Combinatorial Results for Semigroups of Order-Preserving and A-Decreasing Finite Transformations

Abstract

For $n\in \mathbb {N}$, let $O_{n}$ be the semigroup of all order-preserving transformations on the finite chain $X_{n}=\{1,\ldots ,n\}$, under its natural order. For any non-empty subset A of $X_{n}$, let $O_{n}(A)$ and $O_{n}^+(A)$ be the subsemigroups of all order-preserving and A-decreasing, and of all order-preserving and A-increasing transformations on $X_{n}$, respectively. In this paper we obtain formulae for the number of elements and for the number of idempotents in $O_{n}(A)$. Moreover, we show that $O_{n}(A)$ contains a zero element if and only if $1\in A$, and then we obtain the number of nilpotents in $O_{n}(A)$ when $1\in A$.

1 Introduction

For $n\in \mathbb {N}$, let $T_{n}$ be the (full) transformations semigroup (under composition) on the finite chain $X_{n}=\{1,\ldots ,n\}$, under its natural order. A transformation $\alpha \in T_{n}$ is called order-preserving if $x\le y$ implies ($\forall x,y \in X_{n}$) $x\alpha \le y\alpha$, and the semigroup of all order-preserving transformations in $T_{n}$ is denoted by $O_{n}$. The problem to find certain combinatorial properties of subsemigroups of $T_{n}$, in particular of $O_{n}$, is an important problem in Semigroup theory, and has been much studied in a literature, see for example [1, 2, 5, 8,9,10].

A transformation $\alpha \in T_{n}$ is called decreasing (increasing) if $x\alpha \le x$ ($x\alpha \ge x$) for each $x \in X_n$ and the semigroup of all order-preserving and decreasing (increasing) transformations in $T_n$ is denoted by $C_n$ ($C_n^+$). It is a well-known fact from [12, Corollary 2.7] that $C_n$ and $C_n^+$ are isomorphic semigroups. For any non-empty subset A of $X_{n}$, a transformation $\alpha \in T_{n}$ is called A-decreasing (A-increasing) if $x\alpha \le x$ ($x\alpha \ge x$) for each $x \in A$. Then the subsemigroup of all order-preserving and A-decreasing transformations in $T_{n}$ is denoted by $O_{n}(A)$, and the subsemigroup of all order-preserving and A-increasing transformations in $T_{n}$ is denoted by $O_{n}^+(A)$, say

$$\begin{aligned} O_{n}(A)= & {} \left\{ \alpha \in O_{n}: ( \forall x \in A)\quad x\alpha \le x\right\} \text{ and } \\ O_{n}^+(A)= & {} \left\{ \alpha \in O_{n}: ( \forall x \in A)\quad x\alpha \ge x\right\} , \end{aligned}$$

respectively. Notice that

• $O_{n}(A)=O_{n}$ if $A=\{n\}$,

• $O_{n}(A)=C_{n}$ if $A=X_n{\setminus } \{n\}$ or $A=X_n$, and

• $O_{n}(A)= O_{n}(A\cup \{n\})$ for any non-empty subset A of $X_n$.

Similarly, we have

• $O_{n}^+(A)=O_{n}$ if $A=\{1\}$,

• $O_{n}^+(A)=C_{n}^+$ if $A=X_n{\setminus } \{1\}$ or $A=X_n$, and

• $O_{n}^+(A)= O_{n}^+(A\cup \{1\})$ for each non-empty subset A of $X_n$.

Also notice that $C_{n}\subseteq O_{n}(A)\subseteq O_{n}$ and $C_{n}^+\subseteq O_{n}^+(A)\subseteq O_{n}$.

Let S be a semigroup. An element $x\in S$ is called idempotent if $x^2=x$, and the set of all idempotents in S is denoted by E(S). Moreover, if S contains a zero element, denoted in general by 0, that implies $0y=y0=0$ for each $y\in S$, then an element $x\in S$ is called nilpotent if $x^k=0$ for some positive integer k, and the set of all nilpotents in S is denoted by N(S).

