Structural properties and isomorphism theorems for Cayley digraphs of full transformation semigroups with respect to Green's equivalence classes

Abstract

Let $T(X)$ be the full transformation semigroup on a nonempty set X. In this paper, the Cayley digraphs of $T(X)$ with connection sets L and R, the Green's equivalence classes of $T(X)$ according to the Green's relations $\mathcal{L}$ and $\mathcal{R}$, are investigated. Furthermore, their connectedness properties are characterized. In addition, the isomorphism theorems for Cayley digraphs of $T(X)$ are also presented.

1. Introduction and preliminaries

Let A be a nonempty subset of a semigroup S. The Cayley digraph $\mathrm{Cay}(S,A)$ of a semigroup S relative to A is known as a digraph with vertex set S and arc set consisting of ordered pairs $(x,xa)\in S\times S$ for some $a\in A$. The set A will be called a connection set of $\mathrm{Cay}(S,A)$. Cayley digraphs of semigroups have been extensively investigated, see, for example [1], [2], [3], [4], [5]. One of semigroups, which is widely considered, is the transformation semigroup since any semigroup can be embedded into a transformation semigroup on an appropriate set. This would be a general result why the transformation semigroup is interesting to study. Several authors studied structural properties of Cayley graphs of transformation semigroups, see, for example [6], [7], [8], [9], [10], [11].

We now provide important preliminaries for this paper. Throughout the paper, all sets are considered to be finite. Let X be a nonempty set and $T(X)$ the set of all full transformations from X into itself. It is well known that $T(X)$ is a regular semigroup under the composition of functions. Throughout this paper, we write the functions on the right, that is, for a composition αβ, α is applied first. We now introduce the definition of Green's relations $\mathcal{L}$ and $\mathcal{R}$ for which $\mathcal{L}$-classes and $\mathcal{R}$-classes are considered to be connections sets of Cayley digraphs of $T(X)$.

Let S be a semigroup and $S^1$ denote a semigroup obtained from S by adjoining an identity if S has no identity and $S^1=S$ if it already contains an identity. The following definitions are due to Green. For any $a,b\in S$, define

$$a\mathcal{L}b\text{ if and only if }{S}^{1}a=S^{1}b,$$

or equivalently, $a\mathcal{L}b$ if and only if $a=xb,b=ya$ for some $x,y\in S^1$. Dually, define

$$a\mathcal{R}b\text{ if and only if }a{S}^1=b{S}^1,$$

or equivalently, $a\mathcal{R}b$ if and only if $a=bx,b=ay$ for some $x,y\in S^1$. Moreover, define

$$a\mathcal{H}b\text{ if and only if }a\mathcal{L}b\text{ and }a\mathcal{R}b,$$

that is, $\mathcal{H}=\mathcal{L}\cap \mathcal{R}$. Note that the relations $\mathcal{L},\mathcal{R}$ and $\mathcal{H}$ are all equivalence relations on S. Furthermore, for each $a\in S$, denote by $L_a,R_a$ and $H_a$ the $\mathcal{L}$-class, $\mathcal{R}$-class and $\mathcal{H}$-class containing a, respectively.

Let $\alpha \in T(X)$. The notation xα means the image of x under α and $X\alpha =\{x\alpha :x\in X\}$ is called the image of α. In addition, ${\pi }_{\alpha }$ stands for the set $\{x{\alpha }^{-1}:x\in X\alpha \}$. Especially, for the transformation α such that $X\alpha =\{a_1,a_2,\dots ,a_n\}$, we usually write this α as

$$\alpha =\left(\begin{array}{cccc}\hfill A_1\hfill & \hfill A_2\hfill & \hfill \dots \hfill & \hfill A_n\hfill \\ \hfill a_1\hfill & \hfill a_2\hfill & \hfill \dots \hfill & \hfill a_n\hfill \end{array}\right)$$

and take as understood that $a_i{\alpha }^{-1}=A_i$ for all $i=1,2,\dots ,n$, that is, ${\pi }_{\alpha }=\{A_1,A_2,\dots ,A_n\}$ is a partition of X. The following theorem shows the characterizations of Green's relations $\mathcal{L}$ and $\mathcal{R}$ on $T(X)$.

Theorem 1.1

[12] For any $\alpha ,\beta \in T(X)$ ,

• 1. $\alpha \mathcal{L}\beta$ if and only if $X\alpha =X\beta$ ,
• 2. $\alpha \mathcal{R}\beta$ if and only if ${\pi }_{\alpha }={\pi }_{\beta }$ .

In this paper, the Cayley digraphs of the full transformation semigroup $T(X)$, where X is a finite nonempty set, with the connection sets L and R, the Green's equivalence classes of $T(X)$ have been constructed for studying their structural properties. Furthermore, the isomorphism theorems of the Cayley digraphs have been presented.

2. Cayley digraphs of $T(X)$ with respect to $\mathcal{L}$-classes

This section provides structural properties of Cayley digraphs of $T(X)$ with respect to $\mathcal{L}$-classes and X is a nonempty finite set. Hereafter, we denote by $T_n$ the full transformation semigroup $T(X)$ when $X=\{1,2,\dots ,n\}$ for some $n\in \mathbb{N}$. Recall that, for any $\lambda \in T_n$, the $\mathcal{L}$-class containing λ is $L_{\lambda }=\{\alpha :X\alpha =X\lambda \}$. Hence, each $\mathcal{L}$-class of $T_n$ is of the form $L_{\mathcal{A}}=\{\alpha \in T_n:X\alpha =\mathcal{A}\}$, where $\varnothing \ne \mathcal{A}\subseteq X$. Note that a digraph D is said to be totally complete if $E(D)=V(D)\times V(D)$. We now start with some characterizations of Cayley digraphs of $T_n$ with certain $\mathcal{L}$-classes.

Theorem 2.1

Let$\lambda \in T_n$and$L_{\lambda }$be an$\mathcal{L}$-class containing λ. Let$\mathrm{\Gamma }:=\mathrm{Cay}(T_n,L_{\lambda })$. If$\delta \in L_{\lambda }$, then an induced subdigraph$\mathrm{\Gamma }[H_{\delta }]$of Γ is totally complete where $H_{\delta }$ is an $\mathcal{H}$-class of $T_n$ containing δ.

Proof

Let $\delta \in L_{\lambda }$ and $\alpha ,\beta \in H_{\delta }$. Since $\delta \mathcal{L}\lambda$, we have $X\delta =X\lambda$. Then we can write

$$\lambda =\left(\begin{array}{cccc}\hfill A_1\hfill & \hfill A_2\hfill & \hfill \dots \hfill & \hfill A_k\hfill \\ \hfill a_1\hfill & \hfill a_2\hfill & \hfill \dots \hfill & \hfill a_k\hfill \end{array}\right)\text{ and }\delta =\left(\begin{array}{cccc}\hfill B_1\hfill & \hfill B_2\hfill & \hfill \dots \hfill & \hfill B_k\hfill \\ \hfill a_1\hfill & \hfill a_2\hfill & \hfill \dots \hfill & \hfill a_k\hfill \end{array}\right).$$

Further, since $\alpha \mathcal{H}\delta \mathcal{H}\beta$, we obtain that $X\alpha =X\delta =X\beta$ and ${\pi }_{\alpha }={\pi }_{\delta }={\pi }_{\beta }$. Thus we can write

$$\alpha =\left(\begin{array}{cccc}\hfill B_1\hfill & \hfill B_2\hfill & \hfill \dots \hfill & \hfill B_k\hfill \\ \hfill a_{\varphi (1)}\hfill & \hfill a_{\varphi (2)}\hfill & \hfill \dots \hfill & \hfill a_{\varphi (k)}\hfill \end{array}\right)\text{ and }\beta =\left(\begin{array}{cccc}\hfill B_1\hfill & \hfill B_2\hfill & \hfill \dots \hfill & \hfill B_k\hfill \\ \hfill a_{\phi (1)}\hfill & \hfill a_{\phi (2)}\hfill & \hfill \dots \hfill & \hfill a_{\phi (k)}\hfill \end{array}\right),$$

where ϕ and φ are bijections on the set $\{1,2,\dots ,k\}$ and $\{a_{\varphi (i)}:i=1,2,\dots ,k\}=\{a_1,a_2,\dots ,a_k\}=\{a_{\phi (i)}:i=1,2,\dots ,k\}$. Let $\gamma \in T_n$ be defined by

$$\gamma =\left(\begin{array}{cccc}\hfill a_{\varphi (1)}\hfill & \hfill a_{\varphi (2)}\hfill & \hfill \dots \hfill & \hfill a_{\varphi (k)}\cup C\hfill \\ \hfill a_{\phi (1)}\hfill & \hfill a_{\phi (2)}\hfill & \hfill \dots \hfill & \hfill a_{\phi (k)}\hfill \end{array}\right),$$

where $C=X\setminus \{a_{\varphi (i)}:i=1,2,\dots ,k\}$. Clearly, $X\gamma =X\lambda$ which implies that $\gamma \in L_{\lambda }$. Moreover, we obtain that $\beta =\alpha \gamma$. Thus $(\alpha ,\beta )\in E(\mathrm{\Gamma })$. From $\alpha ,\beta \in H_{\delta }$, we have $(\alpha ,\beta )\in E(\mathrm{\Gamma }[H_{\delta }])$. This completes the proof. □

In order to present the next theorem, we need to note that ${\mathcal{S}}_n$ means the group of all permutations in $T_n$ with identity $i{d}_X$ where $|X|=n$. We now illustrate an example of the characterization for the Cayley digraph of $T_n$ with respect to the connection set ${\mathcal{S}}_n$ which is one of $\mathcal{L}$-classes of $T_n$.

