Automorphisms of partition order-decreasing transformation monoids

Abstract

Let T _n be the full transformation semigroup on a finite set X _n ={1,2,…,n}. Let ρ be an equivalence relation on X _n and ⪯ be a total order on the partition set X _n /ρ. We describe all automorphisms of the partition order-decreasing transformation monoid:

$$T(\rho,\preceq)=\bigl\{\alpha\in T_{n}: (x\alpha)\rho\preceq x\rho, \forall x\in X_{n}\bigr\} $$

that generalizes the results of Schreier (Fundam. Math., 28:261–264, 1936) and Šutov (Izv. Vysš. Učebn. Zaved., Mat., 3:177–184, 1961).

1 Introduction

For the standard definitions on semigroups and transformation semigroups we refer the reader to the books [3–5].

Let X _n ={1,2,…,n}. A submonoid M of the semigroup T _n of full transformations on X _n is intransitive if there exist x,y in X _n such that $(x)\varphi\not= y$ for any φ∈M. A submonoid M of T _n said to be half-transitive provided that it is intransitive, and for every ordered pair (x,y)∈X _n ×X _n there is some φ∈M such that either xφ=y or yφ=x. In [10] we showed that, for every half-transitive submonoid M of T _n , there exist a non-universal equivalence relation ρ on X _n and a total order ⪯ on the partition set X _n /ρ such that M lies inside a half-transitive submonoid T(ρ,⪯) of T _n defined by

$$T(\rho,\preceq)=\bigl\{\alpha\in T_{n}: (x\alpha)\rho\preceq x\rho, \forall x\in X_{n}\bigr\}. $$

Here we consider the monoid T(ρ,⪯) for arbitrary equivalence relation ρ on X _n . In particular, if ρ is universal then T(ρ,⪯)=T _n ; T(ρ,⪯) is the order-decreasing finite full transformation monoid if ρ is the identity relation [9]. If X _n /ρ={{1},X _n −{1}} and {1}≺X _n −{1} then T(ρ,⪯) is isomorphic to PT _n−1, the semigroup of partial transformations of X _n −{1}.

Note that automorphisms of T _n and PT _n−1 were described by Schreier and Šutov in [6] and [8] respectively. In present paper, we will describe automorphisms of the monoid T(ρ,⪯). Thus our result generalizes the results of Schreier [6] and Šutov [8].

For the topics of automorphisms of transformation semigroups we refer the reader to the references of [1] and [3]. We point out that T(ρ,⪯) is not the centralizer of any idempotent in T _n [1] and, in general, T(ρ,⪯) does not contain all constant transformations of X _n . Our method is different from any one of the references of [1] and [7]. Here we use a basis fact that an automorphism φ of a monoid S maps units of S to its units, and idempotents of S to its idempotents.

To achieve our aims, the organization of the paper is as follows: In Sect. 2 we investigate the units and the idempotents of T(ρ,⪯), and describe the generators of T(ρ,⪯). In Sect. 3 we describe the Green’s ^∗-relations $\mathcal{L}^{*}$ and $\mathcal{R}^{*}$ on T(ρ,⪯). As a consequence, T(ρ,⪯) is shown to be abundant. Finally, in Sect. 4, we use the results of the previous sections to determine all automorphisms of the monoid T(ρ,⪯).

Throughout the paper, we use the following notations: let |A| denote the cardinality of a set A. 1 _A denote the identity function from A to itself. For a function α:A→B, denote the image of α by $\operatorname{im} \alpha$. $|\operatorname{im} \alpha|$ is said to be the rank of α, and we can write

$$\alpha= \left ( \begin{array}{c@{\quad}c@{\quad}c} A_{1} & \ldots& A_{k}\\ a_{1} & \ldots& a_{k} \end{array} \right ), $$

where $\operatorname{im} \alpha=\{a_{1},\dots,a_{k}\}$, a _i α ⁻¹=A _i (i=1,…,k) and $\{A_{1}, \dots,A_{k}\}=A/\operatorname{ker} \alpha$, where $\operatorname{ker} \alpha$ is the kernel of α (the equivalence relation {(x,y)∈A×A:xα=yα}). Denote the restriction of α to C by α| _C for C⊆A and (C)α denotes the image set of C under α.

Let E _n−1 denote the set of idempotents in T _n of rank n−1. Every element e of E _n−1 has the form

$$e = {b\choose a} $$

for some a,b∈X _n ,a≠b, which maps b to a and x to itself for any x∈X _n −{b} [4].

For the remainder of the paper, ρ will denote an equivalence relation on X _n and ⪯ will denote a total order on the partition set X _n /ρ.

2 Units, idempotents and generators

In this section we describe the units and the idempotents of T(ρ,⪯), and determine the generators of T(ρ,⪯).

We define

$$U_{\rho}=\bigl\{\alpha\in T_{n}: (x\rho)\alpha= x\rho, x\in X_{n}\bigr\}. $$

Clearly, U _ρ ⊆T(ρ,⪯). Let S _n be the symmetric group on X _n . The following lemma gives the description of the group of units of T(ρ,⪯).

Lemma 2.1

Let α∈T _n . Then the following statements are equivalent:

• (1) α is a unit of T(ρ,⪯).

• (2) α∈S _n ∩T(ρ,⪯).