Recall that nth Fibonacci number $f_n$ is defined by the linear recurrence relation

$$\begin{aligned} f_{n}=f_{n-1}+f_{n-2} \text{ for } n\ge 3 \end{aligned}$$

where $f_{1}=f_{2}=1$ (see for example [6]), and nth Catalan number $\mathcal {C}_{n}$ is defined by

$$\begin{aligned} \mathcal {C}_{n}=\frac{1}{n+1}\left( {\begin{array}{c}2n\\ n\end{array}}\right) = \frac{1}{n}\left( {\begin{array}{c}2n\\ n-1\end{array}}\right) \text{ for } n\ge 1 \end{aligned}$$

(see for example [8]). From [6, Theorem 2.1 and Theorem 2.3] we know that

$$\begin{aligned} |O_{n}| = \left( {\begin{array}{c}2n-1\\ n-1\end{array}}\right) \, \text{ and } \, |E(O_{n})| = f_{2n}. \end{aligned}$$

Also, from [8, Theorem 2.1 and Proposition 2.3] and from [5, Theorem 3.19] we know that

$$\begin{aligned} \begin{array}{lllll} |C_{n}| &{}= &{}|C_{n}^+| &{}=&{} \mathcal {C}_{n},\\ |N(C_{n})| &{}=&{}|N(C_{n}^+)| &{}=&{} \mathcal {C}_{n-1} \text{ and } \\ |E(C_{n})| &{}=&{} |E(C_{n}^+)| &{}=&{} 2^{n-1}. \end{array} \end{aligned}$$

In general a transformation $\alpha \in T_{n}$ is represented by the following tabular form:

$$\begin{aligned} \alpha =\left( \begin{array}{cccc} 1 &{} \cdots &{} n \\ 1\alpha &{} \cdots &{} n\alpha \end{array} \right) . \end{aligned}$$

The height, the kernel and the fix of any transformation $\alpha \in T_n$ are defined by

$$\begin{aligned} \mathrm {h\, }(\alpha )= & {} |\mathrm {im\, }(\alpha )|, \\ \ker (\alpha )= & {} \{ (x,y) : x,y\in X_n \text{ and } x\alpha = y\alpha \} \text{ and } \\ \mathrm {Fix\, }(\alpha )= & {} \{ x\in X_n: x\alpha =x\}, \end{aligned}$$

respectively. Recall that $\ker (\alpha )$ is an equivalence relation on $X_{n}$ and the equivalence classes obtained by $\ker (\alpha )$ are all of the pre-image sets of elements in $\mathrm {im\, }(\alpha )$, say $\{ y\alpha ^{-1}: y\in \mathrm {im\, }(\alpha )\}$, that forms a partition of $X_n$. (For the other terms in semigroup theory, which are not explained here, we refer to [7]).

Let $P=\{ I_{1}, \ldots ,I_{p} \}$ be a partition of a set $X_n$ for $1\le p \le n$. P is called an ordered partition, and we write $P=\{I_{1}< \cdots <I_{p}\}$, if $x < y$ for all $x\in I_{i}$ and $y\in I_{i+1}$ ($1\le i\le p-1$), (the idea of ordering a family of sets appeared on p. 335 of [11]). Notice that the kernel classes of $\alpha \in O_{n}$ are convex subsets C of $X_{n}$ in the sense that

$$\begin{aligned} x,y \in C \text{ and } x\le z\le y \Rightarrow z \in C \end{aligned}$$

(see [4, p. 187]). Thus, if $\mathrm {h\, }(\alpha )=p$ ($2\le p\le n$), there exist $x_1,\ldots , x_{p-1}\in X_n$ such that the kernel classes of $\alpha$ are $A_i=\{x_{i-1}+1,\ldots ,x_i\}$ for $1\le i\le p$ where $x_0=0$ and $x_p=n$, then the ordered partition $\{A_1<\cdots < A_k\}$ of $X_n$ is called the kernel partition of $\alpha$. Moreover, there exist $a_1,\ldots ,a_p\in X_n$ such that $\mathrm {im\, }(\alpha )=\{a_1<\cdots < a_p\}$. Then we can also use the following tabular form:

$$\begin{aligned} \alpha =\left( \begin{array}{cccc} A_1 &{} \cdots &{} A_p \\ a_1 &{} \cdots &{} a_p \end{array}\right) \end{aligned}$$

to represent $\alpha \in O_{n}$.