Example 2.2

Let $X=\{1,2,3\}$. Then $|T_3|=27$. For convenience, each element $\left(\begin{array}{ccc}\hfill 1\hfill & \hfill 2\hfill & \hfill 3\hfill \\ \hfill x\hfill & \hfill y\hfill & \hfill z\hfill \end{array}\right)$ in $T_3$ will be written as “xyz”. Then ${\mathcal{S}}_3=\{123,132,213,231,312,321\}$ and thus $\mathrm{Cay}(T_3,{\mathcal{S}}_3)$ is shown in Fig. 1. We can observe that the set of all vertices in the same component forms an $\mathcal{R}$-class of $T_3$. Hence $\mathrm{Cay}(T_3,{\mathcal{S}}_3)$ is the disjoint union of totally complete subdigraphs which each of them is induced by an $\mathcal{R}$-class of $T_3$.

Figure 1.

$\mathrm{Cay}(T_3,{\mathcal{S}}_3)$.

Theorem 2.3

Let L be an$\mathcal{L}$-class of$T_n$and$\mathrm{\Gamma }:=\mathrm{Cay}(T_n,L)$. Then Γ is the disjoint union of totally complete induced subgraphs $\mathrm{\Gamma }[R]$, where R is an $\mathcal{R}$-class of $T_n$, if and only if $L={\mathcal{S}}_n$.

Proof

Let $L={\mathcal{S}}_n$ and R be an $\mathcal{R}$-class of $T_n$. Let $\alpha ,\beta \in R$. Then ${\pi }_{\alpha }={\pi }_{\beta }$ and we can write

$$\alpha =\left(\begin{array}{cccc}\hfill A_1\hfill & \hfill A_2\hfill & \hfill \dots \hfill & \hfill A_k\hfill \\ \hfill a_1\hfill & \hfill a_2\hfill & \hfill \dots \hfill & \hfill a_k\hfill \end{array}\right)\text{ and }\beta =\left(\begin{array}{cccc}\hfill A_1\hfill & \hfill A_2\hfill & \hfill \dots \hfill & \hfill A_k\hfill \\ \hfill b_1\hfill & \hfill b_2\hfill & \hfill \dots \hfill & \hfill b_k\hfill \end{array}\right).$$

We can observe that $|X\alpha |=k=|X\beta |$. It follows that $|X\setminus X\alpha |=n-k=|X\setminus X\beta |$. Let $X\setminus X\alpha =\{c_i:i=1,2,\dots ,n-k\}$ and $X\setminus X\beta =\{d_i:i=1,2,\dots ,n-k\}$. Define $\gamma \in T_n$ by

$$\gamma =\left(\begin{array}{cccccccc}\hfill a_1\hfill & \hfill a_2\hfill & \hfill \dots \hfill & \hfill a_k\hfill & \hfill c_1\hfill & \hfill c_2\hfill & \hfill \dots \hfill & \hfill c_{n-k}\hfill \\ \hfill b_1\hfill & \hfill b_2\hfill & \hfill \dots \hfill & \hfill b_k\hfill & \hfill d_1\hfill & \hfill d_2\hfill & \hfill \dots \hfill & \hfill d_{n-k}\hfill \end{array}\right).$$

Thus $\gamma \in {\mathcal{S}}_n$. Furthermore, $\beta =\alpha \gamma$ which yields that $(\alpha ,\beta )\in E(\mathrm{\Gamma })$. Hence $\mathrm{\Gamma }[R]$ forms a totally complete subdigraph of Γ.

Next, let $(\alpha ,\beta )$ be an arbitrary arc of Γ. Then there exists $\lambda \in L={\mathcal{S}}_n$ in which $\beta =\alpha \lambda$. It follows that $\alpha =\beta {\lambda }^{-1}$ such that ${\lambda }^{-1}\in {\mathcal{S}}_n$. That means $\alpha \mathcal{R}\beta$. Thus α and β belong to the same $\mathcal{R}$-class of $T_n$. Therefore, Γ is the disjoint union of totally complete subdigraphs $\mathrm{\Gamma }[R]$ where R is an $\mathcal{R}$-class of $T_n$.

Conversely, assume that the condition holds. Let $\alpha \in L$. It is easy to verify that $(i{d}_X,\alpha )\in E(\mathrm{\Gamma })$. By the assumption, we have $\alpha \in R_{i{d}_X}$ which implies that ${\pi }_{\alpha }={\pi }_{i{d}_X}$. Thus $\alpha \in {\mathcal{S}}_n$. On the other hand, let $\alpha \in {\mathcal{S}}_n$. Then ${\pi }_{\alpha }={\pi }_{i{d}_X}$ and hence $\alpha \mathcal{R}i{d}_X$. Since a subdigraph of Γ induced by an $\mathcal{R}$-class is totally complete, we have $(i{d}_X,\alpha )\in E(\mathrm{\Gamma }[R])\subseteq E(\mathrm{\Gamma })$. Hence $\alpha =i{d}_X\beta$ for some $\beta \in L={\mathcal{S}}_n$, that is, $\alpha =\beta \in {\mathcal{S}}_n$. Consequently, $L={\mathcal{S}}_n$, as required. □

For each $a\in X$, denote by ${\chi }_a$ the constant map $\left(\begin{array}{c}\hfill X\hfill \\ \hfill a\hfill \end{array}\right)\in T_n$. It is easy to see that $\{{\chi }_a\}=\{\alpha \in T_n:X\alpha =\{a\}\}$ is an $\mathcal{L}$-class of $T_n$. The following theorem is the characterization of Cayley digraphs with respect to $\mathcal{L}$-classes of such type.

Theorem 2.4

Let L be an $\mathcal{L}$ -class of $T_n$ , $\mathrm{\Gamma }:=\mathrm{Cay}(T_n,L)$ and $a\in X$ . Then $L=\{{\chi }_a\}$ if and only if $E(\mathrm{\Gamma })=\{(\alpha ,{\chi }_a):\alpha \in T_n\}$ .

Proof

Let $L=\{{\chi }_a\}$ and $\alpha \in T_n$. Since ${\chi }_a=\alpha {\chi }_a$ where ${\chi }_a\in L$, we have $(\alpha ,{\chi }_a)\in E(\mathrm{\Gamma })$. We next let $(\lambda ,\delta )\in E(\mathrm{\Gamma })$. Then $\delta =\lambda {\chi }_a$ where ${\chi }_a\in L$. Hence $\delta ={\chi }_a$. Therefore, $E(\mathrm{\Gamma })=\{(\alpha ,{\chi }_a):\alpha \in T_n\}$.

Conversely, assume that $E(\mathrm{\Gamma })=\{(\alpha ,{\chi }_a):\alpha \in T_n\}$. Clearly, $(i{d}_X,{\chi }_a)\in E(\mathrm{\Gamma })$ which leads to ${\chi }_a=i{d}_X\lambda$ where $\lambda \in L$. Thus ${\chi }_a=\lambda \in L$. Next, suppose that there exists $\alpha \in L$ in which $\alpha \ne {\chi }_a$. This gives $(i{d}_X,\alpha )\in E(\mathrm{\Gamma })$ which contradicts to the assumption. So we conclude that $L=\{{\chi }_a\}$. □

Let $\mathrm{\Gamma }:=\mathrm{Cay}(T_n,L)$. For each $\alpha \in T_n$, we note that

$$N(\alpha )=\{\beta \in T_n:(\alpha ,\beta )\in E(\mathrm{\Gamma })\}.$$

If $(\alpha ,\beta )\in E(\mathrm{\Gamma })$, then β will be called a neighbour of $\alpha \in T_n$ and the set $N(\alpha )$ is called the neighbourhood of $\alpha \in T_n$. Next, we characterize the neighbourhoods of constant maps and permutations in $T_n$.

