On the Nilpotent Ranks of the Principal Factors of Orientation-Preserving Transformation Semigroups

Abstract

Let $X_n$ be a chain with n elements, and let $\mathcal {OP}_n$ be the monoid of all orientation-preserving transformations of $X_n$. In this paper, we investigate the nilpotent ranks of the principal factors of the semigroup $\mathcal {OP}_n$.

1 Introduction and Preliminaries

Let $X_n$ be a chain with n elements, say $X_n=\{1<2<\cdots <n\}$. As usual, we denote by $\mathcal {T}_n$ the monoid of all full transformations of a finite set $X_n$. We say that a transformation $\alpha \in \mathcal {T}_n$ is order preserving if $x\le y$ implies $x\alpha \le y\alpha$, for all $x,y\in X_n$. Denote by $\mathcal {O}_n$ the submonoid of $\mathcal {T}_n$ of all full order-preserving transformations of $X_n$.

Let $c=(c_1,c_2,\ldots ,c_t)$ be a sequence of t ($t\ge 0$) elements from the chain $X_n$. We say that c is cyclic if there exists not more than one index $i\in \{1,\ldots ,t\}$ such that $c_i>c_{i+1}$, where $c_{t+1}$ denotes $c_1$. We say that $\alpha \in \mathcal {T}_n$ is orientation preserving if the sequence of its image $(1\alpha ,2\alpha ,\ldots ,n\alpha )$ is cyclic. Denote by $\mathcal {OP}_n$ the submonoid of $\mathcal {T}_n$ of all full orientation-preserving transformations of $X_n$.

Let S be a semigroup. Denote by $S^1$ the monoid obtained from S through the adjoining of an identity if S has none and exactly S otherwise. Recall the definition of Green’s equivalence relations ${\mathscr {R}}$, ${\mathscr {L}}$, ${\mathscr {H}}$, and ${\mathscr {J}}$: for all $u, v\in S$,

$$\begin{aligned} \begin{array}{rcl} u{\mathscr {R}} v &{}\text{ if } \text{ and } \text{ only } \text{ if }&{} uS^1=vS^1;\\ u{\mathscr {L}} v &{}\text{ if } \text{ and } \text{ only } \text{ if }&{} S^1u=S^1v;\\ u{\mathscr {H}} v &{}\text{ if } \text{ and } \text{ only } \text{ if }&{} u{\mathscr {R}} v \text{ and } u{\mathscr {L}} v;\\ u{\mathscr {J}} v &{}\text{ if } \text{ and } \text{ only } \text{ if }&{} S^1uS^1=S^1vS^1. \end{array} \end{aligned}$$

Associated to Green’s relation ${\mathscr {J}}$, there is a quasi-order $\le _{\mathscr {J}}$ on S defined by

$$\begin{aligned} u\le _{\mathscr {J}} v \quad \text{ if } \text{ and } \text{ only } \text{ if }\quad S^1uS^1\subseteq S^1vS^1, \end{aligned}$$

for all $u, v\in S$. Notice that, for every $u, v\in S$, we have $u\,{\mathscr {J}}\,v$ if and only if $u\le _{\mathscr {J}}v$ and $v\le _{\mathscr {J}}u$. Denote by $J_{u}^S$ the ${\mathscr {J}}$-class of the element $u\in S$. As usual, a partial order relation $\le _{\mathscr {J}}$ is defined on the quotient set $S/{\mathscr {J}}$ by putting $J_{u}^S\le _{\mathscr {J}}J_{v}^S$ if and only if $u\le _{\mathscr {J}}v$, for all $u, v\in S$. Given a subset A of S and $u\in S$, we denote by E(A) set of idempotents of S belonging to A and by $L^S_u$, $R^S_u$, and $H^S_u$ the ${\mathscr {L}}$-class, ${\mathscr {R}}$-class, and ${\mathscr {H}}$-class of u, respectively. For general background on Semigroup Theory, we refer the reader to Howie’s book [8].

