# Regularity and Green's Relations on a Semigroup of

Authors: Jintana Sanwong; Worachead Sommanee
Journal: International Journal of Mathematics and Mathematical Sciences
 (2008)

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## Abstract

LetT(X) be the full transformation semigroup on the set X and let T(X,Y)={α∈T(X):Xα⊆Y}.
Then T(X,Y) is a sub-semigroup of T(X) determined by a nonempty subset Y of X. In this paper, we give a necessary and sufficient condition for T(X,Y) to be regular. In the case that T(X,Y) is not regular, the largest regular sub-semigroup is obtained and this sub-semigroup
is shown to determine the Green's relations on T(X,Y). Also, a class
of maximal inverse sub-semigroups of T(X,Y) is obtained.


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## Body

## 1. Introduction

LetY be  a nonempty subset of X and let T(X) denote the semigroup of transformations from X into itself. We consider the sub-semigroup of T(X) defined by (1.1)T(X,Y)={α∈T(X):Xα⊆Y}, when Xα denotes the range of α.
In fact, if |Y|=1,
then T(X,Y) contains exactly one element (namely, the
constant map with range Y) and if Y=X, then T(X,Y)=T(X).In 1975, Symons [1] described all the automorphisms of T(X,Y) and found that the most difficult case occurs
when |Y|=2.
He also determined when T(X1,Y1) is isomorphic to T(X2,Y2) and, surprisingly, the answer depends on the
cardinals |Xi| and |Xi∖Yi|,
not on |Yi| for i=1,2.
Here, we study other algebraic properties of this semigroup. Recall that an
element a of a semigroup S is called regular if a=axa for some x in S.
A semigroup S is regular if every element of S is regular. It is already known that T(X) is a regular semigroup (see [[2], page 33]).
But T(X,Y) is not regular in general. So, in Section 2,
we prove that T(X,Y) is regular if and only if |Y|=1 or Y=X.
We also prove that if T(X,Y) is not regular, the set F={α∈T(X,Y):Xα⊆Yα} is the largest regular sub-semigroup of T(X,Y).
In Section 3, we characterize the Green's relations on T(X,Y) and find that its 𝒟 and 𝒥 relations are surprising, but they reduce to
those on T(X) when Y=X.
And in Section 4, we give a class of maximal inverse sub-semigroups of T(X,Y),
of the form Fa={α∈F:aα=a and α is injective on X∖aα−1}.
When Y=X,
the set Fa={α∈T(X):aα=a and α is injective on X∖aα−1} is a class of maximal inverse sub-semigroups
of T(X) given in [4].Note that
throughout the paper, we write functions on the right; in particular, this
means that for a compositionαβ, α is applied first.


