Maximal Regular Subsemibands of the Finite Order-Preserving Partial Transformation Semigroups $\mathcal {PO}(n,r)$

Abstract

Let $\mathcal {PO}_n$ be the semigroup of all order-preserving partial transformations on the finite set $X_n=\{1, 2,\ldots , n\}$. For $1\le r\le n-1$, set $\mathcal {PO}(n, r)=\{\alpha \in \mathcal {PO}_n: |\mathop {\text{ im }}\nolimits (\alpha )|\le r\}$. In this paper, we investigate the maximal regular subsemigroups and the maximal regular subsemibands of the semigroup $\mathcal {PO}(n,r)$. First, we completely describe the maximal regular subsemigroups of the semigroup $\mathcal {PO}(n,r)$, for $1\le r\le n-1$. Secondly, we show that, for $2\le r \le n-2$, any maximal regular subsemigroup of the semigroup $\mathcal {PO}(n,r)$ is a semiband and obtain that the maximal regular subsemigroups and the maximal regular subsemibands of the semigroup $\mathcal {PO}(n,r)$ coincide, for $2\le r\le n-2$. Finally, we obtain the complete classification of maximal regular subsemibands of the semigroup $\mathcal {PO}_n$.

1 Introduction

A subset U of a finite semigroup S is a generating set for S if every element of S may be written as a finite product of elements of U; in this case we write $S = \langle U\rangle$. An element $\varepsilon$ of S is said to be idempotent if $\varepsilon ^2 = \varepsilon$. If S has a generating set consisting of idempotents, then S is said to be idempotent generated or semiband. The latter term was introduced by Pastijn [1]. In the abstract theory of semigroups, idempotents are extremely important in the structure theory of semigroups, both finite and infinite. They help classify different types of semigroups, identify subgroups, determine left or right ideals, and describe the general structure of a given semigroup. Our interest in idempotents stems from the question whether some semigroups are semibands or not.

A large body of work involving semibands arises in the context of the singular (non-invertible) endomorphisms of a structured set. Erdos [2] proved that the semigroup of singular endomorphisms of a finite dimensional vector space is a semiband. Fountain and Lewin [3] proved that the semigroup of singular order-preserving endomorphisms of an independence algebra of finite rank is a semiband, while Oliveira [4] proved a similar result for order-independence algebras.

Let $X_n$ be a finite chain with n elements, say $X_n=\{1<2<\cdots < n\}$. As usual, we denote by $\mathcal {PT}\!_n$ the monoid of all partial transformations of $X_n$ (under composition) and by $\mathcal {T}_n$ the submonoid of $\mathcal {PT}\!_n$ of all full transformations of $X_n$.

We say that a transformation $\alpha \in \mathcal {PT}\!_n$ is order-preserving if $x\le y$ implies $x\alpha \le y\alpha$, for all $x,y\in \mathop {\mathrm {dom}}\nolimits (\alpha )$. We denote by $\mathcal {PO}_n$ the subsemigroup of $\mathcal {PT}\!_n$ of all order-preserving partial transformations (excluding the identity map) and by $\mathcal {O}_n$ the semigroup $\mathcal {PO}_n\cap \mathcal {T}_n$ of all full transformations that preserve the order. Let $\mathcal {M}_n\in \{ \mathcal {O}_n,\mathcal {PO}_n\}$. Set

$$\begin{aligned} \mathcal {M}(n, r)=\{\alpha \in \mathcal {M}_n : |\mathop {\text{ im }}\nolimits (\alpha )|\le r\}. \end{aligned}$$

The sets $\mathcal {M}(n, r)$, with $0\le r\le n-1$, together with the empty set (if necessary), are the two-sided ideals of $\mathcal {M}_n$ and each is a regular subsemigroup of $\mathcal {M}_n$. Various properties of $\mathcal {M}(n, r)$ are known. In particular, Howie [5] showed that the semigroup $\mathcal {O}_n=\mathcal {O}(n,n-1)$ is a semiband and later jointly with Gomes [6] proved that $\mathcal {PO}_n=\mathcal {PO}(n,n-1)$ is also a semiband. Garba [7] proved that, for $2\le r\le n-2$, $\mathcal {M}(n,r)$ is a semiband. We observe that the semigroup $\mathcal {O}_n$ is a semiband but its maximal (regular) subsemigroups in general need not be a semiband (see [8, Theorem 3.1 and Theorem 4.1]). Dimitrova and Koppitz [9] classified completely maximal regular subsemigroups of $\mathcal {O}(n,r)$, for $2\le r\le n-2$. Zhao [10] showed that, for $2\le r \le n-2$, any maximal regular subsemigroup of the semigroup $\mathcal {O}(n,r)$ is a semiband and obtained that the maximal regular subsemigroups and the maximal regular subsemibands of $\mathcal {O}(n,r)$ coincide, for $2\le r\le n-2$ (see [10, Theorem 2.26]). In this paper, we investigate the maximal regular subsemigroups and the maximal regular subsemibands of $\mathcal {PO}(n,r)$. First, we completely describe the maximal regular subsemigroups of the semigroup $\mathcal {PO}(n,r)$, for $1\le r\le n-1$. Secondly, using a similar approach from Zhao [10], we show that, for $2\le r \le n-2$, any maximal regular subsemigroup of the semigroup $\mathcal {PO}(n,r)$ is a semiband and obtain that the maximal regular subsemigroups and the maximal regular subsemibands of $\mathcal {PO}(n,r)$ coincide, for $2\le r\le n-2$. Finally, we obtain the complete classification of maximal regular subsemibands of $\mathcal {PO}_n$.

Let $\alpha \in \mathcal {PO}_n$. As usual, we write $\mathop {\text{ im }}\nolimits (\alpha )$ and $\mathop {\mathrm {dom}}\nolimits (\alpha )$ for the image of $\alpha$ and domain of $\alpha$, respectively. The kernel of $\alpha$ is the equivalence $\mathop {\text{ ker }}\nolimits (\alpha )=\{(x,y)\in \mathop {\mathrm {dom}}\nolimits (\alpha )\times \mathop {\mathrm {dom}}\nolimits (\alpha ): x\alpha =y\alpha \}$. Given a subset U of $\mathcal {PO}_n$, we denote by E(U) its set of idempotents. We denote by $V(\alpha )$ the set of all inverses of $\alpha$, and by $L_{\alpha }$ , $R_{\alpha }$, $H_{\alpha }$, and $J_{\alpha }$ the ${\mathscr {L}}$-class, ${\mathscr {R}}$-class, ${\mathscr {H}}$-class, and ${\mathscr {J}}$-class, respectively, of an element $\alpha \in \mathcal {PO}_n$. For general background on Semigroup Theory, we refer the reader to Howie’s book [11].