Lemma 2.5

Let$L_{\mathcal{A}}$be a connection set of Γ and $a\in X$. Then $N({\chi }_a)=\{{\chi }_b:b\in \mathcal{A}\}$.

Proof

Let $a\in X$. To show that $N({\chi }_a)=\{{\chi }_b:b\in \mathcal{A}\}$, we first let $\alpha \in N({\chi }_a)$. Then $({\chi }_a,\alpha )\in E(\mathrm{\Gamma })$, that is, $\alpha ={\chi }_a\beta$ for some $\beta \in L_{\mathcal{A}}$. It follows that $\alpha ={\chi }_{a\beta }$ where $a\beta \in \mathcal{A}$. On the other hand, let $b\in \mathcal{A}$. We prove that $({\chi }_a,{\chi }_b)\in E(\mathrm{\Gamma })$. For convenience, let $\mathcal{A}=\{b,a_1,a_2,\dots ,a_k\}$ and $X\setminus \mathcal{A}=C$. Let $I=\{1,2,\dots ,k\}$. We have two cases to consider.

Case 1:$a=a_l$ for some $l\in I$. Then define

$$\alpha ={\left(\begin{array}{ccc}\hfill \{a\}\cup C\hfill & \hfill b\hfill & \hfill a_i\hfill \\ \hfill b\hfill & \hfill a\hfill & \hfill a_i\hfill \end{array}\right)}_{i\in I\setminus \{l\}}.$$

Hence $\alpha \in L_{\mathcal{A}}$ and ${\chi }_b={\chi }_a\alpha$ which yields $({\chi }_a,{\chi }_b)\in E(\mathrm{\Gamma })$.

Case 2:$a=b$ or $a\notin \mathcal{A}$. We then define

$$\beta ={\left(\begin{array}{cc}\hfill \{b\}\cup C\hfill & \hfill a_i\hfill \\ \hfill b\hfill & \hfill a_i\hfill \end{array}\right)}_{i\in I}.$$

Thus $\beta \in L_{\mathcal{A}}$ and ${\chi }_b={\chi }_b\beta ={\chi }_a\beta$ and so $({\chi }_a,{\chi }_b)\in E(\mathrm{\Gamma })$.

From the above two cases, we have ${\chi }_b\in N({\chi }_a)$. □

Lemma 2.6

Let$L_{\mathcal{A}}$be a connection set of Γ in which $1\le |\mathcal{A}|\le n-1$. If $\alpha \in {\mathcal{S}}_n$, then $N(\alpha )=L_{\mathcal{A}}$.

Proof

Let $\alpha \in {\mathcal{S}}_n$. We first show that $L_{\mathcal{A}}$ contains $N(\alpha )$. Let $\beta \in N(\alpha )$. Then $(\alpha ,\beta )\in E(\mathrm{\Gamma })$, that is, $\beta =\alpha \gamma$ for some $\gamma \in L_{\mathcal{A}}$. We need to prove that $X\beta =\mathcal{A}$. Clearly, $X\beta \subseteq X\gamma =\mathcal{A}$. Next, let $a\in \mathcal{A}$. Thus there exists $b\in X$ such that $b\gamma =a$. Since α is surjective, there exists $c\in X$ in which $c\alpha =b$. Hence $c\beta =c(\alpha \gamma )=(c\alpha )\gamma =b\gamma =a$ which implies that $a\in X\beta$ and thus $\mathcal{A}\subseteq X\beta$. Therefore, $X\beta =\mathcal{A}$ and then $\beta \in L_{\mathcal{A}}$.

For the other containment, we let $\beta \in L_{\mathcal{A}}$. Since α is surjective, we have $X\alpha =X$. Define γ by $(x\alpha )\gamma =x\beta$ for every $x\in X$. Since α is bijective, γ is well-defined. This implies that $\gamma \in T_n$. We now show that $(\alpha ,\beta )\in E(\mathrm{\Gamma })$. For each $x\in X$, $x\alpha \gamma =(x\alpha )\gamma =x\beta$. Thus $\alpha \gamma =\beta$. From the definition of γ, it is obvious that $X\gamma \subseteq X\beta =\mathcal{A}$. Let $a\in \mathcal{A}$. There exists $x\in X$ in which $x\beta =a$. Hence $(x\alpha )\gamma =x\beta =a$, that is, $a\in X\gamma$. We have $\mathcal{A}\subseteq X\gamma$ which yields that $X\gamma =\mathcal{A}$. Therefore, $\gamma \in L_{\mathcal{A}}$. Consequently, $(\alpha ,\beta )\in E(\mathrm{\Gamma })$ and then $\beta \in N(\alpha )$. □

Next, we provide necessary and sufficient conditions for being arcs of Γ which also give the characterization of neighbours of α where α is neither a permutation nor a constant map in $T_n$. To purpose the results, we need the following definition.

Let $\mathcal{A}$ and $\mathcal{B}$ be families of sets. We say that $\mathcal{A}$ refines $\mathcal{B}$ if for each $A\in \mathcal{A}$, there exists $B\in \mathcal{B}$ in which $A\subseteq B$.

Lemma 2.7

Let $\alpha ,\beta \in T_n$ . Then $\beta \in \alpha T_n$ if and only if ${\pi }_{\alpha }$ refines ${\pi }_{\beta }$ .

Proof

Let $\alpha ,\beta \in T_n$. Assume that $\beta \in \alpha T_n$. Then $\beta =\alpha \gamma$ for some $\gamma \in T_n$. Let $A\in {\pi }_{\alpha }$. Thus $A=x{\alpha }^{-1}$ for some $x\in X\alpha$. For each $y\in A$, we have $y\alpha =x$ and so $x\gamma =(y\alpha )\gamma =y(\alpha \gamma )=y\beta \in X\beta$. Hence $(x\gamma ){\beta }^{-1}\in {\pi }_{\beta }$. We now prove that A is contained in $(x\gamma ){\beta }^{-1}$. Let $a\in A$. It follows that $a\alpha =x$ and thus $a\alpha \gamma =x\gamma$. That means $a\beta =x\gamma$ which implies that $a\in (x\gamma ){\beta }^{-1}$. Therefore, $A\subseteq (x\gamma ){\beta }^{-1}\in {\pi }_{\beta }$. Consequently, ${\pi }_{\alpha }$ refines ${\pi }_{\beta }$.

Conversely, assume that ${\pi }_{\alpha }$ refines ${\pi }_{\beta }$. For each $x\in X\alpha$, there exists $x^{\prime }\in X$ in which $x^{\prime }\alpha =x$. Define $\gamma \in T_n$ as follows:

$$x\gamma =\{\begin{array}{cc}{x}^{\prime }\beta ,\hfill & x\in X\alpha ,\hfill \\ x\alpha ,\hfill & x\notin X\alpha .\hfill \end{array}$$

Hence, for each $x\in X$, we obtain that $x\alpha \gamma =(x\alpha )\gamma ={(x\alpha )}^{\prime }\beta$. Since $x\alpha \in X\alpha$, we have ${(x\alpha )}^{\prime }\alpha =x\alpha$. Then there exists $P\in {\pi }_{\alpha }$ in which $\{{(x\alpha )}^{\prime },x\}\subseteq P$. Since ${\pi }_{\alpha }$ refines ${\pi }_{\beta }$, there exists $P^{\prime }\in {\pi }_{\beta }$ such that $P\subseteq P^{\prime }$ and so $\{{(x\alpha )}^{\prime },x\}\subseteq P^{\prime }$. It follows that ${(x\alpha )}^{\prime }\beta =x\beta$. Hence $\alpha \gamma =\beta$ and thus $\beta \in \alpha T_n$, as required. □

Theorem 2.8

Let$L_{\mathcal{A}}$be a connection set of Γ and $\alpha ,\beta \in T_n$. Then $(\alpha ,\beta )\in E(\mathrm{\Gamma })$ if and only if the following statements hold:

• 1.${\pi }_{\alpha }$refines${\pi }_{\beta }$,
• 2.$X\beta \subseteq \mathcal{A}$,
• 3.$|X\setminus X\alpha |\ge |\mathcal{A}\setminus X\beta |$.

