On Nilpotent Subsemigroups of the Order-Decreasing Transformation Semigroups

Abstract

For $n\in {\mathbb {N}}$, let ${\mathcal {D}}_{n}$ be the semigroup of all order-decreasing transformations on $X_{n}=\{1,\ldots ,n\}$, under its natural order, and let $E({\mathcal {D}}_{n})$ be the set of all idempotents in ${\mathcal {D}}_n$ and for any $\xi \in E({\mathcal {D}}_{n})$, let

$$\begin{aligned} {\mathcal {D}}_{n}(\xi )=\{\ \alpha \in {\mathcal {D}}_n \ :\ \alpha ^m=\xi \,\, \text{ for } \text{ some } \,\, m\in {\mathbb {Z}}^{+}\} . \end{aligned}$$

In this paper, first we show that ${\mathcal {D}}_{n}(\xi )$ is the maximum nilpotent subsemigroup of ${\mathcal {D}}_{n}$ with zero element $\xi$. Moreover, we determine the minimum generating set of ${\mathcal {D}}_{n}(\xi )$, and so the rank of ${\mathcal {D}}_{n}(\xi )$. Also, we find the cardinality of ${\mathcal {D}}_{n}(\xi )$.

1 Introduction

For $n\in {\mathbb {N}}$, let ${\mathcal {T}}_{n}$ be the (full) transformations semigroup (under composition) on $X_{n}=\{1,\ldots ,n\}$, under its natural order. A transformation $\alpha \in {\mathcal {T}}_{n}$ is called order-decreasing (order-increasing) if $x\alpha \le x$ ($x\alpha \ge x$) for each $x\in X_{n}$. The semigroup of all order-decreasing (order-increasing) transformations in ${\mathcal {T}}_{n}$ is denoted by ${\mathcal {D}}_n$ (${\mathcal {D}}_n^+$). It is a well-known fact from [17, Lemma 1.1] that ${\mathcal {D}}_n$ and ${\mathcal {D}}_n^+$ are isomorphic semigroups.

The fix set and image set of any transformation $\alpha \in {\mathcal {T}}_n$ are defined, respectively, by

$$\begin{aligned} \mathrm {fix\, }(\alpha )= & {} \{ x\in X_n: x\alpha =x\},\\ \mathrm {im\, }(\alpha )= & {} \{ x\alpha : x\in X_{n}\}. \end{aligned}$$

An element e of a semigroup S is called idempotent if $e^2=e$, and the set of all idempotents in S is denoted by E(S). It is a well-known fact that, for $\alpha \in {\mathcal {T}}_{n}$, $\alpha$ is an idempotent if and only if $x\alpha =x$ for each $x\in \mathrm {im\, }(\alpha )$. Equivalently, $\alpha$ is an idempotent if and only if $\mathrm {fix\, }(\alpha )=\mathrm {im\, }(\alpha )$. An element a of a semigroup S with the zero element 0, is called nilpotent if $a^m=0$ for some $m \in {\mathbb {Z}}^{+}$. The set of all nilpotents in S is denoted by N(S). A semigroup S with the zero element 0 is called nilpotent semigroup if there exists $m\in {\mathbb {Z}}^{+}$ such that $S^{m}=\{0\}$. It is shown in [2] (see also [7]) that if S is a finite semigroup, then the following statements are equivalent:

• (i) S is nilpotent,

• (ii) Every element $a \in S$ is nilpotent, and

• (iii) The unique idempotent of S is the zero element.

It is well known that the constant map $\varepsilon \in {\mathcal {D}}_{n}$ to 1 is the zero element of ${\mathcal {D}}_{n}$. Also, from [17, Lemma 1.5], we know that an element $\alpha$ in ${\mathcal {D}}_{n}$ is nilpotent if and only if $\mathrm {fix\, }(\alpha )=\{1\}$.

For any non-empty subset A of any semigroup S, the subsemigroup generated by A, is defined as, the smallest subsemigroup of S containing A and denoted by $\langle A\, \rangle$. If there exists a finite subset A of S such that $S=\langle A\, \rangle$, then S is called a finitely generated semigroup. The rank of a finitely generated semigroup S is defined by

$$\begin{aligned} \mathrm {rank\, }(S)=\min \{\, |A|:\langle A\, \rangle =S\, \}. \end{aligned}$$

If there exists a finite subset A of E(S) such that $S=\langle A\, \rangle$, then the idrank of S is defined by

$$\begin{aligned} \mathrm {idrank\, }(S)=\min \{\, |A|: A\subseteq E(S), \langle A\, \rangle =S\, \}. \end{aligned}$$

An element s of a semigroup S is called indecomposable, if $s\ne xy$ for all $x,y\in S$, that is if $s\in S\setminus S^{2}$. It is clear that every generating set of S must contain all indecomposable elements of S. Thus, if A is a non-empty finite set which consists entirely of indecomposable elements and $S=\langle A\rangle$, then A is the minimum generating set of S, and so $\mathrm {rank\, }(S)=|A|$. Moreover, for a finitely generated nilpotent semigroup S, it is proved in [13, Lemma 2.0.2] that $S\setminus S^2$ is the minimum generating set of S, and so

$$\begin{aligned} \mathrm {rank\, }(S)=|S|- |S^{2}|. \end{aligned}$$

(For the other terms in semigroup theory, which are not explained here, we refer to [10]).