• (3) α∈U _ρ .

Proof

Obviously (1)⇒(2).

(2)⇒(3). Suppose that α∈S _n ∩T(ρ,⪯). Then α ⁻¹∈S _n ∩T(ρ,⪯). Given any x∈X _n . Take any y∈xρ, we have

$$y\rho=\bigl(y\alpha\alpha^{-1}\bigr)\rho\preceq(y\alpha)\rho\preceq y \rho=x\rho, $$

and so yα∈xρ. Thus (xρ)α⊆xρ. To prove that xρ⊆(xρ)α consider z∈xρ. Then we have

$$x\rho=z\rho=\bigl[\bigl(z\alpha^{-1}\bigr)\alpha\bigr]\rho\preceq \bigl(z\alpha^{-1}\bigr)\rho\preceq z\rho $$

and so zα ⁻¹∈xρ and z=(zα ⁻¹)α. It follows that (xρ)α=xρ and hence α∈U _ρ .

(3)⇒(1). By the definition of U _ρ , we have that α∈T(ρ,⪯) and the restriction α| _xρ is a bijection from the ρ-class xρ onto itself, and so α∈S _n and α ⁻¹∈T(ρ,⪯). Thus α is a unit of T(ρ,⪯). □

Next we have

Lemma 2.2

Let α∈T(ρ,⪯). Then α is an idempotent if and only if for all $t \in \operatorname{im} \alpha$, tα=t and tρ=min{xρ:x∈tα ⁻¹}.

Proof

First recall that α∈T _n is an idempotent if and only if, for all $t \in \operatorname{im} \alpha$, t∈tα ⁻¹. Hence α∈T(ρ,⪯) is an idempotent if and only if, for all $t\in \operatorname{im} \alpha$,

$$t\alpha= t \quad\mbox{and}\quad t\rho= \min \bigl\{x\rho: x \in t \alpha^{-1}\bigr\} $$

since t∈tα ⁻¹ and x∈tα ⁻¹ implies that t=tα=xα and tρ=(xα)ρ⪯xρ. □

Corollary 2.3

Let a,b∈X _n and a≠b. Then aρ⪯bρ if and only if ${b\choose a}\in T(\rho,\preceq)$.

From Corollary 2.3 we introduce the following idempotent subset of T(ρ,⪯) with rank n−1 (that is often used in the paper):

$$E^{w}_{\rho}=\left\{{b\choose a}\in E_{n-1}: a\rho\preceq b \rho\right\}. $$

Then we have

Lemma 2.4

Let α be an element of T(ρ,⪯)−U _ρ . Then α is a product of elements in $E^{w}_{\rho}\cup U_{\rho}$.

Proof

Let s(α) be the cardinality of the set {x∈X _n :xα≠x}. We will show that α can be written as a product of elements in $E^{w}_{\rho}\cup U_{\rho}$ by using induction on s(α).

Clearly, if s(α)=1 then $\alpha\in E^{w}_{\rho}$. We now assume that s(α)>1 and let

$$\alpha= \left ( \begin{array}{c@{\quad}c@{\quad}c@{\quad}c} A_{1} & A_{2} & \ldots& A_{k}\\ a_{1} & a_{2} & \ldots& a_{k} \end{array} \right )\in T(\rho, \preceq)-U_{\rho}. $$

Then, for every i, a _i ρ⪯min{xρ:x∈A _i } from the definition of T(ρ,⪯). Choose b _i ∈A _i such that b _i ρ=min{xρ:x∈A _i } for every i. Without loss of generality we may assume that

$$a_{1}=b_{1}, \dots, a_{i}=b_{i}, \quad\quad a_{i+1} \ne b_{i+1},\dots, a_{k}\ne b_{k}. $$

Define

$$\beta= \left ( \begin{array}{c@{\quad}c@{\quad}c} A_{1} & \ldots & A_{k}\\ b_{1} & \ldots& b_{k} \end{array} \right ) \quad\mbox{and}\quad \gamma= \left ( \begin{array}{c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c} b_{1} & \ldots& b_{i} & b_{i+1} & \ldots & b_{k} & Y\\ b_{1} & \ldots& b_{i} & a_{i+1} & \ldots & a_{k} & y \end{array} \right ), $$

where Y=X _n −{b ₁,…,b _k }, y∈Y and yρ=min{zρ:z∈Y}. Then β∈T(ρ,⪯)−U _ρ and γ∈T(ρ,⪯). Moreover, γ∈U _ρ if and only if k=n−1.

Note that s(α)=n−i>n−k=s(β) and s(γ)=s(α)−1. Thus both β and γ are products of elements in $E^{w}_{\rho }\cup U_{\rho}$ by inductive supposition. It follows that α=βγ is a product of elements in $E^{w}_{\rho}\cup U_{\rho}$. This completes the proof. □

From Lemma 2.4 we immediately deduce

Proposition 2.5

Let U _ρ and $E^{w}_{\rho}$ be defined above. Then $E^{w}_{\rho}\cup U_{\rho}$ is a generating set of T(ρ,⪯).

Similar to the proof as Lemma 2.4 we also have

Lemma 2.6

Let α be an idempotent of T(ρ,⪯)−U _ρ . Then α is a product of elements in $E^{w}_{\rho}$.