2 Main Result

Recall that the Green’s relations on $\mathcal {PO}_n$ can be characterized as $\alpha {\mathscr {L}} \beta$ if and only if $\mathop {\text{ im }}\nolimits (\alpha )=\mathop {\text{ im }}\nolimits (\beta )$, $\alpha {\mathscr {R}} \beta$ if and only if $\mathop {\text{ ker }}\nolimits (\alpha )=\mathop {\text{ ker }}\nolimits (\beta )$, and $\alpha {\mathscr {J}} \beta$ if and only if $|\mathop {\text{ im }}\nolimits (\alpha )|=|\mathop {\text{ im }}\nolimits (\beta )|$, for every transformations $\alpha$ and $\beta$. The semigroup $\mathcal {PO}_n$ has trivial ${\mathscr {H}}$-classes and has n ${\mathscr {J}}$-classes: $J_0,J_1,\ldots ,J_{n-1}$, where $J_0$ consists of the empty mapping and $J_r=\{\alpha \in \mathcal {PO}_{n}: |\mathop {\text{ im }}\nolimits (\alpha )|=r\}$. Clearly, $\mathcal {PO}(n, r)=J_0\cup J_1\cup \dots \cup J_r$.

It is well known that $\mathcal {PO}_n$ is a regular semiband and $\mathcal {PO}(n,r)$ ($2\le r\le n-1$) is generated by idempotents of rank r (see [7, Lemma 3.3 and Proposition of Page 195] and [6, Lemma 3.14]). Notice that $\mathcal {PO}(n,r)$ is an ideal of $\mathcal {PO}_n$. From the result [12], Corollary 1.4] that every ideal of a regular semiband S is also a regular semiband, we immediately deduce:

Lemma 1

Let $2\le r\le n-1$. Then $\mathcal {PO}(n,r)$ is a regular subsemiband of $\mathcal {PO}_n$ and $\mathcal {PO}(n,r)=\langle E(J_r) \rangle$.

The following lemma is from Clifford and Preston [13, Theorems 2.17 and 2.18].

Lemma 2

(1) For any two elements a, b in a semigroup S, $ab\in R_a\cap L_b$ if and only if $E(R_b\cap L_a)\ne \emptyset$. (2) Let a be a regular element of a semigroup S. (i) Every inverse of a lies in $D_a$. (ii) An ${\mathscr {H}}$-class $H_b$ contains an inverse of a if and only if both of the ${\mathscr {H}}$-classes $R_a \cap L_b$ and $R_b\cap L_a$ contain idempotents.

Let $Q_1, Q_2, \dots , Q_m$ be all subsets of $X_n$ with cardinality r, where $m={n\atopwithdelims ()r}$, and let

$$\begin{aligned} R(Q_i)=\{ \alpha \in J_r: \mathop {\mathrm {dom}}\nolimits (\alpha )=Q_i\}, \end{aligned}$$

where $1\le i\le m$. Then $R(Q_1)$, $R(Q_2)$, $\dots$, $R(Q_m)$ are some of the ${\mathscr {R}}$-classes of $J_r$. Clearly, $|E(R(Q_j))|=1$ for all $1\le j\le m$. We denote the unique idempotent of $R(Q_j)$ by $\zeta _j$. Then $\zeta _j$ is the identity mapping on $Q_j$.

Lemma 3

Let $2\le r\le n-1$. Let S be a regular subsemigroup of $\mathcal {PO}(n,r)$. If $S\cap R_{\alpha }\ne \emptyset$, for all  $\alpha \in J_r$, then $S=\mathcal {PO}(n,r)$.

Proof

We claim that

$$\begin{aligned} S\cap L_{\beta }\ne \emptyset , \ for \ all \ \beta \in J_r. \end{aligned}$$

(2.1)

Otherwise, if there exist $\beta \in J_r$ such that $L_{\beta }\subseteq J_r\backslash S$. Let $\mathop {\text{ im }}\nolimits (\beta )=Q_i$, then $\zeta _i \in E(R(Q_i))\cap L_{\beta }$ and so $\zeta _i\notin S$. Notice that $R_{\zeta _i}=R(Q_i)$ and $|E(R(Q_i))|=1$. This will yield $S\cap R_{\zeta _i}=\emptyset$ (otherwise, since S is regular, we have $\zeta _i \in S$). Notice that S is regular. By condition and (2.1), we have

$$\begin{aligned} S\cap E(R_{\alpha })\ne \emptyset , \quad S\cap E(L_{\beta })\ne \emptyset , {\text{ for } \text{ all }}\, \alpha , \beta \in J_r. \end{aligned}$$

(2.2)

We show that $E(J_r)\subseteq E(S)$ and so $S=\mathcal {PO}(n,r)$ by Lemma 1. Suppose that $e\in E(J_r) \backslash E(S)$. Notice that ${\mathscr {D}} ={\mathscr {J}}$ in every finite semigroup. By (2.2), we can choose $f\in E(S)\cap L_{e}$ and $g\in E(S)\cap R_{e}$. Since $e\notin E(S)$, we have $e\ne f$ and $e\ne g$. By Lemma 2, we have $fg\in S\cap R_f\cap L_g$ (since $e\in E(L_f\cap R_g)$). As usual, by ${\mathscr {L}}^S$, ${\mathscr {R}}^S$ denote Green relations of the subsemigroup S of $\mathcal {PO}(n,r)$. Since S is regular, we have that ${\mathscr {R}}^S={\mathscr {R}}\cap (S \times S)$ and ${\mathscr {L}}^S={\mathscr {L}}\cap (S\times S)$. Then $f{\mathscr {R}}^Sfg{\mathscr {L}}^Sg$ (since $f{\mathscr {R}}fg{\mathscr {L}}g$). Notice that $f,g\in E(S)$ and $fg\in S$. By Lemma 2, there exist $\delta \in V(fg)$ such that $\delta \in L^S_f\cap R^S_g$. Since $f{\mathscr {L}} e{\mathscr {R}} g$, we have $\delta \in L^S_f\cap R^S_g \subseteq S\cap L_f\cap R_g=S\cap L_e\cap R_e=S\cap H_e$. Then $e=\delta \in S$ (since every ${\mathscr {H}}$-class of $\mathcal {PO}(n,r)$ is trivial), which contradicts $e\in E(J_r) \backslash E(S)$. Thus $E(J_r)\subseteq E(S)$. $\square$

Let $2\le r\le n-1$. A proper subsemigroup S of $\mathcal {PO}(n,r)$ is called a maximal regular subsemigroup if S is a regular semigroup, and any regular subsemigroup of $\mathcal {PO}(n, r)$ properly containing S must must be $\mathcal {PO}(n, r)$.

Our first main result is:

Theorem 4

Let $2\le r\le n-1$. Then any maximal regular subsemigroups of the semigroup $\mathcal {PO}(n,r)$ are of the form: $\mathcal {PO}(n,r-1)\cup (J_r\backslash R_{\alpha })$, for some $\alpha \in J_r$.