From [17, Theorem 3.4] and [18, Theorem 5.2.5], we know that

$$\begin{aligned} \begin{array}{lllll} \mathrm {rank\, }({\mathcal {D}}_{n})&{}=&{}\mathrm {idrank\, }({\mathcal {D}}_{n})= \frac{n(n-1)}{2},\\ \mathrm {rank\, }(N({\mathcal {D}}_{n}))&{}=&{} (n-2)\,!\,(n-2). \end{array} \end{aligned}$$

Also, from [17, Lemmas 4.1, 4.2 and 4.3] we know that

$$\begin{aligned} \begin{array}{lllll} |{\mathcal {D}}_{n}| &{}= &{}n!,\\ |N({\mathcal {D}}_{n})| &{}=&{}(n-1)!,\\ |E({\mathcal {D}}_{n})|&{}=&{} \sum _{r=0}^{n}S(n,r). \end{array} \end{aligned}$$

As in other algebraic theories, it is an important problem to find the minimum generating set and the rank of semigroups. It is also an important problem to investigate the structure of nilpotent elements and nilpotent subsemigroups of semigroups. Among recent contributions are [1, 3,4,5,6, 8, 9, 11, 12, 14,15,16, 19, 20]. In this paper, we find the structure of all the nilpotent subsemigroups of ${\mathcal {D}}_n$ and we find the cardinality of ${\mathcal {D}}_{n}(\xi )$. Moreover, we determine the minimum generating set of ${\mathcal {D}}_{n}(\xi )$, and then we show that

$$\begin{aligned} \mathrm {rank\, }({\mathcal {D}}_{n}(\xi ))= \prod _{i=1}^{r} (s_{i}-1)! - \prod _{i=1}^{r} (s_{i}-2)!. \end{aligned}$$

in Theorem 3.7.

2 Nilpotent Subsemigroups of ${\mathcal {D}}_{n}$

In this section, we focus on idempotents in ${\mathcal {D}}_{n}$ inspired by the zero element being an idempotent. For any $\alpha \in {\mathcal {D}}_{n}$, it is shown in [17, Lemma 1.3] that $\alpha$ is an idempotent if and only if, for all $t\in \mathrm {im\, }(\alpha )$, $t=\min \{x:x\in t\alpha ^{-1}\}$. For any $\xi \in E({\mathcal {D}}_{n})$, let

$$\begin{aligned} {\mathcal {D}}_{n}(\xi )=\{\ \alpha \in {\mathcal {D}}_n \ :\ \alpha ^{m}=\xi \,\, \text{ for } \text{ some } \,\, m\in {\mathbb {Z}}^{+}\} . \end{aligned}$$

In particular, ${\mathcal {D}}_{n}(\varepsilon )=N({\mathcal {D}}_{n})$ and ${\mathcal {D}}_{n}(1_{X_{n}})= \{1_{X_{n}}\}$ where $1_{X_{n}}$ is the identity map on $X_{n}$. Hence, for escape from triviality, we suppose that $\xi \in E({\mathcal {D}}_{n})\setminus \{\varepsilon , 1_{X_{n}}\}$, unless otherwise stated.

We would like to draw your attention that the following example shows that there exist $\xi \in E({\mathcal {D}}_{n})$ and $\alpha \in {\mathcal {D}}_{n}$ such that $\mathrm {fix\, }(\alpha )=\mathrm {fix\, }(\xi )$, but $\alpha$ is not an element of ${\mathcal {D}}_{n}(\xi )$ in general as in [19, Proposition 1].

Example 2.1

Let $\xi =\left( \begin{array}{ccccccc} 1&{} 2 &{} 3 &{} 4 &{} 5 &{} 6 &{} 7 \\ 1&{} 1 &{} 3 &{} 1 &{} 3 &{} 6 &{} 1 \end{array}\right)$ in $E({\mathcal {D}}_{7})$, and let $\alpha =\left( \begin{array}{ccccccc} 1&{} 2 &{} 3 &{} 4 &{} 5 &{} 6 &{} 7 \\ 1&{} 1 &{} 3 &{} 2 &{} 3 &{} 6 &{} 6 \end{array}\right)$ in ${\mathcal {D}}_{7}$. Notice that $\mathrm {fix\, }(\alpha )=\mathrm {fix\, }(\xi )$ and that $7\alpha ^{k}=6\not =1=7\xi$ for all positive integer k, and so $\alpha$ is not in ${\mathcal {D}}_{7}(\xi )$. $\square$

It is shown in [12, Lemma 1.1] that for any $\alpha , \beta \in {\mathcal {D}}_n$, $\mathrm {fix\, }( \alpha \beta )=\mathrm {fix\, }(\alpha )\cap \mathrm {fix\, }(\beta )$, and so we are able state the following theorem.

Theorem 2.2

For any $\xi \in E({\mathcal {D}}_{n})$ and any $\alpha \in {\mathcal {D}}_n$, the followings are equivalent:

• (i) $\alpha \in {\mathcal {D}}_{n}(\xi )$,

• (ii) $\mathrm {fix\, }(\alpha )=\mathrm {fix\, }(\xi )$ and $\alpha \xi =\xi$,

• (iii) $\mathrm {fix\, }(\alpha )=\mathrm {fix\, }(\xi )$ and $\alpha \xi =\xi =\xi \alpha$.