Proof

Assume that $(\alpha ,\beta )\in E(\mathrm{\Gamma })$. Then there exists $\gamma \in L_{\mathcal{A}}$ such that $\beta =\alpha \gamma$. By Lemma 2.7, we obtain that ${\pi }_{\alpha }$ refines ${\pi }_{\beta }$. Moreover, we have $X\beta =X\alpha \gamma \subseteq X\gamma =\mathcal{A}$. As the fact that $\beta =\alpha \gamma$, the number of sets in ${\pi }_{\gamma }$ containing elements of Xα is $|X\beta |$. Hence the number of sets in ${\pi }_{\gamma }$ which do not contain elements of Xα is exactly $|\mathcal{A}\setminus X\beta |$. Therefore, $|X\setminus X\alpha |\ge |\mathcal{A}\setminus X\beta |$.

Conversely, assume that the conditions hold. We will show that there exists $\gamma \in L_{\mathcal{A}}$ such that $\beta =\alpha \gamma$. Since $|X\setminus X\alpha |\ge |\mathcal{A}\setminus X\beta |$, there exists a surjective function $\sigma :X\setminus X\alpha \to \mathcal{A}\setminus X\beta$. Moreover, for each $x\in X\alpha$, there will be $x^{\prime }\in X$ such that $x^{\prime }\alpha =x$. Define $\gamma \in T_n$ by

$$x\gamma =\{\begin{array}{cc}{x}^{\prime }\beta ,\hfill & x\in X\alpha ,\hfill \\ x\sigma ,\hfill & x\notin X\alpha .\hfill \end{array}$$

Let $x\in X$. Then $x\alpha \in X\alpha$ and thus there exists ${(x\alpha )}^{\prime }\in X$ such that ${(x\alpha )}^{\prime }\alpha =x\alpha$. Hence ${(x\alpha )}^{\prime },x\in P$ for some $P\in {\pi }_{\alpha }$. Since ${\pi }_{\alpha }$ refines ${\pi }_{\beta }$, we obtain that ${(x\alpha )}^{\prime }\beta =x\beta$. Therefore, $x\alpha \gamma =(x\alpha )\gamma ={(x\alpha )}^{\prime }\beta =x\beta$ which implies that $\beta =\alpha \gamma$ and $(X\alpha )\gamma =X\beta$. Since $X\gamma =(X\alpha )\gamma \cup (X\setminus X\alpha )\gamma =X\beta \cup (\mathcal{A}\setminus X\beta )=\mathcal{A}$, $\gamma \in L_{\mathcal{A}}$. Consequently, $(\alpha ,\beta )\in E(\mathrm{\Gamma })$. □

In order to purpose the following theorem, we need to define certain notation. For a nonempty subset S of $T_n$, the neighbourhood of S in $T_n$ is $N(S)=\bigcup _{\alpha \in S}N(\alpha )$ where $N(\alpha )$ is the neighbourhood of α in $T_n$.

Theorem 2.9

Let$L_{\mathcal{A}}$be a connection set of Γ such that $\varnothing \ne \mathcal{A}⊊X$. If $|\mathcal{A}|=k$ and $|X|=n$, then $N(L_{\mathcal{A}})$ is the union of $\mathcal{L}$-classes $L_B$ where $B\subseteq \mathcal{A}$ and $|B|\ge 2k-n$.

Proof

Let $\varnothing \ne \mathcal{A}⊊X$ be such that $|\mathcal{A}|=k$ and $|X|=n$. Then $k\le n-1$. Let $\alpha \in N(L_{\mathcal{A}})$. Thus $\alpha \in N(\beta )$ for some $\beta \in L_{\mathcal{A}}$ and hence $\alpha =\beta \gamma$ for some $\gamma \in L_{\mathcal{A}}$. We obtain that $X\alpha =X\beta \gamma \subseteq X\gamma =\mathcal{A}$. Again from $\alpha =\beta \gamma$, we can conclude by Theorem 2.8 that $|X\setminus X\beta |\ge |\mathcal{A}\setminus X\alpha |$. This implies that $n-k\ge k-|X\alpha |$ and so $|X\alpha |\ge 2k-n$. Therefore,

$$\alpha \in L_{X\alpha }\subseteq \bigcup _{B\subseteq \mathcal{A}}{L}_B,$$

where $|B|\ge 2k-n$. Next, let $\alpha \in L_D$ for some $D\subseteq \mathcal{A}$ and $|D|\ge 2k-n$. For convenience, we may assume that $\mathcal{A}=\{a_1,a_2,\dots ,a_k\}$. We now consider the following two cases.

Case 1:$D=\mathcal{A}$. Then $X\alpha =\mathcal{A}$. So we can write

$$\alpha =\left(\begin{array}{cccc}\hfill A_1\hfill & \hfill A_2\hfill & \hfill \dots \hfill & \hfill A_k\hfill \\ \hfill a_1\hfill & \hfill a_2\hfill & \hfill \dots \hfill & \hfill a_k\hfill \end{array}\right).$$

Let $C=X\setminus \mathcal{A}$ and define

$$\beta =\left(\begin{array}{cccc}\hfill \{a_1\}\cup C\hfill & \hfill a_2\hfill & \hfill \dots \hfill & \hfill a_k\hfill \\ \hfill a_1\hfill & \hfill a_2\hfill & \hfill \dots \hfill & \hfill a_k\hfill \end{array}\right).$$

Clearly, $\beta \in L_{\mathcal{A}}$ and $\alpha =\alpha \beta$. Thus $(\alpha ,\alpha )\in E(\mathrm{\Gamma })$, that is, $\alpha \in N(\alpha )\subseteq N(L_{\mathcal{A}})$.

Case 2:$D⊊\mathcal{A}$. Let $D=\{d_1,d_2,\dots ,d_t\}$ and $\mathcal{A}\setminus D=\{z_1,z_2,\dots ,z_{k-t}\}$ be such that $t<k$ and $t\ge 2k-n$. We can write α as follows:

$$\alpha =\left(\begin{array}{cccc}\hfill A_1\hfill & \hfill A_2\hfill & \hfill \dots \hfill & \hfill A_t\hfill \\ \hfill d_1\hfill & \hfill d_2\hfill & \hfill \dots \hfill & \hfill d_t\hfill \end{array}\right).$$

Since $t<k<n$, there exists $A_j$ such that $|A_j|\ge 2$ for some $j\in \{1,2,\dots ,t\}$. Choose $a_j\in A_j$. Let $A_j^{\prime }=A_j\setminus \{a_j\}$. Since $|\{d_s\in D:s\ne j\}\cup \{d_j\}\cup (\mathcal{A}\setminus D)|=k$, we can choose $b_1,b_2,\dots ,b_{k-t}\in \bigcup _{j=1}^t{A}_j^{\prime }$ which are all distinct. Further, since $t\ge 2k-n$, we have $n\ge 2k-t$ and thus $n-k\ge k-t$. Now, choose $w_1,w_2,\dots ,w_{k-t}\in X\setminus \mathcal{A}$ and let $Y=X\setminus (\mathcal{A}\cup \{w_1,w_2,\dots ,w_{k-t}\})$. Define β and γ as follows:

$$\begin{gathered} \beta ={\left(\begin{array}{cccccc}\hfill A_s\hfill & \hfill A_j\setminus \{b_1,b_2,\dots ,b_{k-t}\}\hfill & \hfill b_1\hfill & \hfill b_2\hfill & \hfill \dots \hfill & \hfill b_{k-t}\hfill \\ \hfill d_s\hfill & \hfill d_j\hfill & \hfill z_1\hfill & \hfill z_2\hfill & \hfill \dots \hfill & \hfill z_{k-t}\hfill \end{array}\right)}_{s\in \{1,2,\dots t\}\setminus \{j\}}\text{ and } \\ \gamma ={\left(\begin{array}{cccccc}\hfill d_s\hfill & \hfill \{d_j\}\cup T_j\hfill & \hfill w_1\hfill & \hfill w_2\hfill & \hfill \dots \hfill & \hfill \{w_{k-t}\}\cup Y\hfill \\ \hfill d_s\hfill & \hfill d_j\hfill & \hfill z_1\hfill & \hfill z_2\hfill & \hfill \dots \hfill & \hfill z_{k-t}\hfill \end{array}\right)}_{s\in \{1,2,\dots t\}\setminus \{j\}}, \end{gathered}$$

where $T_j=\{z_m:b_m\in A_j\}$. Therefore, $\beta \gamma =\alpha$ where $\gamma \in L_{\mathcal{A}}$. Consequently, $\alpha \in N(\beta )\subseteq N(L_{\mathcal{A}})$. □

Now, we are ready to present characterizations for connectedness of the Cayley digraph $\mathrm{\Gamma }:=\mathrm{Cay}(T_n,L_{\mathcal{A}})$ where $L_{\mathcal{A}}$ is an $\mathcal{L}$-class of $T_n$. If $n=1$, then Γ has only one vertex which is a trivial digraph. Hence, from now on, we will consider the Cayley digraph Γ of $T_n$ where $n>1$. Let $u,v$ be two distinct vertices of a digraph D. Throughout the paper, the notation $[u,v]$-semidipath denotes the semidipath from u to v. Similarly, $[u,v]$-dipath denotes the dipath from u to v. The digraph D is called a strongly connected digraph if D contains a $[u,v]$-dipath. Moreover, D is called a weakly connected digraph if D contains a $[u,v]$-semidipath. The digraph D is called a locally connected digraph whenever a $[u,v]$-dipath exists in D, a $[v,u]$-dipath must exist in D as well. Further, D is called a unilaterally connected digraph if either a $[u,v]$-dipath or a $[v,u]$-dipath exists in D.