Proof

Let $\alpha \in J_r$, and let $M_{\alpha }=\mathcal {PO}(n,r-1)\cup (J_r\backslash R_{\alpha })$. We shall show that $M_\alpha$ is a maximal regular subsemigroup of $\mathcal {PO}(n, r)$. For any $\beta ,\gamma \in J_r$, either $\beta \gamma {\mathscr {R}} \beta$ or $\beta \gamma \in \mathcal {PO}(n,r-1)$ by Lemma 2, so $M_\alpha$ is a subsemigroup of $\mathcal {PO}(n, r)$. From Lemma 1, we know that $\mathcal {PO}(n,r-1)$ is regular. Let $\beta \in J_r\backslash R_{\alpha }$. Suppose that $\mathop {\text{ im }}\nolimits (\beta )=\{b_1<\dots <b_r\}$. Let $e_1$ be the identity mapping on $\mathop {\text{ im }}\nolimits (\beta )$, and let

$$\begin{aligned} e_2 =\left( \begin{array}{llll} B_1 &{}\quad B_2 &{}\quad \cdots &{}\quad B_r\\ b_1 &{}\quad b_2 &{}\quad \cdots &{}\quad b_r \end{array} \right) , \end{aligned}$$

where $B_1=\{1,\dots , b_1\}$, $B_i=\{b_{i-1}+1,\dots , b_i\}$, for $2\le i\le r-1$, and $B_r=\{b_{r-1}+1,\dots ,n\}$. Clearly, $e_1,e_2\in E(L_\beta )$ and $(e_1,e_2)\notin {\mathscr {R}}$. Since $\mathcal {PO}_n$ is regular, we have $|E(R_\beta )|\ge 1$. Let $f\in E(R_\beta )$. Then $e_i{\mathscr {L}} \beta {\mathscr {R}} f$. By Lemma 2, we have that $R_{e_i}\cap L_f$ contains inverse of $\alpha$. Let $\gamma _i\in R_{e_i}\cap L_f\cap V(\alpha )$, $i=1,2$, then $e_i{\mathscr {R}}\gamma _i{\mathscr {L}}f$ and so $(\gamma _1,\gamma _2)\notin {\mathscr {R}}$. It follows that $\gamma _1 \in J_r\backslash R_{\alpha }\subseteq M_{\alpha }$ or $\gamma _2 \in J_r\backslash R_{\alpha }\subseteq M_{\alpha }$. Then $\beta$ is regular and so $M_{\alpha }$ is a regular subsemigroup of $\mathcal {PO}(n,r)$.

Suppose T is a regular subsemigroup of $\mathcal {PO}(n,r)$ properly containing $M_\alpha$. Then $T\cap R_\beta \ne \emptyset$, for all $\beta \in J_r$. Thus, by Lemma 3, $T=\mathcal {PO}(n,r)$ and so $M_\alpha$ is a maximal regular subsemigroup of $\mathcal {PO}(n,r)$.

Conversely, suppose M is a maximal regular subsemigroup of $\mathcal {PO}(n,r)$. Then there exist $\alpha \in J_r$ such that $M\cap R_\alpha =\emptyset$ (otherwise, by Lemma 3, we have $M=\mathcal {PO}(n,r)$) and so $M\subseteq \mathcal {PO}(n,r-1) \cup (J_r\backslash R_{\alpha })=M_\alpha$. Thus, by the maximality of M, $M=M_\alpha$.

$\square$

Let A be a subset of $X_n$. We say that a subset C of A is convex on A if

$$\begin{aligned} x, y \in C , z\in A\ and \ x\le z \le y \Rightarrow z\in C. \end{aligned}$$

We shall refer to an equivalence $\pi$ on the subset A of $X_n$ as convex if its classes are convex subsets of A, and we shall say that $\pi$ is of weight r on A if $|\pi / A|= r$. Now, we consider the top class $J_{r}$ of $\mathcal {PO}(n,r)$. We denote by $\Lambda _r$ the collection of all subsets of $X_n$ of cardinality r. Let $A \in \Lambda _r$. Then a typical ${\mathscr {L}}$-class in $J_r$ may be denoted by $L_A=\{\alpha \in \mathcal {PO}(n,r): \mathop {\text{ im }}\nolimits (\alpha )=A \}$. The number of ${\mathscr {L}}$-classes within $J_r$ is the number of image sets in $X_n$ of cardinality r, namely, $n\atopwithdelims ()r$. We denote by $\Omega _r$ the collection of all convex equivalences of weight r on all subsets of $X_n$. Let $\pi \in \Omega _r$. Then a typical ${\mathscr {R}}$-class in $J_r$ may be denoted by $R_\pi =\{\alpha \in \mathcal {PO}(n,r): \mathop {\text{ ker }}\nolimits (\alpha )=\pi \}$. Thus, $J_r$ has $\Sigma _{k=r}^n{n\atopwithdelims ()k}{k-1\atopwithdelims ()r-1})$ ${\mathscr {R}}$-classes corresponding to the $\Sigma _{k=r}^n{n\atopwithdelims ()k}{k-1\atopwithdelims ()r-1})$ convex equivalences of weight r on all subsets of $X_n$, and $n\atopwithdelims ()r$ ${\mathscr {L}}$-classes corresponding to the $n\atopwithdelims ()r$ subsets of $X_n$ of cardinality r. It follows that $J_r$ has ${n\atopwithdelims ()r}(\Sigma _{k=r}^n{n\atopwithdelims ()k}{k-1\atopwithdelims ()r-1})$ ${\mathscr {H}}$-classes. Also we may write $H_{(\pi ,A)}$ for the ${\mathscr {H}}$-class, which is the intersection of $R_\pi$ and $L_A$. A is said to be a transversal of $\pi$ if $|A\cap \tilde{x}| = 1$ for every equivalence class $\tilde{x}$ of $\pi$. Then $H_{(\pi ,A)}$ is a group ${\mathscr {H}}$-class if and only if A is a transversal of $\pi$. Notice that every ${\mathscr {H}}$-class of $\mathcal {PO}(n,r)$ is trivial. Thus every group $\mathscr {H}$-class consists of an idempotent.

The following lemma is from G.U. Garba [7, Lemma 3.3 and Proposition of Page 195]:

Lemma 5

There is a way of listing the subsets of $X_n$ of cardinality r as $A_1, A_2, \ldots , A_m$ (where $m={n \atopwithdelims ()r}$ and $r\ge 2$) so that there exist distinct convex equivalences $\pi _1, \pi _2, \ldots , \pi _m$ of weight r with the property that $A_{i-1}, A_i$ are both transversals of $\pi _{i}$ ($i=2,\ldots , m$) and $A_m$, $A_1$ are traversals of $\pi _1$. Each ${\mathscr {H}}$-class $H_{(\pi _i,A_i)}$ consists of an idempotent $\varepsilon _i$ $(i=1,\ldots , m)$, ${\mathscr {H}}$-class $H_{(\pi _{i},A_{i-1})}$ consists of an idempotent $\eta _{i}$ $(i=2,\ldots , m)$ and ${\mathscr {H}}$-class $H_{(\pi _{1},A_{m})}$ consists of an idempotent $\eta _1$, and there exist idempotents $\varepsilon _{m+1},\dots , \varepsilon _{p}$ (where $p=\Sigma _{k=r}^n{n\atopwithdelims ()k}{k-1\atopwithdelims ()r-1})$ such that $\varepsilon _{1}, \varepsilon _2, \dots , \varepsilon _{p}$ cover all the ${\mathscr {R}}$-classes in $J_r$ and $\{\varepsilon _{1}, \varepsilon _2, \dots , \varepsilon _{p}\}$ is a set of generators for $\mathcal {PO}(n,r)$.