Recall that, for $k\in \mathbb {N^+}$ and $m\in \mathbb {N}$, the number of integer solutions of the equation

$$\begin{aligned} x_1+ \cdots +x_k=m \quad (x_i\ge 0 \text{ for } 1 \le i \le k) \end{aligned}$$

is $\left( {\begin{array}{c}m+k-1\\ k-1\end{array}}\right) =\left( {\begin{array}{c}m+k-1\\ m\end{array}}\right)$, and throughout the next section we use the notation $x_i=| i\alpha ^{-1}|$ for $1 \le i \le n$. Also we use the equation

$$\begin{aligned} 1+\sum _{l=1}^{t}f_{2l}=f_{2t+1} \end{aligned}$$

that can be shown easily by inductive hypothesis on $t\ge 1$. Moreover, if $A=\{r\}$ for any $1\le r \le n$, we use the notation $O_n(r)$$(O_n^+(r))$ instead of $O_n(\{r\})$$(O_n^+(\{r\}))$.

The rank and the idempotent rank of a finite semigroup S are defined by $\mathrm {rank\, }(S)=\min \{ |A| :A \subseteq S, \langle A \rangle = S \}$ and $\mathrm {idrank\, }(S)=\min \{ |A|:A\subseteq E(S), \langle A \rangle =S \}$, respectively. Zhao, investigated the subsemigroups $O_{n}(A)$ for any non-empty subset A of $X_n$ in [13], and examined in terms of certain algebraic properties for example, the Green’s $*$-Relations and the (idempotent) rank of $O_{n}(A)$. Now we examine the combinatorial properties of $O_{n}(A)$, and obtain formulaes for the numbers of elements, idempotents and (when $1\in A$) nilpotents in $O_{n}(A)$. Moreover, we show that the subsemigroup $O_{n}^+(A)$ is isomorphic to the subsemigroup $O_{n}(\overline{A})$ where $\overline{A}=\{n-a+1:a\in A\}$, and state some properties of $O_{n}^+(A)$ by using this isomorphism.

2 The Number of Elements in $O_n(A)$

Lemma 2.1

For $1\le r \le n$, we have $|O_{n}(r) |=\sum _{s=1}^{r}\left( {\begin{array}{c}r+s-2\\ s-1\end{array}}\right) \left( {\begin{array}{c}2n-(r+s)\\ n-s\end{array}}\right)$.

Proof

Let $1\le r \le n-1$, and let $\alpha \in O_n(r)$. Then there exists $s\in X_n$ such that $1 \le s \le r$ and that $r\alpha =s$. That is $\alpha \in O_n$ with the properties that

• (i) $1\le t\alpha \le s$ for each $1 \le t \le r$,

• (ii) $r\alpha =s$ and

• (iii) $s\le t\alpha \le n$ for each $r\le t \le n$.

Consider and fix an arbitrary s such that $1\le s \le r$. It is easy to see that the number of elements with the properties given above in $O_n$ is equals to the product of the numbers of integer solutions of the equations

$$\begin{aligned} x_1+x_2+\cdots +x_s= & {} r-1 \quad (x_i\ge 0 \text{ for } 1 \le i \le s) \text{ and }\\ x_s+x_{s+1}+\cdots +x_n= & {} n-r \quad (x_i\ge 0 \text{ for } s \le i \le n), \end{aligned}$$

by considering the possible image sets. This product is

$$\begin{aligned} \left( {\begin{array}{c}r+s-2\\ s-1\end{array}}\right) \left( {\begin{array}{c}2n-(r+s)\\ n-s\end{array}}\right) , \end{aligned}$$

and so the result is clear for $1\le r \le n-1$ by considering each $1\le s\le r$ separately. Moreover, $|O_n(n)|=\left( {\begin{array}{c}2n-1\\ n-1\end{array}}\right)$ since $O_n(n)=O_n$, and notice that

$$\begin{aligned} |O_{n}(n)|=\left( {\begin{array}{c}2n-1\\ n-1\end{array}}\right) =\sum _{s=1}^{n}\left( {\begin{array}{c}n+s-2\\ s-1\end{array}}\right) \left( {\begin{array}{c}2n-(n+s)\\ n-s\end{array}}\right) , \end{aligned}$$

by using the well-known basic recurrence that $\left( {\begin{array}{c}m-1\\ k-1\end{array}}\right) +\left( {\begin{array}{c}m-1\\ k\end{array}}\right) =\left( {\begin{array}{c}m\\ k\end{array}}\right) ,$ for $m,k\in \mathbb {Z^+},$ as required. $\square$