Theorem 2.10

The Cayley digraph Γ is never strongly connected.

Proof

As we obtain by Lemma 2.5 that $N({\chi }_a)=\{{\chi }_b:b\in \mathcal{A}\}$, there is no dipath from ${\chi }_a$ through to α in which $|X\alpha |>1$. This implies that Γ is not strongly connected. □

Theorem 2.11

Γ is weakly connected if and only if $|\mathcal{A}|\le n-1$.

Proof

If $|\mathcal{A}|=n$, then $L_{\mathcal{A}}={\mathcal{S}}_n$. By Theorem 2.3, we obtain that Γ is the disjoint union of totally complete induced subdigraphs $\mathrm{\Gamma }[R]$ where R is an $\mathcal{R}$-class of $T_n$. It follows that Γ is not weakly connected.

Conversely, let $|\mathcal{A}|\le n-1$. For convenience, let $\mathcal{A}=\{a_1,a_2,\dots ,a_k\}$ and $b\notin \mathcal{A}$. Further, let $B=X\setminus (\mathcal{A}\cup \{b\})$. Define ${\beta }_1\in T_n$ as follows:

$${\beta }_1=\left(\begin{array}{cccc}\hfill A_1\hfill & \hfill A_2\hfill & \hfill \dots \hfill & \hfill A_k\hfill \\ \hfill a_1\hfill & \hfill a_2\hfill & \hfill \dots \hfill & \hfill a_k\hfill \end{array}\right).$$

Then ${\beta }_1\in L_{\mathcal{A}}$. Moreover, for each $i=2,3,\dots ,k$, we define ${\beta }_i$ by

$${\beta }_i={\left(\begin{array}{ccc}\hfill \{a_1,a_i\}\cup B\hfill & \hfill b\hfill & \hfill x\hfill \\ \hfill a_1\hfill & \hfill a_i\hfill & \hfill x\hfill \end{array}\right)}_{x\in \mathcal{A}\setminus \{a_1,a_i\}}.$$

It is not hard to verify that ${\beta }_1{\beta }_2\cdots {\beta }_k={\chi }_{a_1}$ where ${\beta }_i\in L_{\mathcal{A}}$ for $1\le i\le k$. Now, let $\alpha ,\beta$ be arbitrary vertices of Γ. We obtain that

$$\alpha {\beta }_1{\beta }_2\cdots {\beta }_k=\alpha {\chi }_{a_1}=\beta {\chi }_{a_1}=\beta {\beta }_1{\beta }_2\cdots {\beta }_k.$$

Hence Γ contains an $[\alpha ,{\chi }_{a_1}]$-dipath and a $[\beta ,{\chi }_{a_1}]$-dipath. Therefore, there is a weakly dipath joining between α and β in Γ. Consequently, Γ is weakly connected. □

Let $\beta \in T_n$. Note that $N^{-}(\beta )=\{\alpha \in T_n:(\alpha ,\beta )\in E(\mathrm{\Gamma })\}$.

Theorem 2.12

The Cayley digraph Γ is never unilaterally connected.

Proof

Let $L_{\mathcal{A}}$ be a connection set of Γ. For the case $|\mathcal{A}|=n$, we can conclude by Theorem 2.3 that Γ is the disjoint union of totally complete induced subdigraphs $\mathrm{\Gamma }[R]$ where R is an $\mathcal{R}$-class of $T_n$. It follows that Γ is not unilaterally connected. We next consider $|\mathcal{A}|\le n-1$. Let $\alpha \in T_n$. In fact, $|X(\alpha \mu )|\le |X\mu |\le n-1$ for all $\mu \in L_{\mathcal{A}}$. This implies that $\alpha \mu \notin {\mathcal{S}}_n$ for all $\mu \in L_{\mathcal{A}}$. Hence $N^{-}(\beta )=\varnothing$ for all $\beta \in {\mathcal{S}}_n$. We can conclude that there is no any dipath joining between elements in ${\mathcal{S}}_n$. Therefore, Γ is not unilaterally connected. □

Theorem 2.13

Γ is locally connected if and only if $|\mathcal{A}|=n$.

Proof

Assume that $|\mathcal{A}|=n$. That means $L_{\mathcal{A}}={\mathcal{S}}_n$. By Theorem 2.3, we can conclude that every component of Γ is totally complete. This clearly implies that Γ is locally connected.

Conversely, let Γ be locally connected. Let $\alpha \in L_{\mathcal{A}}$ be a given element. Hence $(i{d}_X,\alpha )\in E(\mathrm{\Gamma })$. By the locally connectedness of Γ, there exists a dipath joining from α back to $i{d}_X$ in Γ, say P. Thus P can be written as a sequence $\alpha ,{\alpha }_1,{\alpha }_2,\dots ,{\alpha }_k,i{d}_X$ for some ${\alpha }_i\in T_n$ where $i\in \{1,2,\dots ,k\}$. Then $i{d}_X={\alpha }_k{\lambda }_1=({\alpha }_{k-1}{\lambda }_2){\lambda }_1=\dots =({\alpha }_1{\lambda }_k){\lambda }_{k-1}\cdots {\lambda }_2{\lambda }_1=\alpha =\lambda {\lambda }_k{\lambda }_{k-1}\cdots {\lambda }_2{\lambda }_1$ where $\lambda ,{\lambda }_i\in L_{\mathcal{A}}$ for all $i\in \{1,2,\dots ,k\}$. Since $\lambda {\lambda }_k{\lambda }_{k-1}\cdots {\lambda }_2{\lambda }_1\in T_n$ and by Lemma 2.7, we obtain that ${\pi }_{\alpha }$ refines ${\pi }_{i{d}_X}$. This implies that $\alpha \in {\mathcal{S}}_n$. Since ${\mathcal{S}}_n$ is an $\mathcal{L}$-class of $T_n$ containing α, we have $L_{\mathcal{A}}={\mathcal{S}}_n$. Hence $|A|=n$. □

The last part of this section, we will present the sufficient condition for an isomorphism theorem of the Cayley digraph Γ with respect to the connection set $L_{\mathcal{A}}$.

Theorem 2.14

Let ${\mathrm{\Gamma }}_1$ and ${\mathrm{\Gamma }}_2$ denote $\mathrm{Cay}(T_n,L_{{\mathcal{A}}_1})$ and $\mathrm{Cay}(T_n,L_{{\mathcal{A}}_2})$ , respectively. If $|{\mathcal{A}}_1|=|{\mathcal{A}}_2|$ , then ${\mathrm{\Gamma }}_1\cong {\mathrm{\Gamma }}_2$ .