Proof

$(i) \Rightarrow (ii)$ Let $\xi \in E({\mathcal {D}}_{n})$ with $\mathrm {fix\, }(\xi )=\{ 1=a_{1}<\cdots < a_{r} \}$ for $2\le r \le n-1$, and let $\alpha \in {\mathcal {D}}_{n}(\xi )$. Then, there exists $m \in {\mathbb {Z}}^{+}$ such that $\alpha ^{m}=\xi$, and so we have $\mathrm {fix\, }(\alpha )=\mathrm {fix\, }(\xi )$ since $\mathrm {fix\, }( \alpha )=\mathrm {fix\, }(\alpha ^{k})$ for any $k\in {\mathbb {Z}}^{+}$. Then, clearly $a_{i}(\alpha \xi ) =(a_{i} \alpha ) \xi = a_{i}\xi =a_{i}$ for each $1\le i \le r$. Now, let $b\in X_{n}\setminus \mathrm {fix\, }( \alpha )$. Then there exists $1\le i \le r$ such that $a_{i}< b <a_{i+1}$ where $a_{r+1}=n+1$, and there exists $1 \le j\le i$ such that $b\xi =a_{j}$ since $\mathrm {fix\, }(\xi )=\mathrm {im\, }(\xi )$, and so $a_{i}=\min \{x: x \in a_{i}\xi ^{-1}\}$. Hence,

$$\begin{aligned} b(\alpha \xi )=(b\alpha )\xi =(b\alpha )\alpha ^{m}=(b\alpha ^{m})\alpha =(b\xi )\alpha =(a_{j}\alpha )=a_{j}=b\xi , \end{aligned}$$

and so $\alpha \xi =\xi$, as required.

$(ii) \Rightarrow (iii)$ Let $\xi \in E({\mathcal {D}}_{n})$ with $\mathrm {fix\, }(\xi )=\{ 1=a_{1}<\cdots <a_{r}\}$ for $2\le r \le n-1$. Suppose that $\mathrm {fix\, }(\alpha )=\mathrm {fix\, }(\xi )$ and that $\alpha \xi =\xi$. Then, clearly $a_{i}(\xi \alpha )= (a_{i}\xi )\alpha =a_{i}\alpha =a_{i}=a_{i}\xi$ for each $1\le i \le r$. Now, let $b\in X_{n}\setminus \mathrm {fix\, }(\alpha )$. Then, $b(\xi \alpha )=(b\xi )\alpha =b\xi$, since $b\xi \in \mathrm {im\, }(\xi )=\mathrm {fix\, }(\xi )=\mathrm {fix\, }(\alpha )$. Hence, $\alpha \xi =\xi =\xi \alpha$, as required.

$(iii) \Rightarrow (i)$ Let $\xi \in E({\mathcal {D}}_{n})$ with $\mathrm {fix\, }(\xi )=\{ 1=a_{1}<\cdots <a_{r} \}$ for $2\le r \le n-1$. Suppose that $\mathrm {fix\, }(\alpha )=\mathrm {fix\, }(\xi )$ and that $\alpha \xi =\xi =\xi \alpha$. It is clear that for each $k\in {\mathbb {Z}}^{+}$,  $\alpha ^{k} \xi =\xi$, and that $a_{i} \alpha ^{k}=a_{i}=a_{i}\xi$ for each $1\le i \le r$. Now, let $b \in X_{n}\setminus \mathrm {fix\, }(\alpha )$. Since $b\alpha \lneqq b$, there exists a positive integer $m_{b}$ such that $b\alpha ^{m_{b}}\in \mathrm {fix\, }(\alpha )$ but $b\alpha ^ {m_{b}-1}\notin \mathrm {fix\, }(\alpha ) \quad (\alpha ^{0}=1_{X_{n}})$. Thus, for every $k\ge m_{b}$, we have $b\alpha ^ {k}=b\alpha ^{m_{b}}$. If we take $m=\max \{ m_{b} : b\in X_{n} \setminus \mathrm {fix\, }(\alpha )\}$, then for any $b\in X_{n}\setminus \mathrm {fix\, }(\alpha )$, we have

$$\begin{aligned} b\alpha ^{m}=b\alpha ^{m_{b}}=(b\alpha ^{m_{b}})\xi =b(\alpha ^{m_{b}}\xi ) =b\xi . \end{aligned}$$

Therefore, $\alpha ^{m}=\xi$, and so $\alpha \in {\mathcal {D}}_{n}(\xi )$, as required. $\square$

Theorem 2.3

For any $\xi \in E({\mathcal {D}}_{n})$, ${\mathcal {D}}_{n}(\xi )$ is the maximum nilpotent subsemigroup of ${\mathcal {D}}_{n}$ with zero element $\xi$.

Proof

For all $\alpha , \beta \in {\mathcal {D}}_{n}(\xi )$, it follows from Theorem 2.2 that

$$\begin{aligned} \mathrm {fix\, }(\alpha \beta ) =\mathrm {fix\, }(\alpha )\cap \mathrm {fix\, }(\beta )=\mathrm {fix\, }(\xi )\cap \mathrm {fix\, }(\xi )=\mathrm {fix\, }(\xi ), \end{aligned}$$

and that $(\alpha \beta )\xi =\alpha (\beta \xi )=\alpha \xi =\xi$, and so $\alpha \beta \in {\mathcal {D}}_{n} (\xi )$. Moreover, from Theorem 2.2, $\xi$ is the zero element of ${\mathcal {D}}_{n}(\xi )$, and so, from the definition of ${\mathcal {D}}_{n}(\xi )$, ${\mathcal {D}}_{n}(\xi )$ is a nilpotent subsemigroup.