Theorem 2.2

For any subset $A=\{r_1,\ldots ,r_k\}$$(k\ge 2)$ of $X_n$, we have

$$\begin{aligned}|O_{n}(A)|= & {} \sum _{s_1=1}^{r_1} \sum _{s_2=s_1}^{r_2}\cdots \sum _{s_k=s_{k-1}}^{r_k} \left( {\begin{array}{c}r_1+s_1-2\\ s_1-1\end{array}}\right) \\&\left[ \prod _{i=2}^{k}\left( {\begin{array}{c}\left( r_i+s_i\right) -\left( r_{i-1}+s_{i-1}\right) -1\\ s_i-s_{i-1}\end{array}}\right) \right] \left( {\begin{array}{c}2n-\left( r_k+s_k\right) \\ n-s_k\end{array}}\right) . \end{aligned}$$

Proof

Let $A=\{r_1,\ldots ,r_k\}\subseteq X_n$$(k\ge 2)$, and let $\alpha \in O_n(A)$. Then there exist $s_1,\ldots , s_k\in X_n$ such that $s_{i-1}\le s_i \le r_i$ and $r_i\alpha =s_i$ for each $1 \le i \le k$ where $s_0=1$. That is $\alpha \in O_n$ with the properties that

• (i) $r_i\alpha =s_i$, for each $1\le i \le k$

• (ii) $1\le t\alpha \le s_1$ for each $1 \le t \le r_1$,

• (iii) $s_{i-1}\le t\alpha \le s_i$ for each $r_{i-1}+1 \le t \le r_i$$(2 \le i \le k)$, and

• (iv) $s_k\le t\alpha \le n$ for $r_k \le t \le n$.

Consider and fix an ordered k-tuple $(s_1,\ldots ,s_k )$ such that $s_{i-1}\le s_i \le r_i$ for each $1\le i \le k$ where $s_0=1$. It is easy to see that the number of elements with the properties given above in $O_n$ equals to the product of the numbers of integer solutions of the equations

$$\begin{aligned} x_1+\cdots +x_{s_{1}}= & {} r_1-1 \quad (x_i\ge 0 \text{ for } 1 \le i \le s_1)\\ x_{s_{i-1}}+ \cdots +x_{s_{i}}= & {} r_{i}-r_{i-1}-1 \quad (x_i\ge 0 \text{ for } 2 \le i \le k) \text{ and }\\ x_{s_{k}}+ \cdots + x_n= & {} n-r_{k}\quad (x_i\ge 0 \text{ for } s_{k} \le i \le n), \end{aligned}$$

by considering the possible image sets. This product is

$$\begin{aligned} \left( {\begin{array}{c}r_1+s_1-2\\ s_1-1\end{array}}\right) \left[ \prod _{i=2}^{k}\left( {\begin{array}{c}\left( r_i+s_i\right) -\left( r_{i-1}+s_{i-1}\right) -1\\ s_i-s_{i-1}\end{array}}\right) \right] \left( {\begin{array}{c}2n-\left( r_k+s_k\right) \\ n-s_k\end{array}}\right) , \end{aligned}$$

and so the result is clear by considering each ordered k-tuple $(s_1,\ldots ,s_k )$ such that $1\le s_1\le \cdots \le s_{k}\le r_k$. $\square$

3 The Number of Idempotents in $O_n(A)$

It is shown in [13, Lemma 2.10] that $\alpha \in O_{n}(A)$ is idempotent if and only if for each $t\in \mathrm {im\, }(\alpha )$, $t\alpha =t$ and $t=\min \{x:x\in t\alpha ^{-1}\cap (A\cup \{t\})\}$. Let $\alpha \in T_n$ and let $\emptyset \ne X \subseteq X_n$. Then the restriction $\alpha$ to the set X is denoted by $\alpha |_{X}$.

Lemma 3.1

For $1\le r \le n$, we have $|E(O_{n}(r))|=f_{2r}\,f_{2(n-r)+1}$.

Proof

Let $1\le r \le n-1$. Since $r \alpha \le r$ for each $\alpha \in E(O_{n}(r))$, we have

$$\begin{aligned} I=\{\alpha |_{\{1,\ldots ,r\}}: \alpha \in E(O_n(r))\}=E(O_r), \end{aligned}$$

and so $|I|=f_{2r}$.