Let $\mathcal {M}_n$ denotes any of the monoids $\mathcal {T}_n$, $\mathcal {O}_n$, or $\mathcal {OP}_n$. Then $\mathcal {M}_n$ is regular. The Green relations ${\mathscr {L}}$ and ${\mathscr {R}}$ of $\mathcal {M}_n$ can be characterized by $\alpha {\mathscr {L}} \beta$ if and only if $\mathop {\mathrm {im}}\nolimits (\alpha )=\mathop {\mathrm {im}}\nolimits (\beta )$, for all $\alpha ,\beta \in \mathcal {M}_n$, and $\alpha {\mathscr {R}} \beta$ if and only if $\mathop {\mathrm {ker}}\nolimits (\alpha )=\mathop {\mathrm {ker}}\nolimits (\beta )$, for all $\alpha ,\beta \in \mathcal {M}_n$. Regarding the Green relation ${\mathscr {J}}$, we have $\alpha \le _{{\mathscr {J}}} \beta$ if and only if $|\mathop {\mathrm {im}}\nolimits (\alpha )|\le |\mathop {\mathrm {im}}\nolimits (\beta )|$ and so $\alpha {{\mathscr {J}}} {\beta }$ if and only if $|\mathop {\mathrm {im}}\nolimits (\alpha )|=|\mathop {\mathrm {im}}\nolimits (\beta )|$, for all $\alpha ,\beta \in \mathcal {M}_n$. It follows that the partial order $\le _{\mathscr {J}}$ on the quotient $\mathcal {M}_n/{\mathscr {J}}$ is linear. More precisely, letting

$$\begin{aligned} J_r^{\mathcal {M}_n}=\{\alpha \in {\mathcal {M}_n}: |\mathop {\mathrm {im}}\nolimits (\alpha )|=r\}, \end{aligned}$$

i.e., the ${\mathscr {J}}$-class of the transformations of image size r (called the rank of the transformations) of $\mathcal {M}_n$, for $1\le r\le n$, we have

$$\begin{aligned} \mathcal {M}_n/{\mathscr {J}}=\{J_1^{\mathcal {M}_n}\le _{\mathscr {J}}J_2^{\mathcal {M}_n}\le _{\mathscr {J}}\dots \le _{\mathscr {J}}J_n^{\mathcal {M}_n}\}. \end{aligned}$$

See [1, 2, 7, 8] for more details.

Since ${\mathcal {M}_n}/ {\mathscr {J}}$ is a chain, the sets

$$\begin{aligned} \mathcal {M}(n,r)=\{\alpha \in \mathcal {M}_n: |\mathop {\mathrm {im}}\nolimits (\alpha )|\le r\}=J_1^{\mathcal {M}_n}\cup J_2^{\mathcal {M}_n}\cup \cdots \cup J_r^{\mathcal {M}_n}, \end{aligned}$$

with $1\le r\le n$, constitute all the non-null ideals of $\mathcal {M}_n$ (see [4, Note of p. 181]). Notice that $\mathcal {M}(n,r)=\mathcal {T}(n,r)$ if $\mathcal {M}_n=\mathcal {T}_n$; $\mathcal {M}(n,r)=\mathcal {O}(n,r)$ if $\mathcal {M}_n=\mathcal {O}_n$; $\mathcal {M}(n,r)=\mathcal {OP}(n,r)$ if $\mathcal {M}_n=\mathcal {OP}_n$.

Notice that every principal factor of $\mathcal {M}_n$ associated with the maximum ${\mathscr {J}}$-class $J_r^{\mathcal {M}_n}$ is the Rees quotient $\mathcal {M}(n,r)/\mathcal {M}(n,r-1)$ ($2\le r\le n$), which we denote by $\mathcal {P}^{\mathcal {M}_n}_r$. It is usually convenient to think of $\mathcal {P}^{\mathcal {M}_n}_r$ as $J_r^{\mathcal {M}_n}\cup \{0\}$, and the product of two elements of $\mathcal {P}^{\mathcal {M}_n}_r$ is taken to be zero if it falls in $\mathcal {M}(n,r-1)$. $\mathcal {P}^{\mathcal {M}_n}_r$ is a completely 0-simple semigroup.

As usual, the rank of a finite semigroup S is defined by $\mathop {\mathrm {rank}}\nolimits S=\min \{|A|: A\subseteq S, \langle A \rangle =S \}$. If S is generated by its set E of idempotents, then the idempotent rank of S is defined by $\mathop {\mathrm {idrank}}\nolimits S=\min \{|A|: A\subseteq E, \langle A \rangle =S \}$. If S is generated by its set N of nilpotents, then the nilpotent rank of S is defined by $\mathop {\mathrm {nilrank}}\nolimits S=\min \{|A|: A\subseteq N, \langle A \rangle =S \}$. Clearly, $\mathop {\mathrm {rank}}\nolimits S\le \mathop {\mathrm {idrank}}\nolimits S$ and $\mathop {\mathrm {rank}}\nolimits S\le \mathop {\mathrm {nilrank}}\nolimits S$.