Now, let T be a nilpotent subsemigroup of ${\mathcal {D}}_{n}$ with zero element $\xi$. It is clear that $T \subseteq {\mathcal {D}}_{n}(\xi )$, and so ${\mathcal {D}}_{n}(\xi )$ is the maximum nilpotent subsemigroup of ${\mathcal {D}}_{n}$ with zero element $\xi$. $\square$

3 Cardinalities and Ranks of Nilpotent Subsemigroups of ${\mathcal {D}}_{n}$

For any $\xi \in E({\mathcal {D}}_{n})$, suppose that $\mathrm {fix\, }(\xi )=\{1=a_{1}<a_{2}<\cdots <a_{r}\}$ with $2\le r\le n-1$. Then, let $B_{i}=a_{i}\xi ^{-1}$ and $|B_{i}|=s_{i}$ for every $1\le i \le r$. Since $a_{i}\in B_{i}$, and $a_{i}=\min \{x:x\in a_{i}\xi ^{-1}\}$ for every $1\le i\le r$, we can take $B_{i}=\{\, b_{i1}< \cdots < b_{is_{i}}\,\}$ with $b_{i1} =a_{i}$. Since there exists at least one $1\le i \le r$ such that $s_{i}\ge 2$, there exists an integer $1\le t \le r$ such that $s_{j(1)},\ldots ,s_{j(t)}$ are at least 2, and that if $s_{i}\notin \{s_{j(1)},\ldots ,s_{j(t)}\}$ ($1\le i\le r$), then $s_{i}=1$. Now, suppose that $|B_{j}|=2$ for every $j \in \{j(1),\ldots ,j(t)\}$. Then, for any $\alpha \in {\mathcal {D}}_{n}(\xi )$, it follows from Theorem 2.2 that there exist $b, c\in X_{n}$ such that $b\alpha =c\ne b$, and so there exists $k \in \{j(1),\ldots ,j(t)\}$ such that $b\in B_{k}\setminus \{a_{k}\}$. Moreover, since

$$\begin{aligned} a_{k}=b\xi =b(\alpha \xi )=(b\alpha )\xi =c\xi , \end{aligned}$$

$c\in B_{j}$. Since $|B_{j}|=2$, we must have $c=a_{k}$. Therefore, for every $b\in X_{n}$, since $b\alpha \in \mathrm {fix\, }(\xi )$, we have $\alpha =\xi$, and so ${\mathcal {D}}_{n}(\xi )=\{\xi \}$. From this fact, if ${\mathcal {D}}_{n}(\xi )\not = \{\xi \}$, then there exists at least one $1\le i \le r$ such that $|B_{i}|=s_{i}\ge 3$.

From Theorem 2.2, we have $b_{i1} \alpha =a_{i} \alpha =a_{i}=b_{i1}\in B_{i}$ for each $1\le i\le r$. Next we consider the case $s_{i}\ge 2$ for each $1\le i\le r$.

Proposition 3.1

With the above notations, for any $\xi \in E({\mathcal {D}}_{n})$, let $\mathrm {fix\, }(\xi )=\{ 1=a_{1}<a_{2}< \cdots <a_{r}\}$ with $2\le r\le n-1$. Then, for any $\alpha$ in ${\mathcal {D}}_{n}(\xi )$, we have $b_{ij}\alpha \in B_{i}$ for each $1\le i\le r$ and $1\le j\le s_{i}$. Moreover, if $s_{i}\ge 2$ some $1\le i\le r$, then we have $b_{ij} \alpha \le b_{i(j-1)}$ for each $2\le j\le s_{i}$, in particular, $b_{i2} \alpha =a_{i}$.

Proof

Let $\alpha \in {\mathcal {D}}_{n}(\xi )$ and consider the element $b_{ij}\in X_{n}$ for any $1\le i\le r$ and $1\le j\le s_{i}$ . Since the set $\{B_{1},\ldots ,B_{r}\}$ is a partition of $X_{n}$, there exist $1\le k\le r$ and $1\le l \le s_{k}$ such that $b_{ij}\alpha =b_{kl}$. Since $b_{ij}\xi =a_{i}$, it follows from Theorem 2.2 that

$$\begin{aligned} a_{i}=b_{ij}\xi =b_{ij}(\alpha \xi )=(b_{ij}\alpha )\xi =b_{kl}\xi =a_{k}, \end{aligned}$$

and so $i=k$, that is $b_{ij}\alpha =b_{il}$ and clearly $b_{il}\in B_{i}$.

Suppose that $s_{i}\ge 2$ for some $1\le i\le r$. Since $a_{i}=b_{i1}< b_{i2}< \cdots < b_{is_{i}}$, it follows from Theorem 2.2 that $b_{ij}\alpha \lneqq b_{ij}$, and so $b_{ij} \alpha \le b_{i(j-1)}$ for each $2\le j\le s_{i}$, in particular, $b_{i2} \alpha =a_{i}$, as required. $\square$

Example 3.2

For $\xi =\left( \begin{array}{ccccccc} 1&{} 2 &{} 3 &{} 4 &{} 5 &{} 6 &{} 7 \\ 1&{} 1 &{} 3 &{} 1 &{} 3 &{} 1 &{} 3 \end{array}\right) \in E({\mathcal {D}}_{7})$, notice that $\mathrm {fix\, }(\xi )= \{ 1,3\}$,  $B_{1}=1\xi ^{-1}= \{1,2,4,6\}$ and $B_{2}=3\xi ^{-1}=\{3,5,7\}$. Then ${\mathcal {D}}_{7}(\xi )$ is

