The rank of the inverse semigroup of partial automorphisms on a finite fence

TL;DR

This paper determines the minimal number of generators needed for the semigroup of all order-preserving partial injections on a finite fence, a specific partial order structure, focusing on odd-sized sets.

Contribution

It provides the exact rank and a minimal generating set for the semigroup of order-preserving partial injections on odd-sized fences, extending previous work to this specific case.

Findings

Calculated the rank of the semigroup for odd n

Provided a minimal generating set for the semigroup

Extended known results to a new class of partial orders

Abstract

A fence is a particular partial order on a (finite) set, close to the linear order. In this paper, we calculate the rank of the semigroup $\mathcal{FI}_{n}$ of all order-preserving partial injections on an $n$-element fence. In particular, we provide a minimal generating set for $\mathcal{FI}_{n}$. In the present paper, $n$ is odd since this problem for even $n$ was already solved by I. Dimitrova and J. Koppitz.

Equations

$$a⪯b\iff a\alpha ⪯b\alpha ,\text{ for all }a,b\in dom\alpha \text{.}$$

$$1$$

$$\text{and }1$$

$$\text{rank}S:=\min \{\left|A\right|:A\subseteq S\text{, }A\text{ generates }S\}.$$

$$J_n:=\{\alpha \in {\mathcal{F}\mathcal{I}}_n:\text{rank}\alpha \ge n-2\}\text{.}$$

$$\gamma_{n}:=\left(\begin{array}[]{ccccc}1&2&\cdots&n-1&n\\ n&n-1&\cdots&2&1\end{array}\right)\text{.}$$

$$r\beta :=r\delta \text{ for all }r\in dom\delta \text{ and }w\beta :=x\text{.}$$

$$a:=(x-1)\delta \text{.}$$

$$w{\beta }_1:=w\delta \text{ for all }w\in dom\delta \text{ and }x{\beta }_1:=a+1\text{.}$$

$$w{\beta }_4:=w\delta \text{ for }w\in dom\delta \text{ and }x{\beta }_4:=a+1\text{.}$$

$$g:=g_{k+1}\cdots g_s\text{.}$$

Full text

††2010 Mathematics Subject Classification: 20M18, 20M20††Key words: partial injections, finite transformation semigroup, fence, rank, generators

1 Introduction

Let $n\in\mathbb{N}$ and denote by $\mathcal{PT}_{n}$ the semigroup (under composition) of all partial transformations on the set $\overline{n}:=\{1,\ldots,n\}$ of the first $n$ natural numbers. The set $\mathcal{I}_{n}$ of all partial injections on $\overline{n}$ forms an inverse subsemigroup of $\mathcal{PT}_{n}$. For more information about the symmetric inverse semigroup $\mathcal{I}_{n}$, we refer the reader to O. Ganyushkin and V. Mazorchuk’s book [9].

Let $\preceq$ be any partial order on $\overline{n}$. The pair $(\overline{n},\preceq)$ can be regarded as a digraph. Let $\alpha\in\mathcal{PT}_{n}$. Then $\alpha$ is called order-preserving on $\overline{n}$ with respect to $\preceq$ if $a\preceq b\Rightarrow a\alpha\preceq b\alpha$, for all $a,b\in dom\alpha$. If $\alpha\in\mathcal{I}_{n}$ is order-preserving then it is a partial injective endomorphisms on the digraph $(\overline{n},\preceq)$. Clearly, the set $\mathcal{I}End(\overline{n},\preceq)$ of all partial injective endomorphisms on $(\overline{n},\preceq)$ forms a submonoid of $\mathcal{I}_{n}$, which has not to be inverse, in general. A regular element $\alpha$ in $\mathcal{I}End(\overline{n},\preceq)$ is characterized by the following property:

[TABLE]

Such regular elements in $\mathcal{I}End(\overline{n},\preceq)$ are called partial automorphisms on $(\overline{n},\preceq)$. The set $\mathcal{P}Aut(\overline{n},\preceq)$ of all strong partial automorphisms on $(\overline{n},\preceq)$ forms an inverse subsemigroup of $\mathcal{I}End(\overline{n},\preceq)$.

