On Maximal Subsemigroups of Partial Baer-Levi Semigroups

Abstract

Suppose that 𝑋 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 𝑀_𝐴 (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 𝑀_𝐴 to obtain maximal subsemigroups of 𝑃𝑆(𝑞) when |𝑋|=𝑞.

1. Introduction

Suppose that 𝑋 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 𝑔(𝛼)=|𝑋⧵dom𝛼|,𝑑(𝛼)=|𝑋⧵ran𝛼|,𝑟(𝛼)=|ran𝛼|, (1.1) 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 𝐵𝐿𝑞=𝛼∈𝐼(𝑋)∶𝑔(𝛼)=0,𝑑(𝛼)=𝑞,𝑃𝑆𝑞=𝛼∈𝐼(𝑋)∶𝑑(𝛼)=𝑞, (1.2) 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 𝑅𝑞=𝛼∈𝑃𝑆𝑞∶𝑔(𝛼)=𝑞. (1.3) 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 subset 𝐴 of 𝑋 such that |𝑋⧵𝐴|≥𝑞, let 𝑀_𝐴=𝛼∈𝐵𝐿𝑞∶𝐴⊈𝑋𝛼or𝐴𝛼⊆𝐴or|𝑋𝛼⧵𝐴|<𝑞. (1.4) In other words, given 𝛼∈𝐵𝐿(𝑞), we have 𝛼∈𝑀_𝐴 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 𝑀_𝐴 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 of 𝑃𝑆(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 𝑆_𝑟 of 𝑃𝑆(𝑞) defined by 𝑆_𝑟=𝛼∈𝑃𝑆𝑞∶𝑔(𝛼)≤𝑟, (1.5) where 𝑞≤𝑟≤𝑝. Moreover, we extend 𝑀_𝐴 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 a disjoint 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 id_𝐴 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 𝑎𝛼=^𝑖𝑥_𝑖 (2.1) and take as understood that the subscript 𝑖 belongs to some (unmentioned) index set 𝐼, that the abbreviation {𝑥_𝑖} denotes {𝑥_𝑖∶𝑖∈𝐼}, and that 𝑋𝛼=ran𝛼={𝑥_𝑖},𝑎_𝑖𝛼=𝑥_𝑖 for each 𝑖 and dom𝛼={𝑎_𝑖}. To simplify notation, if 𝐴⊆𝑋, we sometimes write 𝐴𝛼 in place of (𝐴∩dom𝛼)𝛼.

Let 𝑆 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 any 𝑎,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 of 𝑃𝑆(𝑞).

3. Maximal Subsemigroups of 𝑃𝑆(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. Let 𝑆 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 subsemigroup 𝑀 of 𝑇, 𝑀∪𝐼 is a maximal subsemigroup of 𝑆; (b)for any maximal subsemigroup 𝑁 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, let 𝑀 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), let 𝑁 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⋯𝑐_𝑛 for some natural 𝑛 and some 𝑐_𝑖∈𝑁∪{𝑎} for all 𝑖=1,2,…,𝑛. Since 𝑏∉𝑁, we have 𝑐_𝑖=𝑎 for some 𝑖. Moreover, since 𝑏∉𝐼, we have 𝑐_𝑗∈𝑇∩𝑁 for all 𝑗≠𝑖. It follows that 𝑇⧵𝑁⊆⟨(𝑇∩𝑁)∪{𝑎}⟩, therefore 𝑇=(𝑇⧵𝑁)∪(𝑇∩𝑁)⊆⟨(𝑇∩𝑁)∪{𝑎}⟩, (3.1) that is, 𝑇=⟨(𝑇∩𝑁)∪{𝑎}⟩ and thus 𝑇∩𝑁 is maximal in 𝑇.

Let 𝑢 be a cardinal number. The successor of 𝑢, denoted by 𝑢^′, is defined as 𝑢^′=min{𝑣∶𝑣>𝑢}. (3.2) 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, 𝑆_𝑘=𝛼∈𝑃𝑆𝑞∶𝑔(𝛼)≤𝑘 (3.3) is a subsemigroup of 𝑃𝑆(𝑞). Also, when 𝑝>𝑞, the proper ideals of 𝑃𝑆(𝑞) are precisely the sets: 𝑇_𝑠=𝛼∈𝑃𝑆𝑞∶𝑔(𝛼)≥𝑠, (3.4) where 𝑞<𝑠≤𝑝 (see [3, Theorem  13]). Thus, for any 𝑞≤𝑟<𝑝, it is clear that 𝑃𝑆𝑞=𝑆_𝑟∪̇𝑇_𝑟′, (3.5) that is, 𝑃𝑆(𝑞) can be written as a disjoint union of the semigroup 𝑆_𝑟 and the ideal 𝑇_𝑟′. Hence, the next result follows directly from Lemma 3.1(a).

