of the class L

(U

) together with the respective antitone transformations, then from Proposition 2 and L

(U ) ∩ R

(U

) = ε ∈ E(U

), we have that L

(U )R

(U

)

= U , where U is H - full and its isotone transformations are the same as those of the set U

.

In the set U there are an isotone transformation α and an antitone

transformation β such that X

α is not a cross-section of π

. Then the

product αβ = γ

belongs to the class ˇ J

(see Proposition 1) and from

Lemma 1 we have that h ˆ J

, γ

i = ˜ J

. Continuing in this way, we find

h ˆ J

, γ

i = ˜ J

for all 2 ≤ k ≤ n − 1, i.e. we obtain the ideal ˜ I

.

Hence the semigroup S = hS

, γi = ˜ I

∪ U is H - full.

We have shown that for each α ∈ S

the class H

⊆ S contains exactly two transformations - one isotone and one antitone. Now we will show that the product of any two of the given antitone transformations belongs to the semigroup S

, i.e. S ∩ O

= S

.

We have S

⊆ O

and S

⊆ S, since S = hS

, γi. This shows that S

⊆ S ∩ O

.

Now we will show that S ∩ O

⊆ S

. Let α ∈ S ∩ O

, i.e. α is an isotone transformation of the semigroup S = ˜ I

∪ U . If α ∈ ˜ I

∩ O

= ˆ I

then α ∈ S

, since S

= ˆ I

∪ U

.

We have that S = hS

, γi, thus S

γ, γS

⊆ S and S

γ, γS

⊆ Q

. Then for each isotone transformation α we have α ∈ S

or α = γ

γ

, where γ

, γ

∈ S ∩ Q

.

Let α ∈ U ∩ O

⊆ ˆ J

. Assume that there are antitone transforma- tions γ

, γ

∈ U such that α = γ

γ

. Then from Proposition 1 we have X

γ

# π

. The semigroup S is H - full, and so all H - classes contain one isotone and one antitone transformation. Therefore, there are isotone transformations α

, α

∈ U

such that H

= H

and H

= H

. This shows that

H

= H

=⇒







L

= L

=⇒ X

α

= X

γ

R

= R

=⇒ π

= π

, j = 1, 2.

From X

γ

# π

, it follows that X

α

# π

and α

α

= α ∈ U

⊆ S

. Consequently, we have S ∩ O

⊆ S

and thus S ∩ O

= S

.

By the definition of an H - full semigroup we get:

Corollary 4 . If S

= O

and γ ∈ ˇ J

with γ

∈ S ∩ E

then hO

, γi = M

.

Now, we will study the connection between the maximal subsemigroups of the semigroups O

and M

.

Lemma 7 . Let S

be a maximal subsemigroup of the semigroup O

and

let γ ∈ ˇ J

be an antitone transformation. Then hS

, γi is a maximal

subsemigroup of M

.

P roof. 1) Let S = hS

, γi and α ∈ M

\S. We will show that hS, αi = M

, i.e. for each transformation β ∈ M

\ S there exist transformations η and δ in S, such that β = ηαδ. Since we know the transformations α and β, we can completely define the transformations η and δ by the following conditions (see Corollary 2):

(7) R

= R

, L

= L

and X

α # π

, X

η # π

.

The transformation α belongs to the semigroup M

and hence it can be isotone or antitone. We will consider each of these cases.

a) Let α be an isotone transformation, i.e. α ∈ (M

\ S) ∩ O

. Since S

is a maximal subsemigroup of the semigroup O

we have hS

, αi = O

. Therefore, for each transformation β

∈ O

\ S

there exist transforma- tions η

, δ

∈ S

, such that β

= η

αδ

.

The transformations η

and δ

belong to the semigroup S

⊆ S, but S is an H - full semigroup (by Lemma 6) and thus the antitone transfor- mations η

and δ

of the H - classes of η

and δ

also belong to S.