A very important particular and natural case occurs when a linear order $\leq$, e.g. that one induced by the usual order on the natural numbers, is considered. The monoid $\mathcal{PIO}_{n}$ of all partial order-preserving injections on $(\overline{n},\leq)$ has been extensively studied. Basic information about the monoid of $\mathcal{PIO}_{n}$, one can find in [7]. In [15], the author considers generating sets of the semigroup of all partial injective decreasing maps of $(\overline{n},\leq)$, a submonoid of $\mathcal{PIO}_{n}$. The maximal subsemigroups of the ideals of some semigroups of partial injections on $(\overline{n},\leq)$ were determined by I. Dimitrova and J. Koppitz [3]. In [1], the authors consider distance-preserving injections on $(\overline{n},\leq)$. They study the algebraic structure of such semigroups, e.g. the Green’s relations.

A non-linear order close to a linear order in some sense is the so-called zig-zag order. The pair $(\overline{n},\preceq)$ is called zig-zag poset or fence if

[TABLE]

The definition of the partial order $\preceq$ is self-explanatory. Observe that every element in a fence is either minimal or maximal.

If the domain of an $\alpha\in$ $\mathcal{PT}_{n}$ is $\overline{n}$, i.e. $dom\alpha=\overline{n}$, then $\alpha$ is called (full) transformation on $\overline{n}$. The set $\mathcal{T}_{n}$ of all full transformations on $\overline{n}$ forms a submonoid of $\mathcal{PT}_{n}$. The monoid $\mathcal{TF}_{n}$ of all order-preserving transformations within $\mathcal{T}_{n}$ (with respect to $\preceq$), i.e. of all endomorphisms on $(\overline{n},\preceq)$, was first investigated by J.D. Currie and T.I. Visentin in [2] and by A. Rutkowski [13]. In [2], by using generating functions, the authors calculate the cardinality of $\mathcal{TF}_{n}$ for the case that $n$ is even. On the other hand, an exact formula for the number of endomorphisms on $(\overline{n},\preceq)$ for even as well as odd $n$ was given in [13]. Recently, in [8], the authors determine the rank of $\mathcal{TF}_{n}$. Recall that the rank of a semigroup $S$, denoted by $\text{rank}S$, is the minimal size of a generating set of $S,$

[TABLE]

In particular, a concrete generating set of $\mathcal{TF}_{n}$ of minimal size is given in [8]. Moreover, the authors characterize the transformations on $\overline{n}$ preserving the fence. It is worth mentioning that several other properties of monoids of order-preserving transformations of a fence were also studied. In [12, 14], the authors discussed the regular elements of these monoids. Coregular elements (i.e. elements $\alpha$ with the property $\alpha=\alpha^{3}$) of these monoids were determined in [11]. Some relative ranks of the monoid of all partial transformations preserving an infinite zig-zag order were determined in [5].

In this paper, we will denote the semigroup of all partial automorphisms on $(\overline{n},\preceq)$ by $\mathcal{FI}_{n}$, i.e. $\mathcal{P}Aut(\overline{n},\preceq)=\mathcal{FI}_{n}$. This inverse semigroup was first studied by I. Dimitrova and J. Koppitz in [4]. The authors described the Green’s relations on $\mathcal{FI}_{n}$. In fact, they described only the $\mathcal{J}$- relation since the relations $\mathcal{L},\mathcal{R}$, and $\mathcal{H}$ are clear because $\mathcal{FI}_{n}$ is an inverse semigroup. Moreover, they show that $\mathcal{FI}_{n}$ is generated by the set

[TABLE]

Recall that the rank of a (partial) transformation $\alpha$ (in symbols: $\text{rank}\alpha$) is the size of the range of $\alpha$ (in symbols: $\text{im}\alpha$), i.e. $\text{rank}\alpha=\left|\text{im}\alpha\right|$. For the case that $n$ is even, it is proved that $\text{rank}\mathcal{FI}_{n}=n+1$ and a concrete generating set of $\mathcal{FI}_{n}$ with $n+1$ elements is given in [4]. On the other hand, the rank of $\mathcal{FI}_{n}$ is still an open problem, whenever $n$ is odd. We will solve it in the present paper. We will determine the rank of $\mathcal{FI}_{n}$ and give a concrete generating set of $\mathcal{FI}_{n}$ with minimal size in the case that $n$ is odd.