Corollary 3.2. Suppose that 𝑝>𝑟>𝑞≥ℵ₀. If 𝑀 is a maximal subsemigroup of 𝑆_𝑟, then 𝑀∪𝑇_𝑟′ is a maximal subsemigroup of 𝑃𝑆(𝑞).

Lemma 3.3. Let 𝑝>𝑞≥ℵ₀ and suppose that 𝑀 is a maximal subsemigroup of 𝑃𝑆(𝑞). Then, (a)𝑆_𝑟∩𝑀≠∅ for all 𝑞≤𝑟<𝑝; (b)if there exists 𝛼∉𝑀 with 𝑔(𝛼)<𝑝, then 𝑆_𝑘⧵𝑀≠∅ for some 𝑞≤𝑘<𝑝.

Proof. To show that (a) holds, we first note that 𝑆_𝑞 is contained in 𝑆_𝑟 for all 𝑞≤𝑟<𝑝. If 𝑆_𝑞∩𝑀=∅, then 𝑀⊆𝑇_𝑞′⊊𝑃𝑆(𝑞) and thus 𝑀=𝑇_𝑞′ by the maximality of 𝑀. But 𝑇_𝑞′⊊𝑇_𝑞′∪𝐵𝐿(𝑞)⊊𝑃𝑆(𝑞) where 𝑇_𝑞′∪𝐵𝐿(𝑞) is a subsemigroup of 𝑃𝑆(𝑞) (since 𝑇_𝑞′ is an ideal), so we get a contradiction. Therefore, ∅≠𝑆_𝑞∩𝑀⊆𝑆_𝑟∩𝑀 for all 𝑞≤𝑟<𝑝.
To show that (b) holds, suppose there is 𝛼∉𝑀 with 𝑔(𝛼)=𝑘<𝑝. If 𝑘<𝑞, then 𝛼∈𝑆_𝑟⧵𝑀 for all 𝑞≤𝑟≤𝑝. Otherwise, if 𝑞≤𝑘, then 𝛼∈𝑆_𝑘⧵𝑀. Hence (b) holds.

For what follows, for any cardinal 𝑟≤𝑝, we let 𝐺_𝑟=𝛼∈𝑃𝑆𝑞∶𝑔(𝛼)=𝑟. (3.6)

Then 𝐺₀=𝐵𝐿(𝑞) and 𝐺_𝑞=𝑅(𝑞). Moreover, if 𝑝>𝑞 and 𝑟>𝑞, then 𝐺_𝑟=𝑆_𝑟∩𝑇_𝑟, and so 𝐺_𝑟 is a subsemigroup of 𝑆_𝑟 (since it is the intersection of two semigroups). Also, 𝐺_𝑟 is bisimple and idempotent-free, when 𝑝>𝑞 and 𝑟>𝑞 (see [3, Corollary  3]).

From [3, Theorem  5], if 𝑝≥𝑞, then 𝑆_𝑞=𝛼⋅𝑅(𝑞) for each 𝛼∈𝐵𝐿(𝑞), and by [3, Theorem  6], 𝑆_𝑞=𝐵𝐿(𝑞)⋅𝜇⋅𝐵𝐿(𝑞) for each 𝜇∈𝑅(𝑞) when 𝑝≠𝑞.

This motivates the following result.

Lemma 3.4. Suppose that 𝑝≥𝑟>𝑞≥ℵ₀. Then 𝐺_𝑟=𝐵𝐿(𝑞)⋅𝛼⋅𝐵𝐿(𝑞) for each 𝛼∈𝐺_𝑟.

Proof. Let 𝛼∈𝐺_𝑟 and 𝛽,𝛾∈𝐵𝐿(𝑞). Since 𝑋⧵dom𝛼=𝑋𝛽∩(𝑋⧵dom𝛼)∪𝑋⧵𝑋𝛽∩(𝑋⧵dom𝛼), (3.7) 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 𝑟=𝑋𝛽∩(𝑋⧵dom𝛼)𝛽⁻¹=𝑋𝛽⧵dom𝛼𝛽⁻¹==.dom𝛽⧵dom𝛽𝛼𝑋⧵dom𝛽𝛼=𝑔𝛽𝛼 (3.8) Since dom𝛾=𝑋, we have dom(𝛽𝛼𝛾)=dom(𝛽𝛼), and so 𝑔(𝛽𝛼𝛾)=𝑔(𝛽𝛼)=𝑟. Hence 𝛽𝛼𝛾∈𝐺_𝑟 and therefore 𝐵𝐿(𝑞)⋅𝛼⋅𝐵𝐿(𝑞)⊆𝐺_𝑟.
For the converse, if 𝛼,𝛽∈𝐺_𝑟, then |𝑋⧵dom𝛼|=𝑟=|𝑋⧵dom𝛽|. Since 𝑝>𝑞, every element in 𝑃𝑆(𝑞) has rank 𝑝, so we write 𝑎𝛼=^𝑖𝑥_𝑖𝑏,𝛽=^𝑖𝑦_𝑖where|𝐼|=𝑝. (3.9) Now write 𝑋⧵{𝑦_𝑖}=𝐴∪̇𝐵 and 𝑋⧵{𝑎_𝑖}=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 𝑏𝛿=^𝑖𝑋⧵𝑏_𝑖𝑎_𝑖𝐷𝑥,𝜖=^𝑖𝑋⧵𝑥_𝑖𝑦_𝑖𝐴, (3.10) where 𝛿|(𝑋⧵{𝑏_𝑖}) and 𝜖|(𝑋⧵{𝑥_𝑖}) are bijections. Then 𝛿,𝜖∈𝐵𝐿(𝑞) and 𝛽=𝛿𝛼𝜖, that is, 𝐺_𝑟⊆𝐵𝐿(𝑞)⋅𝛼⋅𝐵𝐿(𝑞) and equality follows.