$$\begin{aligned} {\mathcal {D}}_{7}(\xi )= & {} \Big \{\left( \begin{array}{ccccccc} 1&{} 2 &{} 3 &{} 4 &{} 5 &{} 6 &{} 7 \\ 1&{} 1 &{} 3 &{} 1 &{} 3 &{} 1 &{} 3 \end{array}\right) ,\left( \begin{array}{ccccccc} 1&{} 2 &{} 3 &{} 4 &{} 5 &{} 6 &{} 7 \\ 1&{} 1 &{} 3 &{} 1 &{} 3 &{} 1 &{} 5 \end{array}\right) ,\\{} & {} \left( \begin{array}{ccccccc} 1&{} 2 &{} 3 &{} 4 &{} 5 &{} 6 &{} 7 \\ 1&{} 1 &{} 3 &{} 1 &{} 3 &{} 2 &{} 3 \end{array}\right) ,\left( \begin{array}{ccccccc} 1&{} 2 &{} 3 &{} 4 &{} 5 &{} 6 &{} 7 \\ 1&{} 1 &{} 3 &{} 1 &{} 3 &{} 2 &{} 5 \end{array}\right) ,\\{} & {} \left( \begin{array}{ccccccc} 1&{} 2 &{} 3 &{} 4 &{} 5 &{} 6 &{} 7 \\ 1&{} 1 &{} 3 &{} 1 &{} 3 &{} 4 &{} 3 \end{array}\right) ,\left( \begin{array}{ccccccc} 1&{} 2 &{} 3 &{} 4 &{} 5 &{} 6 &{} 7 \\ 1&{} 1 &{} 3 &{} 1 &{} 3 &{} 4 &{} 5 \end{array}\right) ,\\{} & {} \left( \begin{array}{ccccccc} 1&{} 2 &{} 3 &{} 4 &{} 5 &{} 6 &{} 7 \\ 1&{} 1 &{} 3 &{} 2 &{} 3 &{} 1 &{} 3 \end{array}\right) ,\left( \begin{array}{ccccccc} 1&{} 2 &{} 3 &{} 4 &{} 5 &{} 6 &{} 7 \\ 1&{} 1 &{} 3 &{} 2 &{} 3 &{} 1 &{} 5 \end{array}\right) ,\\{} & {} \left( \begin{array}{ccccccc} 1&{} 2 &{} 3 &{} 4 &{} 5 &{} 6 &{} 7 \\ 1&{} 1 &{} 3 &{} 2 &{} 3 &{} 2 &{} 3 \end{array}\right) ,\left( \begin{array}{ccccccc} 1&{} 2 &{} 3 &{} 4 &{} 5 &{} 6 &{} 7 \\ 1&{} 1 &{} 3 &{} 2 &{} 3 &{} 2 &{} 5 \end{array}\right) ,\\{} & {} \left( \begin{array}{ccccccc} 1&{} 2 &{} 3 &{} 4 &{} 5 &{} 6 &{} 7 \\ 1&{} 1 &{} 3 &{} 2 &{} 3 &{} 4 &{} 3 \end{array}\right) ,\left( \begin{array}{ccccccc} 1&{} 2 &{} 3 &{} 4 &{} 5 &{} 6 &{} 7 \\ 1&{} 1 &{} 3 &{} 2 &{} 3 &{} 4 &{} 5 \end{array}\right) \Big \} . \end{aligned}$$

For any $\xi \in E({\mathcal {D}}_{n})$, let $\mathrm {fix\, }(\xi )=\{ 1=a_{1}<a_{2}< \cdots <a_{r}\}$ with $2\le r\le n-1$. Suppose that there exists at least one $1\le i \le r$ such that $s_{i}\ge 3$. Then, we consider the following lemma.

Lemma 3.3

With the above notations, suppose that $s_{i}\ge 3$ for some $1\le i \le r$.

• (i) For $\alpha \in {\mathcal {D}}_{n}(\xi )$, if there exists at least one $3 \le j\le s_{i}$ such that $b_{ij}\alpha =b_{i(j-1)}$, then $\alpha \in {\mathcal {D}}_{n} (\xi )\setminus {\mathcal {D}}_{n}(\xi )^{2}$.

• (ii) For every $\alpha \in {\mathcal {D}}_{n}(\xi )^{2}$,  $b_{ij}\alpha \le b_{i(j-2)}$ for every $3\le j\le s_{i}$, in particular, we have $b_{i3}\alpha =a_{i}$.

Proof

First of all, for every $\delta \in {\mathcal {D}}_{n}(\xi )$ and every $1\le i\le r$, it follows from Theorem 2.2 that $a_{i}\delta =a_{i}=b_{i1}$. Moreover, if $|B_{i}|=s_{i}\ge 2$ for some $1\le i \le r$, it follows from Proposition 3.1 that $b_{i2}\delta =a_{i}$.

(i) Assume that there are $\beta ,\gamma \in {\mathcal {D}}_{n}(\xi )$ such that $\alpha =\beta \gamma$. Similarly, it follows from Proposition 3.1 that, for every $3\le j\le s_{i}$, there exists $k\le j-1$ such that $b_{ij}\beta =b_{ik}$, and moreover, if $k\ge 2$, then there exists $l\le k-1$ such that $b_{ik}\gamma =b_{il}$, and so

$$\begin{aligned} \begin{array}{ll} b_{ij}\alpha =(b_{ij}\beta )\gamma =b_{i1}\gamma =b_{i1}\le b_{i(j-2)} &{} \text{ if } \, k=1 \\ b_{ij}\alpha =(b_{ij}\beta )\gamma =b_{ik}\gamma =b_{il}\le b_{i(j-2)} &{} \text{ if } \, k\ge 2, \end{array} \end{aligned}$$

which is a contradiction. Therefore, $\alpha \in {\mathcal {D}}_{n} (\xi )\setminus {\mathcal {D}}_{n} (\xi )^{2}$.

