# On Maximal Subsemigroups of Partial Baer-Levi Semigroups

Authors: Boorapa Singha; Jintana Sanwong
Journal: International Journal of Mathematics and Mathematical Sciences
 (2011)

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## Abstract

Suppose thatX is an infinite set with |X|≥q≥ℵ0 and I(X) is the symmetric inverse semigroup defined on X. In 1984, Levi and Wood determined a class of maximal subsemigroups MA (using certain subsets A of X) of the Baer-Levi semigroup BL(q)={α∈I(X): dom α=X and |X∖Xα|=q}. Later, in 1995, Hotzel showed that there are many other classes of maximal subsemigroups of BL(q), but these are far more complicated to describe. It is known that BL(q) is a subsemigroup of the partial Baer-Levi semigroup PS(q)={α∈I(X):|X∖Xα|=q}. In this paper, we characterize all maximal subsemigroups of PS(q) when |X|>q, and we extend MA to obtain maximal subsemigroups of PS(q) when |X|=q.


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## Body

## 1. Introduction

Suppose thatX is a nonempty set, and let P(X) denote the semigroup (under composition) of all partial transformations of X (i.e., all mappings α:A→B, where A,B⊆X). For any α∈P(X), we let domα and ranα (or Xα) denote the domain and the range of α, respectively. We also write
(1.1)g(α)=|X∖dom⁡α|,d(α)=|X∖ranα|,r(α)=|ranα|,
and refer to these cardinals as the gap, the defect, and the rank of α, respectively. Let I(X) denote the symmetric inverse semigroup on X: that is, the set of all injective mappings in P(X). If |X|=p≥q≥ℵ0, we write (1.2)BL(q)={α∈I(X):g(α)=0,d(α)=q},PS(q)={α∈I(X):d(α)=q},
where BL(q) is the Baer-Levi semigroup of type (p,q) defined on X (see [1, 2]). It is wellknown that this semigroup is right simple, right cancellative, and idempotent-free. On the other hand, in [3] the authors showed that PS(q), the partial Baer-Levi semigroup on X, does not have these properties but it is right reductive in the sense that for every α,β∈PS(q), if αγ=βγ for all γ∈PS(q), then α=β. Also, they showed that PS(q) satisfies the dual property, that is, it is left reductive (see [1, 2]). The authors also characterized Green's relations and ideals of PS(q) and, in [3, Corollary  1], they proved that PS(q) contains an inverse subsemigroup: namely, the set R(q) defined by (1.3)R(q)={α∈PS(q):g(α)=q}.
This set consists, in fact, of all regular elements of PS(q), as shown in [3, Theorem  4]. Recently, in [4], the authors studied some properties of Mitsch's natural partial order defined on a semigroup (see [5, Theorem  3]) and some other partial orders defined on PS(q). In particular, they described compatibility and the existence of maximal and minimal elements.For any nonempty subsetA of X such that |X∖A|≥q, let (1.4)MA={α∈BL(q):A⊈Xαor(Aα⊆Aor|Xα∖A|<q)}.
In other words, given α∈BL(q), we have α∈MA if and only if Xα does not contain A, or Xα contains A and either Aα⊆A or |Xα∖A|<q. In [6], Levi and Wood showed that MA is a maximal subsemigroup of BL(q). Later, Hotzel [7] showed that there are many other maximal subsemigroups of BL(q).In this paper, we study maximal subsemigroups ofPS(q). In particular, in Section 3 we describe all maximal subsemigroups of PS(q) when p>q. We also determine some maximal subsemigroups of a subsemigroup Sr of PS(q) defined by (1.5)Sr={α∈PS(q):g(α)≤r},
where q≤r≤p. Moreover, we extend MA to determine maximal subsemigroups of PS(q). In Section 4, we determine some maximal subsemigroups of PS(q) when p=q.