3 Green’s ^∗-relations $\mathcal{L}^{*}$ and $\mathcal{R}^{*}$

We recall some of the basis facts about Green’s *-relations $\mathcal{L}^{*}$ and $\mathcal{R}^{*}$. The relations $\mathcal{L}^{*}(\mathcal{R}^{*})$ is defined on a semigroup S by the rule $a{\mathcal{L}^{*}}b(a{\mathcal{R}^{*}}b)$ if and only if the elements a,b of S are related by the Green’s relation ${\mathcal{L}}(\mathcal{R})$ in some oversemigroup of S [2]. The following lemma, from [2], provides us an alternative description for $\mathcal{L}^{*}(\mathcal{R}^{*})$.

Lemma 3.1

Let S be a semigroup and let a,b be in S. The following conditions are equivalent:

• (1) $a{\mathcal{L}^{*}}b(a{\mathcal{R}^{*}}b)$.

• (2) for all s,t∈S ¹, as=at(sa=ta) if and only if bs=bt(sb=tb).

For α∈T _n , $\operatorname{ker} \alpha=\{(x,y)\in X_{n}\times X_{n}: x\alpha=y\alpha\}$. Then we have

Proposition 3.2

Let α,β be elements of T(ρ,⪯). Then

• (1) $(\alpha, \beta) \in{\mathcal{L}^{*}}$ if and only if $\operatorname{im} \alpha= \operatorname{im} \beta$;

• (2) $(\alpha, \beta) \in{\mathcal{R}^{*}}$ if and only if $\operatorname{ker} \alpha= \operatorname{ker} \beta$.

Proof

(1) Certainly if $\operatorname{im} \alpha = \operatorname{im} \beta$ then $(\alpha, \beta) \in{\mathcal{L}}(T_{n})$ [5] and so $(\alpha, \beta) \in{\mathcal{L}}^{*}$.

Conversely, suppose that $(\alpha, \beta) \in{\mathcal{L}}^{*}$. Then

$$\alpha\gamma= \alpha\delta \quad\mbox{if and only if}\quad \beta\gamma= \beta \delta \quad \bigl(\mbox{for all } \gamma, \delta\in T(\rho,\preceq)\bigr). $$

Let X _n /ρ={Y ₁≺Y ₂≺⋯≺Y _t }. If Y ₁={x}, for some x∈X _n , then, certainly xα=x=xβ; Otherwise, choose y∈Y ₁ and y≠x, then ${x\choose y} \in T(\rho, \preceq)$. Hence

$$x\not\in \operatorname{im} \alpha\Leftrightarrow\alpha\cdot{x\choose y} = \alpha \cdot1_{X_{n}} \Leftrightarrow\beta\cdot{x\choose y} = \beta\cdot 1_{X_{n}}\Leftrightarrow x\not\in \operatorname{im} \beta. $$

We therefore conclude that $\operatorname{im} \alpha= \operatorname{im} \beta$.

(2) Again if $\operatorname{ker} \alpha= \operatorname{ker} \beta$ then $(\alpha, \beta) \in{\mathcal{R}}(T_{n})$ [5] and so $(\alpha, \beta) \in{\mathcal{R}}^{*}$.

Conversely, if $(\alpha,\beta) \in {\mathcal{R}}^{*}$ then

$$\gamma\alpha= \delta\alpha\Leftrightarrow\gamma\beta= \delta\beta \quad\bigl( \mbox{for all } \gamma, \delta\in T(\rho, \preceq)\bigr). $$

For x,y∈X _n with x≠y, we may assume that yρ⪯xρ. Then ${x\choose y} \in T(\rho, \preceq)$. Hence

$$x \alpha= y\alpha\Leftrightarrow{x\choose y}\cdot\alpha= 1_{X_{n}}\cdot \alpha \Leftrightarrow{x\choose y}\cdot\beta= 1_{X_{n}}\cdot\beta \Leftrightarrow x\beta= y\beta. $$

We therefore conclude that $\operatorname{ker} \alpha= \operatorname{ker} \beta$. □

A semigroup S in which each ${\mathcal{L}}^{*}$-class and each ${\mathcal{R}}^{*}$-class contains an idempotent is called abundant [2]. We have

Corollary 3.3

For any equivalence relation ρ on X _n and a total order ⪯ on the partition set X _n /ρ, the semigroup T(ρ,⪯) is abundant.

Proof

For a typical ${\mathcal{R}}^{*}$-class $R^{*}_{\alpha}$ of T(ρ,⪯) with $X_{n}/\operatorname{ker} \alpha= \{ A_{1}, A_{2}, \ldots, A_{k}\}$. Choose b _i ∈A _i such that b _i ρ=min{x _i ρ:x _i ∈A _i } for every i∈{1,…,k}. Then we have that

$$\beta= \left ( \begin{array}{c@{\quad}c@{\quad}c} A_{1} & \ldots& A_{k}\\ b_{1} & \ldots& b_{k} \end{array} \right ) $$

is an idempotent of T(ρ,⪯) by Lemma 2.2. and αR ^∗ β by Proposition 3.2.