Without loss of generality, let $1\prec 2\succ 3\prec\cdots\succ n$. Such fences are also called up-fences. The fence $1\succ 2\prec 3\succ\cdots\prec n$ would be called down-fence. We avoid both notations up-fence and down-fence. In fact, in order to check a fence is an up-fence or down-fence, we need that $1$ and $2$ are comparable with respect to $\preceq$. Recall that $x,y\in\overline{n}$ are comparable with respect to $\preceq$ if $x\prec y$ or $x=y$ or $x\succ y$. Otherwise, $x$ and $y$ are called incomparable. But the restriction that $1$ and $2$ belong to the fence and are comparable is an unnecessary restriction for the concept fence since instead of $\overline{n}$ one could choose another $n$-element set or one could define $\preceq$ on $\overline{n}$ such that $1$ and $2$ are incomparable.

But if the fence $(\overline{n},\preceq)$ is defined as above (it is the most natural way to do it in that way) then we observe that any $x,y\in\overline{n}$ are comparable if and only if $x\in\{y-1,y,y+1\}$. For general background on Semigroup Theory and standard notation, we refer the reader to Howie‘s book [10].

2 Main result

Let us fix now an odd natural number $n$. Clearly, for $U\subseteq\overline{n}$, the partial identity mapping $id_{U}:=id_{\overline{n}}|_{U}$, i.e. the identity mapping $id_{\overline{n}}$ on $\overline{n}$ restricted to $U$, is a partial automorphism on $(\overline{n},\preceq)$. In particular, $id_{\overline{n}}\in\mathcal{FI}_{n}$. There are exactly two automorphisms in  $\mathcal{FI}_{n}$, besides $id_{\overline{n}}$ the reflection

[TABLE]

Note that $\mathcal{FI}_{1}$ consists of the identity mapping on $\{1\}$ and the empty transformation $\emptyset$. Since these both partial transformations do not generate each other, the rank of $\mathcal{FI}_{1}$ is $2$. We suppose now that $n\geq 3$. I. Dimitrova and J. Koppitz proved $\mathcal{FI}_{m}=\left\langle J_{m}\right\rangle$ for all natural numbers $m$ in [4], which comprises several pages and a few of lemmas in [4]. For the case that $n$ is odd, one can shorten the proof. Therefore, and for the sake of completeness, we will give a new proof for the particular case that $n$ is odd. For this, we define a series of partial transformations of $J_{n}$. Let

\alpha_{i}=\left(\begin{array}[]{ccccccc}1&\cdots&i-1&i&i+1&\cdots&n\\ 1&\cdots&i-1&&n&\cdots&i+1\end{array}\right) for even $i\in\{2,\ldots,n-1\}$,

$a_{i}=id_{\overline{n}\setminus\{i\}}$ for odd $i\in\overline{n}$,

\beta_{2}^{odd}=\left(\begin{array}[]{cccccc}1&2&3&4&\cdots&n\\ 2&&&4&\cdots&n\end{array}\right)\text{, }\beta_{n-1}^{odd}=\left(\begin{array}[]{cccccc}1&2&3&\cdots&n-1&n\\ n-1&&1&\cdots&n-3&\end{array}\right),

\beta_{2}^{even}=\left(\begin{array}[]{cccccc}1&2&3&4&\cdots&n\\ &1&&4&\cdots&n\end{array}\right)\text{, and }\beta_{n-1}^{even}=\left(\begin{array}[]{cccccc}1&\cdots&n-3&n-2&n-1&n\\ 3&\cdots&n-1&&1&\end{array}\right).