## 2. Regularity ofT(X,Y)

To give a
necessary and sufficient condition for the semigroupT(X,Y) to be regular, we first note the
following.(1)
If|Y|=1,
say Y={a}, then T(X,Y) contains exactly one element (namely, the
constant map Xa with range {a}), so T(X,Y) is regular.(2)
IfY=X, then T(X,Y)=T(X) which is a regular semigroup.(3)
If|X|≤2,
then |Y|=1 or Y=X,
and T(X,Y) is regular by (1) and (2). Now, we need
some notation. We adopt the convention introduced in ([[3], page 241]),
namely, if α∈T(X,Y),
then we write (2.1)α=(Xiai), and take as understood
that the subscript i belongs to some (unmentioned) index set I,
the abbreviation {ai} denotes {ai:i∈I},
and that Xα = {ai} and aiα−1=Xi.Theorem 2.1.
T(X,Y) is a regular semigroup if and only if |Y|=1 or Y=X.Proof.
Assume that|Y|≠1 and Y≠X.
Let a,b∈Y be such that a≠b and choose c∈X∖Y.
Let (2.2)α=(Xiyi) be any element
in T(X,Y), where Xα={yi}⊆Y and Xi=yiα−1.
We defineβ=(cX∖{c}ab),
and it is clear that β∈T(X,Y).
Sincec∉Y,
so yi≠c for all i, and (2.3)αβ=(Xiyi)(cX∖{c}ab)=(Xb)≠β. So, we conclude
that αβ≠β for all α∈T(X,Y),
this implies that β is a nonregular element in T(X,Y).
Therefore, T(X,Y) is not a regular semigroup. The converse is
clear by the previous note.Now, we
consider the set(2.4)F={α∈T(X,Y):Xα⊆Yα}. It is easy to
see that F={α∈T(X,Y):(X∖Y)α⊆Yα}={α∈T(X,Y):Xα=Yα}.
Since Y≠∅,
there exists a∈Y and we see that the constant map Xa with range {a} satisfies the condition in F,
therefore, Xa∈F and so F≠∅.
And for each α∈F and β∈T(X,Y),
we have Xαβ=(Xα)β⊆(Yα)β=Yαβ,
and thus αβ∈F.
This proves the following.Lemma 2.2.
F is a right ideal of T(X,Y).In general,F is not a left ideal of T(X,Y) as shown in the following example.Example 2.3.
LetX=ℕ denote the set of positive integers, let Y denote the set of all positive even integers,
and define (2.5)α=(n2n),β=(2nX∖Y2n2). Then α∈T(X,Y)∖F and β∈F,
but αβ=α∉F. Thus F is not a left ideal of T(X,Y).Theorem 2.4.
F is the largest regular sub-semigroup of T(X,Y).Proof.
From Lemma2.2, we see that F is a sub-semigroup of T(X,Y).
Let α∈F and write (2.6)α=(xiα−1xi), where ⋃i∈Ixiα−1=X and Xα={xi:i∈I}=Yα.
For each x∈Yα,
choose dx∈xα−1⋂Y,
so dxα=x, and dy≠dz,
for all y,z∈Yα such that y≠z.
Choose k∈I and let J=I∖{k}.
Define (2.7)β=(xjX∖{xj}dxjdxk), where {xj:j∈J}=Yα∖{xk}.
Then β∈T(X,Y) and αβα=α.
Since Xβ={dxi:i∈I}⊆(Yα)β⊆Yβ,
we have β∈F.
Hence F is a regular sub-semigroup of T(X,Y).
Now, let α be any regular element in T(X,Y).
Then αβα=α, for some β∈T(X,Y),
so Xα=Xαβα=(Xαβ)α⊆Yα,
and thus α∈F.
Therefore, F is the largest regular sub-semigroup of T(X,Y) as required.Note that ifY={a}, then for each α∈T(X,Y), we have Xα={a}=Yα which  implies
that   F=T(X,Y) consists of only one element and so is a
regular semigroup, and if X=Y, then F={α∈T(X,Y):Xα⊆Yα}={α∈T(X):Xα⊆Xα}=T(X) which is also a regular semigroup.