In [6], Gomes and Howie showed that both the rank and the idempotent rank of $\mathcal {P}^{\mathcal {T}_n}_{n-1}$ are equal to $n(n-1)/2$. This result was later generalized by Howie and McFadden [9] who showed that the rank and idempotent rank of $\mathcal {P}^{\mathcal {T}_n}_{r}$ are both equal to S(n, r), the Stirling number of the second kind, for $2\le r\le n-1$. Yang [10] showed that the nilpotent rank of $\mathcal {P}^{\mathcal {T}_n}_{r}$ is also S(n, r), $2\le r\le n-1$.

The rank and idempotent rank of $\mathcal {P}^{\mathcal {O}_n}_{n-1}$ were shown to be n and $2n-2$, respectively, by Gomes and Howie [7]. Garba [5] generalized this result by showing that both the rank and the idempotent rank of the principal factor $\mathcal {P}^{\mathcal {O}_n}_{r}$ are equal to ${n\atopwithdelims ()r}$, for $2\le r\le n-2$. Yang [11] showed that the nilpotent rank of the principal factor $\mathcal {P}^{\mathcal {O}_n}_{r}$ are also equal to ${n\atopwithdelims ()r}$, for $3\le r\le n-1$.

Regarding the semigroup $\mathcal {OP}_n$, Zhao [12] showed that both the rank and the idempotent rank of the principal factor $\mathcal {P}^{\mathcal {OP}_n}_{n-1}$ are equal to ${n\atopwithdelims ()2}$. Recently, Zhao and Fernandes [13] showed that the rank and the idempotent rank of the principal factor $\mathcal {P}^{\mathcal {OP}_n}_{r}$ are equal to ${n\atopwithdelims ()r}$, for $2\le r\le n-1$. In this paper, we investigate the nilpotent rank of the principal factor $\mathcal {P}^{\mathcal {OP}_n}_{r}$, for $2\le r\le n-1$. In Sect. 2, we characterize the structure of the minimal generating sets of $\mathcal {OP}(n,r)$. As applications, we prove that the number of distinct minimal generating sets is $r^n n!$. In Sect. 3, we show that the nilpotent rank of the principal factor $\mathcal {P}^{\mathcal {OP}_n}_{r}$ is equal to ${n\atopwithdelims ()r}$, for $2\le r\le n-1$.

Remark 1

In this paper, it will always be clear from context when additions are taken modulo n (or modulo t where t is the number of elements of any sequence).

Throughout this paper, for simplicity, we always assume that $n\ge 3$.

2 The Minimal Generating Sets of $\mathcal {OP}(n,r)$

Let $\alpha \in \mathcal {T}_n$. As usual, we write $\mathop {\mathrm {im}}\nolimits (\alpha )$ for the image of $\alpha$. The kernel of $\alpha$ is the equivalence $\mathop {\mathrm {ker}}\nolimits (\alpha )=\{(x,y)\in X_n \times X_n: x\alpha =y\alpha \}$. The equivalence classes of $X_n$ with respect to $\mathop {\mathrm {ker}}\nolimits (\alpha )$ are called the kernel classes of $\alpha$.

We denote by [i, k] the set $\{i,i+1, \dots , k-1,k\}$ for $i,k\in X_n$. A set $K\subseteq X_n$ is convex if K has the form $[i,i+t]$, for some $i\in X_n$ and $0\le t\le n-1$. We shall refer to an equivalence $\pi$ on $X_n$ as convex if its classes are convex subsets of $X_n$, and we shall say that $\pi$ is of weight r if $|X_n/\pi |=r$.

It is known that that every kernel $\mathop {\mathrm {ker}}\nolimits (\alpha )$ of $\alpha \in \mathcal {OP}_n$ is convex (see [2]). Let $2\le r\le n-1$. Then, given a transformation $\alpha \in \mathcal {OP}_n$ of rank r, the kernel $\mathop {\mathrm {ker}}\nolimits (\alpha )$ is convex. Therefore, we may establish a one-to-one correspondence between the collection of all subsets of $X_n$ of cardinality r (which consists of all possible images of elements of rank r of $\mathcal {OP}_n$) and the collection of all possible convex kernels of elements of rank r of $\mathcal {OP}_n$ as follows: associate to each r-set $\{a_1, a_2,\ldots , a_r\}$, with $1\le a_1<a_2<\cdots <a_r\le n$, the r-partition $\{A_1,\ldots , A_r\}$ of $X_n$ defined by

$$\begin{aligned} A_i=\{a_i,a_i+1,\ldots ,a_{i+1}-1\},~\hbox { for}\ 1\le i\le r \end{aligned}$$