Before presenting our next lemma, we introduce the following notation.

Let $A_i$, $\pi _i$, $\varepsilon _i$, and $\eta _i$ be as defined in Lemma 5, and let $\pi _i=\mathop {\text{ ker }}\nolimits (\varepsilon _i)$, for $i=m+1,\dots , p$. Then $\Omega _r=\{\pi _1, \pi _2, \ldots , \pi _p\}$ $(p=\Sigma _{k=r}^n{n\atopwithdelims ()k}{k-1\atopwithdelims ()r-1})$ and $\Lambda _r=\{ A_{1}, A_{2}, \dots , A_m\}$ ($m={n \atopwithdelims ()r}$). Let $\Sigma =\cup ^m_{i,j=1}H_{(\pi _i,A_j)}$, $G^+(\Sigma )=\{\varepsilon _1,\varepsilon _2,\ldots , \varepsilon _m\}$, $G^-(\Sigma )=\{\eta _1,\eta _2,\ldots , \eta _m\}$, and $G(\Sigma )=G^+(\Sigma )\cup G^-(\Sigma )$.

As in [14], let $k\in \{0,1,2,\ldots ,m\}$ and define a total order $\le _k$ on the set $\{1,\ldots , m\}$ by

$$\begin{aligned} k+1\le _{k}k+2\le _{k}\cdots \le _{k}m\le _{k}1\le _{k}\cdots \le _{k} k, \end{aligned}$$

where $i<_{k}j$ if $i\le _{k}j$ and $i\ne j$. In the following lemma, it will always be clear from context when additions are modular m.

Lemma 6

$\Sigma \subseteq \langle G(\Sigma )\rangle$.

Proof

Let

$$\begin{aligned} \alpha _{ij}= & {} \eta _i\eta _{i-1}\cdots \eta _{j+1}, 1\le j<i\le m,\\ \beta _{ij}= & {} \varepsilon _i\varepsilon _{i+1}\cdots \varepsilon _j, 1\le i\le j\le m. \quad \end{aligned}$$

Then, by Lemmas 2 and 5, $\alpha _{ij}\in \langle G^-(\Sigma )\rangle \cap H_{(\pi _i,A_j)}$ and $\beta _{ij}\in \langle G^+(\Sigma )\rangle \cap H_{(\pi _i,A_j)}$. Since $\mathcal {PO}_n$ is ${\mathscr {H}}$-trivial, we have

$$\begin{aligned} H_{(\pi _i,A_j)}= & {} \{\alpha _{ij}\}\subseteq \langle G^-(\Sigma )\rangle , 1\le j<i\le m,\\ H_{(\pi _i,A_j)}= & {} \{\beta _{ij}\}\subseteq \langle G^+(\Sigma )\rangle , 1\le i\le j\le m. \end{aligned}$$

Notice that $\Sigma =\cup ^m_{i,j=1}H_{(\pi _i,A_j)}$. It follows immediately that $\Sigma \subseteq \langle G^+(\Sigma )\cup G^-(\Sigma )\rangle =\langle G(\Sigma )\rangle$. $\square$

Lemma 7

$\Sigma \backslash R_{\pi _k}\subseteq \langle E(J_r\backslash R_{\pi _k})\rangle$, for all $1\le k\le m$.

Proof

Let $\pi _i$, $A_i$, $\varepsilon _i$, and $\eta _i$ be defined as before. Let

$$\begin{aligned} \alpha ^{[k]}_{i,j}= & {} \eta _i\eta _{i-1}\cdots \eta _{j}, \quad j\le _k i <_k k,\\ \beta ^{[k]}_{i,j}= & {} \varepsilon _i\varepsilon _{i+1}\cdots \varepsilon _{j}, \quad i\le _k j<_k k. \end{aligned}$$

Notice that $G(\Sigma )\backslash \{\varepsilon _k, \eta _{k}\}\subseteq E(J_r\backslash R_{\pi _k})$. Then, by Lemmas 2 and 5,

$$\begin{aligned}&\alpha ^{[k]}_{i,j} \in \langle G^-(\Sigma )\backslash \{\eta _{k}\}\rangle \cap H_{(\pi _i,A_{j-1})}\subseteq \langle E(J_r\backslash R_{\pi _k})\rangle \cap H_{(\pi _i,A_{j-1})},\\&\beta ^{[k]}_{i,j} \in \langle G^+(\Sigma )\backslash \{\varepsilon _{k}\}\rangle \cap H_{(\pi _i,A_j)}\subseteq \langle E(J_r\backslash R_{\pi _k})\rangle \cap H_{(\pi _i,A_j)}. \qquad \end{aligned}$$

Since $\mathcal {PO}_n$ is ${\mathscr {H}}$-trivial, we have

$$\begin{aligned} H_{(\pi _i, A_{j-1})}= & {} \{\alpha ^{[k]}_{i,j}\}\subseteq \langle E(J_r\backslash R_{\pi _k})\rangle , j\le _k i <_k k,\\ H_{(\pi _i, A_{j})}= & {} \{\beta ^{[k]}_{i,j}\}\subseteq \langle E(J_r\backslash R_{\pi _k})\rangle , i\le _k j <_k k. \end{aligned}$$

It follows immediately that $\Sigma \backslash R_{\pi _k} = \mathop {\mathop {\cup }\nolimits _{i,j=1}}\limits _{i\ne k}^m H_{(\pi _i,A_j)}= [(\cup _{j\le _k i <_k k}H_{(\pi _i,A_{j-1})})\cup (\cup _{i\le _k j <_k k}H_{(\pi _i,A_j)})] \subseteq \langle E(J_r\backslash R_{\pi _k}) \rangle$. $\square$

Lemma 8

Let $2\le r\le n-2$. Then $J_r\backslash R_{\alpha }=\langle E(J_r\backslash R_{\alpha })\rangle \cap J_r$, for all $\alpha \in J_r$.

