Semigroup Forum

, Volume 85, Issue 3, pp 513–524

# Automorphisms of partition order-decreasing transformation monoids

Open Access
RESEARCH ARTICLE

## 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).

### Keywords

Transformation semigroup Partition order-decreasing Automorphism group

## 1 Introduction

For the standard definitions on semigroups and transformation semigroups we refer the reader to the books [3, 4, 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 φM. 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 φM such that either =y or =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:AB, 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:=}). Denote the restriction of α to C by α| C for CA 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,bX n ,ab, which maps b to a and x to itself for any xX 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 αT n . Then the following statements are equivalent:
1. (1)

α is a unit of T(ρ,⪯).

2. (2)

αS n T(ρ,⪯).

3. (3)

αU ρ .

### Proof

Obviously (1)⇒(2).

(2)⇒(3). Suppose that αS n T(ρ,⪯). Then α −1S n T(ρ,⪯). Given any xX n . Take any y, we have
$$y\rho=\bigl(y\alpha\alpha^{-1}\bigr)\rho\preceq(y\alpha)\rho\preceq y \rho=x\rho,$$
and so . Thus ()α. To prove that ⊆()α consider z. 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 −1 and z=( −1)α. It follows that ()α= and hence αU ρ .

(3)⇒(1). By the definition of U ρ , we have that αT(ρ,⪯) and the restriction α| is a bijection from the ρ-class onto itself, and so αS n and α −1T(ρ,⪯). Thus α is a unit of T(ρ,⪯). □

Next we have

### Lemma 2.2

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

### Proof

First recall that αT n is an idempotent if and only if, for all $$t \in \operatorname{im} \alpha$$, t −1. Hence αT(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 −1 and x −1 implies that t== and =()ρ. □

### Corollary 2.3

Let a,bX n and ab. Then 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 {xX n :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{:xA i } from the definition of T(ρ,⪯). Choose b i A i such that b i ρ=min{:xA 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 }, yY and =min{:zY}. Then βT(ρ,⪯)−U ρ and γT(ρ,⪯). Moreover, γU ρ if and only if k=n−1.

Note that s(α)=ni>nk=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. (1)

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

2. (2)

for all s,tS 1, as=at(sa=ta) if and only if bs=bt(sb=tb).

For αT 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. (1)

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

2. (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 1Y 2≺⋯≺Y t }. If Y 1={x}, for some xX n , then, certainly =x=; Otherwise, choose yY 1 and yx, 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,yX n with xy, we may assume that . 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 orderon 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≤kn. 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 εT(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 1b 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 xX 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 1b 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 ρb 2 ρ. We have
1. (1)

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

2. (2)

If b 1b 2 then there exists δT(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 ρb 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 2ε 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: bb 1; b 1 ρ(Similar argument for b 1 ρ). By (2) of Lemma 4.3, there exists δT(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 1Y 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 xX 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 zt and Y 1, 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 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 = 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 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 φ =y . It follows that if then ( φ )ρ⪯( φ )ρ, and so μ φ U ρ .
Case 2: |Y 1|≥2. In this case, for any xX n , there exists yX n ∖{x} such that , 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 2X n . Since |Y 1|≥2, there exist y 1,y 2X 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 ρx 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′= φ .

Since $${x\choose y}\in E^{w}_{\rho}$$,we have , 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 yz. Then or 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 tX 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 φ = φ and so x=z, which contradicts xz.
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′= 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 μU ρ we have that
$$\mu^{-1}{x\choose y}\mu={x\mu\choose y\mu}.$$

### Proof

For every zX 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 ={μS n :()μ=,xX n }. Then U ρ is the unit group of T(ρ,⪯) and, for any $$\varphi\in \operatorname{Aut} T(\rho,\preceq)$$, there exists μU ρ such that (α)φ=μ −1 αμ for all αT(ρ,⪯).

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

### Proof

Let $$\varphi\in \operatorname{Aut} T(\rho,\preceq)$$. Then, first by Lemmas 4.5 and 4.6, there exists μU 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 αU ρ . For αU ρ , we must have (α)φU ρ , and by Lemma 4.6, for any $${x\choose y}\in E^{w}_{\rho}$$, we have
Hence, xαμ=(α)φ and yαμ=(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 αT(ρ,⪯). Now, given any xX n , we can choose an element yX 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 1= 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 $$\operatorname{Aut} S_{n}^{-} \cong\{1_{X_{n}}\}$$.

## Notes

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