Now we can describe all maximal subsemigroups of 𝑃𝑆(𝑞) when 𝑝>𝑞.

Theorem 3.5. Suppose that 𝑝>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)⧵𝐺_𝑝={𝛼∈𝑃𝑆(𝑞)∶𝑔(𝛼)<𝑝};(b)𝑁∪𝑇_𝑟′, where 𝑞≤𝑟<𝑝 and 𝑁 is a maximal subsemigroup of 𝑆_𝑟.

Proof. Let 𝛼,𝛽∈𝑃𝑆(𝑞) be such that 𝑔(𝛼)<𝑝 and 𝑔(𝛽)<𝑝. Clearly |𝑋𝛼⧵dom𝛽|≤|𝑋⧵dom𝛽|=𝑔(𝛽)<𝑝. Then dom𝛼⧵dom𝛼𝛽=𝑋𝛼⧵𝑋𝛼∩dom𝛽𝛼⁻¹=𝑋𝛼⧵dom𝛽𝛼⁻¹=𝑋𝛼⧵dom𝛽<𝑝. (3.11) Hence, 𝑋⧵dom𝛼𝛽=|𝑋⧵dom𝛼|+dom𝛼⧵dom𝛼𝛽<𝑝, (3.12) and this shows that 𝑃𝑆(𝑞)⧵𝐺_𝑝 is a subsemigroup of 𝑃𝑆(𝑞). To show that 𝑃𝑆(𝑞)⧵𝐺_𝑝 is maximal in 𝑃𝑆(𝑞), we let 𝛼,𝛽∈𝑃𝑆(𝑞)⧵(𝑃𝑆(𝑞)⧵𝐺_𝑝)=𝐺_𝑝. By Lemma 3.4, 𝛼=𝜆𝛽𝜇 for some 𝜆,𝜇∈𝐵𝐿(𝑞)⊆𝑃𝑆(𝑞)⧵𝐺_𝑝. Thus, 𝛼 can be written as a finite product of elements of (𝑃𝑆(𝑞)⧵𝐺_𝑝)∪{𝛽}, and hence 𝑃𝑆(𝑞)⧵𝐺_𝑝 is maximal in 𝑃𝑆(𝑞). Also, if 𝑞≤𝑟<𝑝 and 𝑁 is a maximal subsemigroup of 𝑆_𝑟, then 𝑁∪𝑇_𝑟′ is maximal in 𝑃𝑆(𝑞) by Corollary 3.2.
We now suppose that 𝑀 is a maximal subsemigroup of 𝑃𝑆(𝑞) such that 𝑀≠𝑃𝑆(𝑞)⧵𝐺_𝑝. Then there exists 𝛼∉𝑀 with 𝑔(𝛼)<𝑝. Thus, Lemma 3.3 implies that 𝑆_𝑘⧵𝑀≠∅ and 𝑆_𝑘∩𝑀≠∅ for some 𝑘, where 𝑞≤𝑘<𝑝. Since 𝑃𝑆(𝑞)=𝑆_𝑘∪̇𝑇_𝑘′, Lemma 3.1(b) implies that 𝑆_𝑘∩𝑀 is maximal in 𝑆_𝑘. We also see that 𝑀=𝑆_𝑘∩𝑀∪𝑇_𝑘′∩𝑀⊆𝑆_𝑘∩𝑀∪𝑇_𝑘′, (3.13) where (𝑆_𝑘∩𝑀)∪𝑇_𝑘′ is maximal in 𝑃𝑆(𝑞) by Corollary 3.2. This means that 𝑀=(𝑆_𝑘∩𝑀)∪𝑇_𝑘′ by the maximality of 𝑀.

By the previous theorem, when 𝑝>𝑞, most of the maximal subsemigroups of 𝑃𝑆(𝑞) are induced by maximal subsemigroups of 𝑆_𝑟 where 𝑞≤𝑟<𝑝. Hence we now determine some maximal subsemigroups of 𝑆_𝑟.