Next, consider that a typical ${\mathcal{L}}^{*}$-class $L^{*}_{\alpha}$ of T(ρ,⪯) with $\operatorname{im} \alpha= \{ a_{1},\ldots, a_{k}\}$, where 1≤k≤n. We will use induction on k to show that $L^{*}_{\alpha}$ contains an idempotent of T(ρ,⪯). If k=1 then, clearly α is an idempotent. Suppose now that the conclusion holds for k−1. Consider that

$$\alpha= \left ( \begin{array}{c@{\quad}c@{\quad}c@{\quad}c} A_{1} & \ldots& A_{k-1} & A_{k}\\ a_{1} & \ldots& a_{k-1} & a_{k} \end{array} \right ) \in T(\rho, \preceq). $$

Without loss of generality, we may assume that

$$a_{1}\rho\preceq \cdots\preceq a_{k-1}\rho\preceq a_{k}\rho. $$

Then ${a_{k}\choose a_{k-1}}\in T(\rho,\preceq)$ by Corollary 2.3, and so

$$\beta= \left ( \begin{array}{c@{\quad}c@{\quad}c@{\quad}c} A_{1} & \ldots& A_{k-2} & A_{k-1}\cup A_{k}\\ a_{1} & \ldots& a_{k-2} & a_{k-1} \end{array} \right ) = \alpha{a_{k}\choose a_{k-1}}\in T(\rho,\preceq). $$

By induction supposition, there exists an idempotent ε∈T(ρ,⪯) such that $\beta{\mathcal{L}}^{*} \varepsilon$. Hence, by Lemma 2.2, we can let

$$\varepsilon= \left ( \begin{array}{c@{\quad}c@{\quad}c@{\quad}c} B_{1} & \ldots& B_{k-2} & B_{k-1}\\ a_{1} & \ldots& a_{k-2} & a_{k-1} \end{array} \right ) $$

with a _i ∈B _i ,i=1,…,k−1 and a _i ρ=min{x _i ρ:x _i ∈B _i } for every i∈{1…,k−1}. Since a _k ∈X _n =B ₁∪⋯∪B _k−1, there exist a unique B _j such that a _k ∈B _j . Note that a _k ≠a _j , so a _j ∈B _j ∖{a _k }. It follows, by Lemma 2.2, that

$$\varepsilon^{*} = \left ( \begin{array}{c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c} B_{1} & \ldots& B_{j-1} & B_{j}\backslash\{a_{k}\} & B_{j+1} & \ldots & B_{k-1} & a_{k}\\ a_{1} & \ldots& a_{j-1} & a_{j} & a_{j+1} & \ldots& a_{k-1} & a_{k} \end{array} \right ) $$

is an idempotent in T(ρ,⪯), so that $\alpha{\mathcal{L}}^{*}\varepsilon^{*}$ as required. □

4 Automorphisms of T(ρ,⪯)

After the preliminaries of the previous two sections, to determine every automorphism of T(ρ,⪯), we need the following lemmas.

Lemma 4.1

Let ${b_{1}\choose a_{1}}, {b_{2}\choose a_{2}} \in E_{n-1}$. Then ${b_{1}\choose a_{1}}{b_{2}\choose a_{2}} ={b_{1}\choose a_{1}}$ if and only if b ₁=b ₂.

Proof

Assume that b ₁=b ₂. Then, from a ₁≠b ₁ we have

$$b_{1}{b_{1}\choose a_{1}} {b_{1}\choose a_{2}}=a_{1}{b_{1}\choose a_{2}}=a_{1}=b_{1}{b_{1} \choose a_{1}}, $$

and for every x∈X _n ∖{b ₁},

$$x{b_{1}\choose a_{1}} {b_{1}\choose a_{2}}=x{b_{1}\choose a_{2}}= x = x{b_{1}\choose a_{1}} $$

and so ${b_{1}\choose a_{1}}{b_{1}\choose a_{2}} ={b_{1}\choose a_{1}}$.

Assume now that b ₁≠b ₂. Then

$$b_{2}{b_{1}\choose a_{1}} {b_{2} \choose a_{2}} = b_{2}{b_{2}\choose a_{2}} = a_{2} \neq b_{2} =b_{2}{b_{1} \choose a_{1}}, $$

and so ${b_{1}\choose a_{1}}{b_{2}\choose a_{2}} \neq{b_{1}\choose a_{1}}$. □

Similar argument as in Lemma 4.1, we have

Lemma 4.2

Let ${x\choose y}, {y\choose w} \in E_{n-1}$. Then ${y\choose w}{x\choose y}$ is an idempotent if and only if x=w.

Lemma 4.3

Let ${b_{1}\choose a_{1}}, {b_{2}\choose a_{2}} \in T(\rho,\preceq)$ and b ₁ ρ⪯b ₂ ρ. We have

• (1) If b ₁=b ₂ then ${b_{1}\choose a_{1}}{\mathcal{L}}^{*} {b_{2}\choose a_{2}}$.

• (2) If b ₁≠b ₂ then there exists δ∈T(ρ,⪯) such that ${b_{1}\choose a_{1}}{\mathcal{R}}^{*}\delta{\mathcal{L}}^{*}{b_{2}\choose a_{2}}$.

Proof

Since $\operatorname{im} {b_{1}\choose a_{1}} = X_{n}\backslash\{b_{1}\} = \operatorname{im} {b_{1}\choose a_{2}}$, so (1) holds by Lemma 3.1.