(1)

(notice that $a_{r+1}=a_1$ by Remark 1). Thus, the ${\mathscr {J}}$-class $J^{\mathcal {OP}_n}_r$ of $\mathcal {OP}_n$ (and of $\mathcal {OP}(n,r)$) contains $\left( {\begin{array}{c}n\\ r\end{array}}\right)$ ${\mathscr {R}}$-classes and $\left( {\begin{array}{c}n\\ r\end{array}}\right)$ ${\mathscr {L}}$-classes. See [2] for more details.

Now, for $1\le a_1<a_2<\cdots <a_r\le n$, define

$$\begin{aligned} \varepsilon _{a_1, a_2,\ldots , a_r} = \begin{pmatrix} A_{1} &{} A_{2} &{} \cdots &{} A_{r} \\ a_{1} &{} a_{2} &{} \cdots &{} a_{r} \end{pmatrix}, \end{aligned}$$

where $\{A_1,\ldots , A_r\}$ is the r-partition of $X_n$ associated to $\{a_1, a_2,\ldots , a_r\}$ as in (1). Clearly, $\varepsilon _{a_1, a_2,\ldots , a_r}\in E(J^{\mathcal {OP}_n}_r)$. Moreover, the set

$$\begin{aligned} E_r=\left\{ \varepsilon _{a_1, a_2,\ldots , a_r}: 1\le a_1<a_2<\cdots <a_r\le n\right\} \end{aligned}$$

contains exactly one (idempotent) element from each ${\mathscr {R}}$-class and from each ${\mathscr {L}}$-class of $\mathcal {OP}_n$ of rank r.

The following lemma was proved by Zhao and Fernandes [13, Proposition 2.4].

Lemma 2.1

For $2\le r \le n-1$, the set $E_r$ generates $\mathcal {OP}(n,r)$.

Now, we record a well-known result, due to Miller and Clifford ([3, Theorem 2.17]).

Lemma 2.2

For any two elements a, b in a semigroup S, $ab\in R_a\cap L_b$ if and only if $E(R_b\cap L_a)\ne \emptyset$.

Let U be a subset of $J^{\mathcal {OP}_n}_r$. We say that U satisfies Condition $(\mathbf {R\sim L})$ if U contains exactly one element from each ${\mathscr {R}}$-class and from each ${\mathscr {L}}$-class of $J^{\mathcal {OP}_n}_r$. Notice that the set $E_r$ satisfies Condition $(\mathbf {R\sim L})$. From Lemma 2.1, we know that $E_r$ is a generating set of $\mathcal {OP}(n,r)$. In fact, we have the following.

Lemma 2.3

Let G be a subset of $J^{\mathcal {OP}_n}_r$. If G satisfies Condition $(\mathbf {R\sim L})$, then $\mathcal {OP}(n,r)=\langle G \rangle$.

