Semigroup Forum

, Volume 85, Issue 3, pp 513–524

Automorphisms of partition order-decreasing transformation monoids

Open Access


Let Tn be the full transformation semigroup on a finite set Xn={1,2,…,n}. Let ρ be an equivalence relation on Xn and ⪯ be a total order on the partition set Xn/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).


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 Xn={1,2,…,n}. A submonoid M of the semigroup Tn of full transformations on Xn is intransitive if there exist x,y in Xn such that \((x)\varphi\not= y\) for any φM. A submonoid M of Tn said to be half-transitive provided that it is intransitive, and for every ordered pair (x,y)∈Xn×Xn there is some φM such that either =y or =x. In [10] we showed that, for every half-transitive submonoid M of Tn, there exist a non-universal equivalence relation ρ on Xn and a total order ⪯ on the partition set Xn/ρ such that M lies inside a half-transitive submonoid T(ρ,⪯) of Tn 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 Xn. In particular, if ρ is universal then T(ρ,⪯)=Tn; T(ρ,⪯) is the order-decreasing finite full transformation monoid if ρ is the identity relation [9]. If Xn/ρ={{1},Xn−{1}} and {1}≺Xn−{1} then T(ρ,⪯) is isomorphic to PTn−1, the semigroup of partial transformations of Xn−{1}.

Note that automorphisms of Tn and PTn−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 Tn [1] and, in general, T(ρ,⪯) does not contain all constant transformations of Xn. 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. 1A 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 \(\operatorname{im} \alpha=\{a_{1},\dots,a_{k}\}\), aiα−1=Ai (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 En−1 denote the set of idempotents in Tn of rank n−1. Every element e of En−1 has the form
$$e = {b\choose a} $$
for some a,bXn,ab, which maps b to a and x to itself for any xXn−{b} [4].

For the remainder of the paper, ρ will denote an equivalence relation on Xn and ⪯ will denote a total order on the partition set Xn/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, UρT(ρ,⪯). Let Sn be the symmetric group on Xn. The following lemma gives the description of the group of units of T(ρ,⪯).

Lemma 2.1

LetαTn. Then the following statements are equivalent:
  1. (1)

    αis a unit ofT(ρ,⪯).

  2. (2)


  3. (3)




Obviously (1)⇒(2).

(2)⇒(3). Suppose that αSnT(ρ,⪯). Then α−1SnT(ρ,⪯). Given any xXn. 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 αUρ.

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

Next we have

Lemma 2.2

LetαT(ρ,⪯). Thenαis an idempotent if and only if for all\(t \in \operatorname{im} \alpha\), =tand=min{:x−1}.


First recall that αTn 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

Leta,bXnandab. Thenif 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

Letαbe an element ofT(ρ,⪯)−Uρ. Thenαis a product of elements in\(E^{w}_{\rho}\cup U_{\rho}\).


Let s(α) be the cardinality of the set {xXn: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, aiρ⪯min{:xAi} from the definition of T(ρ,⪯). Choose biAi such that biρ=min{:xAi} for every i. Without loss of generality we may assume that
$$a_{1}=b_{1}, \dots, a_{i}=b_{i}, \quad\quad a_{i+1} \ne b_{i+1},\dots, a_{k}\ne b_{k}. $$
$$\beta= \left ( \begin{array}{c@{\quad}c@{\quad}c} A_{1} & \ldots & A_{k}\\ b_{1} & \ldots& b_{k} \end{array} \right ) \quad\mbox{and}\quad \gamma= \left ( \begin{array}{c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c} b_{1} & \ldots& b_{i} & b_{i+1} & \ldots & b_{k} & Y\\ b_{1} & \ldots& b_{i} & a_{i+1} & \ldots & a_{k} & y \end{array} \right ), $$
where Y=Xn−{b1,…,bk}, yY and =min{:zY}. Then βT(ρ,⪯)−Uρ and γT(ρ,⪯). Moreover, γUρ 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

LetUρand\(E^{w}_{\rho}\)be defined above. Then\(E^{w}_{\rho}\cup U_{\rho}\)is a generating set ofT(ρ,⪯).

Similar to the proof as Lemma 2.4 we also have

Lemma 2.6

Letαbe an idempotent ofT(ρ,⪯)−Uρ. Thenαis 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

LetSbe a semigroup and leta,bbe inS. The following conditions are equivalent:
  1. (1)


  2. (2)

    for alls,tS1, as=at(sa=ta) if and only ifbs=bt(sb=tb).


For αTn, \(\operatorname{ker} \alpha=\{(x,y)\in X_{n}\times X_{n}: x\alpha=y\alpha\}\). Then we have

Proposition 3.2

Letα,βbe elements ofT(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\(\operatorname{ker} \alpha= \operatorname{ker} \beta\).



(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 Xn/ρ={Y1Y2≺⋯≺Yt}. If Y1={x}, for some xXn, then, certainly =x=; Otherwise, choose yY1 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,yXn 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 relationρonXnand a total orderon the partition setXn/ρ, the semigroupT(ρ,⪯) is abundant.