For (2), note that ${b_{2}\choose b_{1}} \in T(\rho,\preceq)$ by Corollary 2.3, since b ₁ ρ⪯b ₂ ρ. It follows that $\delta= {b_{1}\choose a_{1}}{b_{2}\choose b_{1}} \in T(\rho,\preceq)$ with $\operatorname{ker} \delta= \operatorname{ker} {b_{1}\choose a_{1}}$ and $\operatorname{im} \delta= \operatorname{im} {b_{2}\choose b_{1}}= \operatorname{im} {b_{2}\choose a_{2}}$. Hence, by Proposition 3.2, we have ${b_{1}\choose a_{1}}{\mathcal{R}}^{*}\delta{\mathcal{L}}^{*}{b_{2}\choose a_{2}}$. □

Let $\operatorname{Aut} T(\rho,\preceq)$ be the automorphism group of T(ρ,⪯). Then we have

Lemma 4.4

Let $\varphi\in \operatorname{Aut} T(\rho,\preceq)$. Then $(E^{w}_{\rho})\varphi= E^{w}_{\rho}$.

Proof

First, we claim that

$$\bigl(E^{w}_{\rho}\bigr)\varphi\cap E^{w}_{\rho} \ne\emptyset. $$

Assume that it is not such case. Note that (ε _i )φ is idempotent for every $\varepsilon_{i} \in E^{w}_{\rho}$, and so we certainly have $|\operatorname{im} (\varepsilon_{i})\varphi |\leq n-2$. Next, given an element $\varepsilon\in E^{w}_{\rho}$ then there exists an idempotent $\omega\in T(\rho,\preceq)-\{1_{X_{n}}\}$ such that (ω)φ=ε. Thus, by Lemma 2.6, there exist elements ε ₁,ε ₂,…,ε _k of $E^{w}_{\rho}$ such that ω=ε ₁ ε ₂⋯ε _k . Hence we have

$$\varepsilon=(\varepsilon_{1} \varepsilon_{2} \cdots \varepsilon_{k})\varphi=(\varepsilon_{1})\varphi ( \varepsilon_{2})\varphi\cdots(\varepsilon_{k})\varphi. $$

It follows that $n-1 = |\operatorname{im} \varepsilon| \leq|\operatorname{im} (\varepsilon_{k})\varphi| \leq n-2$ (since $(\varepsilon_{k})\varphi\not\in E^{w}_{\rho}$), a contradiction.

We have shown that $(E^{w}_{\rho})\varphi\cap E^{w}_{\rho}\neq \emptyset$. Next, let $\varepsilon_{0}= {b_{0}\choose a_{0}} = \bigl({b_{1}\choose a_{1}}\bigr)\varphi\in(E^{w}_{\rho})\varphi\cap E^{w}_{\rho}$. For any $\varepsilon={b\choose a}\in E^{w}_{\rho}$, we distinguish two cases as follows.

Case 1: b=b ₁. In this case, by (1) of Lemma 4.3 we have $\varepsilon{\mathcal{L}}^{*} {b_{1}\choose a_{1}}$. Further, by Lemma 3.1, we immediately deduce $(\varepsilon)\varphi{\mathcal{L}}^{*}\varepsilon_{0}$. Hence, by (1) of Proposition 3.2, $|\operatorname{im}(\varepsilon)\varphi| = n-1$ and so $(\varepsilon)\varphi\in E^{w}_{\rho}$.

Case 2: b≠b ₁; bρ⪯b ₁ ρ(Similar argument for b ₁ ρ⪯bρ). By (2) of Lemma 4.3, there exists δ∈T(ρ,⪯) such that $\varepsilon{\mathcal{R}}^{*} \delta {\mathcal{L}}^{*} {b_{1}\choose a_{1}}$. Hence, by Lemma 3.1 we have $(\varepsilon)\varphi{\mathcal{R}}^{*}(\delta)\varphi{\mathcal{L}}^{*} \varepsilon_{0}$. That is, $\operatorname{ker} (\varepsilon)\varphi= \operatorname{ker} (\delta)\varphi$ and $\operatorname{im} (\delta)\varphi= \operatorname{im} \varepsilon_{0}$. It follows that

$$\big|\operatorname{im} (\varepsilon)\varphi\big| = \big|X_{n}/ \operatorname{ker} (\varepsilon)\varphi\big| = \big|X_{n}/ \operatorname{ker} (\delta) \varphi\big| = \big|\operatorname{im} (\delta) \varphi\big| = |\operatorname{im} \varepsilon_{0} | = n-1, $$

and so $(\varepsilon)\varphi \in E^{w}_{\rho}$.

By Case 1 and Case 2 above, we have shown that $(E^{w}_{\rho})\varphi\subseteq E^{w}_{\rho}$. By using the foregoing argument for the automorphism φ ⁻¹, we have $(E^{w}_{\rho})\varphi^{-1}\subseteq E^{w}_{\rho}$. It follows that

$$E^{w}_{\rho}=\bigl(E^{w}_{\rho}\bigr) \varphi^{-1}\varphi\subseteq\bigl(E^{w}_{\rho }\bigr) \varphi\subseteq E^{w}_{\rho}, $$

and so $(E^{w}_{\rho})\varphi= E^{w}_{\rho}$. □

Lemma 4.5

Let $\varphi\in \operatorname{Aut} T(\rho,\preceq)$. Then there exists μ _φ ∈U _ρ such that $\big({x\choose y}\big)\varphi= {x\mu_{\varphi} \choose y\mu_{\varphi}}$ for every ${x\choose y}\in E^{w}_{\rho}$.