Proof

We shall show that $E_r\subseteq \langle G \rangle$ and so $\mathcal {OP}(n,r)=\langle E_r\rangle$ by Lemma 2.1. Let $\varepsilon \in E_r$. Since G satisfies Condition $(\mathbf {R\sim L})$, there exists unique $\alpha \in G$ such that $\alpha {{\mathscr {R}}} \varepsilon$. If $(\alpha , \varepsilon )\in {{\mathscr {L}}}$, then $\alpha {{\mathscr {H}}} \varepsilon$. Notice that each ${\mathscr {H}}$-class of $J^{\mathcal {OP}_n}_r$ that contains an idempotent is a cyclic group of order r (by [2, Corollary 3.6]). Thus $\varepsilon =\alpha ^r\in \langle G \rangle$. If $(\alpha , \varepsilon )\notin {{\mathscr {L}}}$, then since $E_r$ satisfies Condition $(\mathbf {R\sim L})$, there exists unique $\varepsilon _1\in E_r\backslash \{\varepsilon \}$ such that $(\alpha ,\varepsilon _1)\in {\mathscr {L}}$. Since G satisfies Condition $(\mathbf {R\sim L})$, there exists unique $\alpha _1\in G\backslash \{\alpha \}$ such that $(\alpha _1, \varepsilon _{1})\in {{\mathscr {R}}}$. Notice that $\varepsilon _1\in E(L_\alpha \cap R_{\alpha _1})$. Then, by Lemma 2.2, $\alpha \alpha _1\in R_{\alpha }\cap L_{\alpha _1}=R_{\varepsilon }\cap L_{\alpha _1}$. If $(\alpha _1, \varepsilon )\in {\mathscr {L}}$, then $(\alpha \alpha _1) {\mathscr {H}} \varepsilon$ and so $\varepsilon =(\alpha \alpha _1) ^r\in \langle G \rangle$. If $(\alpha _1, \varepsilon )\notin {{\mathscr {L}}}$, then since $E_r$ satisfies Condition $(\mathbf {R\sim L})$, there exists unique $\varepsilon _2\in E_r\backslash \{\varepsilon , \varepsilon _1 \}$ such that $(\alpha _1,\varepsilon _2)\in {\mathscr {L}}$. Since G satisfies Condition $(\mathbf {R\sim L})$, there exists unique $\alpha _2 \in G\backslash \{\alpha , \alpha _1\}$ such that $(\alpha _2, \varepsilon _{2})\in {{\mathscr {R}}}$. Notice that $\varepsilon _2\in E(L_{\alpha _1}\cap R_{\alpha _2})=E(L_{\alpha \alpha _1}\cap R_{\alpha _2})$. Then, by Lemma 2.2, $(\alpha \alpha _1)\alpha _2\in R_{\alpha \alpha _1}\cap L_{\alpha _2}=R_{\varepsilon }\cap L_{\alpha _2}$. If $(\alpha _2, \varepsilon )\in {\mathscr {L}}$, then $(\alpha \alpha _1\alpha _2) {\mathscr {H}} \varepsilon$ and so $\varepsilon =(\alpha \alpha _1\alpha _2) ^r\in \langle G \rangle$. If $(\alpha _2, \varepsilon )\notin {{\mathscr {L}}}$, then since $E_r$ satisfies Condition $(\mathbf {R\sim L})$, there exists unique $\varepsilon _3\in E_r\backslash \{\varepsilon , \varepsilon _1, \varepsilon _2 \}$ such that $(\alpha _2,\varepsilon _3)\in {\mathscr {L}}$. Since G satisfies Condition $(\mathbf {R\sim L})$, there exists unique $\alpha _3 \in G\backslash \{\alpha , \alpha _1,\alpha _2\}$ such that $(\alpha _3,\varepsilon _{3})\in {{\mathscr {R}}}$. Notice that $\varepsilon _3\in E(L_{\alpha _2}\cap R_{\alpha _3})=E(L_{\alpha \alpha _1\alpha _2}\cap R_{\alpha _3})$. Then, by Lemma 2.2, $(\alpha \alpha _1\alpha _2)\alpha _3\in R_{\alpha \alpha _1\alpha _2}\cap L_{\alpha _3}=R_{\varepsilon }\cap L_{\alpha _3}$. Continuing this demonstration, since G and $E_r$ satisfy Condition $(\mathbf {R\sim L})$, there must exist $k\le m$ ($m={n\atopwithdelims ()r}$) such that $\alpha _k\in G\backslash \{\alpha , \alpha _1, \alpha _{k-1}\}$, $(\alpha \dots \alpha _{k-1})\alpha _k\in R_{\varepsilon }\cap L_{\alpha _k}$ and $\alpha _k {\mathscr {L}} \varepsilon$. Then $(\alpha \alpha _1\dots \alpha _k) {\mathscr {H}} \varepsilon$ and so $\varepsilon =(\alpha \alpha _1\dots \alpha _k) ^r\in \langle G \rangle$. $\square$

Since $\mathcal {OP}(n,r)$ has rank ${n\atopwithdelims ()r}$ (see [13, Theorem 2.7]), a generating set of $\mathcal {OP}(n,r)$ with ${n\atopwithdelims ()r}$ elements is a minimal generating set. Moreover, if $\alpha$ is an element of $\mathcal {OP}(n,r)$ of rank r and $\beta$ and $\gamma$ are two elements of $\mathcal {OP}(n,r)$ such that $\alpha =\beta \gamma$, then $\mathop {\mathrm {ker}}\nolimits (\alpha )=\mathop {\mathrm {ker}}\nolimits (\beta )$ and $\mathop {\mathrm {im}}\nolimits (\alpha )=\mathop {\mathrm {im}}\nolimits (\gamma )$. Then any generating set of $\mathcal {OP}(n,r)$ with ${n\atopwithdelims ()r}$ elements be the subset having exactly one element from each ${\mathscr {R}}$-class and from each ${\mathscr {L}}$-class of rank r. These observations, together with the Lemma 2.3, prove the following result:

Theorem 2.4

Let M be a subset of $\mathcal {OP}(n,r)$ with ${n\atopwithdelims ()r}$ elements. Then M is a minimal generating set of $\mathcal {OP}(n,r)$ if and only if M be the subset having exactly one element from each ${\mathscr {R}}$-class and from each ${\mathscr {L}}$-class of $\mathcal {OP}_n$ of rank r.