As mentioned in Section 1, for every nonempty subset 𝐴 of 𝑋 with |𝑋⧵𝐴|≥𝑞, 𝑀_𝐴 is a maximal subsemigroup of 𝐵𝐿(𝑞). Here we extend the definition of 𝑀_𝐴 and consider the set 𝑀_𝐴 defined as 𝑀_𝐴=𝛼∈𝑃𝑆𝑞∶𝐴⊈𝑋𝛼or𝐴𝛼⊆𝐴⊆dom𝛼or|𝑋𝛼⧵𝐴|<𝑞, (3.14) 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 on 𝑀_𝐴.

Lemma 3.6. Suppose that 𝑝≥𝑞≥ℵ₀, and let 𝐴 be a nonempty subset of 𝑋 such that |𝑋⧵𝐴|≥𝑞. Then, (a)for any cardinal 𝑘 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 each 𝛾∉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 find 𝛽∈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 {𝑎}{𝑎}{𝑎}{𝑎}𝛽=𝑃∪𝑄∪𝐴⧵𝑃∪𝐾∪𝐴⧵, (3.15) 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 ()∪{𝑎}𝐴𝛽=𝑋⧵𝐴^′⧵{𝑎}𝑅∪{𝑎}𝐴⧵{𝑎}, (3.16) 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 is 𝛾∉𝑀_𝐴, then 𝐴⊆𝑋𝛾 and |𝑋𝛾⧵𝐴|≥𝑞. So |𝐴𝛾⁻¹|=|𝐴| since 𝛾 is injective. Also, 𝑋⧵𝐴=𝑋⧵𝑋𝛾∪̇𝑋𝛾⧵𝐴, (3.17) 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 𝑀_𝐴 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 that 𝑝≥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. Let 𝛼,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 𝑋𝛼𝛽⧵𝐴≤𝑋𝛽⧵𝐴<𝑞 (3.18) 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 𝑋𝛼𝛽⧵𝐴≤𝑋𝛼𝛽⧵𝐴𝛽=(𝑋𝛼⧵𝐴)𝛽≤|𝑋𝛼⧵𝐴|<𝑞, (3.19) 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 cardinal 𝑟 such that 𝑞≤𝑟≤𝑝, 𝑆_𝑟∩𝑀_𝐴 is a proper subsemigroup of 𝑆_𝑟 but it is not maximal when 𝑞<𝑟. To see this, suppose 𝑆_𝑟∩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∈𝑆_𝑟⧵𝑀_𝐴 where dom𝛽=𝑋. Moreover ⟨(𝑆_𝑟∩𝑀_𝐴)∪{𝛼}⟩=𝑆_𝑟, and so 𝛽=𝛾₁𝛾₂⋯𝛾_𝑛𝛼𝜆₁𝜆₂⋯𝜆_𝑚 (3.20) for some 𝑛,𝑚∈ℕ₀ and 𝛾_𝑖,𝜆_𝑗∈(𝑆_𝑟∩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⋯𝛾_𝑛), it follows that 𝛾=𝛾₁𝛾₂⋯𝛾_𝑛∈𝐵𝐿(𝑞). Moreover, 𝑋𝛾⊆dom𝛼, and this implies, 𝑞≤𝑟=|𝑋⧵dom𝛼|≤𝑋⧵𝑋𝛾=𝑞, (3.21) and hence 𝑟=𝑞.
Since 𝑀_𝐴 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 𝑆_𝑝=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 that 𝑝≥𝑟≥𝑞≥ℵ₀, and let 𝐴 be a nonempty subset of 𝑋 such that |𝑋⧵𝐴|≥𝑞. Then 𝑆_𝑟∩𝑀_𝐴 is a maximal subsemigroup of 𝑆_𝑟 precisely when 𝑟=𝑞.