## 3. Green's Relations onT(X,Y)

LetS be a semigroup. Then we define S1 to be a semigroup of adding an identity to S if S does not already have an identity element in
it and S1=S if S contains an identity. The following
definitions are due to J. A. Green. For any a,b∈S,
we define (3.1)aℒbiffS1a=S1b, or
equivalently; aℒb if and only if a=xb,b=ya for some x,y∈S1.Dually, we
define(3.2)aℛbtomeansaS1=bS1, or equivalently; aℛb if and only if a=bx,b=ay for some x,y∈S1.And we define(3.3)a𝒥btomeansS1aS1=S1bS1, or
equivalently; a𝒥b if and only if a=xby,b=uav for some x,y,u,v∈S1.Finally, we
defineℋ=ℒ∩ℛ and 𝒟=ℒ∘ℛ.In [2, 3],
Clifford and Preston characterized Green's relations on the full transformation
semigroup T(X), where X is an arbitrary set. They proved that (3.4)αℒβiffXα=Xβ,αℛβiffπα=πβ. Here, we do the
same for the semigroup T(X,Y) and we obtain results which generalize the
same results on T(X).Lemma 3.1.
Letα,β∈T(X,Y).
If β∈F,
then Xα⊆Xβ if and only if α=γβ for some γ∈T(X,Y).Proof.
Letβ be an element of F.
It is clear that if α=γβ for some γ∈T(X,Y),
then Xα⊆Xβ.
Now, we assume that Xα⊆Xβ and write (3.5)α=(aiα−1ai),β=(bjβ−1bj), where {ai}⊆{bj}.
For each a∈Xα⊆Xβ⊆Yβ (since β∈F), we get a=yβ for some y∈Y which implies y∈aβ−1 and thus y∈Y∩aβ−1≠∅.
Choose da∈Y∩aβ−1,
so da∈Y and daβ=a.
Since X=⋃ai∈Xαaiα−1 is the disjoint union of the aiα−1,
we can define (3.6)γ=(aiα−1dai). Then γ∈T(X,Y) and γβ=α.From now on,
the notationsLα(Rα,Hα,Dα) denote the set of all elements of T(X,Y) which are ℒ-related (ℛ-related, ℋ-related, 𝒟-related) to α, where α∈T(X,Y).Theorem 3.2.
Forα∈T(X,Y),
the following statements hold.
(1)
Ifα∈F,
then Lα={β∈F:Xα=Xβ}.(2)
Ifα∈T(X,Y)∖F,
then Lα={α}.Proof.
Letα be any element in T(X,Y) and let β∈Lα.
Then αℒβ which implies that α=α′β and β=β′α, for some α′,β′∈T(X,Y)1.
(1) Assume
thatα∈F.
If β=α,
then β∈F and Xα=Xβ.
If β≠α,
then α′ and β′ both belong to T(X,Y).
Thus Xβ=(Xβ′α′)β⊆Yβ,
and hence β∈F.
From α∈F and β=β′α,
we get Xβ⊆Xα by Lemma 3.1. Similarly, from β∈F and α=α′β, we get Xα⊆Xβ.
Therefore, Xα=Xβ.
Now, if γ∈F and Xα=Xγ, then it is clear by Lemma 3.1 that γ∈Lα.
(2) Assume
thatα∈T(X,Y)∖F.
If α′,β′∈T(X,Y),
then Xα=X(α′β)=X(α′(β′α))=(Xα′β′)α⊆Yα.
Thus α∈F which is a contradiction, so α′=1 or β′=1 and β=α.We note that
for anyα∈T(X,Y),  πα={(a,b)∈X×X:aα=bα} is an equivalence on X and |X/πα|=|Xα|. The relation πα is usually called the kernel of α.Theorem 3.3.
Letα,β∈T(X,Y).
Then πβ⊆πα if and only if α=βγ for some γ∈T(X,Y).
Hence αℛβ if and only if πα=πβ.Proof.
It is clear that ifα=βγ for some γ∈T(X,Y),
then πβ⊆πα. Now, suppose that πβ⊆πα. If x∈Xβ,
then x=zβ for some z∈X,