Now consider and fix an arbitrary element $\sigma \in I$, and let $r\sigma =s$ for any $1 \le s \le r$. Then let

$$\begin{aligned} I_{\sigma }=\left\{ \alpha |_{ \{r+1,\ldots ,n\} }:\alpha \in E(O_n(r)) \text{ and } \alpha |_{ \{1,\ldots ,r\} }=\sigma \right\} . \end{aligned}$$

Notice that $I_\sigma =\bigcup _{i=r}^{n}I_{\sigma }^i$ where

for each $r \le i \le n-1$, and

$$\begin{aligned} I_{\sigma }^n= & {} \left\{ \alpha |_{ \{r+1,\ldots , n\}}:\alpha \in E(O_n(r)), \,\alpha |_{ \{1,\ldots ,r\} }=\sigma , \, t\alpha =s \text{ for } r\le t \le n \right\} \\= & {} \left\{ \left( \begin{array}{ccc} r+1 &{} \cdots &{} n\\ s &{} \cdots &{} s \end{array}\right) \right\} . \end{aligned}$$

Also notice that for each $\alpha |_{\{r+1,\ldots , n\}}\in I_{\sigma }^i$ we have

$$\begin{aligned} (i+1)\alpha |_{ \{r+1,\ldots ,n\} }\ge i+1, \end{aligned}$$

since $\alpha \in E(O_n(r))$, and so

for each $r\le i\le n-1$. Thus we have

$$\begin{aligned} \left| I_{\sigma }^i\right| =\left| E(O_{n-i})\right| =f_{2(n-i)} \end{aligned}$$

for $r\le i \le n-1$, and so

$$\begin{aligned} |I_{\sigma }|=\left| \bigcup _{i=r}^{n}I_{\sigma }^i\right| =1+\sum _{l=1}^{n-r}f_{2l}=f_{2(n-r)+1}. \end{aligned}$$

It is easy to see that $|I_{\sigma }|=f_{2(n-r)+1}$ for each $\sigma \in I$, and so we have

$$\begin{aligned} |E(O_{n}(r))|=|I|\, f_{2(n-r)+1}=f_{2r}\,f_{2(n-r)+1} \end{aligned}$$

for $1\le r \le n-1$.

Moreover, $|E(O_n(n))|=f_{2n}$ since $O_n(n)=O_n$, and notice that $|E(O_{n}(n))|= f_{2n} = f_{2n} \, f_{2(n-n)+1}$, as required. $\square$

Theorem 3.2

For any subset $A=\{r_1,\ldots ,r_k\}$$(k\ge 2)$ of $X_n$, we have

$$\begin{aligned} \left| E(O_{n}(A)\right| =f_{2r_1} \,\prod _{j=1}^{k}f_{2(r_{j+1}-r_{j})+1}, \end{aligned}$$

where $r_{k+1}=n$.

Proof

Let $A=\{r_1,\ldots ,r_k\}\subseteq X_n$$(k\ge 2)$. Since $r_1\alpha \le r_1$, for each $\alpha \in E(O_{n}(A))$, we have

$$\begin{aligned} I=\left\{ \alpha |_{\{1,\ldots ,r_1\}}: \alpha \in E(O_n(A))\right\} =E(O_{r_1}), \end{aligned}$$

and so $|I|=f_{2r_1}$.

Now consider and fix an arbitrary element $\sigma _1\in I$ and let $r_1\sigma _1=s_1$ for any $1 \le s_1 \le r_1$. Then let

$$\begin{aligned} I_{\sigma _1}=\left\{ \alpha |_{ \{r_1+1,\ldots ,r_2\} }:\alpha \in E(O_n(r)) \text{ and } \alpha |_{ \{1,\ldots ,r_1\} }=\sigma _1 \right\} . \end{aligned}$$

As in the proof of Lemma 3.1 it is clear that $I_{\sigma _1}=\bigcup _{i=r_{1}}^{r_{2}}I_{\sigma _1}^{i}$ where

for $r_{1}\le i\le r_{2}-1$ and

and that

$$\begin{aligned} \left| I_{\sigma _1}^i\right| =|E(O_{r_2-i})|=f_{2(r_2-i)} \end{aligned}$$

for $r_1\le i \le r_2-1$. Thus

$$\begin{aligned} \left| I_{\sigma _1}\right| =1+\sum _{l=1}^{r_2-r_1}f_{2l}=f_{2(r_2-r_1)+1}. \end{aligned}$$

Also it is easy to see that $|I_{\sigma }|=f_{2(r_2-r_1)+1}$ for each $\sigma \in I$.