Proof

Let $\pi _i$, $A_i$, $\varepsilon _i$, and $\eta _i$ be defined as before. Notice that $\Omega _r=\{\pi _1, \pi _2, \ldots , \pi _p \}$ ($p=\Sigma _{k=r}^n{n\atopwithdelims ()k}{k-1\atopwithdelims ()r-1}$) and $\Lambda _r=\{ A_{1}, A_{2}, \dots , A_m \}$ ($m={n\atopwithdelims ()r}$). Let $\widetilde{\Sigma }=J_r\backslash \Sigma$. Then $\widetilde{\Sigma }=\cup ^{p}_{i=m+1}\cup ^m_{j=1}H_{(\pi _i,A_j)}$. Let $\alpha \in J_r$ and $\mathop {\text{ ker }}\nolimits (\alpha )=\pi$, then there exist $k\in \{1,2,\dots , p\}$ such that $\pi =\pi _k$. Clearly, $R_{\alpha }=R_{\pi _k}$ and $J_r \backslash R_{\alpha }=J_r \backslash R_{\pi _k}$. We first prove that $J_r \backslash R_{\alpha }\subseteq \langle E(J_r \backslash R_{\alpha })\rangle \cap J_r$, i.e., $J_r \backslash R_{\pi _k}\subseteq \langle E(J_r \backslash R_{\pi _k})\rangle \cap J_r$. We distinguish two cases:

Case 1 $1\le k\le m$. Clearly $J_r \backslash R_{\pi _k}=(\Sigma \backslash R_{\pi _k})\cup \widetilde{\Sigma }$. We shall show that $\widetilde{\Sigma }\subseteq \langle E(J_r \backslash R_{\pi _k})\rangle$. Let $\alpha \in \widetilde{\Sigma }$. Then $\alpha \in H_{(\pi _i,A_j)}$ for some $i\in \{m+1,\dots , p\}$, $j\in \{1, \dots , m\}$. Notice that $\alpha {\mathscr {R}} \varepsilon _i$. Suppose that $\varepsilon _i\in H_{(\pi _i,A_s)}$ for some $s\in \{1,\dots ,m\}$. Let

$$\begin{aligned} \sigma _s=\left\{ \begin{array}{lll} \eta _{s+1} &{}\quad s=k\\ \varepsilon _s &{}\quad s\ne k. \end{array} \right. \end{aligned}$$

Then $\varepsilon _i {\mathscr {L}} \sigma _s$ and $\sigma _s\in \Sigma \backslash R_{\pi _k}$. Let $\beta \in R_{\sigma _s}\cap L_\alpha$. Clearly $\sigma _s\in E(L_{\varepsilon _i}\cap R_{\beta })$ and $\beta \in \Sigma \backslash R_{\pi _k}$. Notice that $\mathcal {PO}_n$ is ${\mathscr {H}}$-trivial and $\varepsilon _i\in E(\widetilde{\Sigma })\subseteq E(J_r \backslash R_{\pi _k})$. Then, by Lemmas 2 and 7,

$$\begin{aligned} \alpha =\varepsilon _i\beta \in E(J_r \backslash R_{\pi _k})\cdot \Sigma \backslash R_{\pi _k}\subseteq E(J_r \backslash R_{\pi _k})\cdot \langle E(J_r \backslash R_{\pi _k})\rangle \subseteq \langle E(J_r \backslash R_{\pi _k})\rangle \end{aligned}$$

and so $\widetilde{\Sigma }\subseteq \langle E(J_r \backslash R_{\pi _k})\rangle$. Thus, by Lemma 7, $J_r \backslash R_{\pi _k}=(\Sigma \backslash R_{\pi _k})\cup \widetilde{\Sigma }\subseteq \langle E(J_r \backslash R_{\pi _k})\rangle$ and so $J_r \backslash R_{\pi _k}=(J_r \backslash R_{\pi _k})\cap J_r\subseteq \langle E(J_r \backslash R_{\pi _k})\rangle \cap J_r$.

Case 2 $m+1\le k\le p$. Clearly $J_r \backslash R_{\pi _k}=\Sigma \cup (\widetilde{\Sigma }\backslash R_{\pi _k})$. We shall show that $\widetilde{\Sigma }\backslash R_{\pi _k}\subseteq \langle E(J_r \backslash R_{\pi _k})\rangle$. Let $\alpha \in \widetilde{\Sigma }\backslash R_{\pi _k}$. Then $\alpha \in H_{(\pi _i,A_j)}$ for some $i\in \{m+1,\dots , p\}\backslash \{k\}$, $j\in \{1, \dots , m\}$. Notice that $\alpha {\mathscr {R}} \varepsilon _i$. Suppose that $\varepsilon _i\in H_{(\pi _i,A_s)}$ for some $s\in \{1,\dots ,m\}$. Then $\varepsilon _i{\mathscr {L}}\varepsilon _s$ and $\varepsilon _s\in \Sigma$. Let $\beta \in R_{\varepsilon _s}\cap L_\alpha$. Clearly $\varepsilon _s\in E(L_{\varepsilon _i}\cap R_{\beta })$ and $\beta \in \Sigma$. Notice that $\varepsilon _i\in E(\widetilde{\Sigma }\backslash R_{\pi _k})\subseteq E(J_r\backslash R_{\pi _k})$, $G(\Sigma )\subseteq E(J_r\backslash R_{\pi _k})$, and $\mathcal {PO}_n$ is ${\mathscr {H}}$-trivial. Then, by Lemmas 2 and 7,

$$\begin{aligned} \alpha =\varepsilon _i\beta \in E(J_r \backslash R_{\pi _k})\cdot \Sigma \subseteq E(J_r \backslash R_{\pi _k})\cdot \langle G(\Sigma )\rangle \subseteq \langle E(J_r \backslash R_{\pi _k})\rangle \end{aligned}$$

and so $\widetilde{\Sigma }\backslash R_{\pi _k}\subseteq \langle E(J_r \backslash R_{\pi _k})\rangle$. Thus, by Lemma 6, $J_r \backslash R_{\pi _k}=\Sigma \cup (\widetilde{\Sigma }\backslash R_{\pi _k})\subseteq \langle G(\Sigma )\rangle \cup \langle E(J_r \backslash R_{\pi _k})\rangle \subseteq \langle E(J_r \backslash R_{\pi _k})\rangle$ and so $J_r \backslash R_{\pi _k}=(J_r \backslash R_{\pi _k})\cap J_r\subseteq \langle E(J_r \backslash R_{\pi _k})\rangle \cap J_r$.

It remains to prove that $\langle E(J_r \backslash R_{\alpha } )\rangle \cap J_r\subseteq J_r \backslash R_{\alpha }$. Let $S=\mathcal {PO}(n,r-1)\cup (J_r \backslash R_{\alpha })$. From Theorem 4, we know that S is a subsemigroup of $\mathcal {PO}(n,r)$. It is obvious that $E(J_r \backslash R_{\alpha })\subseteq S$. Then $\langle E(J_r \backslash R_{\alpha }) \rangle \subseteq S$ and so $\langle E(J_r \backslash R_{\alpha })\rangle \cap J_{r}\subseteq S\cap J_{r}=J_r \backslash R_{\alpha }$. $\square$

Our next main result is:

Theorem 9

Let $2\le r\le n-2$. Then any maximal regular subsemigroup of $\mathcal {PO}(n,r)$ is a semiband.