Proof. In Remark 3.8, we have shown that 𝑆_𝑟∩𝑀_𝐴 is not maximal in 𝑆_𝑟 when 𝑟>𝑞. It remains to show 𝑆_𝑞∩𝑀_𝐴 is maximal in 𝑆_𝑞. Let 𝛼,𝛽∈𝑆_𝑞⧵𝑀_𝐴. Then 𝑔(𝛼),𝑔(𝛽)≤𝑞 and Lemma 3.6(b) implies that 𝐴𝛼⁻¹=|𝐴|=𝐴𝛽⁻¹,dom𝛼⧵𝐴𝛼⁻¹=dom𝛽⧵𝐴𝛽⁻¹=𝑋𝛽⧵𝐴=|𝑋𝛼⧵𝐴|=|𝑋⧵𝐴|=𝑠say≥𝑞. (3.22) 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 dom𝛽⧵𝐴𝛽⁻¹∪𝑦=𝑐_𝑗∪̇𝑑_𝑘, (3.23) where |𝐽|=𝑠 and |𝐾|=𝑞. Also, since 𝛼,𝛽∉𝑀_𝐴, we have 𝐴⊆𝑋𝛼 and 𝐴⊆𝑋𝛽. Thus, for convenience, write 𝐴={𝑎_𝑖}, let 𝑦_𝑖,𝑧_𝑖∈𝑋 be such that 𝑦_𝑖𝛼=𝑎_𝑖=𝑧_𝑖𝛽 for each 𝑖, and let dom𝛼⧵𝐴𝛼⁻¹={𝑏_𝑗}. Hence, we can write 𝑧𝛽=^𝑖𝑐_𝑗𝑑_𝑘𝑦𝑎_𝑖𝑐_𝑗𝛽𝑑_𝑘𝛽𝑦𝛽. (3.24)
Now define 𝛾∈𝑃(𝑋) by 𝑦𝛾=^𝑖𝑏_𝑗𝑧_𝑖𝑐_𝑗. (3.25) Then, 𝑑(𝛾)=|{𝑑_𝑘}∪{𝑦}|+𝑔(𝛽)=𝑞, that is, 𝛾∈𝑃𝑆(𝑞). Also, since dom𝛾=dom𝛼, we have 𝑔(𝛾)=𝑔(𝛼)≤𝑞 and so 𝛾∈𝑆_𝑞. 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={𝑚_𝑘}∪̇{𝑛_𝑘}∪̇{𝑧} and define 𝜇 in 𝑃(𝑋) by 𝑎𝜇=^𝑖𝑐_𝑗𝛽𝑑_𝑘𝑎𝛽𝑦𝛽_𝑖𝑏_𝑗𝛼𝑚_𝑘𝑧. (3.26) Then 𝑑(𝜇)=|{𝑛_𝑘}|=𝑞=𝑑(𝛽)=𝑔(𝜇), that is, 𝜇∈𝑆_𝑞. Moreover, 𝜇∈𝑀_𝐴 since 𝐴𝜇=𝐴⊆dom𝜇. Finally, we can see that 𝛼=𝛾𝛽𝜇 where 𝛾,𝜇∈𝑆_𝑞∩𝑀_𝐴.
On the other hand, if 𝐴⊈dom𝛽, then there exists 𝑤∈𝐴∩(𝑋⧵dom𝛽). In this case, we rewrite dom𝛽⧵𝐴𝛽⁻¹={𝑐_𝑗}∪̇{𝑑_𝑘} and 𝑋⧵𝑋𝛼={𝑚_𝑘}∪̇{𝑛_𝑘} where |𝐽|=𝑠,|𝐾|=𝑞. Like before, we write 𝐴={𝑎_𝑖} and dom𝛼={𝑦_𝑖}∪̇{𝑏_𝑗} where {𝑏_𝑗}=dom𝛼⧵𝐴𝛼⁻¹, then 𝑧𝛽=^𝑖𝑐_𝑗𝑑_𝑘𝑎_𝑖𝑐_𝑗𝛽𝑑_𝑘𝛽. (3.27) Define 𝛾,𝜇∈𝑃(𝑋) by 𝑦𝛾=^𝑖𝑏_𝑗𝑧_𝑖𝑐_𝑗𝑎,𝜇=^𝑖𝑐_𝑗𝛽𝑑_𝑘𝛽𝑎_𝑖𝑏_𝑗𝛼𝑚_𝑘. (3.28) Then, 𝑑(𝛾)=|{𝑑_𝑘}|+𝑔(𝛽)=𝑞, 𝑔(𝛾)=𝑔(𝛼)≤𝑞, 𝑑(𝜇)=|{𝑛_𝑘}|=𝑞=𝑑(𝛽)=𝑔(𝜇), and so 𝛾,𝜇∈𝑆_𝑞. 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∈𝑆_𝑞⧵𝑀_𝐴, 𝛼 can be written as a finite product of elements of (𝑆_𝑞∩𝑀_𝐴)∪{𝛽}. Therefore, 𝑆_𝑞∩𝑀_𝐴 is maximal in 𝑆_𝑞.

We now determine some other classes of maximal subsemigroups of 𝑆_𝑟.

Lemma 3.10. Suppose that 𝑝≥𝑟≥𝑞≥ℵ₀. Let 𝑘 be a cardinal such that 𝑘=0 or 𝑞≤𝑘≤𝑟. Then 𝑆_𝑟⧵𝐺_𝑘=𝛼∈𝑃𝑆𝑞∶𝑘≠𝑔(𝛼)≤𝑟 (3.29) is a proper subsemigroup of 𝑆_𝑟.