With similar steps, for each $2\le j \le k$ consider and fix an arbitrary element $\sigma _j\in I_{\sigma _{j-1}}$ and let $r_j\sigma _j=s_j$ for any $s_{j-1}\le s_j \le r_j$. Then let

$$\begin{aligned} I_{\sigma _j}=\left\{ \alpha |_{ \{r_j+1,\ldots , r_{j+1}\} }:\alpha \in E(O_n(r)) \text{ and } \alpha |_{ \{r_{j-1}+1,\ldots , r_{j}\} }=\sigma _{j}\right\} \end{aligned}$$

where $r_{k+1}=n$. Similarly, we have $I_{\sigma _j}=\bigcup _{i=r_{j}}^{r_{j+1}}I_{\sigma _{j}}^{i}$ where

for $r_{j}\le i\le r_{j+1}-1$ and

and

$$\begin{aligned} \left| I_{\sigma _j}^i\right| =\left| E\left( O_{r_{j+1}-i}\right) \right| =f_{2\left( r_{j+1}-i\right) } \end{aligned}$$

for each $r_j\le i \le r_{j+1}-1$. Thus we have

$$\begin{aligned} \left| I_{\sigma _j}\right| =1+\sum _{l=1}^{r_{j+1}-r_{j}}f_{2l}=f_{2\left( r_{j+1}-r_{j}\right) +1}. \end{aligned}$$

Also it is easy to see that $|I_{\sigma }|=f_{2(r_{j+1}-r_{j})+1}$ for each $\sigma \in I_{\sigma _{j-1}}$ ($2\le j \le k$). Thus we have

$$\begin{aligned} |E(O_{n}(A))|= |I\times I_{\sigma _1}\times \cdots \times I_{\sigma _k}|=f_{2r_1}\prod _{j=1}^{k}f_{2\left( r_{j+1}-r_{j}\right) +1} \end{aligned}$$

where $r_{k+1}=n$, as required. $\square$

4 The Number of Nilpotents in $O_n(A)$

It is well known that $\varepsilon =\left( \begin{array}{ccc} 1 &{} \cdots &{} n\\ 1 &{} \cdots &{} 1 \end{array}\right)$ and $\varepsilon ^+=\left( \begin{array}{ccc} 1 &{} \cdots &{} n\\ n &{} \cdots &{} n \end{array}\right)$ are the zero elements of $C_{n}$ and $C_n^+$, respectively. Also, from [8, Lemma 2.2], we know that an element $\alpha$ in $C_{n}$$(C_n^+)$ is nilpotent if and only if $\mathrm {Fix\, }(\alpha )=\{1\}$$(\mathrm {Fix\, }(\alpha )=\{n\})$.

Theorem 4.1

For any non-empty subset A of $X_{n}$$(n\ge 2)$, $O_{n}(A)$ contains a zero element if and only if $1\in A$.

Proof

($\Leftarrow$) Let A be a subset of $X_{n}$ contains 1. Then clearly

$$\begin{aligned} \varepsilon =\left( \begin{array}{ccc} 1 &{} \cdots &{} n\\ 1 &{} \cdots &{} 1 \end{array}\right) \end{aligned}$$

is a zero element in $O_{n}(A)$, as required.

($\Rightarrow$) Suppose that $O_{n}(A)$ contains a zero element denoted by $\xi$. Since $O_{n}(A)$ contains at least one constant transformation, and since $\alpha \xi =\xi \alpha =\xi$ for each $\alpha \in O_{n}(A)$, there exists $1\le i\le n$ such that $\xi =\left( \begin{array}{ccc} 1 &{} \cdots &{} n\\ i &{} \cdots &{} i \end{array}\right) .$ Now assume that $i\ne 1$. Then it is clear that there exist at least one element $\alpha \in O_{n}(A)$ such that $i\alpha \ne i$, and so we have

$$\begin{aligned} i=i\xi =i\xi \alpha =i\alpha \ne i \end{aligned}$$

which is a contradiction. Hence $i=1$, and so $\xi =\varepsilon$.

Next assume that $1\notin A$. Then there exist at least one element $\beta \in O_{n}(A)$ such that $1\beta \ge 2$, and so we have $\xi \beta \ne \xi$ which is a contradiction. Thereby $1\in A$, as required. $\square$

Lemma 4.2

For any subset A of $X_n$ contains 1, an element $\alpha \in O_{n}(A)$ is nilpotent if and only if $x\alpha \ne x$ for each $2\le x\le n$.