## 2. Preliminaries

In this paper,Y=A∪̇B means Y is adisjoint union of sets A and B. As usual, ∅ denotes the empty (one-to-one) mapping which acts as a zero for P(X). For each nonempty A⊆X, we write idA for the identity transformation on A: these mappings constitute all the idempotents in I(X) and belong to PS(q) precisely when |X∖A|=q.We modify the convention introduced in [1, 2]: namely, if α∈I(X) is non-zero, then we write
(2.1)α=(aixi)
and take as understood that the subscript i belongs to some (unmentioned) index set I, that the abbreviation {xi} denotes {xi:i∈I}, and that Xα=ranα={xi},aiα=xi for each i and domα={ai}. To simplify notation, if A⊆X, we sometimes write Aα in place of (A∩domα)α.LetS be a semigroup and ∅≠A⊆S. Then 〈A〉 denotes the subsemigroup of S generated by A. Recall that a proper subsemigroup M of S is maximal in S if, whenever M⊆N⊊S and N is a subsemigroup of S, then M=N. Note that this is equivalent to each one of the following:(a)
〈M∪{a}〉=S for all a∈S∖M;(b)
for anya,b∈S∖M, a can be written as a finite product of elements of M∪{b} (note that a is not expressible as a product of elements of M since a∉M).Throughout this paper, we will use this fact to show the maximality of subsemigroups ofPS(q).


## 3. Maximal Subsemigroups ofPS(q) When p>q

The characterisation of maximal subsemigroups of a given semigroup is a natural topic to consider when studying its structure. Sometimes, it is difficult to describe all of them (see [6, 7], e.g.), but for a semigroup with some special properties, we can easily describe some of its maximal subsemigroups.Lemma 3.1.
LetS be a semigroup and suppose that S is a disjoint union of a subsemigroup T and an ideal I of S. Then, (a)
for any maximal subsemigroupM of T, M∪I is a maximal subsemigroup of S;(b)
for any maximal subsemigroupN of S such that T∖N≠∅ and T∩N≠∅, the set T∩N is a maximal subsemigroup of T.Proof.
To see that (a) holds, letM be a maximal subsemigroup of T. Since I is an ideal, we have M∪I is a subsemigroup of S. Clearly, M∪I⊊T∪I=S. If a∈S∖(M∪I), then a∈T∖M and thus T=〈M∪{a}〉⊆〈M∪I∪{a}〉. Since 〈M∪I∪{a}〉 contains I, we have S=T∪I=〈M∪I∪{a}〉 and so M∪I is maximal in S as required.
To prove (b), letN be a maximal subsemigroup of S, where T∖N≠∅ and T∩N≠∅, and let a∈T∖N. Since N is maximal in S, we have 〈N∪{a}〉=S. Thus, for each b∈T∖N, b=c1c2⋯cn for some natural n and some ci∈N∪{a} for all i=1,2,…,n. Since b∉N, we have ci=a for some i. Moreover, since b∉I, we have cj∈T∩N for all j≠i. It follows that T∖N⊆〈(T∩N)∪{a}〉, therefore
(3.1)T=(T∖N)∪(T∩N)⊆〈(T∩N)∪{a}〉,
that is, T=〈(T∩N)∪{a}〉 and thus T∩N is maximal in T.Letu be a cardinal number. The successor of u, denoted by u′, is defined as
(3.2)u′=min⁡{v:v>u}.
Note that u′ always exists since the cardinals are wellordered, and when u is finite we have u′=u+1.From [3, p 95], for ℵ0≤k≤p,
(3.3)Sk={α∈PS(q):g(α)≤k}
is a subsemigroup of PS(q). Also, when p>q, the proper ideals of PS(q) are precisely the sets:
(3.4)Ts={α∈PS(q):g(α)≥s},
where q<s≤p (see [3, Theorem  13]). Thus, for any q≤r<p, it is clear that (3.5)PS(q)=Sr∪̇Tr′,
that is, PS(q) can be written as a disjoint union of the semigroup Sr and the ideal Tr′. Hence, the next result follows directly from Lemma 3.1(a).Corollary 3.2.
Suppose thatp>r>q≥ℵ0. If M is a maximal subsemigroup of Sr, then M∪Tr′ is a maximal subsemigroup of PS(q).Lemma 3.3.
Letp>q≥ℵ0 and suppose that M is a maximal subsemigroup of PS(q). Then, (a)
Sr∩M≠∅ for all q≤r<p;(b)
if there existsα∉M with g(α)<p, then Sk∖M≠∅ for some q≤k<p.Proof.