(ii) From the proof of the first item, it is clear that $b_{ij}\alpha \le b_{i(j-2)}$ for every $3\le j\le s_{i}$, in particular, we have $b_{i3}\alpha =a_{i}$ for every $\alpha \in {\mathcal {D}}_{n}(\xi )^{2}$. $\square$

Since ${\mathcal {D}}_{n}(\xi )$ is a nilpotent semigroup with zero $\xi$, it follows from [13, Lemma 2.0.2] that ${\mathcal {D}}_{n} (\xi ) \setminus {\mathcal {D}}_{n}(\xi )^{2}$ is the minimum generating set of ${\mathcal {D}}_{n}(\xi )$. If we let

$$\begin{aligned} Y= & {} \{\, \alpha \in {\mathcal {D}}_{n}(\xi )\, : \text{ there } \text{ exist } \text{ at } \text{ least } \text{ one } \, 1\le i\le r\, \text{ and } \text{ at } \text{ least } \text{ one }\\{} & {} \qquad \,\,\, 3 \le j\le s_{i}\, \text{ such } \text{ that } \, s_{i}\ge 3\, \text{ and } \, b_{ij}\alpha =b_{i(j-1)} \, \}, \end{aligned}$$

then from Lemma 3.3, we have $Y \subseteq {\mathcal {D}}_{n} (\xi )\setminus {\mathcal {D}}_{n} (\xi )^{2}$. Since one of our aim is to characterize the indecomposable elements of ${\mathcal {D}}_{n} (\xi )$, we need to show that $Y= {\mathcal {D}}_{n} (\xi ) \setminus {\mathcal {D}}_{n} (\xi )^{2}$, that is ${\mathcal {D}}_{n} (\xi )=\langle Y\rangle$ in the following theorem. However, we determine the rank of ${\mathcal {D}}_{n} (\xi )$ without using this equality.

Theorem 3.4

With the above notations, if $s_{k}\ge 3$ for some $1\le k \le r$, then Y is the minimum generating set of ${\mathcal {D}}_{n}(\xi )$. Moreover, ${\mathcal {D}}_{n}(\xi )= Y\cup Y^{2}$.

Proof

It is enough to show that for all $\alpha \in {\mathcal {D}}_{n}(\xi )^{2}$,  $\alpha \in \langle Y\rangle$. Suppose that $s_{k}\ge 3$ for some $1\le k\le r$. Then, there exists an integer $1\le t \le r$ such that $s_{k_{1}},\ldots ,s_{k_{t}}$ are at least 3, and that if $s_{i}\notin \{ s_{k_{1}} ,\ldots ,s_{k_{t}}\}$ ($1\le i\le r$), then $s_{i}=1$ or 2. For any $\alpha \in {\mathcal {D}}_{n}(\xi )^2$, consider the transformations $\beta$ and $\gamma$ which are defined as follows:

$$\begin{aligned} x\beta =\left\{ \begin{array}{ll} a_{i} &{} \text{ if } x=b_{i1}=a_{i} \text{ or } x=b_{i2}\,\, (\text{ if } \text{ exists}) \text{ for } 1\le i \le r\\ b_{k_{u}(j-1)} &{} \text{ if } x=b_{k_{u}j}\in B_{k_{u}}\setminus \{b_{k_{u}1},b_{k_{u}2}\} \text{ for } 3\le j\le s_{k_{u}} \text{ and } 1\le u\le t, \end{array} \right. \end{aligned}$$

and

$$\begin{aligned} x\gamma =\left\{ \begin{array}{ll} a_{i} &{} \, \text{ if } x=b_{i1}=a_{i} \text{ or } x=b_{i2}\,\, (\text{ if } \text{ exists}) \text{ for } 1\le i \le r\\ b_{k_{u}(s_{k_{u}}-1)} &{}\, \text{ if } x=b_{k_{u}s_{k_{u}}} \, \text{ for } 1\le u\le t\\ (b_{k_{u}(j+1)})\alpha &{} \begin{array}{l}\text{ if } x=b_{k_{u}j}\in B_{k_{u}}\setminus \{ b_{k_{u}1}, b_{k_{u}2}, b_{k_{u}s_{k_{u}}} \} \text{ for } 3\le j\le s_{k_{u}}-1 \\ \text{ and } 1\le u\le t \end{array} \end{array} \right. \end{aligned}$$

for all $x\in X_{n}$. Then, it is clear that $\mathrm {fix\, }(\beta )=\mathrm {fix\, }(\gamma )=\mathrm {fix\, }(\xi )$,  $\beta \xi =\xi$. From Proposition 3.1, since $(b_{k_{u}(j+1)})\alpha \in B_{k_{u}}$, and $B_{k_{u}}\xi =a_{k_{u}}$, we also have $\gamma \xi =\xi$, and so it follows from Theorem 2.2 (ii) that $\beta , \gamma \in {\mathcal {D}}_{n} (\xi )$. In fact, since $b_{k_{u}j}\beta =b_{k_{u}(j-1)}$ for all $3\le j\le s_{k_{u}}$ and all $1\le u\le t$, and $b_{k_{u}s_{k_{u}}}\gamma =b_{k_{u} (s_{k_{u}}-1)}$ for all $1\le u\le t$, we have $\beta , \gamma \in Y$.