In the case $n\geq 5$, we define

\alpha_{i,j}=\left(\begin{array}[]{ccccccccccc}1&\cdots&i-1&i&i+1&\cdots&j-1&j&j+1&\cdots&n\\ 1&\cdots&i-1&&j-1&\cdots&i+1&&j+1&\cdots&n\end{array}\right) for $2\leq i<j\leq n-1$, where $i$ and $j$ have the same parity,

\alpha_{1,j}=\left(\begin{array}[]{cccccccc}1&2&\cdots&j-1&j&j+1&\cdots&n\\ &j-1&\cdots&2&&j+1&\cdots&n\end{array}\right) and

\alpha_{j,n}=\left(\begin{array}[]{cccccccc}1&\cdots&j-1&j&j+1&\cdots&n-1&n\\ 1&\cdots&j-1&&n-1&\cdots&j+1&\end{array}\right) for odd $j\in\{3,\ldots,n-2\}$, and

\alpha_{1,n}=\left(\begin{array}[]{ccccc}1&2&\cdots&n-1&n\\ &n-1&\cdots&2&\end{array}\right).

In the case $n\geq 7$, we define

\beta_{i}^{odd}=\left(\begin{array}[]{ccccccccc}1&2&3&\cdots&i&i+1&i+2&\cdots&n\\ i&&1&\cdots&i-2&&i+2&\cdots&n\end{array}\right) and

\beta_{i}^{even}=\left(\begin{array}[]{ccccccccc}1&\cdots&i-2&i-1&i&i+1&i+2&\cdots&n\\ 3&\cdots&i&&1&&i+2&\cdots&n\end{array}\right) for even $i\in\{4,\ldots,n-3\}$.

Note that $\beta_{i}^{even}\beta_{i}^{odd}=id_{\overline{n}\setminus\{i-1,i+1\}}$ for each even $i\in\{2,\ldots,n-1\}$, $\alpha_{i}\alpha_{i}=id_{\overline{n}\setminus\{i\}}$ for each $i\in\overline{n}$, and $\alpha_{i,j}\alpha_{i,j}=id_{\overline{n}\setminus\{i,j\}}$ for all $i<j\in\overline{n}$ having the same parity. Further, let $Par_{n}$ be the set of all $\delta\in\mathcal{FI}_{n}$ such that $x$ and $x\delta$ have different parity for some $x\in dom\delta$. First, we observe that each $\delta\in Par_{n}$ is generated by elements of $\mathcal{FI}_{n}\setminus Par_{n}$ and $\{id_{\overline{n}}\}\cup\{\beta_{i}^{odd},\beta_{i}^{even}:i\in\{2,4,\ldots,n-1\}\}\subseteq J_{n}$.

Lemma 2.1**.**

Let $\delta\in Par_{n}$. Then there are $\beta\in\mathcal{FI}_{n}\setminus Par_{n}$ and $l_{1},\ldots,l_{p},r_{1}\ldots,r_{p}\in\{id_{\overline{n}}\}\cup\{\beta_{i}^{odd},\beta_{i}^{even}:i\in\{2,4,\ldots,n-1\}\}$ with $\delta=l_{1}\cdots l_{p}\beta r_{1}\cdots r_{p}$.

Proof.

Let $x\in dom\delta$ such that $x$ and $x\delta$ have different parity. If $x$ is odd then $x\delta$ is even, where $x\delta-1,x\delta+1\notin\text{im}\delta$. Let $\beta:=id_{\overline{n}}\delta\beta_{x\delta}^{even}$. Then it is easy to verify that $\left|\left\{w\in dom\delta:w\text{ and }w\delta\text{ have different parity}\right\}\right|>\left|\left\{w\in dom\beta:w\text{ and }w\beta\text{ have different parity}\right\}\right|$ and $\beta\beta_{x\delta}^{odd}=\delta\beta_{x\delta}^{even}\beta_{x\delta}^{odd}=\delta$ since $\text{im}\delta\subseteq dom\beta_{x\delta}^{even}$. If $x$ is even then $x\delta$ is odd and $x-1,x+1\notin dom\delta$. In this case, let $\beta:=\beta_{x}^{odd}\delta id_{\overline{n}}$ and we observe that $\left|\left\{w\in dom\delta:w\text{ and }w\delta\text{ have different parity}\right\}\right|>\left|\left\{w\in dom\beta:w\text{ and }w\beta\text{ have different parity}\right\}\right|$, where $\beta_{x}^{even}\beta=\beta_{x}^{even}\beta_{x}^{odd}\delta=\delta$ since $\text{im}\beta_{x}^{odd}\supseteq dom\delta$.  Now, we can consider $\beta$ instead of $\delta$. Since the domain of $\delta$ is finite, we obtain successively after $p$ steps $l_{1},\ldots,l_{p},r_{1}\ldots,r_{p}\in\{id_{\overline{n}}\}\cup\{\beta_{i}^{odd},\beta_{i}^{even}:i\in\{2,4,\ldots,n-1\}\}$ and a $\widetilde{\beta}\in\mathcal{FI}_{n}\setminus Par_{n}$ such that $\delta=l_{1}\cdots l_{p}\widetilde{\beta}r_{1}\cdots r_{p}$. ∎