so we define γ:X→X by (3.7)xγ={zα,ifx∈Xβ,xβ,ifx∈X∖Xβ. Then γ is well defined (since πβ⊆πα) and γ∈T(X,Y).
For each x∈X,
let y=xβ∈Xβ,
so xβγ=(xβ)γ=yγ=xα by the definition of γ.
Thus α=βγ as required, and the remaining assertion is
clear.Lemma 3.4.
Letα,β∈T(X,Y).
If πα=πβ then either both α and β are in F,
or neither is in F.Proof.
Assume thatπα=πβ and suppose that α,β∈F is false. So one of α or β is not in F,
we suppose that α∉F.
Thus (X∖Y)α⊈Yα,
so there is x0∈X∖Y such that x0α≠yα for all y∈Y.
Thus (x0,y)∉πα for all y∈Y.
If β∈F,
then Xβ=Yβ,
so (x0,y)∈πβ  for some y∈Y which contradicts πα=πβ. Therefore, β∉F.Using Theorem3.3 and Lemma 3.4, we have the following corollary.Corollary 3.5.
Forα∈T(X,Y),
the following statements hold.
(1)
Ifα∈F,
then Rα={β∈F:πα=πβ}.(2)
Ifα∈T(X,Y)∖F,
then Rα={β∈T(X,Y)∖F:πα=πβ}.As a direct consequence of Theorems3.2 and 3.3, we
have the following.Theorem 3.6.
Forα∈T(X,Y),
the following statements hold.
(1)
Ifα∈F,
then Hα={β∈F:Xα=Xβandπα=πβ}.(2)
Ifα∈T(X,Y)∖F,
then Hα={α}.In [2, 3], volume 1, Clifford and Preston proved that two
elements of T(X) are 𝒟-related if and only if they have the same
rank (i.e., the ranges of the two elements have the same cardinality). But
for T(X,Y), we have the
following.Theorem 3.7.
Forα∈T(X,Y),
the following statements hold.
(1)
Ifα∈F,
then Dα={β∈F:|Xα|=|Xβ|}.(2)
Ifα∈T(X,Y)∖F,
then Dα={β∈T(X,Y)∖F:πα=πβ}.Proof.
Letα be any element in T(X,Y) and let β∈Dα.
Then αℒγ and γℛβ for some γ∈T(X,Y).
(1) Ifα∈F,
then since αℒγ, we must have γ∈F and Xα=Xγ.
From γℛβ,
we get πγ=πβ and β=γλ for some λ∈T(X,Y)1.
Since F is a right ideal of T(X,Y),
so β∈F.
And |Xα|=|Xγ|=|X/πγ|=|X/πβ|=|Xβ|. Conversely, assume that λ∈F and |Xα|=|Xλ|.
Then there is a bijection θ:Xλ→Xα.
We let μ=λθ,
then μ∈T(X,Y) and Xμ=Xλθ=(Xλ)θ=Xα.
Since λ∈F implies Xλ⊆Yλ,
so Xμ=Xλθ⊆Yλθ=Yμ,
hence μ∈F.
Since α,μ∈F and Xα=Xμ,
so αℒμ by Theorem 3.2. Now, since μ=λθ and θ is injective on Xλ, we get πμ=πλ, so μℛλ.
Therefore, α and λ are 𝒟-related and λ∈Dα.
(2) Ifα∈T(X,Y)∖F,
then γ=α (since αℒγ) and thus αℛβ which implies that πα=πβ. So by Lemma 3.4 we must have β∈T(X,Y)∖F.
The other containment is clear since ℛ⊆𝒟.In order to characterize the𝒥-relation on T(X,Y),
the following lemma is needed.Lemma 3.8.
Letα,β∈T(X,Y).
If α=λβμ for some λ∈T(X,Y) and μ∈T(X,Y)1,
then |Xα|≤|Yβ|.Proof.
Ifα=λβμ for some λ∈T(X,Y) and μ∈T(X,Y)1.
Then Xλ⊆Y,
which implies that (Xλ)β⊆Yβ and so |Xλβ|≤|Yβ|.
If μ=1,
then α=λβ and so |Xα|=|Xλβ|≤|Yβ|.
If μ∈T(X,Y),
then |Xα|=|X(λβμ)|=|(Xλβ)μ|≤|Xλβ|≤|Yβ|.
Thus |Xα|≤|Yβ| as required.Theorem 3.9.
Letα,β∈T(X,Y).
Then (3.8)α𝒥βiffπα=πβor|Xα|=|Yα|=|Yβ|=|Xβ|.Proof.