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 biAi such that biρ=min{xiρ:xiAi} 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 αRβ 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 aiBi,i=1,…,k−1 and aiρ=min{xiρ:xiBi} for every i∈{1…,k−1}. Since akXn=B1∪⋯∪Bk−1, there exist a unique Bj such that akBj. Note that akaj, so ajBj∖{ak}. 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 ifb1=b2.


Assume that b1=b2. Then, from a1b1 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 xXn∖{b1},
$$x{b_{1}\choose a_{1}} {b_{1}\choose a_{2}}=x{b_{1}\choose a_{2}}= x = x{b_{1}\choose a_{1}} $$
and so \({b_{1}\choose a_{1}}{b_{1}\choose a_{2}} ={b_{1}\choose a_{1}}\).
Assume now that b1b2. 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 ifx=w.

Lemma 4.3

Let\({b_{1}\choose a_{1}}, {b_{2}\choose a_{2}} \in T(\rho,\preceq)\)andb1ρb2ρ. We have
  1. (1)

    Ifb1=b2then\({b_{1}\choose a_{1}}{\mathcal{L}}^{*} {b_{2}\choose a_{2}}\).

  2. (2)

    Ifb1b2then there existsδT(ρ,⪯) such that\({b_{1}\choose a_{1}}{\mathcal{R}}^{*}\delta{\mathcal{L}}^{*}{b_{2}\choose a_{2}}\).



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 b1ρb2ρ. 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\((E^{w}_{\rho})\varphi= E^{w}_{\rho}\).


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 (εi)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 ε1,ε2,…,εk of \(E^{w}_{\rho}\) such that ω=ε1ε2εk. 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=b1. 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: bb1; b1ρ(Similar argument for b1ρ). 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 existsμφUρsuch that\(\big({x\choose y}\big)\varphi= {x\mu_{\varphi} \choose y\mu_{\varphi}}\)for every\({x\choose y}\in E^{w}_{\rho}\).


Let Xn/ρ={Y1Y2≺⋯≺Yt}. We distinguish two cases: |Y1|=1 or |Y1|≥2.

Case 1: |Y1|=1. Let Y1={1}. For any xXn∖{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 Y1, 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 μφ:XnXn by 1μφ=1 and φ=x′ (defined the above), x≠1. Obviously, μφ 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 φy 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 μφUρ.
Case 2: |Y1|≥2. In this case, for any xXn, there exists yXn∖{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 μφ from Xn 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, μφ is surjective.

Fact 1

μφ is injective.

Indeed, assume that t=x1μφ=x2μφ, for x1,x2Xn. Since |Y1|≥2, there exist y1,y2Xn 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 x1=x2 from Lemma 4.1.

Fact 2

If x1ρx2ρ. Then (x1μφ)ρ⪯(x2μφ)ρ.

In fact, assume that (x1μφ)ρ≻(x2μφ)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=x2 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 (x1μφ)ρ=(x2μφ)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 tXn. 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′∈Xn. 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 μφUρ, 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 anyμUρwe have that
$$\mu^{-1}{x\choose y}\mu={x\mu\choose y\mu}. $$


For every zXn 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

LetSnthe symmetric group ofXnandUρ={μSn:()μ=,xXn}. ThenUρis the unit group ofT(ρ,⪯) and, for any\(\varphi\in \operatorname{Aut} T(\rho,\preceq)\), there existsμUρsuch that (α)φ=μ−1αμfor allαT(ρ,⪯).

Conversely, letμUρ. Thenμ−1αμT(ρ,⪯) for anyαT(ρ,⪯), and the mapφ:T(ρ,⪯)→T(ρ,⪯) defined by (α)φ=μ−1αμ,αT(ρ,⪯) is an automorphism.


Let \(\varphi\in \operatorname{Aut} T(\rho,\preceq)\). Then, first by Lemmas 4.5 and 4.6, there exists μUρ 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−1αμ for all αUρ. For αUρ, we must have (α)φUρ, 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−1αμ.

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−1αμ as required.

The converse part is obvious. □


The μ is unique in Theorem 4.7. Indeed, let μ1,μ2 be elements of Uρ such that \(\mu_{1}^{-1}\alpha\mu_{1}=\mu_{2}^{-1}\alpha\mu_{2}\) for all αT(ρ,⪯). Now, given any xXn, we can choose an element yXn 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 μ1=μ2.
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 PTn be the semigroup of partial transformations on Xn. 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\(\operatorname{Aut} S_{n}^{-} \cong\{1_{X_{n}}\}\).


Open Access

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Copyright information

© The Author(s) 2012

Authors and Affiliations

  1. 1.Qiangjiang CollegeHangzhou Normal UniversityHangzhou CityP.R. China
  2. 2.Department of MathematicsHangzhou Normal UniversityHangzhou CityP.R. China

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