Proof

From Theorem 4, we know that each maximal regular subsemigroup of $\mathcal {PO}(n,r)$ must be in the following form: $\mathcal {PO}(n,r-1) \cup (J_r\backslash R_\alpha )$, for some $\alpha \in J_r$. By Lemmas 1 and 8, we have

$$\begin{aligned} \mathcal {PO}(n,r-1) \cup (J_r\backslash R_\alpha )=\langle E(J_{r-1})\rangle \cup (\langle E(J_r\backslash R_\alpha )\rangle \cap J_r)=\langle E(J_{r-1}) \cup E(J_r\backslash R_\alpha )\rangle . \end{aligned}$$

Then any maximal regular subsemigroup of $\mathcal {PO}(n,r)$ is a semiband. $\square$

As in [7], we shall refer to an element $\alpha$ in $\mathcal {PO}_n$ as being of type (k, r) or belonging to the set [k, r] if $|\mathop {\mathrm {dom}}\nolimits (\alpha )|=k$, $|\mathop {\text{ im }}\nolimits (\alpha )|=r$. Clearly $J_r=\cup ^n_{i=r}[i,r]$ and $J_{n-1}=[n,n-1]\cup [n-1,n-1]$. We draw attention to the ${\mathcal {J}}$-class $J_{n-1}$. Gomes and Howie [6] used the notation $[i\rightarrow i+1]$ for the increasing idempotent $\varepsilon$ defined by $i\varepsilon =i+1$, $x\varepsilon =x$ $(x\ne i)$ and the notation $[i\rightarrow i-1]$ for the decreasing idempotent $\eta$ defined by $i\eta =i-1$, $x\eta =x$ $(x\ne i)$. They also used the notation $\delta _{k}$ for the identity mapping on $X_n\backslash \{k\}$. Let $E^+_{n-1}=\{[i\rightarrow i+1]: 1\le i\le n-1\}$ and $E^-_{n-1}=\{[i\rightarrow i-1]: 2\le i\le n\}$ be the increasing and decreasing idempotent sets of $[n,n-1]$, respectively. Let $E_{n-1}=E^+_{n-1}\cup E^-_{n-1}$, and let $F_{n-1}=\{\delta _1, \ldots , \delta _n \}$. Then $E([n,n-1])=E_{n-1}$ and $E([n-1,n-1])=F_{n-1}$. For convenience, we use $[n\rightarrow n+1]$ or $[1\rightarrow 0]$ to denote $\theta$ (the empty mapping).

For any $i,j\in X_n$, let

$$\begin{aligned} M_{i,j}= & {} \{\alpha \in \mathcal {O}_n : (\forall x,y \in X_n ) \ x \le i \Longrightarrow x\alpha \le i, \ y \ge j \Longrightarrow y\alpha \ge j\}, \end{aligned}$$

(2.3)

$$\begin{aligned} PM_{i,j}= & {} \{ \alpha \in \mathcal {PO}_n : (\forall x,y \in \mathop {\mathrm {dom}}\nolimits (\alpha ) ) \ x \le i \Longrightarrow x\alpha \le i, \ y \ge j \Longrightarrow y\alpha \ge j \} \cup J_0.\nonumber \\ \end{aligned}$$

(2.4)

Let $\alpha \in \mathcal {O}_n$ and fix some $i\in X_n$. It is easy to prove that

$$\begin{aligned}&i\alpha \le i\Leftrightarrow (x\in X_n) x\le i \Rightarrow x\alpha \le i,\\&i\alpha \ge i\Leftrightarrow (y\in X_n) y\ge i \Rightarrow y\alpha \ge i. \end{aligned}$$

Notice that $[n,n-1]=\mathcal {O}_n\cap J_{n-1}$. From the above fact and [15, Lemma 2.5], we easily deduce the following:

Lemma 10

Let $n\ge 3$. Then $M_{i,j}=\langle E([n,n-1])\backslash \{[i\rightarrow i+1], [j\rightarrow j-1]\}\rangle$, for all $i,j \in X_n$.

Let $1_{X_n}$ be the identity mapping on $X_n$, and let $\mathcal {O}^1_n=\mathcal {O}_n\cup \{1_{X_n}\}$. For a strictly partial transformation $\beta =\left( \begin{array}{l} A \\ a \end{array} \right) \in \mathcal {PO}_n$, let $c=min (A\cup \{a\})$ and $d=max(A\cup \{a\})$. Now, we define the completion $\beta ^*$ of $\beta$ in $\mathcal {O}^1_n$ as follows:

$$\begin{aligned} x\beta ^*=\left\{ \begin{array}{ll} a, &{} \quad c \le x \le d, \\ x, &{} \quad otherwise. \end{array} \right. \end{aligned}$$

Clearly $\beta ^*=1_{X_n}$ if and only if $A=\{a\}$.

Lemma 11

Let $n\ge 3$. Then $PM_{i,j}=\langle E(J_{n-1})\backslash \{[i\rightarrow i+1], [j\rightarrow j-1]\}\rangle$, for all $i,j \in X_n$.

Proof

Let $\tilde{E}=E(J_{n-1})\backslash \{[i\rightarrow i+1], [j\rightarrow j-1]\}$. We first show that $PM_{i,j} \subseteq \langle \tilde{E} \rangle$. It is obvious that $\theta =\delta _1\delta _2\cdots \delta _n$. Notice that $F_{n-1} \subseteq \tilde{E}$. Thus $J_0=\{\theta \}\subseteq \langle \tilde{E}\rangle$. Let $\alpha \in PM_{i,j} \backslash J_0$. We distinguish three cases.

Case 1 $\alpha \in \mathcal {O}_n$. Clearly $\alpha \in M_{i,j}$. By Lemma 10, we have

$$\begin{aligned} \alpha \in M_{i,j}=\langle E([n,n-1])\backslash \{ [i \rightarrow i+1], [j\rightarrow j-1]\} \rangle \subseteq \langle \tilde{E} \rangle . \end{aligned}$$

Case 2 $\alpha$ is an identity mapping in [r, r], where $1 \le r \le n-1$. Now, put

$$\begin{aligned} X_n\backslash \mathop {\mathrm {dom}}\nolimits (\alpha )=\{x_1,x_2,\ldots ,x_{n-r}\}, \end{aligned}$$

and define $e_k$ to be the identity mapping on $X_n\backslash \{x_k\}$, then $e_k\in F_{n-1}\subseteq \tilde{E}$, for all $k\in \{1,\dots , n-r\}$. It is obvious that $\alpha =e_1e_2\cdots e_{n-r}$. Thus $\alpha \in \langle \tilde{E}\rangle$.