First, assume
thatα𝒥β.
Then α=γβλ and β=γ′αλ′ for some γ,λ,γ′,λ′∈T(X,Y)1. If γ=1=γ′,
then α=βλ and β=αλ′ which imply αℛβ and thus πα=πβ. If γ∈T(X,Y) or γ′∈T(X,Y),
then we conclude that α=σβδ and β=σ′αδ′ for some σ,σ′∈T(X,Y) and δ,δ′∈T(X,Y)1.
For example, if γ=1 and γ′∈T(X,Y),
then α=βλ and β=γ′αλ′ imply α=βλ=(γ′αλ′)λ=γ′α(λ′λ)=γ′(βλ)λ′λ=γ′β(λλ′λ).
By using Lemma 3.8, we get that |Yβ|≥|Xα|≥|Yα|≥|Xβ|≥|Yβ|,
so it follows that |Xα|=|Yα|=|Yβ|=|Xβ|.
Conversely, ifπα=πβ, then αℛβ which implies that α𝒥β since ℛ⊆𝒥. If |Xα|=|Yα|=|Yβ|=|Xβ|,
then by applying Lemma 2.7 in [2, 3], to |X/πα|=|Xα|=|Yβ|  and |X/πβ|=|Xβ|=|Yα|, we get there are γ,λ∈T(X,Y) such that πγ=πα, Xγ=Yβ;
and πλ=πβ, Xλ=Yα.
From πγ=πα  and πλ=πβ, we get γℛα and λℛβ,
so α=γγ′ and β=λλ′ for some γ′,λ′∈T(X,Y)1.
And from Xγ=Yβ,
we write Yβ={yi:i∈I},
so (3.9)γ=(yiγ−1yi). For each i∈I,
choose ai∈yiβ−1∩Y and define β′:X→X by (3.10)β′=(yiγ−1ai). Then β′∈T(X,Y) and γ=β′β.
Similarly, from Xλ=Yα, we can prove that λ=α′α for some α′∈T(X,Y).
Therefore,α=γγ′=β′βγ′ and β=λλ′=α′αλ′ which implies that α𝒥β as required.Recall that𝒟⊆𝒥 on any semigroup and 𝒟=𝒥 on T(X);
but in T(X,Y), this is not always true, as shown in the
following example.Example 3.10.
LetX=ℕ denote the set of positive integers and
let Y denote the set of all positive even integers.
Then we define (3.11)α=(n2n),β=(2nX∖Y4n2). Hence α,β∈T(X,Y)∖F and |Xα|=|Yα|=ℵ0=|Yβ|=|Xβ|,
so α𝒥β.
Since πα≠πβ, we have α and β are not 𝒟-related on T(X,Y).
As a consequence of Theorems3.7 and 3.9, we see
that 𝒟=𝒥 on the sub-semigroup F of T(X,Y).Corollary 3.11.
Ifα,β∈F,
then α𝒥β on T(X,Y) if and only if α𝒟β on T(X,Y).Proof.
In general, we have𝒟⊆𝒥.
Let α,β∈F and α𝒥β on T(X,Y). Then πα=πβ or |Xα|=|Yα|=|Yβ|=|Xβ|. If πα=πβ, then |Xα|=|X/πα|=|X/πβ|=|Xβ|. Thus, both cases imply |Xα|=|Xβ| and α𝒟β on T(X,Y) by Theorem 3.7.If we replaceY with X in the above corollary, we then get 𝒟=𝒥 on T(X,Y)=T(X).
Next, we will consider the case when Y is a finite subset of X.Theorem 3.12.
IfY is a finite subset of X,
then 𝒟=𝒥 on T(X,Y).Proof.
LetY be a finite subset of X and let α,β∈T(X,Y) be such that α and β are 𝒥-related. Then πα=πβ or |Xα|=|Yα|=|Yβ|=|Xβ|. We note that if α∉F,
then |Xα|>|Yα|.
For if |Xα|=|Yα|, then Xα is a finite set (since Y is finite) which implies
that Xα=Yα and thus α∈F.
Now, if α∈F and β∉F, then |Xα|=|Yα| but |Yβ|<|Xβ|,
so πα=πβ which contradicts Lemma 3.4. Therefore, either both α and β are in F,
or neither is in F.
If α,β∈F,
then α𝒟β by Corollary 3.11. If α,β∉F,
then |Xα|>|Yα| which implies that πα=πβ and thus α𝒟β by Theorem 3.7. Therefore, 𝒥⊆𝒟 and the other containment is clear.