Proof

Let X _n /ρ={Y ₁≺Y ₂≺⋯≺Y _t }. We distinguish two cases: |Y ₁|=1 or |Y ₁|≥2.

Case 1: |Y ₁|=1. Let Y ₁={1}. For any x∈X _n ∖{1}, clearly ${x\choose1}\in E^{w}_{\rho}$ and by Lemma 4.4 we can let $\bigl({x\choose1}\bigr)\varphi={x'\choose1_{x}'} \in E^{w}_{\rho}$. We will prove that $1=1_{x}'$. Indeed, if $1 \neq 1_{x}'$ then ${1_{x}'\choose1}\in E^{w}_{\rho}$, and by Lemma 4.4 we can let $\bigl({z\choose t}\bigr)\varphi={1_{x}'\choose1}$, where ${z\choose t}\in E^{w}_{\rho}$. Since z≠t and Y ₁⪯tρ⪯zρ, we have z≠1 and so

$${z\choose t} {x\choose1}= \left \{ \begin{array}{l@{\quad}l} {x\choose1}, & t=1,x=z\\[6pt] {z\choose t}, & t\neq1,x=z\\[6pt] \left ( \begin{array}{c@{\quad}c@{\quad}c} \{z,x,1\} & k & \dots\\ 1 & k & \dots \end{array} \right ), & t=1,x\neq z \\[12pt] \left ( \begin{array}{c@{\quad}c@{\quad}c@{\quad}c} \{z,t\} & \{x, 1\}& k &\dots\\ t & 1 & k & \dots \end{array} \right ), & t\neq1,x\neq z \end{array} \right . $$

is an idempotent, but ${1_{x}'\choose 1}{x'\choose1_{x}'}$ is not idempotent by Lemma 4.2, a contradiction. Thus $1=1_{x}'$ and we have proved that

$$\left({x\choose1}\right)\varphi={x'\choose1} \quad\mbox{for any } x\in X_{n}\backslash\{1\}. $$

Now we define a mapping μ _φ :X _n →X _n by 1μ _φ =1 and xμ _φ =x′ (defined the above), x≠1. Obviously, μ _φ is a bijection.

Next, we will prove that $\bigl({x\choose y}\bigr)\varphi={x\mu_{\varphi}\choose y\mu_{\varphi}}$ for any ${x\choose y}\in E^{w}_{\rho}, x\neq1, y\neq1$. By Lemma 4.4 we can suppose that $\bigl({x\choose y}\bigr)\varphi={x^{*}\choose y^{*}}\in E^{w}_{\rho}$. Since

$${x\mu_{\varphi}\choose1}=\left({x\choose1}\right)\varphi=\left({x\choose1} {x\choose y}\right) \varphi ={x\mu_{\varphi}\choose1} {x^{*}\choose y^{*}}, $$

we immediately deduce that x ^∗=xμ _φ by Lemma 4.1. Moreover,from $\bigl({y\choose1}{x\choose y}\bigr)\varphi={y\mu_{\varphi}\choose 1}{x\mu_{\varphi}\choose y^{*}}$ we see that if yμ _φ ≠y ^∗ then ${y\mu_{\varphi}\choose 1}{x\mu_{\varphi}\choose y^{*}}$ is an idempotent, and so ${y\choose1}{x\choose y}$ is idempotent (since φ is an automorphism). But this is impossible since ${y\choose1}{x\choose y}$ is not an idempotent. Hence yμ _φ =y ^∗. It follows that if yρ⪯xρ then (yμ _φ )ρ⪯(xμ _φ )ρ, and so μ _φ ∈U _ρ .

Case 2: |Y ₁|≥2. In this case, for any x∈X _n , there exists y∈X _n ∖{x} such that yρ⪯xρ, and so ${x\choose y}\in E^{w}_{\rho}$. By Lemma 4.4 we can let

$$\left({x\choose y}\right)\varphi= {x'\choose y'}\in E^{w}_{\rho} \quad\mbox{and}\quad \left({x\choose z}\right)\varphi= {x'_{z}\choose z'}\in E^{w}_{\rho}. $$

Note that

$${x'\choose y'} = \left({x\choose y}\right)\varphi=\left[\left({x \choose y} {x\choose z}\right)\right]\varphi ={x'\choose y'} {x'_{z}\choose z'}, $$

and so $x' = x'_{z}$ by Lemma 4.1. Hence, φ induces a map μ _φ from X _n to itself, defined by

$$\left({x\choose y}\right)\varphi= {x\mu_{\varphi}\choose y'},\quad \hbox{for every } {x\choose y}\in E^{w}_{\rho}. $$

Obviously, μ _φ is surjective.

Fact 1

μ _φ is injective.