Hence both transformations η

and δ

also satisfy the conditions in (7). Then we have

β

= η

αδ

= η

αδ

for each isotone transformation β

∈ M

\ S and

β

= η

αδ

= η

αδ

for each antitone transformation β

∈ M

\ S. Hence hS, αi = M

for each isotone transformation α ∈ (M

\ S) ∩ ˆ J

.

b) Let α be an antitone transformation, i.e. α ∈ (M

\ S) ∩ ˇ J

. Since

S

is a maximal subsemigroup of the semigroup O

by the results of

Yang Xiuliang ([10]) each L - and each R - class of S

contains an

idempotent. Therefore, either L

or R

contains an idempotent ε ∈ S.

Since S is an H - full semigroup, the antitone transformation η

of the H - class of ε also belongs to S. The transformation α does not belong to S and thus the isotone transformation α

∈ H

also does not belong to S.If L

= L

then from the conditions R

= R

and L

= L

= L

we have αη

= α

∈ hS, αi. If R

= R

then from the conditions R

= R

= R

and L

= L

we have η

α = α

∈ hS, αi.

From the case a), hS, α

i = M

and so hS, αi = M

for each antitone transformation α ∈ (M

\ S) ∩ ˇ J

.

Lemma 8 . Let S be a maximal subsemigroup of the semigroup M

and S

= S ∩ O

. Then either S

= O

or S

is maximal in O

.

P roof. Let S = ˜ I

∪ U for some U ⊆ ˜ J

. If S is not an H - full semigroup then S does not contain any antitone transformations with defect 1 and U = ˆ J

(see Lemma 5). Therefore, for the semigroup S

we have S

= ( ˆ I

∪ ˆ J

) ∩ O

= ˆ I

∪ ˆ J

= O

.

If S contains at least one antitone transformation, then U 6= ˆ J

and it is an H - full semigroup, also by Lemma 5.

Now we will show that in this case the semigroup S

= S ∩ O

is a max- imal subsemigroup of the semigroup O

. Assume that S

is not a maximal subsemigroup of O

. Then there is a semigroup T

such that S

⊂ T

⊂ O

. Since S is maximal in M

, it contains an idempotent ε ∈ S

, and so ε ∈ T

. The semigroup S is H - full and so it contains the antitone transformation γ ∈ H

. From Lemma 6, it follows that hT

, γi = T is an H - full semi- group for which S ⊂ T ⊂ M

. This contradicts the condition that S is a maximal subsemigroup of M

. Therefore, we obtain that S

is a maximal subsemigroup of O

.

Now we are able to give a description of all maximal subsemigroups of M

.

Theorem 2 . Let S be a subsemigroup of the semigroup M

. Then S is a maximal subsemigroup iff one of the following statements holds:

1) S := ˜ I

∪ ˆ J

.

2) S := hS

, γi for some maximal subsemigroup S

of O

and some

γ ∈ ˇ J

.

P roof. 1) Let γ be an antitone transformation which belongs to (M

\ S) ∩ ˇ J

. By Lemma 1, we have that ˆ J

∪ γ generates the class J ˜

, and ˜ I

∪ ˜ J

= M

. Therefore, hS, γi = M

for each γ ∈ ˇ J

, i.e.

S is a maximal subsemigroup of the semigroup M

.

2) The semigroup S

can be only one of the semigroups given in [10], i.e., S

= ˆ I

∪ U

for some U

⊆ ˆ J

. Hence S is maximal by Lemma 7.

Conversely, let S be a maximal subsemigroup of the semigroup M

. By Lemma 8 the semigroup S

= S ∩ O

coincides with the semigroup O

, i.e., S

= O

or S

is a maximal subsemigroup of O

.

Let S

= O

and let γ ∈ ˇ J

be an antitone transformation, then from Corollary 4, it follows that hO

, γi = M

. Hence S ⊆ O

∪ ˇ I

= I ˜

∪ ˆ J

, but since S is maximal, we have S = ˜ I

∪ ˆ J

. Therefore, the semigroup S is of type 1).