Proof

We prove this claim as in the proof of [3, Lemma 2.1].

$(\Rightarrow )$ Let $\alpha \in N(O_{n}(A))$. If there exists $2\le x\le n$ such that $x\alpha =x$, it is clear that

$$\begin{aligned} 1\ne x=x\alpha =x\alpha ^2=x\alpha ^3=\cdots , \end{aligned}$$

and so $\alpha ^k\ne \varepsilon$ for all $k\in \mathbb {Z^+}$, which is a contradiction.

$(\Leftarrow )$ Let $\alpha \in O_{n}(A)$, and let $x\alpha \ne x$ for each $2\le x\le n$. Then we have

$$\begin{aligned} \left\{ \begin{array}{ll} x\alpha<x \text{ or } x\alpha >x &{} \text{ if } x\in X_n{\backslash } A,\\ x\alpha <x &{} \text{ if } x \in A{\setminus }\{1\} \end{array} \right. , \end{aligned}$$

and so for each $k\ge 2$

$$\begin{aligned} \left\{ \begin{array}{ll} x\alpha ^k<x \text{ or } x\alpha ^k>x &{} \text{ if } x\in X_n{\backslash } A,\\ x\alpha ^k<x &{} \text{ if } x \in A{\setminus }\{1\} \end{array} \right. . \end{aligned}$$

Consequently we have $x\alpha ^k\ne x$ for each $2\le x\le n$ and for each $k\in \mathbb {Z}^+$.

If $\mathrm {im\, }(\alpha )=\{1\}$ then we have $\alpha =\varepsilon$, and so the result is clear. Now let $\{1\}\subset \mathrm {im\, }(\alpha )$, and let

$$\begin{aligned} \alpha =\left( \begin{array}{cccc} A_1 &{} A_2 &{}\cdots &{} A_r \\ 1 &{} a_2 &{} \cdots &{} a_r \end{array} \right) \end{aligned}$$

where $\{A_1<\cdots <A_r\}$ is the kernel partition of $\alpha$ and $\mathrm {im\, }(\alpha )=\{1<a_2<\cdots <a_r\}\subseteq X_n$ for any $2\le r\le n$. It is clear that $a_r\notin A_r$ since $a_r\alpha \ne a_r$, and so $a_r < \min \{A_r\}$. Thus we have $\mathrm {im\, }(\alpha )\cap A_r=\emptyset$ and so $a_r\notin \mathrm {im\, }(\alpha ^2)$, that is

$$\begin{aligned} \mathrm {im\, }(\alpha ^2)\subset \mathrm {im\, }(\alpha ). \end{aligned}$$

If $\mathrm {im\, }(\alpha ^2)=\{1\}$ then clearly $\alpha ^2=\varepsilon$, and so the result is clear. Now let $\{1\}\ne \mathrm {im\, }(\alpha ^2)\subset \mathrm {im\, }(\alpha )$, and let

$$\begin{aligned} \alpha ^2=\left( \begin{array}{cccc} B_1 &{} B_2 &{} \cdots &{} B_t \\ 1 &{} b_2 &{} \cdots &{} b_t \end{array} \right) \end{aligned}$$

where $\{B_1<\cdots <B_t\}$ is the kernel partition of $\alpha ^2$ and $\mathrm {im\, }(\alpha ^2)=\{1< b_2<\cdots<b_t\}\subset \{1<a_2<\cdots <a_r\}$ for $2\le t< r\le n$. Similarly we have

$$\begin{aligned} \mathrm {im\, }(\alpha ^2)^2\subset \mathrm {im\, }(\alpha ^2) \end{aligned}$$

since $x\alpha ^2\ne x$ for each $2\le x\le n$. If we continue with this procedure, we will obtain a strict descent sequence:

$$\begin{aligned} \mathrm {im\, }(\alpha )\supset \mathrm {im\, }(\alpha ^2)\supset \mathrm {im\, }(\alpha ^2)^2\supset \cdots . \end{aligned}$$

Since $|\mathrm {im\, }(\alpha )|$ is finite, there exists $m>2$ such that

$$\begin{aligned} \alpha ^{m}=\left( \begin{array}{ccc} 1 &{}\cdots &{} n \\ 1 &{} \cdots &{} 1 \end{array} \right) =\varepsilon , \end{aligned}$$

as required. $\square$

Lemma 4.3

For any subset A of $X_n$ contains 1, $N(O_n(A))=N(C_n)$.