For all $1\le i \le r$, it is clear that $a_{i}\beta \gamma =a_{i}= a_{i}\alpha$, and from Proposition 3.1, $b_{i2}\beta \gamma =a_{i}\gamma =a_{i}=b_{i2}\alpha$ (if $b_{i2}$ exists). Moreover, for every $b_{k_{u}j}\in B_{k_{u}}\setminus \{ b_{k_{u}1},b_{k_{u}2} \}$ where $3\le j\le s_{k_{u}}$ and $1\le u\le t$, it follows from Proposition 3.1 and Lemma 3.3 that

$$\begin{aligned} b_{k_{u}3}\beta \gamma= & {} b_{k_{u}2}\gamma =a_{k_{u}}=b_{k_{u}3}\alpha \, \text{ and }\\ b_{k_{u}j}\beta \gamma= & {} b_{k_{u}(j-1)}\gamma =b_{k_{u}j}\alpha \, \text{ for } j\ge 4, \end{aligned}$$

and so $\alpha =\beta \gamma$. Therefore, we have $\alpha \in Y^{2}\subseteq \langle Y\rangle$, as required. $\square$

Notice that we may have $b_{k_{u}j}\gamma \not = b_{k_{u}(j-1)}$ for some $3\le j\le s_{k_{u}}-1$ in the above proof.

Example 3.5

Let $\xi \in E({\mathcal {D}}_{7})$ be the map defined in Example 3.2. Then, the set

$$\begin{aligned} Y= & {} \Big \{\left( \begin{array}{ccccccc} 1&{} 2 &{} 3 &{} 4 &{} 5 &{} 6 &{} 7 \\ 1&{} 1 &{} 3 &{} 1 &{} 3 &{} 1 &{} 5 \end{array}\right) ,\left( \begin{array}{ccccccc} 1&{} 2 &{} 3 &{} 4 &{} 5 &{} 6 &{} 7 \\ 1&{} 1 &{} 3 &{} 1 &{} 3 &{} 4 &{} 3 \end{array}\right) ,\\{} & {} \left( \begin{array}{ccccccc} 1&{} 2 &{} 3 &{} 4 &{} 5 &{} 6 &{} 7 \\ 1&{} 1 &{} 3 &{} 1 &{} 3 &{} 4 &{} 5 \end{array}\right) ,\left( \begin{array}{ccccccc} 1&{} 2 &{} 3 &{} 4 &{} 5 &{} 6 &{} 7 \\ 1&{} 1 &{} 3 &{} 1 &{} 3 &{} 2 &{} 5 \end{array}\right) ,\\{} & {} \left( \begin{array}{ccccccc} 1&{} 2 &{} 3 &{} 4 &{} 5 &{} 6 &{} 7 \\ 1&{} 1 &{} 3 &{} 2 &{} 3 &{} 1 &{} 5 \end{array}\right) ,\left( \begin{array}{ccccccc} 1&{} 2 &{} 3 &{} 4 &{} 5 &{} 6 &{} 7 \\ 1&{} 1 &{} 3 &{} 2 &{} 3 &{} 2 &{} 3 \end{array}\right) ,\\{} & {} \left( \begin{array}{ccccccc} 1&{} 2 &{} 3 &{} 4 &{} 5 &{} 6 &{} 7 \\ 1&{} 1 &{} 3 &{} 2 &{} 3 &{} 2 &{} 5 \end{array}\right) ,\left( \begin{array}{ccccccc} 1&{} 2 &{} 3 &{} 4 &{} 5 &{} 6 &{} 7 \\ 1&{} 1 &{} 3 &{} 2 &{} 3 &{} 4 &{} 3 \end{array}\right) ,\\{} & {} \left( \begin{array}{ccccccc} 1&{} 2 &{} 3 &{} 4 &{} 5 &{} 6 &{} 7 \\ 1&{} 1 &{} 3 &{} 2 &{} 3 &{} 4 &{} 5 \end{array}\right) ,\left( \begin{array}{ccccccccc} 1&{} 2 &{} 3 &{} 4 &{} 5 &{} 6 &{} 7 \\ 1&{} 1 &{} 3 &{} 2 &{} 3 &{} 1 &{} 3 \end{array}\right) \Big \} \end{aligned}$$

is the minimum generating set of ${\mathcal {D}}_{7}(\xi )$ from Theorem 3.4.

For any $\xi \in E({\mathcal {D}}_{n})$, if $\xi =\varepsilon$, then it is shown in [17, Lemma 4.2] that $|{\mathcal {D}}_{n}(\varepsilon )|=|N({\mathcal {D}}_{n})|=(n-1)!$, and clearly $|{\mathcal {D}}_{n}(1_{X_{n}})|=|\{1_{X_{n}}\}|=1$. From now on we suppose that $2\le r\le n-1$.

Theorem 3.6

With the above notations, for any $\xi \in E({\mathcal {D}}_{n})$, let $\mathrm {fix\, }(\xi )=\{1=a_{1}<a_{2}< \cdots <a_{r}\}$ with $2\le r\le n-1$. Then,

$$\begin{aligned} |{\mathcal {D}}_{n}(\xi )|= \prod _{i=1}^{r}(s_{i}-1)! \end{aligned}$$

where $|a_{i}\xi ^{-1}|=s_{i}$ for every $1\le i \le r$.