The following fact will be used frequently without to refer it. If $U$ is a convex subset of the domain of a $\delta\in\mathcal{FI}_{n}$ then $U\delta=\{x\delta:x\in U\}$ is also a convex set. Next, we show that any $\delta\in\mathcal{FI}_{n}\setminus J_{n}$ with a convex domain can be generated by elements of $J_{n}$ and a transformation $\beta\in\mathcal{FI}_{n}$ with $\text{rank}\beta>\text{rank}\delta$.

Lemma 2.2**.**

Let $\delta\in\mathcal{FI}_{n}\setminus J_{n}$. If $dom\delta$ is a convex set then there are $\beta\in\mathcal{FI}_{n}$ with $\text{rank}\beta=\text{rank}\delta+1$ and $\alpha\in J_{n}$ such that $\delta=\alpha\beta$.

Proof.

Suppose that $dom\delta$ is a convex set.

Since both intervals $dom\delta$ and $\text{im}\delta$ have a lenght less than or equal $n-3$, there are $x,w\in\bar{n}$ such that $w-1,w,w+1\in\{0,1,\dots,n,n+1\}\setminus dom\delta$ and $x-1,x,x+1\in\{0,1,\dots,n,n+1\}\setminus\text{im}\delta$. We define a transformation $\beta$ by

[TABLE]

Clearly, $\beta\in\mathcal{FI}_{n}$ with $\text{rank}\beta=\text{rank}\delta+1$ and $\delta=id_{\overline{n}\setminus\{w\}}\beta$. ∎

We note that the empty set is convex. So, the empty transformation $\emptyset$ is a product of two transformations with rank $\geq 1$. Now, we can prove the following proposition, which is a particular case of Theorem 3.15 in [4] for $n$ is odd.

Proposition 2.3**.**

$\mathcal{FI}_{n}$* is generated by $J_{n}$.*

Proof.

Let $3\leq r\leq n$ and suppose that $\beta\in\left\langle J_{n}\right\rangle$ for all $\beta\in\mathcal{FI}_{n}$ with $\text{rank}\beta>n-r$. Let now $\delta\in\mathcal{FI}_{n}$ with $\text{rank}\delta=n-r$. By Lemma 2.1, we can restrict us to the case $\delta\in\mathcal{FI}_{n}\setminus Par_{n}$. Now, we choose $x\in\overline{n}\setminus dom\delta$ in the following way: There is $z<x$ such that $w\notin dom\delta$ for all $w<z$ and $\{z,\ldots,x-1\}\subseteq dom\delta$. Clearly, if $dom\delta\neq\{r+1,\ldots,n\}$ then such $x$ exists. Note, by Lemma 2.2, we can skip the case that $dom\delta=\{r+1,\ldots,n\}$.