Indeed, assume that t=x ₁ μ _φ =x ₂ μ _φ , for x ₁,x ₂∈X _n . Since |Y ₁|≥2, there exist y ₁,y ₂∈X _n such that ${x_{1}\choose y_{1}},{x_{2}\choose y_{2}} \in E^{w}_{\rho}$. By Lemma 4.4 we let $\bigl({x_{1}\choose y_{1}}\bigr)\varphi= {t\choose y_{1}'}$ and $\bigl({x_{2}\choose y_{2}}\bigr)\varphi= {t\choose y_{2}'}$. Then

$$\left({x_{1}\choose y_{1}} {x_{2}\choose y_{2}}\right)\varphi={t\choose y_{1}'} {t\choose y_{2}'} ={t\choose y_{1}'} =\left({x_{1}\choose y_{1}}\right)\varphi. $$

As φ is a bijection it follows that ${x_{1}\choose y_{1}}{x_{2}\choose y_{2}}={x_{1}\choose y_{1}}$, and so x ₁=x ₂ from Lemma 4.1.

Fact 2

If x ₁ ρ⪯x ₂ ρ. Then (x ₁ μ _φ )ρ⪯(x ₂ μ _φ )ρ.

In fact, assume that (x ₁ μ _φ )ρ≻(x ₂ μ _φ )ρ. Then ${x_{2}\choose x_{1}}, {x_{1}\mu_{\varphi}\choose x_{2}\mu_{\varphi}} \in E^{w}_{\rho}$. By Lemma 4.4 we can let

$$\left({x_{2}\choose x_{1}}\right)\varphi= {x_{2} \mu_{\varphi}\choose x_{1}'}, \quad\quad\left({x_{1}\choose z}\right)\varphi={x_{1}\mu_{\varphi}\choose x_{2} \mu_{\varphi}}. $$

Note that ${x_{1}\mu_{\varphi}\choose x_{2}\mu_{\varphi}}{x_{2}\mu_{\varphi}\choose x_{1}'}$ is an idempotent. Hence ${x_{1}\choose z}{x_{2}\choose x_{1}}$ is also an idempotent, and so z=x ₂ by Lemma 4.2. Further,

$${x_{1}\mu_{\varphi}\choose x_{2}\mu_{\varphi}}=\left({x_{1} \choose x_{2}}\right)\varphi= \left({x_{2}\choose x_{1}} {x_{1}\choose x_{2}}\right)\varphi={x_{2} \mu_{\varphi}\choose x_{1}'} {x_{1} \mu_{\varphi} \choose x_{2}\mu_{\varphi}} $$

is an idempotent. Hence, $x_{1}' = x_{1}\mu_{\varphi}$ by Lemma 4.2. Since now both ${x_{2}\mu_{\varphi}\choose x_{1}\mu_{\varphi}}$ and ${x_{1}\mu_{\varphi}\choose x_{2}\mu_{\varphi}}$ belong to $E^{w}_{\rho}$, we have (x ₁ μ _φ )ρ=(x ₂ μ _φ )ρ, which contradicts the assumption.

Fact 3

Let $\bigl({x\choose y}\bigr)\varphi= {x\mu_{\varphi}\choose y'}$ for ${x\choose y}\in E^{w}_{\rho}$. Then y′=yμ _φ .

Since ${x\choose y}\in E^{w}_{\rho}$,we have yρ⪯xρ, and so ${x\mu_{\varphi}\choose y\mu_{\varphi}}\in E^{w}_{\rho}$ by Facts 2 and 1. By Lemma 4.4 we can let

$$\left({x\choose z}\right)\varphi= {x\mu_{\varphi}\choose y\mu_{\varphi}} \quad{\mbox{for some}} z\in X_{n}\backslash\{x\}. $$

It is sufficient to show that y=z. Assume now that y≠z. Then yρ⪯zρ or zρ⪯yρ since ⪯ is a totally order.

For the former, by Facts 2 and 1 we have ${z\mu_{\varphi}\choose y\mu_{\varphi}}\in E^{w}_{\rho}$ and by Lemma 4.4 we can let ${z\mu_{\varphi}\choose y\mu_{\varphi}}=\bigl({z\choose t}\bigr)\varphi$, for some t∈X _n . Note that ${z\mu_{\varphi}\choose y\mu_{\varphi}}{x\mu_{\varphi}\choose y\mu_{\varphi}}$ is idempotent. Then ${z\choose t}{x\choose z}$ is also idempotent, and so t=x by Lemma 4.2. Hence

$${z\mu_{\varphi}\choose y\mu_{\varphi}}=\left({z\choose x}\right)\varphi= \left({x \choose z} {z\choose x}\right)\varphi={x\mu_{\varphi}\choose y\mu_{\varphi}} {z\mu_{\varphi}\choose y\mu_{\varphi}}. $$

It follows that xμ _φ =zμ _φ and so x=z, which contradicts x≠z.

For the latter,then ${y\choose z}\in E^{w}_{\rho}$ and by Lemma 4.4 we can let $\bigl({y\choose z}\bigr)\varphi= {y\mu_{\varphi}\choose z'}$ for some z′∈X _n . Note that ${y\choose z}{x\choose z}$ is idempotent. We have that ${y\mu_{\varphi}\choose z'}{x\mu_{\varphi}\choose y\mu_{\varphi}}$ is also idempotent, and so z′=xμ _φ by Lemma 4.2. Hence, from

$$\left({x\choose z} {y\choose z}\right)\varphi={x\mu_{\varphi}\choose y \mu_{\varphi}} {y\mu_{\varphi}\choose x\mu_{\varphi}}={y \mu_{\varphi}\choose x\mu_{\varphi}}=\left({y\choose z}\right)\varphi $$

and φ is a bijection, we have that ${x\choose z}{y\choose z}={y\choose z}$ and so x=y, a contradiction.