Proof. Since 𝑘≤𝑟, we have 𝑆_𝑟⧵𝐺_𝑘⊊𝑆_𝑟. If 𝑘=0, then 𝑆_𝑟⧵𝐺₀=𝑆_𝑟⧵𝐵𝐿(𝑞), and this is a subsemigroup of 𝑆_𝑟 since, for 𝛼,𝛽∈𝑆_𝑟⧵𝐵𝐿(𝑞), dom(𝛼𝛽)⊆dom𝛼⊊𝑋, and this implies 𝛼𝛽∈𝑆_𝑟⧵𝐵𝐿(𝑞). Now suppose 𝑞≤𝑘≤𝑟 and let 𝛼,𝛽∈𝑆_𝑟 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 𝑘=𝑋⧵dom𝛼𝛽=|𝑋⧵dom𝛼|+dom𝛼⧵dom𝛼𝛽, (3.30) we have |𝑋⧵dom𝛼|<𝑘, thus 𝑘=dom𝛼⧵dom𝛼𝛽=𝑋𝛼⧵𝑋𝛼∩dom𝛽𝛼⁻¹=𝑋𝛼⧵dom𝛽𝛼⁻¹=𝑋𝛼⧵dom𝛽. (3.31)
Note that 𝑋⧵dom𝛽=𝑋𝛼⧵dom𝛽∪̇(𝑋⧵𝑋𝛼)∩𝑋⧵dom𝛽, (3.32) where the intersection on the right has cardinal at most 𝑞. Hence, 𝑔(𝛽)=|𝑋⧵dom𝛽|=𝑘 and we have shown that 𝑆_𝑟⧵𝐺_𝑘 is a subsemigroup of 𝑆_𝑟.

Remark 3.11. Observe that, if 0<𝑘<𝑞 then 𝑆_𝑟⧵𝐺_𝑘 is not a semigroup for all 𝑞≤𝑟≤𝑝. To see this, let 𝛼∈𝐵𝐿(𝑞) and 𝛽=id_{𝑋𝛼⧵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∈𝑆_𝑟⧵𝐺_𝑘. But dom𝛼𝛽=𝑋𝛼∩dom𝛽𝛼⁻¹=(𝑋𝛼⧵𝐾)𝛼⁻¹=𝑋⧵𝐾𝛼⁻¹, (3.33) thus 𝑔(𝛼𝛽)=|𝐾𝛼⁻¹|=𝑘, that is, 𝛼𝛽∈𝐺_𝑘.

Theorem 3.12. Suppose that 𝑝≥𝑟≥𝑞≥ℵ₀. Then the following statements hold: (a)𝑆_𝑟⧵𝐺₀ is a maximal subsemigroup of 𝑆_𝑟;(b)if 𝑝>𝑞, then for each cardinal 𝑘 such that 𝑞≤𝑘≤𝑟, 𝑆_𝑟⧵𝐺_𝑘 is a maximal subsemigroup of 𝑆_𝑟.

Proof. By Lemma 3.10, 𝑆_𝑟⧵𝐺₀ is a subsemigroup of 𝑆_𝑟. To see that it is maximal, let 𝛼,𝛽∈𝐺₀=𝐵𝐿(𝑞)⊆𝑆_𝑞. By [3, Theorem  5], 𝑆_𝑞=𝛽⋅𝑅(𝑞), and this implies that 𝛼=𝛽𝛾 for some 𝛾∈𝑅(𝑞)⊆𝑆_𝑟⧵𝐺₀. Hence 𝑆_𝑟⧵𝐺₀ is maximal in 𝑆_𝑟.
Now suppose that 𝑝>𝑞 and let 𝑞≤𝑘≤𝑟. Let 𝛼,𝛽∈𝐺_𝑘. If 𝑘=𝑞, then 𝐺_𝑘=𝑅(𝑞)⊆𝑆_𝑞 and, by [3, Theorem  6], 𝑆_𝑞=𝐵𝐿(𝑞)⋅𝛽⋅𝐵𝐿(𝑞). If 𝑘>𝑞, then 𝐺_𝑘=𝐵𝐿(𝑞)⋅𝛽⋅𝐵𝐿(𝑞) (by Lemma 3.4). Therefore, 𝛼=𝛾𝛽𝜇 for some 𝛾,𝜇∈𝐵𝐿(𝑞)⊆𝑆_𝑟⧵𝐺_𝑘, and so 𝑆_𝑟⧵𝐺_𝑘 is maximal in 𝑆_𝑟.

Corollary 3.13. Suppose that 𝑝>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∪𝑇_𝑞′;(b)𝑁_𝑘={𝛼∈𝑃𝑆(𝑞)∶𝑔(𝛼)≠𝑘} where 𝑘=0 or 𝑞≤𝑘≤𝑝.