Let S

be a maximal subsemigroup of the semigroup O

. Then by Lemma 7 the semigroup hS

, γi, where γ ∈ ˇ J

, is maximal in M

. The semigroup S contains all elements of the semigroup S

and the respective antitone transformations. Therefore, S ⊆ hS

, γi, and so S = hS

, γi, since S is maximal. Consequently, we conclude that S is of type 2).

In the last part of this section we consider the class ˜ J

and subsemigroups of M

which are contained in ˜ J

. We are interested in subsemigroups S of M

with S ⊆ ˜ J

for which hS, γi is not a subset of the class ˜ J

for any γ ∈ ˜ J

\ S. Such semigroups are called maximal in ˜ J

.

Let us put:

1) S

(i) := H

∪ H

for 1 ≤ i ≤ n − 1.

2) S

(i) := H

∪ H

for 2 ≤ i ≤ n − 1.

3) S

(1) := H

. 4) S

(n) := H

.

Theorem 3 . A subsemigroup S of M

is maximal in ˜ J

iff either S =

S

(i) for some 1 ≤ i ≤ n − 1 or S = S

(i) for some 1 ≤ i ≤ n.

P roof. Let S = S

(i) for some 1 ≤ i ≤ n − 1. Then we have S = {α

, γ

, α

, γ

} contains two isotone transformations and two antitone transformations. Moreover, the isotone transformations α

and α

are idempotents. It is easy to verify that the product of any two elements α, β of the set S also belongs to S. Therefore, S is a semigroup and since the classes H

and H

belong to ˜ J

, we have that S is contained in the class J ˜

.

Now we will show that S is maximal in ˜ J

.

Let us assume that S is not maximal, i.e. there is a transformation β ∈ J ˜

\ S such that hS, βi ⊆ ˜ J

. Then for each transformation α ∈ S we have βα, αβ ∈ ˜ J

, i.e. X

β # π

and X

α # π

(see Proposition 1).

From Corollary 3 we have that L

= L

or L

= L

, since R

= R

. If α ∈ H

then L

= L

and from Corollary 3, it follows that R

= R

or R

= R

. If α ∈ H

then L

= L

and so R

= R

or R

= R

. Since X

α # π

has to be satisfied for all α ∈ S, it follows that R

= R

.

Finally, we obtain that β ∈ L

∩ R

= H

or β ∈ L

∩ R

= H

, i.e. β ∈ S. This contradicts the condition β ∈ ˜ J

\ S and so S is maximal in ˜ J

.

The proof of the case S = S

(i) is similar.

Conversely, let S be maximal in ˜ J

. Then βα, αβ ∈ ˜ J

for all α, β ∈ S. This implies X

α # π

and X

β # π

(see Proposition 1).

Obviously, αα ∈ S ⊆ ˜ J

and thus αα ∈ R

∩ L

for all α ∈ S. This shows that αα = α ∈ E

or αα = α ∈ Q

and H

contains an idempotent.

Therefore, we have α ∈ O

∩ (H

∪ H

) or α ∈ Q

∩ (H

∪ H

) for some 1 ≤ i ≤ n − 1.

Let α ∈ H

. Then from the conditions X

α # π

and X

β # π

, it follows β ∈ {R

, R

} and β ∈ {L

, L

} for all β ∈ S (see Corollary 3).

Hence we get

β ∈ n

H

, H

, H

o .

If L

= L

and R

= R

for all β, γ ∈ S then X

β is not a cross-section

of π

. Therefore, we obtain S = H

∪H

= S

(i) or S = H

∪H

= S

(i)

for all 1 ≤ i ≤ n−1. In the second case, if i = 1 then the H - class H

= H

does not exist and S = H

= S

(1).

Now let α ∈ H

. Then in the same way we obtain that S = H

∪ H

= S

(i) or S = H

∪ H

= S

(i + 1) for all 1 ≤ i ≤ n − 1. In the second case, if i = n − 1 then the H - class H

= H

does not exist and S = H

= S

(n).

Acknowledgement

The authors would like to thank Prof. Kalcho Todorov for his helpful suggestions and encouragement.

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Received 19 January 2006

Revised 25 March 2006