Notice that each ${\mathscr {H}}$-class of $J^{\mathcal {OP}_n}_r$ that contains an idempotent is a cyclic group of order r (by [2, Corollary 3.6]). Thus, we have the following corollary from Theorem 2.4:

Corollary 2.5

Let M be a minimal generating set of $\mathcal {OP}(n,r)$. Then the number of distinct sets M is $r^nn!$.

3 The Nilpotent Rank of $\mathcal {P}^{\mathcal {OP}_n}_{r}$

Recall that Zhao [12] showed that both the rank and the idempotent rank of the principal factor $\mathcal {P}^{\mathcal {OP}_n}_{n-1}$ are equal to ${n\atopwithdelims ()2}$. Recently, Zhao and Fernandes [13] showed that the rank and the idempotent rank of the principal factor $\mathcal {P}^{\mathcal {OP}_n}_{r}$ are equal to ${n\atopwithdelims ()r}$, for $2\le r\le n-1$. In this section, we show that the nilpotent rank of the principal factor $\mathcal {P}^{\mathcal {OP}_n}_{r}$ are also equal to ${n\atopwithdelims ()r}$, for $2\le r\le n-1$.

Let A be a subset of $X_n$ of cardinality r and let $\pi$ be a convex equivalence of weight r on $X_n$. We may write $H_{(\pi ,A)}$ for the ${\mathscr {H}}$-class of $J^{\mathcal {OP}_n}_{r}$, which is the intersection of $R_{\pi }=\{\alpha \in J^{\mathcal {OP}_n}_{r}: \mathop {\mathrm {ker}}\nolimits (\alpha )=\pi \}$ and $L_A=\{\alpha \in J^{\mathcal {OP}_n}_{r}: \mathop {\mathrm {im}}\nolimits (\alpha )=A\}$. The subset A of $X_n$ of cardinality r is said to be a transversal of the convex equivalence $\pi$ of weight r on $X_n$ if each convex equivalence $\pi$-class contains exactly one element of A. The following lemma is obvious:

Lemma 3.1

Let $\alpha \in \mathcal {P}^{\mathcal {OP}_n}_{r}$. Then $\alpha$ is nilpotent if and only if $\mathop {\mathrm {im}}\nolimits (\alpha )$ is not a transversal of $\mathop {\mathrm {ker}}\nolimits (\alpha )$.

Our main result of this section is as follows:

Theorem 3.2

Let $n\ge 3$ and $2\le r\le n-1$. Then

$$\begin{aligned}\mathop {\mathrm {nilrank}}\nolimits \mathcal {P}^{\mathcal {OP}_n}_{r}={n\atopwithdelims ()r}. \end{aligned}$$

The proof depends on the following lemma:

Lemma 3.3

Let $A_1, A_2,\dots , A_m$ (where $m={n\atopwithdelims ()r}$) be a list of all the subsets of $X_n$ with cardinality r. Suppose that there exist distinct convex equivalences $\pi _1,\pi _2,\dots , \pi _m$ of weight r on $X_n$ with the property that $A_i$ is not a transversal of $\pi _i$, for $1\le i\le m$. Then there exist nilpotent $\gamma _i$ in the ${\mathscr {H}}$-class $H_{(\pi _i, A_i)}$ ($1\le i\le m$) such that the set $\{ \gamma _1,\gamma _2,\dots , \gamma _m \}$ is a minimal generating set of $\mathcal {P}^{\mathcal {OP}_n}_{r}$.