## 4. Maximal Inverse Sub-Semigroups onT(X,Y)

We first recall that a semigroup is said to be aninverse
semigroup if it is regular and any two idempotents commute. In this
section, we give one class of maximal inverse sub-semigroups on T(X,Y).
If |Y|=1,
then there is only one element in T(X,Y),
the constant map. Hence, in this case, there is no maximal inverse
sub-semigroup on T(X,Y).
Therefore, from now on, we assume that |Y|≥2.In 1976,
Nichols [4] gave a class of maximal inverse sub-semigroups of T(X).
Later in 1978, Reilly [5] generalized Nichols' result. Here, with some mild
modifications of the proof given in [4], we get one class of maximal inverse
sub-semigroups of T(X,Y) which generalizes Nichols' result.LetX be a set and Y a nonempty subset of X.
For each a∈Y,
define (4.1)Fa={α∈F:aα=aandαisinjectiveonX∖aα−1}. We see that Fa≠∅,
since the constant map Xa∈Fa.
To describe maximal inverse sub-semigroups on T(X,Y),
we first prove the following.Theorem 4.1.
Letα∈T(X,Y) and a∈Y.
Then α∈Fa if and only if {a}∪(X∖Y)⊆aα−1 and α is injective on Y∖aα−1.Proof.
IfY=X,
then X∖Y=∅ and F=T(X),
thus α∈Fa if and only if α∈T(X), {a}⊆aα−1 and α is injective on X∖aα−1.
Now, we prove for the caseY≠X.
Assume that α∈Fa.
So α∈F,aα=a and α is injective on X∖aα−1.
We show that (X∖Y)α={a}.
Let b∈(X∖Y)α,
so there exists x∈X∖Y such that xα=b.
Thus b∈(X∖Y)α⊆Yα since α∈F.
Hence b=yα for some y∈Y.
So, x,y∈bα−1 and x≠y.
By the definition of Fa,
we must have b=a.
Therefore, {a}∪(X∖Y)⊆aα−1 and α is injective on Y∖aα−1.
Conversely, assume that the conditions hold. Since a∈Y and {a}⊆aα−1,
we get a=aα∈Yα, and thus {a}⊆Yα.
So (X∖Y)α={a}⊆Yα and therefore α∈F.
Since α is injective on Y∖aα−1 and Y∖aα−1=X∖aα−1,
it follows that α is injective on X∖aα−1,
and so α∈Fa.Recall that for eachα∈T(X), α is an idempotent in T(X) if and only if xα=x for all x∈Xα.
Since T(X,Y) is a sub-semigroup of T(X),
we conclude that α is an idempotent in T(X,Y) if and only if xα=x for all x∈Xα.
And by Theorem 4.1, if α is an idempotent in Fa, then xα=x for all x∈Y∖aα−1.Lemma 4.2.
LetL be a regular sub-semigroup of T(X,Y) such that Fa⊊L and suppose that α∈L∖Fa.
Then
(i)
αℒβ on L for some idempotent β∈Fa.(ii)
Ifaα=a,
then α is not ℛ-related on L to any element in Fa.Proof.
(i) We write(4.2)α=(aα−1Aiaai), where ⋃Ai=X∖aα−1,
and define β∈T(X,Y) by (4.3)β=(aiX∖Baia), where {ai}=B.
Then β:X→Y,
is an identity map on B and X∖B=aβ−1.
So B=X∖aβ−1 and β is injective on X∖aβ−1=B.
Since a∈X∖B,
so a=aβ∈Yβ and {a}⊆Yβ.
From B={ai}⊆Y,
we get (X∖Y)β⊆(X∖B)β={a}⊆Yβ,
so β∈F.
Thus β∈Fa and it is also an idempotent. From the fact
that L⊆F is a regular sub-semigroup of T=T(X,Y),
it follows from Hall's theorem that ℒL=ℒT∩(L×L)⊆ℒT∩(F×F), where (a,b)∈ℒS means there exist s,t in S1 such that a=sb,b=ta.
Since Xα=Xβ,
we must have by Theorem 3.2 that αℒβ on L.
(ii) Assume thataα=a and suppose that αℛβ on L for some β∈Fa.
Thus πα=πβ by Theorem 3.3, and hence aα−1=aβ−1.
Now, let x1,x2∈X∖aα−1 be such that x1α=x2α.
Then (x1,x2)∈πα implies (x1,x2)∈πβ (since πα=πβ), so x1β=x2β and since β is injective on X∖aβ−1=X∖aα−1,
we get x1=x2 which implies that α is injective on X∖aα−1,
which contradicts α∉Fa. Therefore, α is not ℛ-related to any element in Fa.Theorem 4.3.
Fa is a maximal inverse sub-semigroup of T(X,Y).Proof.
First, we prove thatFa is a sub-semigroup of T(X,Y).
Letα,β be elements in Fa.
Then α,β∈F,aα=a=aβ,
and α,β are injective on X∖aα−1 and X∖aβ−1,
respectively. Since F is a right ideal of T(X,Y),
it follows that αβ∈F.
Clearly a(αβ)=a,
and αβ is injective on X∖a(αβ)−1.
Therefore, αβ∈Fa.