Proof

Assume that $|{\mathcal{A}}_1|=|{\mathcal{A}}_2|$. Let ${\mathcal{A}}_1\cap {\mathcal{A}}_2=\mathcal{B}$. Then $|{\mathcal{A}}_1\setminus \mathcal{B}|=|{\mathcal{A}}_2\setminus \mathcal{B}|$. Hence there is a bijection $g:{\mathcal{A}}_1\setminus \mathcal{B}\to {\mathcal{A}}_2\setminus \mathcal{B}$. Let $\alpha :X\to X$ be defined by

$$x\alpha =\{\begin{array}{cc}g(x),\hfill & x\in {\mathcal{A}}_1\setminus \mathcal{B},\hfill \\ g^{-1}(x),\hfill & {\mathcal{A}}_2\setminus \mathcal{B},\hfill \\ x,\hfill & \text{ otherwise},\hfill \end{array}$$

for all $x\in X$. Clearly, $\alpha \in T_n$ is also a bijection. Then we define $\phi :T_n\to T_n$ as follows:

$$\phi \left(\left(\begin{array}{cccc}\hfill A_1\hfill & \hfill A_2\hfill & \hfill \dots \hfill & \hfill A_k\hfill \\ \hfill a_1\hfill & \hfill a_2\hfill & \hfill \dots \hfill & \hfill a_k\hfill \end{array}\right)\right)=\left(\begin{array}{cccc}\hfill A_1\hfill & \hfill A_2\hfill & \hfill \dots \hfill & \hfill A_k\hfill \\ \hfill a_1\alpha \hfill & \hfill a_2\alpha \hfill & \hfill \dots \hfill & \hfill a_k\alpha \hfill \end{array}\right),$$

for all $\left(\begin{array}{cccc}\hfill A_1\hfill & \hfill A_2\hfill & \hfill \dots \hfill & \hfill A_k\hfill \\ \hfill a_1\hfill & \hfill a_2\hfill & \hfill \dots \hfill & \hfill a_k\hfill \end{array}\right)\in T_n$. It is not hard to verify that φ is a bijection. For each $\beta \in T_n$, we can see that $X\beta \subseteq {\mathcal{A}}_1$ if and only if $X\phi (\beta )\subseteq {\mathcal{A}}_2$. Further, we have ${\pi }_{\phi (\beta )}={\pi }_{\beta }$ which implies that $|X\beta |=|X\phi (\beta )|$. Moreover, if $X\beta \subseteq {\mathcal{A}}_1$, then $|{\mathcal{A}}_1\setminus X\beta |+|X\beta |=|{\mathcal{A}}_1|=|{\mathcal{A}}_2|=|{\mathcal{A}}_2|=|{\mathcal{A}}_2\setminus X\phi (\beta )|+|X\phi (\beta )|$. Thus $|{\mathcal{A}}_1\setminus X\beta |=|{\mathcal{A}}_2\setminus X\phi (\beta )|$. Therefore,

$$(\alpha ,\beta )\in E({\mathrm{\Gamma }}_1)\iff {\pi }_{\alpha }\text{ refines }{\pi }_{\beta },X\beta \subseteq {\mathcal{A}}_1\text{ and }|X\setminus X\alpha |\ge |{\mathcal{A}}_1\setminus X\beta |\iff {\pi }_{\phi (\alpha )}\text{ refines }{\pi }_{\phi (\beta )},X\phi (\beta )\subseteq {\mathcal{A}}_2\text{ and }|X\setminus X\phi (\alpha )|\ge |{\mathcal{A}}_2\setminus X\phi (\beta )|\iff (\phi (\alpha ),\phi (\beta ))\in E({\mathrm{\Gamma }}_2).$$

Consequently, ${\mathrm{\Gamma }}_1\cong {\mathrm{\Gamma }}_2$. □

3. Cayley digraphs of $T(X)$ with respect to $\mathcal{R}$-classes

In this section, we study structural properties of Cayley digraphs Γ of $T_n$ where their connection sets are precisely $\mathcal{R}$-classes of $T_n$. For each $\alpha \in T_n$, we recall that ${\pi }_{\alpha }=\{x{\alpha }^{-1}:x\in X\alpha \}$. Indeed, ${\pi }_{\alpha }$ forms a partition of X. Note that two elements $\alpha ,\beta \in T_n$ will be $\mathcal{R}$-related if and only if ${\pi }_{\alpha }={\pi }_{\beta }$, that is, the partitions of X induced by those two elements coincide. Given the partition $\mathcal{P}=\{A_1,A_2,\dots ,A_k\}$ of X where $1\le k\le n$ and let $R_{\mathcal{P}}=\{\alpha \in T_n:{\pi }_{\alpha }=\mathcal{P}\}$. Clearly, $R_{\mathcal{P}}$ is an $\mathcal{R}$-class of $T_n$. From now on, the set $R_{\mathcal{P}}$ stands for the connection set of Γ.

Lemma 3.1

Let$\mathcal{P}$be any partition of X and$R_{\mathcal{P}}$a connection set of Γ. If $K=\{{\chi }_a:a\in X\}$, then an induced subdigraph $\mathrm{\Gamma }[K]$ is totally complete.

Proof

Let $K=\{{\chi }_a:a\in X\}$ and ${\chi }_a,{\chi }_b\in K$. For convenience, assume that $\mathcal{P}=\{A_1,A_2,\dots ,A_t\}$ for some $t\in \mathbb{N}$. Without loss of generality, let $a\in A_1$. We then have the following two cases to consider.

Case 1:$b\in A_1$. We define $\alpha \in T_n$ as follows:

$$\alpha =\left(\begin{array}{cc}\hfill A_1\hfill & \hfill A_i\hfill \\ \hfill b\hfill & \hfill a_i\hfill \end{array}\right),$$

where $a_i\in A_i$ for all $i\in \{2,3,\dots ,t\}$ and $a_i\notin \{a,b\}$. Then $\alpha \in R_{\mathcal{P}}$. Furthermore, we obtain that ${\chi }_a\alpha ={\chi }_b$ which leads to $({\chi }_a,{\chi }_b)\in E(\mathrm{\Gamma })$.

Case 2:$b\notin A_1$. Assume, without loss of generality, that $b\in A_2$. We define $\alpha \in T_n$ as follows:

$$\alpha =\left(\begin{array}{ccc}\hfill A_1\hfill & \hfill A_2\hfill & \hfill A_i\hfill \\ \hfill b\hfill & \hfill a\hfill & \hfill a_i\hfill \end{array}\right),$$

where $a_i\in A_i$ for all $i\in \{3,4,\dots ,t\}$ and $a_i\notin \{a,b\}$. Then $\alpha \in R_{\mathcal{P}}$. We can see that ${\chi }_a\alpha ={\chi }_b$ which implies that $({\chi }_a,{\chi }_b)\in E(\mathrm{\Gamma })$.

From the above two cases, we conclude that $\mathrm{\Gamma }[K]$ is totally complete. □

Lemma 3.2

If $\mathcal{P}=\{A_1,A_2,\dots ,A_k\}$ such that $1\le k\le n-1$ , then ${\chi }_a\in \langle R_{\mathcal{P}}\rangle$ for some $a\in X$ where $\langle R_{\mathcal{P}}\rangle$ is a subsemigroup of $T_n$ generated by $R_{\mathcal{P}}$ .

Proof

Let $\mathcal{P}=\{A_1,A_2,\dots ,A_k\}$ be such that $1\le k\le n-1$. Then there exists $A_i\in \mathcal{P}$ in which $|A_i|\ge 2$. Without loss of generality, assume that $|A_1|\ge 2$. Let $a,b\in A_1$ and $a_i\in A_i$ for all $i\in \{2,3,\dots ,k\}$. Now, for each $j\in \{2,3,\dots ,k\}$, we define ${\alpha }_j\in T_n$ as follows:

$${\alpha }_j=\left(\begin{array}{cccccccc}\hfill A_1\hfill & \hfill A_2\hfill & \hfill A_3\hfill & \hfill \dots \hfill & \hfill A_j\hfill & \hfill A_{j+1}\hfill & \hfill \dots \hfill & \hfill A_k\hfill \\ \hfill a\hfill & \hfill a_2\hfill & \hfill a_3\hfill & \hfill \dots \hfill & \hfill b\hfill & \hfill a_{j+1}\hfill & \hfill \dots \hfill & \hfill a_k\hfill \end{array}\right).$$

Hence ${\alpha }_j\in R_{\mathcal{P}}$. Moreover, we see that

$${\alpha }_j^2=\left(\begin{array}{ccccc}\hfill A_1\cup A_j\hfill & \hfill A_2\hfill & \hfill A_3\hfill & \hfill \dots \hfill & \hfill A_k\hfill \\ \hfill a\hfill & \hfill a_2\hfill & \hfill a_3\hfill & \hfill \dots \hfill & \hfill a_k\hfill \end{array}\right).$$

It is not hard to verify that ${\alpha }_2^2{\alpha }_3^2\cdots {\alpha }_k^2={\chi }_a$, that is, ${\chi }_a\in \langle R_{\mathcal{P}}\rangle$. □

We next present characterizations of Cayley digraphs Γ of $T_n$ with respect to connection sets $R_{\mathcal{P}}$ where $|\mathcal{P}|=1$ and $|\mathcal{P}|=n$.

Theorem 3.3

Let$\mathcal{P}=\{X\}$and$R_{\mathcal{P}}$be a connection set of Γ. Then $E(\mathrm{\Gamma })=\{(\alpha ,{\chi }_a):\alpha \in T_n\mathit{\text{and}}a\in X\}$.