To show that (a) holds, we first note thatSq is contained in Sr for all q≤r<p. If Sq∩M=∅, then M⊆Tq′⊊PS(q) and thus M=Tq′ by the maximality of M. But Tq′⊊Tq′∪BL(q)⊊PS(q) where Tq′∪BL(q) is a subsemigroup of PS(q) (since Tq′ is an ideal), so we get a contradiction. Therefore, ∅≠Sq∩M⊆Sr∩M for all q≤r<p.
To show that (b) holds, suppose there isα∉M with g(α)=k<p. If k<q, then α∈Sr∖M for all q≤r≤p. Otherwise, if q≤k, then α∈Sk∖M. Hence (b) holds.For what follows, for any cardinalr≤p, we let (3.6)Gr={α∈PS(q):g(α)=r}.ThenG0=BL(q) and Gq=R(q). Moreover, if p>q and r>q, then Gr=Sr∩Tr, and so Gr is a subsemigroup of Sr (since it is the intersection of two semigroups). Also, Gr is bisimple and idempotent-free, when p>q and r>q (see [3, Corollary  3]).From [3, Theorem  5], if p≥q, then Sq=α·R(q) for each α∈BL(q), and by [3, Theorem  6], Sq=BL(q)·μ·BL(q) for each μ∈R(q) when p≠q.This motivates the following result.Lemma 3.4.
Suppose thatp≥r>q≥ℵ0. Then Gr=BL(q)·α·BL(q) for each α∈Gr.Proof.
Letα∈Gr and β,γ∈BL(q). Since
(3.7)X∖dom⁡α=[Xβ∩(X∖dom⁡α)]∪[(X∖Xβ)∩(X∖dom⁡α)],
where g(α)=|X∖domα|=r>q and the second intersection on the right has cardinal at most q (since |X∖Xβ|=q), we have |Xβ∩(X∖domα)|=r. This means that
(3.8)r=|[Xβ∩(X∖dom⁡α)]β-1|=|(Xβ∖dom⁡α)β-1|=|dom⁡β∖dom⁡(βα)|=|X∖dom⁡(βα)|=g(βα).
Since domγ=X, we have dom(βαγ)=dom(βα), and so g(βαγ)=g(βα)=r. Hence βαγ∈Gr and therefore BL(q)·α·BL(q)⊆Gr.
For the converse, ifα,β∈Gr, then |X∖domα|=r=|X∖domβ|. Since p>q, every element in PS(q) has rank p, so we write
(3.9)α=(aixi),β=(biyi)where|I|=p.
Now write X∖{yi}=A∪̇B and X∖{ai}=C∪̇D where |A|=|B|=|C|=q and |D|=r (note that this is possible since d(β)=q≥ℵ0 and g(α)=r>q≥ℵ0). Define
(3.10)δ=(biX∖{bi}aiD),ϵ=(xiX∖{xi}yiA),
where δ|(X∖{bi}) and ϵ|(X∖{xi}) are bijections. Then δ,ϵ∈BL(q) and β=δαϵ, that is, Gr⊆BL(q)·α·BL(q) and equality follows.Now we can describe all maximal subsemigroups ofPS(q) when p>q.Theorem 3.5.
Suppose thatp>q≥ℵ0. Then M is a maximal subsemigroup of PS(q) if and only if M equals one of the following sets: (a)
PS(q)∖Gp={α∈PS(q):g(α)<p};(b)
N∪Tr′, where q≤r<p and N is a maximal subsemigroup of Sr.Proof.
Letα,β∈PS(q) be such that g(α)<p and g(β)<p. Clearly |Xα∖domβ|≤|X∖domβ|=g(β)<p. Then
(3.11)|dom⁡α∖dom⁡(αβ)|=|[Xα∖(Xα∩dom⁡β)]α-1|=|(Xα∖dom⁡β)α-1|=|Xα∖dom⁡β|<p.
Hence,
(3.12)|X∖dom⁡(αβ)|=|X∖dom⁡α|+|dom⁡α∖dom⁡(αβ)|<p,
and this shows that PS(q)∖Gp is a subsemigroup of PS(q). To show that PS(q)∖Gp is maximal in PS(q), we let α,β∈PS(q)∖(PS(q)∖Gp)=Gp. By Lemma 3.4, α=λβμ for some λ,μ∈BL(q)⊆PS(q)∖Gp. Thus, α can be written as a finite product of elements of (PS(q)∖Gp)∪{β}, and hence PS(q)∖Gp is maximal in PS(q). Also, if q≤r<p and N is a maximal subsemigroup of Sr, then N∪Tr′ is maximal in PS(q) by Corollary 3.2.