Proof

Let A be a subset of $X_{n}$ contains 1. First recall that $C_n \subseteq O_n(A)$, and that the zero element $\varepsilon$ of $C_n$ is also the zero element of $O_n(A)$. Hence, we immediately have $N(C_n) \subseteq N(O_n(A))$.

Conversely, let $\alpha \in N(O_n(A))$. Then clearly $1\alpha =1$. If $2\in A$ we immediately have $2\alpha =1$ from Lemma 4.2. Now let $2\notin A$ and assume that $2\alpha >2$. Then, by using Lemma 4.2 in each step, we have

$$\begin{aligned} 2\alpha>2\Rightarrow 3\alpha\ge & {} 2\alpha>2,\\ \Rightarrow 4\alpha\ge & {} 3\alpha>3, \\&\vdots&\,\\ \Rightarrow n\alpha\ge & {} (n-1)\alpha >n-1, \end{aligned}$$

and so $n\alpha =n$, which is in a contradict with Lemma 4.2. Thus, even $2\notin A$ we also have $2\alpha =1$.

If $3\in A$, then we immediately have $3\alpha =1$ or $3\alpha =2$ from Lemma 4.2. Now let $3\notin A$. Similarly, it can be shown that to be $3\alpha >3$ implies that $n\alpha =n$, which is in a contradict with Lemma 4.2. Thus even $3\notin A$ we also have $3\alpha =1$ or $3\alpha =2$.

If we continue with this procedure, we will obtain that $1\le x\alpha <x$ for each $1\le x\le n$, and so $\alpha \in C_n$. Moreover, from Lemma 4.2 and [8, Lemma 2.2], we have $\alpha \in N(C_n)$, as required. $\square$

Corollary 4.4

For any subset A of $X_n$ contains 1, we have

$$\begin{aligned} |N(O_n(A))|=|N(C_n)|=\mathcal {C}_{n-1}. \end{aligned}$$

$\square$

5 Remark

First we state the following theorem which can be proved easily and will be used throughout this section.

Theorem 5.1

For any non-empty subset A of $X_{n}$, let $\overline{A}=\{n-a+1:a\in A\}$. Then the map

$$\begin{aligned} \varphi : O_{n}^+(A)\rightarrow & {} O_{n}(\overline{A})\\ \alpha\mapsto & {} \alpha ^* \end{aligned}$$

is an isomorphism where $x\alpha ^*=n-(n-x+1)\alpha +1$ for each $x\in X_n$. $\square$

This isomorphism proves to be powerful tool in translating results on $O_{n}(\overline{A})$ to $O_{n}^+(A)$. For example, we can immediately translate all results, stated in [13], to $O_{n}^+(A)$, more precisely, for $n\ge 3$ we have

$$\begin{aligned} \mathrm {rank\, }(O_{n}^+(A))= & {} \left\{ \begin{array}{ll} n-1 &{} \text{ if } n\in A \\ n &{} \text{ if } n\notin A \end{array} \right. \\ \mathrm {idrank\, }(O_{n}^+(A))= & {} 2n-2-|A{\setminus }\{1\}|. \end{aligned}$$

Moreover, we have

• $\alpha \in E(O_{n}^+(A))$ if and only if   $\alpha ^*\in E(O_{n}(\overline{A}))$,

• $O_{n}^+(A)$ has a zero element if and only if $n\in A$, and

• $\alpha \in N(O_{n}^+(A))$ if and only if   $\alpha ^*\in N(O_{n}(\overline{A}))$ when $n\in A$.

Thereby we can state certain combinatorial results on $O_{n}^+(A)$, for any non-empty subset A of $X_{n}$, as in the following theorem.

Theorem 5.2

For any non-empty subset A of $X_{n}$, we have

• (i) $|O_{n}^+(A)|=|O_{n}(\overline{A})|$,

• (ii) $|E(O_{n}^+(A))|=|E(O_{n}(\overline{A}))|$, and

• (iii) $|N(O_{n}^+(A))|=|N(O_{n}(\overline{A}))|=|N(C_n)|=\mathcal {C}_{n-1}$ when $n\in A$. $\square$