Proof

In this case, $R_{\mathcal{P}}=\{{\chi }_a:a\in X\}$. For each $\alpha \in T_n$ and $a\in X$, we have ${\chi }_a=\alpha {\chi }_a$ where ${\chi }_a\in R_{\mathcal{P}}$. Thus $(\alpha ,{\chi }_a)\in E(\mathrm{\Gamma })$. On the other hand, let $(\alpha ,\beta )\in E(\mathrm{\Gamma })$. Hence $\beta =\alpha \gamma$ for some $\gamma \in R_{\mathcal{P}}$. That means $\gamma ={\chi }_a$ for some $a\in X$. Therefore, $\beta =\alpha {\chi }_a={\chi }_a$. The proof is done. □

Note that $|\mathcal{P}|=n$ is equivalent to $R_{\mathcal{P}}={\mathcal{S}}_n$ where ${\mathcal{S}}_n$ is the group of all permutations in $T_n$. By Theorem 2.3, we have already proved the characterization of Γ in which the connection set is ${\mathcal{S}}_n$. Consequently, the following result is obtained, directly.

Theorem 3.4

Let$R_{\mathcal{P}}={\mathcal{S}}_n$be a connection set of Γ. Then Γ is the disjoint union of totally complete induced subdigraphs $\mathrm{\Gamma }[R]$ where R is an $\mathcal{R}$-class of $T_n$.

Now, we provide the necessary and sufficient conditions of two elements in $T_n$ for being adjacent in Γ.

Theorem 3.5

Let$\mathcal{P}=\{A_1,A_2,\dots ,A_k\}$be such that$1\le k\le n$and$R_{\mathcal{P}}$a connection set of Γ. Let $\alpha ,\beta \in T_n$. Then $(\alpha ,\beta )\in E(\mathrm{\Gamma })$ if and only if the following statements hold:

• 1.${\pi }_{\alpha }$refines${\pi }_{\beta }$,
• 2.for each$x,y\in X$,$x\beta =y\beta$if and only if$x\alpha ,y\alpha \in A_i$for some$i\in \{1,2,\dots ,k\}$.

Proof

Let $(\alpha ,\beta )\in E(\mathrm{\Gamma })$. Then $\beta =\alpha \lambda$ for some $\lambda \in R_{\mathcal{P}}$. By Lemma 2.7, we obtain that ${\pi }_{\alpha }$ refines ${\pi }_{\beta }$. Let $x,y\in X$. We conclude that

$$x\beta =y\beta \iff x(\alpha \lambda )=y(\alpha \lambda )\iff (x\alpha )\lambda =(y\alpha )\lambda \iff x\alpha ,y\alpha \in A_i,$$

for some $i\in \{1,2,\dots ,k\}$.

Conversely, assume that the conditions hold. Since $\mathcal{P}$ is a partition of X, there exists $i\in \{1,2,\dots ,k\}$ such that $A_i\cap X\alpha \ne \varnothing$. Let $J=\{j\in \{1,2,\dots ,k\}:A_j\cap X\alpha \ne \varnothing \}$. For each $j\in J$, choose $c_j\alpha \in A_j\cap X\alpha$. Further, for each $t\in \{1,2,\dots ,k\}\setminus J$, choose $c_t\in X\setminus \{c_j\beta :j\in J\}$. Define $\lambda :X\to X$ as follows:

$$x\lambda =\{\begin{array}{cc}{c}_j\beta ,\hfill & x\in A_j;j\in J,\hfill \\ c_t,\hfill & x\in A_t;t\in \{1,2,\dots ,k\}\setminus J.\hfill \end{array}$$

Consider $c_j\alpha ,c_j^{\prime }\alpha \in A_j\cap X\alpha$, by the second condition in our assumption, we have $c_j\beta =c_j^{\prime }\beta$. Hence λ is well-defined, that is, $\lambda \in T_n$. We next show that $\lambda \in R_{\mathcal{P}}$. Let $x,y\in X$. If $x,y\in A_i$ for some $i\in \{1,2,\dots ,k\}$, then $x\lambda =y\lambda$ by the definition of λ. Suppose that $x\in A_i$ and $y\in A_j$ for some $i,j\in J$ and $i\ne j$. Thus $x\lambda =c_i\beta \ne c_j\beta =y\lambda$ by the choice of $c_i$ and $c_j$. This implies that $\lambda \in R_{\mathcal{P}}$. We now prove that $\beta =\alpha \lambda$. Let $x\in X$. We have $x\alpha \in A_j$ for some $j\in J$ and hence $x(\alpha \lambda )=(x\alpha )\lambda =c_j\beta$. Therefore, $x\alpha ,c_j\alpha \in A_j$ and by assumption, we conclude that $x\beta =c_j\beta$. Consequently, $\beta =\alpha \lambda$ which implies that $(\alpha ,\beta )\in E(\mathrm{\Gamma })$, as required. □

The following proposition provides characterizations for neighbourhoods of constant maps and permutations in $T_n$.

Proposition 3.6

Let$\mathcal{P}=\{A_1,A_2,\dots ,A_k\}$and$R_{\mathcal{P}}$be a connection set of Γ.

• 1.If$a\in X$, then$N({\chi }_a)=\{{\chi }_b:b\in X\}$.
• 2.If$\alpha \in {\mathcal{S}}_n$, then$N(\alpha )=\{\beta \in T_n:{\pi }_{\beta }=\{A_1{\alpha }^{-1},A_2{\alpha }^{-1},\dots ,A_k{\alpha }^{-1}\}\}$.

Proof

To prove 1, let $a\in X$ and $\beta \in N({\chi }_a)$. Then $({\chi }_a,\beta )\in E(\mathrm{\Gamma })$. By Theorem 3.5, we obtain that ${\pi }_{{\chi }_a}$ refines ${\pi }_{\beta }$. Thus ${\pi }_{\beta }=\{X\}={\pi }_{{\chi }_a}$. Hence $\beta \in \{{\chi }_b:b\in X\}$ which implies that $N({\chi }_a)\subseteq \{{\chi }_b:b\in X\}$. On the other hand, we have by Lemma 3.1 that $\mathrm{\Gamma }[K]$ is totally complete, where $K=\{{\chi }_b:b\in X\}$, which yields $\{{\chi }_b:b\in X\}\subseteq N({\chi }_a)$. Therefore, the statement 1 is completely proved.

For proving 2, we let $\alpha \in {\mathcal{S}}_n$. Further, let $\beta \in N(\alpha )$. Thus there exists $\gamma \in R_{\mathcal{P}}$ in which $\beta =\alpha \gamma$. For each $x,y\in X$, we obtain that

$$x,y\in A_i{\alpha }^{-1}\text{ for some }{A}_i\in \mathcal{P}\iff x\alpha ,y\alpha \in A_i\text{ for some }{A}_i\in \mathcal{P}\iff x\beta =y\beta \text{ (by Theorem 3.5)}.$$

So we can conclude that ${\pi }_{\beta }=\{A_1{\alpha }^{-1},A_2{\alpha }^{-1},\dots ,A_k{\alpha }^{-1}\}$. It is obvious that ${\pi }_{\alpha }$ refines ${\pi }_{\beta }$ since $\alpha \in {\mathcal{S}}_n$. Moreover, for each $x,y\in X$, we have

$$x\beta =y\beta \iff x,y\in A_i{\alpha }^{-1}\text{ for some }i\in \{1,2,\dots ,k\}\iff x\alpha ,y\alpha \in A_i\text{ for some }i\in \{1,2,\dots ,k\}.$$

By Theorem 3.5, we conclude that $(\alpha ,\beta )\in E(\mathrm{\Gamma })$, that is, $\beta \in N(\alpha )$. □

Next, we characterize the connectedness for Cayley digraphs Γ of $T_n$ ($n>1$) with respect to $\mathcal{R}$-classes of $T_n$.

Theorem 3.7

Let$\mathcal{P}$be a partition of X and$R_{\mathcal{P}}$a connection set of Γ. The Cayley digraph Γ is never unilaterally connected.

Proof

We first consider the case $|\mathcal{P}|=n$, that is, $R_{\mathcal{P}}={\mathcal{S}}_n$. By Theorem 2.3, we obtain that Γ is the disjoint union of totally complete induced subdigraphs $\mathrm{\Gamma }[R]$ where R is an $\mathcal{R}$-class of $T_n$. Hence there is no any dipath joining between elements from different $\mathcal{R}$ classes. It follows that Γ is not unilaterally connected. We next consider $|\mathcal{P}|\le n-1$. For each $\delta \in T_n$, we have $|X(\delta \lambda )|\le |X\lambda |<n$ for all $\lambda \in R_{\mathcal{P}}$. Therefore, $(\delta ,\alpha )\notin E(\mathrm{\Gamma })$ for all $\alpha \in {\mathcal{S}}_n$. We can conclude that there is no any dipath joining between elements in ${\mathcal{S}}_n$. Consequently, Γ is not unilaterally connected. □

As in the proof of Theorem 3.7, we know that Γ does not contain any dipath joining between two permutations in ${\mathcal{S}}_n$. The following corollary is obtained, evidently.