Proof. By Theorem 3.9, 𝑆_𝑞∩𝑀_𝐴 is maximal in 𝑆_𝑞. Then Corollary 3.2 implies that (𝑆_𝑞∩𝑀_𝐴)∪𝑇_𝑞′ is maximal in 𝑃𝑆(𝑞). But 𝑆_𝑞∩𝑀_𝐴∪𝑇_𝑞′=𝑆_𝑞∪𝑇_𝑞′∩𝑀_𝐴∪𝑇_𝑞′=𝑃𝑆𝑞∩𝑀_𝐴∪𝑇_𝑞′=𝑀_𝐴∪𝑇_𝑞′, (3.34) and so (a) holds. To show that (b) holds, let 𝑟=𝑝 in Theorem 3.12. Then 𝑆_𝑝=𝑃𝑆(𝑞) and thus 𝑁_𝑘=𝑆_𝑝⧵𝐺_𝑘 is maximal in 𝑃𝑆(𝑞).

Theorem 3.14. Suppose that 𝑝>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 𝑆_𝑞∩𝑀_𝐴 and 𝑆_𝑞⧵𝐺_𝑘 of 𝑆_𝑞 are always disjoint.

Proof. By Theorems 3.9 and 3.12, 𝑆_𝑞∩𝑀_𝐴 and 𝑆_𝑞⧵𝐺_𝑘 are maximal subsemigroups of 𝑆_𝑞. By Lemma 3.6(a), there exists 𝛼∈𝑀_𝐴 with 𝑔(𝛼)=𝑘. Then 𝛼∈𝑆_𝑘∩𝑀_𝐴⊆𝑆_𝑞∩𝑀_𝐴 but 𝛼∉𝑆_𝑞⧵𝐺_𝑘, that is, 𝑆_𝑞∩𝑀_𝐴⊈𝑆_𝑞⧵𝐺_𝑘. Also, 𝑆_𝑞⧵𝐺_𝑘⊈𝑆_𝑞∩𝑀_𝐴 by the maximality of 𝑆_𝑞∩𝑀_𝐴 and 𝑆_𝑞⧵𝐺_𝑘. Therefore, 𝑆_𝑞∩𝑀_𝐴 is not equal to 𝑆_𝑞⧵𝐺_𝑘.

4. Maximal Subsemigroups of 𝑃𝑆(𝑞) When 𝑝=𝑞

We first recall that, when 𝑝=𝑞, the empty transformation ∅ belongs to 𝑃𝑆(𝑞) since 𝑑(∅)=𝑝=𝑞. In this case, the ideals of 𝑃𝑆(𝑞) are precisely the sets: 𝐽_𝑟=𝛼∈𝑃𝑆𝑞∶𝑟(𝛼)<𝑟, (4.1) where 1≤𝑟≤𝑝^′ (see [3, Theorem  14]). Clearly, 𝐽_𝑝′=𝑃𝑆(𝑞) and 𝐽_𝑝={𝛼∈𝑃𝑆(𝑞)∶𝑟(𝛼)<𝑝} is the largest proper ideal. In this case, the complement of each 𝐽_𝑟 in 𝑃𝑆(𝑞) is not a semigroup. To see this, write 𝑋=𝐴∪̇𝐵∪̇𝐶 where |𝐴|=𝑝 and |𝐵|=𝑟=|𝐶|. Then id_𝐵,id_𝐶∈𝑃𝑆(𝑞)⧵𝐽_𝑟 whereas id_𝐵.id_𝐶=∅∈𝐽_𝑟. 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 that 𝑝=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)if 𝑅(𝑞)⊆𝑀, then 𝑀∩B𝐿(𝑞)=∅.

Proof. Suppose that there exists 𝛼∉𝑀 with 𝑟(𝛼)=𝑘<𝑝. Then 𝑔(𝛼)=𝑝, and we write in the usual way 𝑎𝛼=^𝑖𝑥_𝑖. (4.2) Also, write 𝑋⧵{𝑎_𝑖}=𝑃∪̇𝑄 and 𝑋⧵{𝑥_𝑖}=𝑅∪̇𝑆 where |𝑃|=|𝑄|=𝑝=|𝑅|=|𝑆|, and define 𝛽,𝛾 in 𝑃(𝑋) by 𝑎𝛽=^𝑖𝑃𝑎_𝑖𝑃𝑎,𝛾=^𝑖𝑄𝑥_𝑖𝑅, (4.3) where 𝛽|𝑃 and 𝛾|𝑄 are bijections. Then 𝛽,𝛾∈𝑃𝑆(𝑞). Also, 𝛼=𝛽⋅𝛼⋅id_𝑋𝛼∈𝑃𝑆𝑞⋅𝛼⋅𝑃𝑆𝑞, (4.4) 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 that 𝑅(𝑞)⊆𝑀. If there exists 𝛼∈𝑀∩𝐵𝐿(𝑞), then [3, Theorem  5] implies that 𝑃𝑆(𝑞)=𝛼⋅𝑅(𝑞)⊆𝑀 (note that 𝑆_𝑞=𝑃𝑆(𝑞) when 𝑝=𝑞), so 𝑀=𝑃𝑆(𝑞), contrary to the maximality of 𝑀. Thus 𝑀∩𝐵𝐿(𝑞)=∅.