We now suppose thatM is a maximal subsemigroup of PS(q) such that M≠PS(q)∖Gp. Then there exists α∉M with g(α)<p. Thus, Lemma 3.3 implies that Sk∖M≠∅ and Sk∩M≠∅ for some k, where q≤k<p. Since PS(q)=Sk∪̇Tk′, Lemma 3.1(b) implies that Sk∩M is maximal in Sk. We also see that
(3.13)M=(Sk∩M)∪(Tk′∩M)⊆(Sk∩M)∪Tk′,
where (Sk∩M)∪Tk′ is maximal in PS(q) by Corollary 3.2. This means that M=(Sk∩M)∪Tk′ by the maximality of M.By the previous theorem, whenp>q, most of the maximal subsemigroups of PS(q) are induced by maximal subsemigroups of Sr where q≤r<p. Hence we now determine some maximal subsemigroups of Sr.As mentioned in Section1, for every nonempty subset A of X with |X∖A|≥q, MA is a maximal subsemigroup of BL(q). Here we extend the definition of MA and consider the set M¯A defined as (3.14)M¯A={α∈PS(q):A⊈Xαor(Aα⊆A⊆dom⁡αor|Xα∖A|<q)},
that is, α in PS(q) belongs to M¯A if and only if(a)
A⊈Xα, or(b)
A⊆Xα and either Aα⊆A⊆domα, or |Xα∖A|<q.The next result gives more detail onM¯A.Lemma 3.6.
Suppose thatp≥q≥ℵ0, and let A be a nonempty subset of X such that |X∖A|≥q. Then, (a)
for any cardinalk such that 0≤k≤p, there exist α,β∈PS(q) such that g(α)=k=g(β) and α∈M¯A,β∉M¯A;(b)
for eachγ∉M¯A, |domγ∖Aγ-1|=|X∖A|=|Xγ∖A| and |Aγ-1|=|A|.Proof.
To show that (a) holds, let|X∖A|=r≥q, and let k be a cardinal such that 0≤k≤p. We write X∖A=R∪̇Q where |R|=r and |Q|=q. If r=p, then |A∪R|≥r=p; if not, then |X∖A|<p, and this implies |A|=p, and so |A∪R|=p. Fix a∈A and let B=(A∖{a})∪R. Then, |B|=p and |X∖B|=|Q∪{a}|=q. We write X=K∪̇L where |K|=k and |L|=p. Then there exists a bijection α:L→B and so g(α)=k, d(α)=q. Also, since A⊈B=Xα, we have α∈M¯A.
To findβ∈PS(q)∖M¯A with g(β)=k, we consider two cases. First, if r=p, we write X∖A=P∪̇Q∪̇K where |P|=p,|Q|=q,|K|=k. Fix a∈A and define
(3.15)β=(P∪Q∪{a}A∖{a}P∪K∪{a}A∖{a}),
where β|(P∪Q∪{a}) and β|(A∖{a}) are bijections and aβ≠a. On the other hand, if r<p, then |A|=p. In this case we write A=A′∪̇K′ and X∖A=R∪̇Q where |A′|=p,|K′|=k,|R|=r and |Q|=q. Fix a∈A′ and redefine
(3.16)β=((X∖A)∪{a}A′∖{a}R∪{a}A∖{a}),
where β|((X∖A)∪{a}) and β|(A′∖{a}) are bijections and aβ≠a. In both cases, we have d(β)=q, g(β)=k, A⊆Xβ, Aβ⊈A, and |Xβ∖A|≥q, that is β∈PS(q)∖M¯A.
To see that (b) holds, suppose that there isγ∉M¯A, then A⊆Xγ and |Xγ∖A|≥q. So |Aγ-1|=|A| since γ is injective. Also,
(3.17)X∖A=(X∖Xγ)∪̇(Xγ∖A),
where |X∖Xγ|=q. Since |X∖A|≥q and by our assumption |Xγ∖A|≥q, we have |X∖A|=|Xγ∖A|=|(Xγ∖A)γ-1|=|domγ∖Aγ-1| as required.In [6, Theorem  1], the authors proved that MA is a maximal subsemigroup of BL(q) for every nonempty subset A of X such that |X∖A|≥q. Using a similar argument, we show that M¯A is a subsemigroup of PS(q).Lemma 3.7.