Now, we will show that one can restrict oneself to the case that $w\delta>\{z,\ldots,x-1\}\delta$ for all $w\in dom\delta\setminus\{z,\ldots,x-1\}$ and $(x-2)\delta<(x-1)\delta$, whenever $x-2\in dom\delta$. For this let $\{r_{1}<\cdots<r_{s}\}$ be a convex interval in $dom\delta$ with $r_{1}-1,r_{s}+1\in\{0,1,\ldots,n,n+1\}\setminus dom\delta$ such that $w\delta>\{r_{1},\ldots,r_{s}\}\delta$ for all $w\in dom\delta\setminus\{r_{1},\ldots,r_{s}\}$. Suppose that $x-1\notin\{r_{1},\ldots,r_{s}\}$, i.e. $r_{1}>2$. If $r_{s}$ is odd then we define $\sigma:=\gamma_{n}\alpha_{n-r_{s}}\gamma_{n}\delta$, where $\delta=\gamma_{n}\alpha_{n-r_{s}}\gamma_{n}\gamma_{n}\alpha_{n-r_{s}}\gamma_{n}\delta$ and $\alpha_{0}=\gamma_{n}$. If $r_{s}$ is even and $r_{1}$ is odd then we define $\sigma:=\alpha_{r_{1}-1}\delta\gamma_{n}$, where $\delta=\alpha_{r_{1}-1}\alpha_{r_{1}-1}\delta\gamma_{n}\gamma_{n}$. If both $r_{1}$ and $r_{s}$ are even and $p:=\max\{z\delta,\ldots,(x-1)\delta\}$ is odd then we define $\sigma:=\delta\gamma_{n}\alpha_{n-p}\gamma_{n}$, where $\delta=\delta\gamma_{n}\alpha_{n-p}\gamma_{n}\gamma_{n}\alpha_{n-p}\gamma_{n}$. If all $r_{1}$, $r_{s}$, and $p$ are even then $1\notin im\delta$ and we define $\sigma:=\delta\alpha_{1,p+1}$, where $\delta=\delta\alpha_{1,p+1}\alpha_{1,p+1}$. Hence, $\delta$ is the product of $\sigma$ and transformations in $J_{n}$. Straightforward calculations show that there are $\widetilde{z}<\widetilde{x}\in\overline{n}$ such that $\{\widetilde{z},\ldots,\widetilde{x}-1\}\subseteq dom\sigma$ is a convex set and $\widetilde{x},w\notin dom\sigma$ for all $w<\widetilde{z}$, where $w\sigma>\{\widetilde{z},\ldots,\widetilde{x}-1\}\sigma$ for all $w\in dom\sigma\setminus\{\widetilde{z},\ldots,\widetilde{x}-1\}$. Suppose that $\widetilde{x}-2\in dom\sigma$ with $(\tilde{x}-2)\sigma>(\tilde{x}-1)\sigma$. If $\widetilde{z}$ is odd then we define $\upsilon:=\sigma\gamma_{n}\alpha_{n-\widetilde{z}\delta}\gamma_{n}$, where $\sigma=\sigma\gamma_{n}\alpha_{n-\widetilde{z}\delta}\gamma_{n}\gamma_{n}\alpha_{n-\widetilde{z}\delta}\gamma_{n}$. If $\widetilde{z}$ is even and $\widetilde{x}-1$ is odd then we define $\upsilon:=\gamma_{n}\alpha_{n-\widetilde{x}+1}\gamma_{n}\sigma$, where $\sigma=\gamma_{n}\alpha_{n-\widetilde{x}+1}\gamma_{n}\gamma_{n}\alpha_{n-\widetilde{x}+1}\gamma_{n}\sigma$. If both $\widetilde{z}$ and $\widetilde{x}-1$ are even then we define $\upsilon:=\alpha_{\widetilde{z}-1,\widetilde{x}}\sigma$, where $\sigma=$ $\alpha_{\widetilde{z}-1,\widetilde{x}}\alpha_{\widetilde{z}-1,\widetilde{x}}\sigma$. Hence, $\sigma$ is the product of $\upsilon$ and transformations in $J_{n}$. Simple calculations show that there are $\widehat{z}<\widehat{x}\in\overline{n}$ such that $\{\widehat{z},\ldots,\widehat{x}-1\}\subseteq dom\upsilon$ is a convex set and $\widehat{x},w\notin dom\upsilon$ for all $w<\widehat{z}$, where $w\upsilon>\{\widehat{z},\ldots,\widehat{x}-1\}\upsilon$ for all $w\in dom\upsilon\setminus\{\widehat{z},\ldots,\widehat{x}-1\}$ and $(\widehat{x}-2)\upsilon<(\widehat{x}-1)\upsilon$, whenever $\widetilde{x}-2\in dom\nu$. This proves the desired restriction.