We have proved Fact 3, and from Fact 2 we see that μ _φ ∈U _ρ , and so the proof of Lemma 4.5 is now completes.  □

Lemma 4.6

For any ${x\choose y}\in E^{w}_{\rho}$ and for any μ∈U _ρ we have that

$$\mu^{-1}{x\choose y}\mu={x\mu\choose y\mu}. $$

Proof

For every z∈X _n we have

$$z\mu^{-1}{x\choose y}\mu=\left \{ \begin{array}{c@{\quad}c} y\mu, & z\mu^{-1}=x\\ z, & z\mu^{-1}\neq x \end{array} \right .=\left \{ \begin{array}{c@{\quad}c} y\mu, & z=x\mu\\ z, & z \neq x\mu \end{array} \right .=z{x\mu\choose y \mu}, $$

and so $\mu^{-1}{x\choose y}\mu={x\mu\choose y\mu}$. □

We now can state and prove the main result of the paper as follows:

Theorem 4.7

Let S _n the symmetric group of X _n and U _ρ ={μ∈S _n :(xρ)μ=xρ,x∈X _n }. Then U _ρ is the unit group of T(ρ,⪯) and, for any $\varphi\in \operatorname{Aut} T(\rho,\preceq)$, there exists μ∈U _ρ such that (α)φ=μ ⁻¹ αμ for all α∈T(ρ,⪯).

Conversely, let μ∈U _ρ . Then μ ⁻¹ αμ∈T(ρ,⪯) for any α∈T(ρ,⪯), and the map φ:T(ρ,⪯)→T(ρ,⪯) defined by (α)φ=μ ⁻¹ αμ,α∈T(ρ,⪯) is an automorphism.

Proof

Let $\varphi\in \operatorname{Aut} T(\rho,\preceq)$. Then, first by Lemmas 4.5 and 4.6, there exists μ∈U _ρ such that

$$\left({x\choose y}\right)\varphi={x\mu\choose y\mu}=\mu^{-1}{x\choose y}\mu, \quad \hbox{for any } {x\choose y}\in E^{w}_{\rho}. $$

Secondly, we will prove that (α)φ=μ ⁻¹ αμ for all α∈U _ρ . For α∈U _ρ , we must have (α)φ∈U _ρ , and by Lemma 4.6, for any ${x\choose y}\in E^{w}_{\rho}$, we have

Hence, xαμ=xμ(α)φ and yαμ=yμ(α)φ. It follows that (α)φ=μ ⁻¹ αμ.

Finally, for any $\alpha\in T(\rho,\preceq)-E^{w}_{\rho}-U_{\rho}$, we have that α is a product of elements in $E^{w}_{\rho}\cup U_{\rho }$ by Proposition 2.5. It follows that (α)φ=μ ⁻¹ αμ as required.

The converse part is obvious. □

Remark

The μ is unique in Theorem 4.7. Indeed, let μ ₁,μ ₂ be elements of U _ρ such that $\mu_{1}^{-1}\alpha\mu_{1}=\mu_{2}^{-1}\alpha\mu_{2}$ for all α∈T(ρ,⪯). Now, given any x∈X _n , we can choose an element y∈X _n such that ${x\choose y}\in E^{w}_{\rho}$ or ${y\choose x}\in E^{w}_{\rho}$. For the former (Similarly for the latter), we have

$${x\mu_{1}\choose y\mu_{1}}=\mu_{1}^{-1}{x \choose y}\mu_{1}=\mu_{2}^{-1}{x\choose y} \mu_{2}={x\mu_{2}\choose y\mu_{2}}. $$

Hence xμ ₁=xμ ₂, and so μ ₁=μ ₂.

The map $\varPsi: \operatorname{Aut} T(\rho,\preceq)\rightarrow U_{\rho}$ defined by

$$(\varphi)\varPsi=\mu_{\varphi},\!\quad \varphi\in \operatorname{Aut} T(\rho,\preceq),\!\quad \mu_{\varphi}\in U_{\rho} \!\quad \hbox{with } (\alpha)\varphi= \mu_{\varphi}^{-1}\alpha\mu_{\varphi}, \!\quad\forall\alpha \in T(\rho,\preceq) $$

is an isomorphism. We have

Corollary 4.8

$\operatorname{Aut} T(\rho,\preceq) \cong U_{\rho}$.

Let PT _n be the semigroup of partial transformations on X _n . Let $S_{n}^{-}=\{\alpha\in T_{n}: x\alpha\leq x, \forall x\in X_{n}\}$ [9]. Then, by Corollary 4.8, we have

Corollary 4.9

$\operatorname{Aut} T_{n}\cong S_{n}$; $\operatorname{Aut} \mathit{PT}_{n} \cong S_{n}$ and $\operatorname{Aut} S_{n}^{-} \cong\{1_{X_{n}}\}$.