Suppose thatp≥q≥ℵ0, and let A be a nonempty subset of X such that |X∖A|≥q. Then M¯A is a proper subsemigroup of PS(q).Proof.
Letα,β∈M¯A. If A⊈Xαβ, then αβ∈M¯A. Now we suppose that A⊆Xαβ. Then, A⊆Xβ and since β∈M¯A, we either have Aβ⊆A⊆domβ, or |Xβ∖A|<q. If |Xβ∖A|<q, then
(3.18)|Xαβ∖A|≤|Xβ∖A|<q
and so αβ∈M¯A. Otherwise, we have Aβ⊆A⊆Xαβ and hence A⊆Xα since β is injective. Since α∈M¯A, we either have Aα⊆A⊆domα, or |Xα∖A|<q. If the latter occurs, then
(3.19)|Xαβ∖A|≤|Xαβ∖Aβ|=|(Xα∖A)β|≤|Xα∖A|<q,
therefore αβ∈M¯A. On the other hand, if Aα⊆A⊆domα, we have Aαβ⊆Aβ⊆A. Moreover, Aα⊆Xα∩domβ, that is, A⊆(Xα∩domβ)α-1=dom(αβ). Therefore αβ∈M¯A, and hence M¯A is a subsemigroup of PS(q). Finally, this subsemigroup is properly contained in PS(q) by Lemma 3.6(a).Remark 3.8.
For any cardinalr such that q≤r≤p, Sr∩M¯A is a proper subsemigroup of Sr but it is not maximal when q<r. To see this, suppose Sr∩M¯A is maximal and choose α,β∉M¯A such that g(α)=r and g(β)=0 (possible by Lemma 3.6(a)). Then α,β∈Sr∖M¯A where domβ=X. Moreover 〈(Sr∩M¯A)∪{α}〉=Sr, and so
(3.20)β=γ1γ2⋯γnαλ1λ2⋯λm
for some n,m∈ℕ0 and γi,λj∈(Sr∩M¯A)∪{α}, i=1,…,n, j=1,…,m. If n=0 or γ1=α, then domβ⊆domα and so g(α)=0, a contradiction. Thus, n≠0 and γ1≠α. Since X=domβ⊆dom(γ1γ2⋯γn), it follows that γ=γ1γ2⋯γn∈BL(q). Moreover, Xγ⊆domα, and this implies,
(3.21)q≤r=|X∖dom⁡α|≤|X∖Xγ|=q,
and hence r=q.
SinceMA is maximal in BL(q), a subsemigroup of PS(q), it is natural to think that M¯A is maximal in PS(q). But when p>q, by taking r=p, the above observation shows that this claim is false since Sp=PS(q). Thus, M¯A is not always a maximal subsemigroup of PS(q).The proof of the next result follows some ideas from [6, Theorem  1].Theorem 3.9.
Suppose thatp≥r≥q≥ℵ0, and let A be a nonempty subset of X such that |X∖A|≥q. Then Sr∩M¯A is a maximal subsemigroup of Sr precisely when r=q.Proof.
In Remark3.8, we have shown that Sr∩M¯A is not maximal in Sr when r>q. It remains to show Sq∩M¯A is maximal in Sq. Let α,β∈Sq∖M¯A. Then g(α),g(β)≤q and Lemma 3.6(b) implies that
(3.22)|Aα-1|=|A|=|Aβ-1|,|dom⁡α∖Aα-1|=|dom⁡β∖Aβ-1|=|Xβ∖A|=|Xα∖A|=|X∖A|=s(say)≥q.
We also have Aβ⊈A or A⊈domβ. In the case that Aβ⊈A, we have Aβ∩(X∖A)≠∅. Thus, there exists y∈A∩(X∖A)β-1, so y∉Aβ-1. Since |domβ∖(Aβ-1∪{y})|=s, we can write
(3.23)dom⁡β∖(Aβ-1∪{y})={cj}∪̇{dk},
where |J|=s and |K|=q. Also, since α,β∉M¯A, we have A⊆Xα and A⊆Xβ. Thus, for convenience, write A={ai}, let yi,zi∈X be such that yiα=ai=ziβ for each i, and let domα∖Aα-1={bj}. Hence, we can write
(3.24)β=(zicjdkyaicjβdkβyβ).