Case 3 $\alpha$ is in [r, s] ($1 \le s\le r \le n-1$), but $\alpha$ is not an identity mapping. Suppose that

$$\begin{aligned} \alpha =\left( \begin{array}{llll} A_1 &{}A_2 &{} \cdots &{} A_s \\ a_1 &{} a_2 &{}\cdots &{} a_s \end{array} \right) \in [r,s]. \end{aligned}$$

Let $\alpha ^*_k$ be a completion of $\left( \begin{array}{l} A_k \\ a_k \end{array} \right)$ in $\mathcal {O}^1_n$. We shall prove that

$$\begin{aligned} \alpha ^*_1, \alpha ^*_2, \ldots , \alpha ^*_s\in M_{i,j} \cup \{1_{X_n}\}. \end{aligned}$$

Suppose that $\alpha ^*_k\ne 1_{X_n}$. Notice that ($\forall x\in X_n$) $x\alpha ^*_k\in \{x,a_k\}$. Let $x\in X_n$ such that $x\ge j$. (i) If $a_k\ge j$, then $x\alpha ^*_k \ge j$. (ii) If $a_k< j$, then $x>a_k$. Suppose that $x \alpha _k^*= a_k$. By the definition of $\alpha _k^*$ and $x>a_k$, we have $x \le max A_k$ and so $j\le max A_k$. Since $max A_k\in A_k\subseteq \mathop {\mathrm {dom}}\nolimits (\alpha )$ and $\alpha \in PM_{i,j}\backslash J_0$, we have $j\le (max A_k)\alpha =a_k$, a contradiction. Then $x \alpha _k^*= x$ and so $x \alpha _k^* \ge j$. Similarly, we can prove that $x\le i\Rightarrow x \alpha _k^*\le i$. Thus $\alpha _k^*\in M_{i,j}$.

Let $\alpha _0$ be an identity mapping on $\mathop {\mathrm {dom}}\nolimits (\alpha )$. It is easy to prove that

$$\begin{aligned} \alpha =\alpha _0\alpha ^*_1\cdots \alpha ^*_s . \end{aligned}$$

From Case 2 above, we know that $\alpha _0\in \langle \tilde{E}\rangle$. By Lemma 10, we have

$$\begin{aligned} \alpha ^*_k\in M_{i,j}= \langle E([n,n-1]\}\backslash \{[i \rightarrow i+1],[j \rightarrow j-1]\}\rangle \subseteq \langle \tilde{E}\rangle . \end{aligned}$$

Thus $\alpha \in \langle \tilde{E}\rangle$.

It remains to prove that $\langle \tilde{E}\rangle \subseteq PM_{i,j}$. Notice that $F_{n-1}\subseteq PM_{i,j}$ and $\tilde{E}= (E([n,n-1])\backslash \{[i \rightarrow i+1],[j \rightarrow j-1]\})\cup F_{n-1}$. By Lemma 10, we have

$$\begin{aligned} E([n,n-1])\backslash \{ [i \rightarrow i+1],[j \rightarrow j-1]\} \subseteq M_{i,j}\subseteq PM_{i,j}. \end{aligned}$$

Then $\tilde{E} \subseteq PM_{i,j}$. It is easy to prove that $PM_{i,j}$ is a subsemigroup of $\mathcal {PO}_n$. Thus $\langle \tilde{E} \rangle \subseteq PM_{i,j}$. $\square$

Lemma 12

Let $1\le i\le n-1$. Then $PM_{i,i+1}$ is a regular subsemiband of $\mathcal {PO}_n$.

Proof

From Lemma 11, we know that $PM_{i,i+1}$ is a subsemiband of $\mathcal {PO}_n$. Let $\alpha \in PM_{i,i+1}$. If $|\mathop {\text{ im }}\nolimits (\alpha )|=0$, then clearly $\alpha =\theta$. Then $\alpha =\alpha ^2$ and so $\alpha$ is regular. If $|\mathop {\text{ im }}\nolimits (\alpha )|\ge 1$, suppose that

$$\begin{aligned} \alpha = \left( \begin{array}{llll} A_{1} &{}\quad A_{2} &{}\quad \cdots &{}\quad A_{r} \\ a_{1} &{}\quad a_{2} &{}\quad \cdots &{}\quad a_{r} \end{array} \right) . \end{aligned}$$

Let $c_k\in A_k$, $1\le k\le r$. Notice that if $a_k\le i$, then $c_k\le i$ (otherwise, since $\alpha \in PM_{i,i+1}$, we have $a_k=A_k\alpha =c_k\alpha \ge i+1$, a contradiction); if $a_k\ge i+1$, then $c_k\ge i+1$ (otherwise, since $\alpha \in PM_{i,i+1}$, we have $a_k=A_k\alpha =c_k\alpha \le i$, a contradiction). Let

$$\begin{aligned} \beta = \left( \begin{array}{llll} A_{1} &{}\quad A_{2} &{}\quad \cdots &{}\quad A_{r} \\ c_{1} &{}\quad c_{2} &{}\quad \cdots &{}\quad c_{r} \end{array} \right) . \end{aligned}$$

Then clearly $\beta \in PM_{i,i+1}$ and $\alpha =\alpha \beta \alpha$. Thus $\alpha$ is regular. $\square$

Let $2\le r\le n-1$. A proper subsemigroup S of $\mathcal {PO}(n,r)$ is called a maximal regular subsemiband of $\mathcal {PO}(n,r)$ if S is a regular semiband, and any regular subsemiband of $\mathcal {PO}(n,r)$ properly containing S must be $\mathcal {PO}(n,r)$.

Lemma 13

Let $1\le i\le n-1$. Then $\mathcal {PO}(n,n-2)\cup PM_{i,i+1}$ is a maximal regular subsemiband of $\mathcal {PO}_n$.

Proof

Let $B_i=\mathcal {PO}(n,n-2)\cup PM_{i,i+1}$. From Lemmas 1 and 12, we easily deduce that $B_i$ is a regular subsemigroup of $\mathcal {PO}_n$. By Lemmas 1 and 10, we have $B_i=\mathcal {PO}(n,n-2)\cup PM_{i,i+1}=\langle E(J_{n-2})\rangle \cup \langle E(J_{n-1})\backslash \{[i\rightarrow i+1], [i+1\rightarrow i]\}\rangle =\langle E(J_{n-2}) \cup (E(J_{n-1})\backslash \{[i\rightarrow i+1], [i+1\rightarrow i]\})\rangle$. Then $B_i$ is a regular subsemiband of $\mathcal {PO}_n$.