Proof

From Lemma 3.1, we know that the ${\mathscr {H}}$-classes $H_{(\pi _1, A_1)}, \dots , H_{(\pi _m,A_m)}$ are non-group ${\mathscr {H}}$-classes, whose elements are nilpotents of $\mathcal {P}^{\mathcal {OP}_n}_{r}$. Put

$$\begin{aligned} \gamma _i\in H_{(\pi _i,A_i)}, \ {\text{ for }} \ 1\le i\le m. \end{aligned}$$

Then the set $\{ \gamma _1,\gamma _2,\dots , \gamma _m \}$ is the subset having exactly one element from each ${\mathscr {R}}$-class and from each ${\mathscr {L}}$-class of $\mathcal {OP}_n$ of rank r. It follows immediately from Theorem 2.4 that the set $\{ \gamma _1,\gamma _2,\dots , \gamma _m \}$ is a minimal generating set of $\mathcal {P}^{\mathcal {OP}_n}_{r}$. $\square$

It remains to prove that the listing of images and convex equivalences postulated in the statement of Lemma 3.3 can actually be carried out. Let $n\ge 3$ and $2\le r\le n-1$, and consider the statement:

$\mathbf{P}(n,r)$: There is a way of listing all the subsets of $X_n$ of cardinality r as $A_1,A_2,\dots , A_m$ (with $m={n\atopwithdelims ()r}$) so that there exist distinct convex equivalences $\pi _1, \dots , \pi _m$ of weight r on $X_n$ with the properties that $A_i$ is not a transversal of $\pi _i$ ($i=1,\dots , m$).

We shall prove this by a double induction on n and r, the key step being a kind of Pascal’s Triangle implication

$$\begin{aligned} \mathbf{P}(n-1,r-1) \ and \ \mathbf{P}(n-1,r)\Rightarrow \mathbf{P}(n,r). \end{aligned}$$

First, however, we anchor the induction with two lemmas:

Lemma 3.4

$\mathbf{P}(n,n-1)$ holds for every $n\ge 3$.

Proof

Consider the list $A_1, A_2, \dots , A_{n}$ of $X_n$ of cardinality $n-1$, where $A_i=X_n\backslash \{i\}$. For $1\le i\le n-2$, let $\pi _i$ be the convex equivalence with a unique non-singleton class $\{i+1,i+2\}$ and all other classes being singletons. Let $\pi _{n-1}$ have classes $\{2\}, \{3\}, \dots , \{n-1\}, \{n,1\}$; and let $\pi _{n}$ have classes $\{1,2\}, \{3\}, \dots , \{n-1\}, \{n\}$. It is easy to verify that $A_1, A_2, \dots , A_{n}$ and $\pi _1, \pi _2, \dots , \pi _{n}$ have the required property. $\square$

Lemma 3.5

$\mathbf{P}(n,2)$ holds for every $n\ge 3$.

Proof

We prove this by the induction n. Notice first that P(3, 2) holds by Lemma 3.4. For $n=4$, we arrange the subsets and convex equivalences as follows:

$$\begin{aligned}&\{1,2\}, \quad \{1,3\}, \quad \{1,4\},\quad \{2,3\},\quad \{2,4\},\quad \{3,4\}\\&12/34, \quad 123/4, \quad 3/412, \quad 23/41, \quad 1/234, \quad 2/341. \end{aligned}$$

Then, it is easy to verify that the subsets and convex equivalences as arranged above satisfy P(4, 2).

Suppose inductively that $P(n-1,2)$ holds $(n\ge 5)$. Thus, we have a list $A_1, A_2,\dots , A_t$ (with $t={n-1\atopwithdelims ()2}$) of all the subsets of $X_{n-1}$ of cardinality 2, and a list $\pi _1,\pi _2,\dots , \pi _t$ of distinct convex equivalences of weight 2 on $X_{n-1}$ such that $A_i$ is not a transversal of $\pi _i$ ($i=1,2,\dots , t$). Notice that $t+n-1={n\atopwithdelims ()2}$. All subsets of $X_{n}$ of cardinality 2 are

$$\begin{aligned} B_{1},B_{2}, \dots , B_{n-1}; A_1, A_2, \dots , A_t, \end{aligned}$$

where $B_{j}=\{j,n\}$, $1\le j\le n-1$. Let $\pi _i'$ ($i=1,2, \dots , t$) be the convex equivalence on $X_n$ obtained from $\pi _i$ by adjoining n to the $\pi _i$-class containing $n-1$. Then $\pi _1', \pi _i', \dots , \pi _t'$ are all distinct, and $A_i$ is not a transversal of $\pi _i'$ ($i=1,2,\dots , t$).