Now defineγ∈P(X) by
(3.25)γ=(yibjzicj).
Then, d(γ)=|{dk}∪{y}|+g(β)=q, that is, γ∈PS(q). Also, since domγ=domα, we have g(γ)=g(α)≤q and so γ∈Sq. Moreover, since y∈A and y∉Xγ, we have A⊈Xγ, that is, γ∈M¯A. Also, since d(α)=q, we can write X∖Xα={mk}∪̇{nk}∪̇{z} and define μ in P(X) by
(3.26)μ=(aicjβdkβyβaibjαmkz).
Then d(μ)=|{nk}|=q=d(β)=g(μ), that is, μ∈Sq. Moreover, μ∈M¯A since Aμ=A⊆domμ. Finally, we can see that α=γβμ where γ,μ∈Sq∩M¯A.
On the other hand, ifA⊈domβ, then there exists w∈A∩(X∖domβ). In this case, we rewrite domβ∖Aβ-1={cj}∪̇{dk} and X∖Xα={mk}∪̇{nk} where |J|=s,|K|=q. Like before, we write A={ai} and domα={yi}∪̇{bj} where {bj}=domα∖Aα-1, then
(3.27)β=(zicjdkaicjβdkβ).
Define γ,μ∈P(X) by
(3.28)γ=(yibjzicj),μ=(aicjβdkβaibjαmk).
Then, d(γ)=|{dk}|+g(β)=q, g(γ)=g(α)≤q, d(μ)=|{nk}|=q=d(β)=g(μ), and so γ,μ∈Sq. Also, γ,μ∈M¯A since A⊈Xγ (note that w∈A∖domβ⊆A∖Xγ) and Aμ=A⊆domμ. Moreover, α=γβμ. In other words, we have shown that for every α,β∈Sq∖M¯A, α can be written as a finite product of elements of (Sq∩M¯A)∪{β}. Therefore, Sq∩M¯A is maximal in Sq.We now determine some other classes of maximal subsemigroups ofSr.Lemma 3.10.
Suppose thatp≥r≥q≥ℵ0. Let k be a cardinal such that k=0 or q≤k≤r. Then
(3.29)Sr∖Gk={α∈PS(q):k≠g(α)≤r}
is a proper subsemigroup of Sr.Proof.
Sincek≤r, we have Sr∖Gk⊊Sr. If k=0, then Sr∖G0=Sr∖BL(q), and this is a subsemigroup of Sr since, for α,β∈Sr∖BL(q), dom(αβ)⊆domα⊊X, and this implies αβ∈Sr∖BL(q). Now suppose q≤k≤r and let α,β∈Sr be such that g(αβ)=k. We claim that g(α)=k or g(β)=k. To see this, assume that g(α)≠k. Since
(3.30)k=|X∖dom⁡(αβ)|=|X∖dom⁡α|+|dom⁡α∖dom⁡(αβ)|,
we have |X∖domα|<k, thus
(3.31)k=|dom⁡α∖dom⁡(αβ)|=|[Xα∖(Xα∩dom⁡β)]α-1|=|(Xα∖dom⁡β)α-1|=|Xα∖dom⁡β|.
Note that(3.32)X∖dom⁡β=[Xα∖dom⁡β]∪̇[(X∖Xα)∩(X∖dom⁡β)],
where the intersection on the right has cardinal at most q. Hence, g(β)=|X∖domβ|=k and we have shown that Sr∖Gk is a subsemigroup of Sr.Remark 3.11.
Observe that, if0<k<q then Sr∖Gk is not a semigroup for all q≤r≤p. To see this, let α∈BL(q) and β=idXα∖K for some subset K of Xα such that |K|=k (possible since |Xα|=p>k), then α,β∈PS(q) since d(β)=d(α)+k=q. Moreover, since g(α)=0 and g(β)=q≠k, we have α,β∈Sr∖Gk. But
(3.33)dom⁡(αβ)=(Xα∩dom⁡β)α-1=(Xα∖K)α-1=X∖Kα-1,
thus g(αβ)=|Kα-1|=k, that is, αβ∈Gk.Theorem 3.12.