Proof

Let $B_{i}=a_{i}\xi ^{-1}$ and $|B_{i}|=s_{i}$ for every $1\le i \le r$. For any $\alpha \in {\mathcal {D}}_{n}(\xi )$, it follows from Proposition 3.1 that $B_{i}\alpha \subseteq B_{i}=\{\, b_{i1}< \cdots < b_{is_{i}}\,\}$,  $b_{i1}\alpha =b_{i1}$, and $b_{ij} \alpha \le b_{i(j-1)}<b_{ij}$ for every $1\le i\le r$ and $2\le j\le s_{i}$ if $s_{i}\ge 2$. Thus the map $\alpha _{\mid _{B_{i}}}: B_{i}\rightarrow B_{i}$ is a nilpotent order-decreasing transformation on $B_{i}$ for every $1\le i\le r$. If we consider the map $\alpha _{i} : X_{s_{i}} \rightarrow X_{s_{i}}$ defined by the rule $j\alpha _{i}=k$ if $b_{ij}\alpha =b_{ik}$ for every $1\le i\le r$ and $1\le j\le s_{i}$, then it is clear that $\alpha _{i} \in N({\mathcal {D}}_{s_i}).$ Moreover, if we consider the bijection

$$\begin{aligned} f:{\mathcal {D}}_{n}(\xi )\rightarrow N({\mathcal {D}}_{s_1})\times N({\mathcal {D}}_{s_2}) \times \cdots \times N({\mathcal {D}}_{s_r}) \end{aligned}$$

defined by the rule $\alpha f=(\alpha _1,\alpha _2,\ldots ,\alpha _r)$, then we have

$$\begin{aligned} |{\mathcal {D}}_{n}(\xi )|= \prod _{i=1}^{r}(s_{i}-1)! , \end{aligned}$$

as required. $\square$

By convention, we take ${\mathcal {D}}_{s_{i}-1}={\mathcal {D}}_{1}$ and $(s_{i}-2)!=1$ for $s_{i}=1$ in the following theorem.

Theorem 3.7

With the above notations, for any $\xi \in E({\mathcal {D}}_{n})$, let $\mathrm {fix\, }(\xi )= \{ 1=a_{1}< a_{2}< \cdots <a_{r} \}$ with $2\le r\le n-1$. Then

$$\begin{aligned} \mathrm {rank\, }({\mathcal {D}}_{n}(\xi ))= \prod _{i=1}^{r} (s_{i}-1)! - \prod _{i=1}^{r} (s_{i}-2)! \end{aligned}$$

where $|a_{i}\xi ^{-1}|=s_{i}$ for every $1\le i \le r$.

Proof

Since $\mathrm {rank\, }({\mathcal {D}}_{n}(\xi ))=|{\mathcal {D}}_{n}(\xi )|-|{\mathcal {D}}_{n} (\xi )^{2}|$, from Theorem 3.6, it is enough to find the cardinality of ${\mathcal {D}}_{n} (\xi )^{2}$.

Let $B_{i}=a_{i}\xi ^{-1}$ and $|B_{i}|=s_{i}$ for every $1\le i \le r$. For any $\alpha \in {\mathcal {D}}_{n}(\xi )^{2}$, similarly, it follows from Proposition 3.1 that the map $\alpha _{\mid _{B_{i}}}: B_{i}\rightarrow B_{i}$ is a nilpotent order-decreasing transformation on $B_{i}$ for every $1\le i\le r$. Moreover, if $s_{i}=1,\, 2$ ($1\le i\le r$), then $\alpha _{ \mid _{B_{i}}}$ is the constant map to $b_{i1}$. Then, we consider the trivial map $\alpha _{i} : \{1\} \rightarrow \{1\}$ in the case $s_{i}=1,\, 2$ ($1\le i\le r$). If $s_{i}\ge 3$ $(1\le i \le r)$, then it follows from Lemma 3.3 that $b_{i3}\alpha =b_{i1}$ and $b_{ij}\alpha \le b_{i(j-2)}$ for every $4\le j\le s_{i}$ (if $s_{i}\ge 4$). Then we consider the map $\alpha _{i} : X_{s_{i}-1} \rightarrow X_{s_{i}-1}$ defined by the rule $j\alpha _{i}=k$ if $b_{i(j+1)}\alpha =b_{i(k+1)}$ for every $1\le j\le s_{i}-1$, similarly, it is clear that $\alpha _{i} \in N({\mathcal {D}}_{s_{i}-1})$ in the case $s_{i}\ge 3$ $(1\le i \le r)$. Thus, if we consider the function

$$\begin{aligned} f:{\mathcal {D}}_{n}(\xi )^2\rightarrow N({\mathcal {D}}_{s_1-1})\times N({\mathcal {D}}_{s_2-1})\times \cdots \times N({\mathcal {D}}_{s_r-1}), \end{aligned}$$

defined by the rule $\alpha f=(\alpha _1,\alpha _2,\ldots ,\alpha _r)$, then it is easy to see that f is also a bijection. Therefore,

$$\begin{aligned} |{\mathcal {D}}_{n}(\xi )^2|=\prod _{i=1}^{r} (s_{i}-2)!, \end{aligned}$$

and so we have

$$\begin{aligned} \mathrm {rank\, }({\mathcal {D}}_{n}(\xi ))=\prod _{i=1}^{r} (s_{i}-1)! - \prod _{i=1}^{r} (s_{i}-2)! \end{aligned}$$

as required. $\square$

Example 3.8

For the given $\xi$ in Example 3.2, we have

$$\begin{aligned} |{\mathcal {D}}_{7}(\xi )|\;&=12 \\ \mathrm {rank\, }({\mathcal {D}}_{7}(\xi ))\;&=10. \end{aligned}$$

from Theorems 3.6 and 3.7.