Without loss of generality, we can assume that $z\delta\in\{1,2\}$. Otherwise, we consider the transformation \widehat{\delta}:=\delta\left(\begin{array}[]{cccc}3&4&\cdots&n\\ 1&2&\cdots&n-2\end{array}\right)^{\left\lfloor\frac{z\delta-1}{2}\right\rfloor}, where \delta=\widehat{\delta}\left(\begin{array}[]{cccc}1&2&\cdots&n-2\\ 3&4&\cdots&n\end{array}\right)^{\left\lfloor\frac{z\delta-1}{2}\right\rfloor} and $z\widehat{\delta}\in\{1,2\}$. We put now

[TABLE]

Since $x\notin dom\delta$ and $w\delta>(x-1)\delta$ for all $w\in dom\delta\setminus\{z,\ldots,x-1\}$, either $a+1\notin\text{im}\delta$ or $dom\delta$ is a convex set. In the latter case, by Lemma 2.2, there are $\beta\in\mathcal{FI}_{n}$ with $\text{rank}\beta=\text{rank}\delta+1$ and $\alpha\in J_{n}$ such that $\delta=\alpha\beta$. Since $\beta\in\left\langle J_{n}\right\rangle$ by the inductive assumption, we obtain $\delta=\alpha\beta\in\left\langle J_{n}\right\rangle$. So, we have to consider the case $a+1\notin\text{im}\delta$, where either $x+1\in dom\delta$ or $x+1\notin dom\delta.$

Suppose that $x+1\in dom\delta$. If $(x+1)\delta=a+2$ then we define a partial transformation $\beta_{1}$ by

[TABLE]

Clearly, $\beta_{1}\in\mathcal{FI}_{n}$ with $\text{rank}\beta_{1}>n-r.$ We have $\delta=\alpha_{a}\beta_{1}\in\left\langle J_{n}\right\rangle$. Now, we show that there is $\widehat{\delta}\in\mathcal{FI}_{n}$ with $w\widehat{\delta}=w\delta$ for $w\in dom\delta\cap\{1,\ldots,x-1\}$ and $(x+1)\widehat{\delta}=a+2$ such that $\delta\in\langle J_{n},\widehat{\delta}\rangle$. If it is the case then we have $\delta\in\left\langle J_{n}\right\rangle$ since $\widehat{\delta}\in\left\langle J_{n}\right\rangle$ by the previous considerations.

Suppose that $a+2\in\text{im}\delta$. Then there is $y\in dom\delta$ with $y\delta=a+2$. Since $w\delta\leq a$ for all $w\in dom\delta$ with $w<x$, we conclude that $y\geq x+1$. If $y+1\notin dom\delta$ then we put $\beta_{2}:=\alpha_{x,y+1}$, whenever $y<n$ and $\beta_{2}:=\alpha_{x}$, whenever $y=n$. Clearly, $\beta_{2}\in\mathcal{FI}_{n}$. Then $\delta=\beta_{2}\beta_{2}\delta$, where $w\beta_{2}\delta=w\delta$ for $w\in dom\delta\cap\{1,\ldots,x-1\}$ and $(x+1)\beta_{2}\delta=a+2$. Now, we consider the case that $y+1\in dom\delta$. This provides $(y+1)\delta=a+3$ and $y-1\notin dom\delta$ since $a+1\notin\text{im}\delta$. If $y-1$ is even then $\delta=\alpha_{y-1}\alpha_{y-1}\delta$, where $w\alpha_{y-1}\delta=w\delta$ for $w\in dom\delta\cap\{1,\ldots,x-1\}$, $n\alpha_{y-1}\delta=a+2$, and $n+1\notin dom(\alpha_{y-1}\delta)$ (which provides a previous case). If $y-1$ is odd and there is an odd  $u>y+1$ with $u\notin dom\delta$ then we have $\delta=\alpha_{y-1,u}\alpha_{y-1,u}\delta$, where $w\alpha_{y-1,u}\delta=w\delta$ for $w\in dom\delta\cap\{1,\ldots,x-1\}$, $(u-1)\alpha_{y-1,u}\delta=a+2$, and $u\notin dom(\alpha_{y-1,u}\delta)$ (which provides a previous case). Let now $y-1$ be odd and let each $u\in\overline{n}\setminus dom\delta$ with $u>y+1$ be even. Then we observe that $w-1,w+1\in\{y,\ldots,n\}\delta$, for each even $w\in\{y,\ldots,n\}\delta$. Hence, there is an odd $k\in\{y,\ldots,n\}\delta$ with $k+1\notin\{y,\ldots,n\}\delta$. Because $(x+1)\delta$ ($>y\delta$) is even, this implies the existence of an even $u<y$ with $u\notin dom\delta$.