Remark 4.2. If 𝑝>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, 𝑁_𝑘 is a maximal subsemigroup of 𝑃𝑆(𝑞) containing 𝑅(𝑞)∪𝐵𝐿(𝑞), this contrasts with Lemma 4.1(b).

As in Section 3, for any cardinal 𝑘, we let 𝑁_𝑘=𝛼∈𝑃𝑆𝑞∶𝑔(𝛼)≠𝑘. (4.5)

By Lemma 3.10 and Remark 3.11, if 𝑝=𝑞, then 𝑁_𝑘 is a subsemigroup of 𝑃𝑆(𝑞) exactly when 𝑘=0 or 𝑘=𝑝. From Corollary 3.13(b), when 𝑝>𝑞, 𝑁_𝑝 is a maximal subsemigroup of 𝑃𝑆(𝑞). But when 𝑝=𝑞, Lemma 4.1(a) implies that 𝑁_𝑝 is not maximal since ∅∉𝑁_𝑝. Moreover, Lemma 4.1(a) implies that every maximal subsemigroup of 𝑃𝑆(𝑞) must contain the largest proper ideal 𝐽_𝑝=𝛼∈𝑃𝑆𝑞∶𝑟(𝛼)<𝑝. (4.6) Note that 𝐽_𝑝 itself is a subsemigroup of 𝑃𝑆(𝑞), but it is not maximal since 𝐽_𝑝⊊𝑅(𝑞) (in case 𝑝=𝑞, 𝑟(𝛼)<𝑝 implies 𝑔(𝛼)=𝑝).

Theorem 4.3. Suppose that 𝑝=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)𝑁_𝑝∪𝐽_𝑝.

Proof. If 𝑝=𝑞, then 𝑆_𝑞=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 𝑁_𝑝=𝑆_𝑝⧵𝐺_𝑝 is a subsemigroup of 𝑃𝑆(𝑞). Moreover, 𝑁_𝑝∪𝐽_𝑝 is also a subsemigroup of 𝑃𝑆(𝑞) since 𝐽_𝑝 is an ideal. To show the maximality of 𝑁_𝑝∪𝐽_𝑝, let 𝛼,𝛽∈𝑃𝑆(𝑞)⧵(𝑁_𝑝∪𝐽_𝑝). Then 𝑔(𝛼)=𝑔(𝛽)=𝑝=𝑟(𝛼)=𝑟(𝛽). Write in the usual way 𝑎𝛼=^𝑖𝑥_𝑖𝑏,𝛽=^𝑖𝑦_𝑖, (4.7) where |𝐼|=𝑝, and let 𝑋⧵𝑎_𝑖=𝐴∪̇𝐵,𝑋⧵𝑦_𝑖=𝐶∪̇𝐷, (4.8) where |𝐴|=|𝐵|=|𝐶|=|𝐷|=𝑝. Then define 𝛾,𝜇∈𝑃(𝑋) by 𝑏𝛾=^𝑖𝑋⧵𝑏_𝑖𝑎_𝑖𝐴𝑥,𝜇=^𝑖𝑋⧵𝑥_𝑖𝑦_𝑖𝐶, (4.9) where 𝛾|(𝑋⧵{𝑏_𝑖}) and 𝜇|(𝑋⧵{𝑥_𝑖}) are bijections. Thus 𝛾,𝜇∈𝑃𝑆(𝑞) since 𝑑(𝛾)=|𝐵|=𝑝=|𝐷|=𝑑(𝜇). Moreover 𝛾,𝜇∈𝑁_𝑝∪𝐽_𝑝 since 𝑔(𝛾)=𝑔(𝜇)=0<𝑝. It is clear that 𝛽=𝛾𝛼𝜇 and therefore 𝑁_𝑝∪𝐽_𝑝 is maximal in 𝑃𝑆(𝑞).

Remark 4.4. When 𝑝=𝑞, if 𝑀 is a maximal subsemigroup containing 𝑅(𝑞), then 𝑀⊆𝑃𝑆𝑞⧵𝐵𝐿𝑞=𝑁₀ (4.10) 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 Section 3, to see all maximal subsemigroups of 𝑃𝑆(𝑞) when 𝑝>𝑞, it is necessary to describe all maximal subsemigroups of 𝑆_𝑟 where 𝑞≤𝑟<𝑝. So we leave this as a direction for future research.

Acknowledgments

The authors wish to thank the referees for the careful review and the valuable comments, which helped to improve the readability of this paper. B. Singha thanks the Office of the Higher Education Commission, Thailand, for its support by a grant. He also thanks the Graduate School, Chiang Mai University, Chiangmai, Thailand, for its financial support during the preparation of this paper. J. Sanwong thanks the National Research University Project under the Office of the Higher Education Commission, Thailand, for its financial support.