Suppose thatp≥r≥q≥ℵ0. Then the following statements hold: (a)
Sr∖G0 is a maximal subsemigroup of Sr;(b)
ifp>q, then for each cardinal k such that q≤k≤r, Sr∖Gk is a maximal subsemigroup of Sr.Proof.
By Lemma3.10, Sr∖G0 is a subsemigroup of Sr. To see that it is maximal, let α,β∈G0=BL(q)⊆Sq. By [3, Theorem  5], Sq=β·R(q), and this implies that α=βγ for some γ∈R(q)⊆Sr∖G0. Hence Sr∖G0 is maximal in Sr.
Now suppose thatp>q and let q≤k≤r. Let α,β∈Gk. If k=q, then Gk=R(q)⊆Sq and, by [3, Theorem  6], Sq=BL(q)·β·BL(q). If k>q, then Gk=BL(q)·β·BL(q) (by Lemma 3.4). Therefore, α=γβμ for some γ,μ∈BL(q)⊆Sr∖Gk, and so Sr∖Gk is maximal in Sr.Corollary 3.13.
Suppose thatp>q≥ℵ0 and let A be a nonempty subset of X such that |X∖A|≥q. Then the following sets are maximal subsemigroups of PS(q): (a)
M¯A∪Tq′;(b)
Nk={α∈PS(q):g(α)≠k} where k=0 or q≤k≤p.Proof.
By Theorem3.9, Sq∩M¯A is maximal in Sq. Then Corollary 3.2 implies that (Sq∩M¯A)∪Tq′ is maximal in PS(q). But
(3.34)(Sq∩M¯A)∪Tq′=(Sq∪Tq′)∩(M¯A∪Tq′)=PS(q)∩(M¯A∪Tq′)=M¯A∪Tq′,
and so (a) holds. To show that (b) holds, let r=p in Theorem 3.12. Then Sp=PS(q) and thus Nk=Sp∖Gk is maximal in PS(q).Theorem 3.14.
Suppose thatp>q≥ℵ0 and k equals 0 or q. Let A be a nonempty subset of X such that |X∖A|≥q. Then the two classes of maximal subsemigroups Sq∩M¯A and Sq∖Gk of Sq are always disjoint.Proof.
By Theorems3.9 and 3.12, Sq∩M¯A and Sq∖Gk are maximal subsemigroups of Sq. By Lemma 3.6(a), there exists α∈M¯A with g(α)=k. Then α∈Sk∩M¯A⊆Sq∩M¯A but α∉Sq∖Gk, that is, Sq∩M¯A⊈Sq∖Gk. Also, Sq∖Gk⊈Sq∩M¯A by the maximality of Sq∩M¯A and Sq∖Gk. Therefore, Sq∩M¯A is not equal to Sq∖Gk.


## 4. Maximal Subsemigroups ofPS(q) When p=q

We first recall that, whenp=q, the empty transformation ∅ belongs to PS(q) since d(∅)=p=q. In this case, the ideals of PS(q) are precisely the sets: (4.1)Jr={α∈PS(q):r(α)<r},
where 1≤r≤p′ (see [3, Theorem  14]). Clearly, Jp′=PS(q) and Jp={α∈PS(q):r(α)<p} is the largest proper ideal. In this case, the complement of each Jr in PS(q) is not a semigroup. To see this, write X=A∪̇B∪̇C where |A|=p and |B|=r=|C|. Then idB,idC∈PS(q)∖Jr whereas idB.idC=∅∈Jr. Hence, unlike what was done in Section 3, we cannot use Lemma 3.1 to find maximal subsemigroups of PS(q) when p=q. In this section, we determine some maximal subsemigroups of PS(q), for p=q, using a different approach. We first describe some properties of each maximal subsemigroup in this case.Lemma 4.1.
Suppose thatp=q≥ℵ0 and M is a maximal subsemigroup of PS(q). Then the following statements hold: (a)
M contains all α∈PS(q) with r(α)<p,(b)
ifR(q)⊆M, then M∩BL(q)=∅.Proof.
Suppose that there existsα∉M with r(α)=k<p. Then g(α)=p, and we write in the usual way
(4.2)α=(aixi).