Next, we show thatFa is a regular sub-semigroup of T(X,Y).
For each α∈Fa, aα=a and |xα−1|=1 for all x∈Yα∖{a} (see Theorem 4.1). Let {xi}=Yα∖{a} and write xiα−1=yi for all i,
thus (4.4)α=(yiaα−1xia), where ⋃{yi}=Y∖aα−1, X=Y⋃aα−1 and define β∈T(X,Y) by (4.5)β=(xiAyia), where {xi}=Yα∖{a} and A=X∖{xi}.
Since a∈A,
we get a=aβ∈Yβ,
and so {a}⊆Yβ.
And for each x∈X∖Y, x≠xi for all i since xi∈Y. Thus (X∖Y)β={a}⊆Yβ,
and so β∈F.
Since aβ=a and β is injective on {xi}=X∖aβ−1,
it follows that β∈Fa.
And, it is clear that α=αβα.
Now, we prove that any two idempotents inFa commute, which is enough to show that Fa is an inverse semigroup. Assume that α,β are idempotents in Fa.
Then xα=x for all x∈X∖aα−1 and xβ=x for all x∈X∖aβ−1.
Let x be any element in X.
Case 1.
x∈X∖aα−1. Then xα=x.
So, if x∈X∖aβ−1,
we get xβ=x and x(αβ)=(xα)β=xβ=x=xα=(xβ)α=x(βα).
But if x∈aβ−1,
then xβ=a,
and x(αβ)=(xα)β=xβ=a=aα=(xβ)α=x(βα).
Thus in this case αβ=βα.
Case 2.
x∈aα−1. Then xα=a.
So by the same proof as given in Case 1, we get αβ=βα.
Therefore,Fa is an inverse sub-semigroup of T(X,Y).
To prove the maximality, we suppose thatFa is properly contained in an inverse
sub-semigroup L of T(X,Y), where L⊆F⊆T(X,Y) and let α∈L∖Fa.
Let β be the constant map with range {a},
so β is an idempotent in Fa,
and thus βα is an idempotent in L.
Since βαβ=β and every two idempotents in L commute, it follows that β=βαβ=ββα=βα and aα=(aβ)α=a(βα)=aβ=a.
Since L is regular, α=αα′α for some α′∈L and αℛαα′ on L such that αα′ is an idempotent in L.
Let γ=αα′,
then by Lemma 4.2 we must have γ∈L∖Fa and γℒσ for some idempotent σ∈Fa.
Since every idempotent e in a semigroup is a right identity for Le,
we have γ=γσ=σγ=σ which is a contradiction since γ∉Fa but σ∈Fa.
Therefore, L=Fa as required.As an application of Theorem4.3, we get the following
corollary which first appeared in [4].Corollary 4.4.
Fa={α∈T(X):aα=a and α is injective on X∖aα−1} is a maximal inverse sub-semigroup of T(X).Proof.
By takingY=X in Theorem 4.3, we get T(X,Y)=T(X)=F and Fa={α∈T(X):aα=a and α is injective on X∖aα−1} which is a maximal inverse sub-semigroup of T(X).Recall that the number of combinations ofn distinct things taken r at a time written (nr) is given
by (4.6)(nr)=n!(n−r)!r!. That is, (nr) is the number
of ways that r objects can be chosen from n distinct objects.In the next
result, we use the above information to find the number of elements inFa when Y is a finite subset of X.Theorem 4.5.
Suppose thatX is an arbitrary set and Y is a nonempty subset of X such that |Y|=n.
Then for each a∈Y, |Fa|=∑r=0n−1r!(n−1r)2.Proof.
Leta∈Y and α∈Fa.
Then by Theorem 4.1 we see that (4.7)α=((X∖Y)⋃Y1Y2aY3), where Y=Y1⋃˙Y2,a∈Y1,Y3⊆Y∖{a},
and |Y2|=|Y3|.
If Y2=∅,
then α can have only one form, the constant map Xa.
If Y2 has t elements, where 1≤t≤n−1,
then Y2 can have (n−1t) choices and for each choice of Y2,Y3 can have (n−1t) choices, thus there are (n−1t)2 ways to choose Y2 and Y3.
Since the restriction of α to Y2 is a permutation, for each choice of Y2 and Y3,
the map α has t! possible forms. Hence in this case α can have t!(n−1t)2 forms.
Therefore,|Fa|=1+∑r=1n−1r!(n−1r)2=∑r=0n−1r!(n−1r)2 as required.We observe that the number of elements inFa depends only on the elements of Y,
and when we replace Y with X in Theorem 4.5, we have the following corollary
which first appeared in [4].Corollary 4.6.
IfX is a finite set with |X|=n,
then the number of elements in Fa={α∈T(X):aα=a and α is injective on X∖aα−1} equals ∑r=0n−1r!(n−1r)2.



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*Source: 794013-2008-11-05.xml*