Corollary 3.8

Let$\mathcal{P}$be a partition of X and$R_{\mathcal{P}}$a connection set of Γ. The Cayley digraph Γ is never strongly connected.

Theorem 3.9

Let$\mathcal{P}$be a partition of X and$R_{\mathcal{P}}$a connection set of Γ. Then Γ is weakly connected if and only if $|\mathcal{P}|\le n-1$.

Proof

If $|\mathcal{P}|=n$, then $R_{\mathcal{P}}={\mathcal{S}}_n$. We can conclude by Theorem 3.4 that Γ is not weakly connected. Conversely, let $|\mathcal{P}|\le n-1$. For the case $|\mathcal{P}|=1$, that is, $R_{\mathcal{P}}=\{{\chi }_a:a\in X\}$, we have shown in Theorem 3.3 that $E(\mathrm{\Gamma })=\{(\alpha ,{\chi }_a):\alpha \in T_n\text{ and }a\in X\}$. This implies that Γ is weakly connected. Next, let $2\le |\mathcal{P}|\le n-1$. Let $\alpha \in T_n$. We will prove that Γ contains an $[\alpha ,{\chi }_a]$-dipath for some $a\in X$. By Lemma 3.2, we have ${\chi }_a\in \langle R_{\mathcal{P}}\rangle$ for some $a\in X$. Thus there exists ${\alpha }_1,{\alpha }_2,\dots ,{\alpha }_t\in R_{\mathcal{P}}$ for some $t\in \mathbb{N}$ such that ${\chi }_a={\alpha }_1{\alpha }_2\cdots {\alpha }_t$. Consider $\alpha {\alpha }_1{\alpha }_2\cdots {\alpha }_t=\alpha {\chi }_a={\chi }_a$, the sequence $\alpha ,\alpha {\alpha }_1,\alpha {\alpha }_1{\alpha }_2,\dots ,\alpha {\alpha }_1{\alpha }_2\cdots {\alpha }_{t-1},{\chi }_a$ forms an $[\alpha ,{\chi }_a]$-dipath in Γ. Therefore, for each $\beta ,\gamma \in T_n$, there exist a $[\beta ,{\chi }_a]$-dipath and a $[\gamma ,{\chi }_b]$-dipath in Γ for some $a,b\in X$. Furthermore, we have by Lemma 3.1 that $\mathrm{\Gamma }[K]$ is totally complete where $K=\{{\chi }_a:a\in X\}$ which leads to $({\chi }_a,{\chi }_b)\in E(\mathrm{\Gamma })$. Hence the $[\beta ,{\chi }_a]$-dipath, the arc $({\chi }_a,{\chi }_b)$ and the $[\gamma ,{\chi }_b]$-dipath form a $[\beta ,\gamma ]$-semidipath in Γ. Therefore, Γ is weakly connected, as required. □

Note that $|\mathcal{A}|=n\iff L_{\mathcal{A}}={\mathcal{S}}_n=R_{\mathcal{P}}\iff |\mathcal{P}|=n$. Hence the following theorem can be obtained, directly, by Theorem 2.13.

Theorem 3.10

Let$\mathcal{P}$be a partition of X and$R_{\mathcal{P}}$a connection set of Γ. Then Γ is locally connected if and only if $|\mathcal{P}|=n$.

Finally, we present the sufficient condition for an isomorphism theorem of Cayley digraphs of $T_n$ with respect to $\mathcal{R}$-classes. Let $\mathcal{P}=\{A_1,A_2,\dots ,A_k\}$ be a partition of X such that $|A_i|=m_i$ for all $i=1,2,\dots ,k$ and $m_1\le m_2\le \dots \le m_k$. The sequence $m_1,m_2,\dots ,m_k$ will be called the degree sequence of $\mathcal{P}$.

Theorem 3.11

Let ${\mathcal{P}}_1=\{A_1,A_2,\dots ,A_k\}$ and ${\mathcal{P}}_2=\{B_1,B_2,\dots ,B_k\}$ be partitions of X. If ${\mathcal{P}}_1$ and ${\mathcal{P}}_2$ have the same degree sequence, then $\mathrm{Cay}(T_n,R_{{\mathcal{P}}_1})\cong \mathrm{Cay}(T_n,R_{{\mathcal{P}}_2})$ .

Proof

Let ${\mathrm{\Gamma }}_1$ and ${\mathrm{\Gamma }}_2$ denote $\mathrm{Cay}(T_n,R_{{\mathcal{P}}_1})$ and $\mathrm{Cay}(T_n,R_{{\mathcal{P}}_2})$, respectively. Further, assume that ${\mathcal{P}}_1$ and ${\mathcal{P}}_2$ have the same degree sequence. Thus $|A_i|=|B_i|$ for all $i=1,2,\dots ,k$. Then, for each $i=1,2,\dots ,k$, there exists a bijective function $f_i:A_i\to B_i$. Let $\lambda :X\to X$ be defined by $x\lambda =f_i(x)$ if $x\in A_i$. Hence $\lambda \in T_n$ is also a bijection. For each $\alpha \in T_n$, we define ${\alpha }^{⁎}:X\to X$ by $x{\alpha }^{⁎}=x(\alpha \lambda )$ for all $x\in X$, that is, ${\alpha }^{⁎}=\alpha \lambda$. So, for each $x,y\in X$, we obtain that

$$x\alpha =y\alpha \iff x(\alpha \lambda )=y(\alpha \lambda )\iff x{\alpha }^{⁎}=y{\alpha }^{⁎}.$$

This also implies that ${\pi }_{\alpha }={\pi }_{{\alpha }^{⁎}}$. Let $\phi :T_n\to T_n$ be defined by $\phi (\alpha )={\alpha }^{⁎}$ for all $\alpha \in T_n$. Since λ is a bijection, it follows that φ is a bijection. Moreover, for each $\alpha ,\beta \in T_n$, we have

$$\begin{array}{ccc}(\alpha ,\beta )\in E({\mathrm{\Gamma }}_1)\hfill & \iff \hfill & {\pi }_{\alpha }\text{ refines }{\pi }_{\beta }\text{ and for each }x,y\in X,\hfill \\ \hfill & \hfill & (x\beta =y\beta \iff x\alpha ,y\alpha \in A_i\text{ for some }i\in \{1,2,\dots ,k\})\hfill \\ \hfill & \iff \hfill & {\pi }_{{\alpha }^{⁎}}\text{ refines }{\pi }_{{\beta }^{⁎}}\text{ and for each }x,y\in X,\hfill \\ \hfill & \hfill & (x{\beta }^{⁎}=y{\beta }^{⁎}\iff x(\alpha \lambda ),y(\alpha \lambda )\in B_i\text{ for some }i\in \{1,2,\dots ,k\})\hfill \\ \hfill & \iff \hfill & {\pi }_{\phi (\alpha )}\text{ refines }{\pi }_{\phi (\beta )}\text{ and for each }x,y\in X,\hfill \\ \hfill & \hfill & (x\phi (\beta )=y\phi (\beta )\iff x\phi (\alpha ),y\phi (\alpha )\in B_i\text{ for some }i\in \{1,2,\dots ,k\})\hfill \\ \hfill & \iff \hfill & (\phi (\alpha ),\phi (\beta ))\in E({\mathrm{\Gamma }}_2).\hfill \end{array}$$

Consequently, ${\mathrm{\Gamma }}_1\cong {\mathrm{\Gamma }}_2$. □

4. Conclusions

In this research, we have constructed the Cayley digraphs of full transformation semigroups whose connection sets are Green's equivalence classes $\mathcal{L}$ and $\mathcal{R}$, respectively. To obtain their structural properties, the characterizations of being arcs in Cayley digraphs have been provided. Moreover, their connectedness have been investigated. Indeed, the connectedness property of Cayley digraphs with respect to $\mathcal{L}$-classes and $\mathcal{R}$-classes are rather resembled. Furthermore, their isomorphism theorems have also been presented.

CRediT authorship contribution statement

Yanisa Chaiya, Nuttawoot Nupo: Conceived and designed the experiments; Performed the experiments; Analyzed and interpreted the data; Contributed reagents, materials, analysis tools or data; Wrote the paper.

Declaration of Competing Interest

The authors declare no conflict of interest.

Data availability

No data was used for the research described in the article.