Let $\sigma _{1}, \dots , \sigma _{n-1}$ be the list of convex equivalences of weight 2 on $X_n$, where

$$\begin{aligned}&\sigma _{j} \ \text{ has } \text{ classes } \ \{n,1,\dots ,j\}, \{j+1, \dots , n-1\}, 1\le j\le n-2, \\&\quad \sigma _{n-1} \ \text{ has } \text{ classes } \ \{n\}, \{1, 2, \dots ,n-1\}. \end{aligned}$$

Then $\sigma _{1}, \dots , \sigma _{n-1}$ are all distinct, $B_j$ is not a transversal of $\sigma _j$, for $1\le j\le n-2$, and each $\sigma _{j}$ is distinct from every $\pi '_i$, since $(n-1,n)\in \pi '_i$ and $(n-1,n)\notin \sigma _{j}$. Notice that $t+n-1={n\atopwithdelims ()2}$. Hence $\{ \pi _1',\dots ,\pi _t', \sigma _1,\dots , \sigma _{n-1},\}$ is a complete list of all the convex equivalences of weight 2 on $X_n$. Notice that $A_1$ is a subset of $X_{n-1}$ and $(n,n-1)\in \pi _1'$. Then $A_1$ is not a transversal of $\sigma _{n-1}$ and $B_{n-1}=\{n-1,n\}$ is not a transversal of $\pi _1'$. Arrange the subsets and the convex equivalence as follows:

$$\begin{aligned}&A_{2}, \quad \dots , \quad A_{t}; B_1, \quad B_2, \quad \dots , \quad B_{n-2}, \quad A_{1}, \quad B_{n-1}, \\&\quad \pi _{2}',\quad \dots , \quad \pi _{t}'; \quad \sigma _1,\quad \sigma _2, \quad \dots ,\quad \sigma _{n-2},\quad \sigma _{n-1},\quad \pi _{1}'. \end{aligned}$$

Then, it is easy to verify that the subsets and convex equivalences as arranged above satisfy P(n, 2). $\square$

Lemma 3.6

Let $n\ge 5$ and $3\le r\le n-2$. Then $\mathbf{P}(n-1,r-1)$ and $\mathbf{P}(n-1,r)$ together imply $\mathbf{P}(n,r)$.

Proof

From the assumption $P(n-1,r)$, we have a list $A_1,\dots , A_m$ (where $m={n-1\atopwithdelims ()r}$) of the subsets of $X_{n-1}$ with cardinality r and a list $\sigma _1,\dots ,\sigma _m$ of the convex equivalences of weight r on $X_{n-1}$ such that $A_i$ is not a transversal of $\sigma _i$, for $1\le i\le m$.

From the assumption $P(n-1,r-1)$, we have a list $B_1,\dots , B_t$ (where $t={n-1\atopwithdelims ()r-1}$) of the subsets of $X_{n-1}$ with cardinality $r-1$ and a list $\pi _1,\dots ,\pi _t$ of the convex equivalences of weight $r-1$ on $X_{n-1}$ such that $B_i$ is not a transversal of $\pi _i$, for $1\le i\le t$.

Let $\sigma _i'$ be the convex equivalence obtained from $\sigma _i$ by adjoining n to the $\sigma _i$-class containing $n-1$, and define $\pi _i'=\pi _i\cup \{(n,n)\}$. Then $\sigma _1',\dots , \sigma _m', \pi _1',\dots ,\pi _t'$ are all distinct. Notice that $m+t={n\atopwithdelims ()r}$. Hence $\{\sigma _1',\dots , \sigma _m', \pi _1',\dots ,\pi _t'\}$ is a complete list of all the convex equivalences of weight r on $X_n$. Next we define

$$\begin{aligned} B_i'=B_i\cup \{n\}, \ \text {for} \ 1\le i\le t. \end{aligned}$$

Then $A_1,\dots , A_m, B_1',B_2',\dots ,B_t'$ are all distinct. Moreover, $A_i$ is not a transversal of $\sigma _i'$, for $1\le i\le m$ and $B_i'$ is not a transversal of $\pi _i'$, for $1\le i\le t$. Arrange the subsets and convex equivalences as follows:

$$\begin{aligned}&A_1,\dots , A_m, \ \ B_1', \ \ B_2',\dots ,B_t' \\&\quad \sigma _1', \ \ \dots , \sigma _m', \ \ \pi _1', \ \ \pi _2', \dots ,\pi _t' \end{aligned}$$

They satisfy all the properties necessary and the inductive step is complete. We have shown that P(n, r) holds for all $n\ge 5$ and $3\le r\le n-2$. $\square$

The pattern of deductions is