Let T be a regular semiband of $\mathcal {PO}_n$ properly containing $B_i$. Then $E(B_i)\subset E(T)$. Notice that $\mathcal {PO}(n,n-2)\subseteq B_i \subset T$ and $E(PM_{i,i+1}\cap J_{n-1})=E(J_{n-1}) \backslash \{[i\rightarrow i+1], [i+1\rightarrow i]\}$. Thus

$$\begin{aligned} E(J_{n-1}) \backslash \{[i\rightarrow i+1], [i+1\rightarrow i]\}= & {} E(PM_{i,i+1}\cap J_{n-1})\\= & {} E(B_i\cap J_{n-1}) \subset E(T\cap J_{n-1}), \end{aligned}$$

whence $E^-_{n-1}\cup F_{n-1}\subseteq E(J_{n-1}) \backslash \{[i\rightarrow i+1]\}\subseteq T$ or $E^+_{n-1}\cup F_{n-1}\subseteq E(J_{n-1}) \backslash \{[i+1\rightarrow i]\}\subseteq T$. Notice that $E^-_{n-1}\cup F_{n-1}$ ($E^+_{n-1}\cup F_{n-1}$) contains exactly one (idempotent) element from each ${\mathscr {R}}$-class of $\mathcal {PO}_n$ of rank $n-1$. It follows that $T\cap E(R_\alpha )\ne \emptyset$, for all $\alpha \in J_{n-1}$. Thus, by Lemma 3, $T=\mathcal {PO}_n$. $\square$

Lemma 14

Let $1\le i\le n-1$. Then $\mathcal {PO}(n,n-2)\cup \langle E(J_{n-1}\backslash \{\delta _i\}\rangle =\mathcal {PO}(n,n-2)\cup \{\alpha \in \mathcal {PO}_n: i\in \mathop {\mathrm {dom}}\nolimits (\alpha )\}$.

Proof

Let $S_i=\{\alpha \in \mathcal {PO}_n: i\in \mathop {\mathrm {dom}}\nolimits (\alpha )\}$, and let $W_i=\mathcal {PO}(n,n-2)\cup S_i$. It is obvious that $W_i=\mathcal {PO}(n,n-2)\cup (S_i\cap J_{n-1})$. For any $\alpha ,\beta \in W_i$, either $\alpha \beta \in \mathcal {PO}(n,n-2)$ or $\alpha \beta \in J_{n-1}$. If $\alpha \beta \in J_{n-1}$, then $\alpha ,\beta \in W_i\cap J_{n-1}=S_i\cap J_{n-1}\subseteq S_i$ and $\alpha {\mathscr {R}}\alpha \beta$. Thus $\mathop {\text{ ker }}\nolimits (\alpha )=\mathop {\text{ ker }}\nolimits (\alpha \beta )$ and so $\alpha \beta \in S_i$. Hence $W_i$ is a subsemigroup of $\mathcal {PO}_n$. It is obvious that $E(J_{n-1})\backslash \{\delta _i\}\subseteq S_i\subseteq W_i$. Then $\langle E(J_{n-1})\backslash \{\delta _i\}\rangle \subseteq W_i$ and so $\mathcal {PO}(n,n-2)\cup \langle E(J_{n-1})\backslash \{\delta _i\}\rangle \subseteq W_i$. From [16, Lemma 3.6], we know that $\langle S_i\rangle = \langle E(J_{n-1})\backslash \{\delta _i\}\rangle$. It follows that $W_i\subseteq \mathcal {PO}(n,n-2)\cup \langle S_i \rangle =\mathcal {PO}(n,n-2)\cup \langle E(J_{n-1}\backslash \{\delta _i\}\rangle$. $\square$

Our final main result is:

Theorem 15

Let $n\ge 3$. Then each maximal regular subsemiband of $\mathcal {PO}_n$ must be in one of the following forms:

(A) $A_i=\mathcal {PO}(n,n-2) \cup \{ \alpha \in \mathcal {PO}_n: i\in \mathop {\mathrm {dom}}\nolimits (\alpha ) \}$, $1\le i\le n$.

(B) $B_i=\mathcal {PO}(n,n-2)\cup PM_{i,i+1}$, $1\le i\le n-1$.

Proof

Let $S_i=\{\alpha \in \mathcal {PO}_n: i\in \mathop {\mathrm {dom}}\nolimits (\alpha ) \}$. From Lemma 13, we know that $B_i$ is a maximal regular subsemiband of $\mathcal {PO}_n$. By the definition of $S_i$, we easily deduce that $S_i\cap J_{n-1}=J_{n-1}\backslash R_{\delta _i}$ and so $A_i=\mathcal {PO}(n,n-2) \cup (J_{n-1}\backslash R_{\delta _i})$. Then, by Theorem 4, $A_i$ is a maximal regular subsemigroup of $\mathcal {PO}_n$. To show that $A_i$ is a maximal regular subsemiband of $\mathcal {PO}_n$, it is enough to verify that $A_i$ is a semiband. By Lemmas 1 and 14, we have

$$\begin{aligned} A_i=\mathcal {PO}(n,n-2)\cup S_i=\langle E(J_{n-2})\rangle \cup \langle E(J_{n-1})\backslash \{\delta _i\}\rangle =\langle E(J_{n-2})\cup (E(J_{n-1})\backslash \{\delta _i\}) \rangle . \end{aligned}$$

Conversely, let S be an arbitrary maximal regular subsemiband of $\mathcal {PO}_n$ not of the form $A_i$ or $B_i$. Notice that

$$\begin{aligned} E(A_i\cap J_{n-1})= & {} E(S_i\cap J_{n-1})=E(J_{n-1})\backslash \{\delta _i\},\\ E(B_i\cap J_{n-1})= & {} E(PM_{i,i+1}\cap J_{n-1})=E(J_{n-1})\backslash \{[i\rightarrow i+1],[i+1\rightarrow i]\}. \end{aligned}$$

We claim that S satisfies

$$\begin{aligned} S\cap E(R_{\alpha })\ne \emptyset , \ for \ all \ \alpha \in J_{n-1} . \end{aligned}$$

(2.5)

Otherwise, there exists $A_i$ or $B_i$ for some $i\in X_n$ such that $E(S\cap J_{n-1}) \subseteq E(A_i\cap J_{n-1})$ or $E(S\cap J_{n-1}) \subseteq E(B_i\cap J_{n-1})$. Notice that $\mathcal {PO}(n,n-2)\subseteq A_i, B_i$. Then $E(S) \subseteq E(A_i)$ or $E(S) \subseteq E(B_i)$. It follows that $S\subseteq A_i$ or $S\subseteq B_i$, since S, $A_i$, and $B_i$ are semibands. Thus, by the maximality of S, $S=A_i$ or $S=B_i$

By Lemma 3 and (2.5), We have $S=\mathcal {PO}_n$ and this is a contradiction. $\square$

From Theorem 4 and Theorem 15, we know that the maximal regular and the maximal regular subsemibands of $\mathcal {PO}_n$ do not coincide. As an immediate consequence of Theorem 4 and Theorem 9, we have the following:

Corollary 16

Let $2\le r\le n-2$. Then the maximal regular subsemigroups and maximal regular subsemibands of $\mathcal {PO}(n,r)$ coincide.