Also, write X∖{ai}=P∪̇Q and X∖{xi}=R∪̇S where |P|=|Q|=p=|R|=|S|, and define β,γ in P(X) by
(4.3)β=(aiPaiP),γ=(aiQxiR),
where β|P and γ|Q are bijections. Then β,γ∈PS(q). Also,
(4.4)α=β⋅α⋅idXα∈PS(q)⋅α⋅PS(q),
thus M⊊M∪(PS(q)·α·PS(q)). But M∪(PS(q)·α·PS(q)) is a subsemigroup of PS(q) and this means that M∪(PS(q)·α·PS(q))=PS(q) by the maximality of M. Since all mappings in PS(q)·α·PS(q) have rank at most k, it follows that M contains all mappings with rank greater than k. Therefore β,γ∈M and thus α=βγ∈M, a contradiction.
To show that (b) holds, suppose thatR(q)⊆M. If there exists α∈M∩BL(q), then [3, Theorem  5] implies that PS(q)=α·R(q)⊆M (note that Sq=PS(q) when p=q), so M=PS(q), contrary to the maximality of M. Thus M∩BL(q)=∅.Remark 4.2.
Ifp>q, then every α∈PS(q) has rank p. This contrasts with Lemma 4.1(a). Also, by Corollary 3.13, if p>q and q<k≤p, Nk is a maximal subsemigroup of PS(q) containing R(q)∪BL(q), this contrasts with Lemma 4.1(b).As in Section3, for any cardinal k, we let (4.5)Nk={α∈PS(q):g(α)≠k}.By Lemma3.10 and Remark 3.11, if p=q, then Nk is a subsemigroup of PS(q) exactly when k=0 or k=p. From Corollary 3.13(b), when p>q, Np is a maximal subsemigroup of PS(q). But when p=q, Lemma 4.1(a) implies that Np is not maximal since ∅∉Np. Moreover, Lemma 4.1(a) implies that every maximal subsemigroup of PS(q) must contain the largest proper ideal (4.6)Jp={α∈PS(q):r(α)<p}.
Note that Jp itself is a subsemigroup of PS(q), but it is not maximal since Jp⊊R(q) (in case p=q, r(α)<p implies g(α)=p).Theorem 4.3.
Suppose thatp=q≥ℵ0, and let A be a nonempty subset of X such that |X∖A|≥q. The following are maximal subsemigroups of PS(q):(a)
M¯A;(b)
N0;(c)
Np∪Jp.Proof.
Ifp=q, then Sq=PS(q), and so (a) holds by Theorem 3.9. Also, by taking r=p in Theorem 3.12(a), we see that (b) holds. To show that (c) holds, take r=p=k in Lemma 3.10, we have Np=Sp∖Gp is a subsemigroup of PS(q). Moreover, Np∪Jp is also a subsemigroup of PS(q) since Jp is an ideal. To show the maximality of Np∪Jp, let α,β∈PS(q)∖(Np∪Jp). Then g(α)=g(β)=p=r(α)=r(β). Write in the usual way
(4.7)α=(aixi),β=(biyi),
where |I|=p, and let
(4.8)X∖{ai}=A∪̇B,X∖{yi}=C∪̇D,
where |A|=|B|=|C|=|D|=p. Then define γ,μ∈P(X) by
(4.9)γ=(biX∖{bi}aiA),μ=(xiX∖{xi}yiC),
where γ|(X∖{bi}) and μ|(X∖{xi}) are bijections. Thus γ,μ∈PS(q) since d(γ)=|B|=p=|D|=d(μ). Moreover γ,μ∈Np∪Jp since g(γ)=g(μ)=0<p. It is clear that β=γαμ and therefore Np∪Jp is maximal in PS(q).Remark 4.4.
Whenp=q, if M is a maximal subsemigroup containing R(q), then
(4.10)M⊆(PS(q)∖BL(q))=N0
by Lemma 4.1(b). Thus, M=N0 by the maximality of M. So we conclude that N0 is the only maximal subsemigroup of PS(q) containing R(q).Remark 4.5.
As we showed in Section3, to see all maximal subsemigroups of PS(q) when p>q, it is necessary to describe all maximal subsemigroups of Sr where q≤r<p. So we leave this as a direction for future research.



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