If $u>x$ then we have $\delta=\alpha_{u}\alpha_{u}\delta$, where $w\alpha_{u}\delta=w\delta$ for $w\in dom\delta\cap\{1,\ldots,u-1\}$ with $u>x$ and $(n-y+1)\alpha_{w}\delta=a+2$, where $(n-y+1+1)\notin dom(\alpha_{u}\delta)$. We have again a previous case. Finally, we have to consider the case $u<x$. In particular, $1,2\notin dom\delta$. We have $\alpha_{2}\gamma_{n}\gamma_{n}\alpha_{2}\delta=\delta$, where $w\gamma_{n}\alpha_{2}\delta=(w+2)\delta$, for all $w\in\{1,\ldots,n-2\}$ with $w+2\in dom\delta$. Then $(y-2)\gamma_{n}\alpha_{2}\delta=a+1$, $y-3$ is odd and $n\notin dom(\gamma_{n}\alpha_{2}\delta)$, i.e. $\gamma_{n}\alpha_{2}\delta$ is a previous case.

Admit now $a+2\notin\text{im}\delta$. If $(x+1)\delta+1\notin\text{im}\delta$ then we have $\delta=\delta\alpha_{a+1,(x+1)\delta+1}\alpha_{a+1,(x+1)\delta+1}$, where $w\delta\alpha_{a+1,(x+1)\delta+1}=w\delta$, for $w\in dom\delta\cap\{1,\ldots,x-1\}$, and $(x+1)\delta\alpha_{a+1,(x+1)\delta+1}=a+2$. Suppose now that $(x+1)\delta+1\in\text{im}\delta$, i.e. $(x+1)\delta-1\notin im\delta$. If $(x+1)\delta$ is even then $a+2$ is even and we put $\beta_{3}:=\alpha_{a+2}.$ If $(x+1)\delta$ is odd then we put $\beta_{3}:=\alpha_{(x+1)\delta-1}$. Clearly, $\beta_{3}\in\mathcal{FI}_{n}$, $\delta=\delta\beta_{3}\beta_{3}$, where $w\delta\beta_{3}=w\delta$ for $w\in dom\delta\cap\{1,\ldots,x-1\}$, and $a+2\notin im\delta\beta_{3}$. Since $(x+1)\delta-1\notin\text{im}\delta$, we can conclude that $(x+1)\delta\beta_{3}+1\notin\text{im}\delta\beta_{3}$ (the previous case for $\delta\beta_{3}$).

We finish the proof with the case that $x+1\notin dom\delta$. If $a+2\notin\text{im}\delta$ we define a partial transformation $\beta_{4}$ by

[TABLE]

It is easy to verify that $\beta_{4}\in\mathcal{FI}_{n}$. Then we have $\delta=id_{\overline{n}\setminus\{x\}}\beta_{4}$. Suppose that there is $y\in dom\delta$ with $y\delta=a+2$. We can assume that $y+1\notin dom\delta$. Otherwise, $y-1\notin dom\delta$ and there is $i\in\{0,1\}$ such that $x+i$ is even and $\delta=\alpha_{x+i}\alpha_{x+i}\delta$, where $w\alpha_{x+i}\delta=w\delta$ for $w\in dom\delta\cap\{1,\ldots,x-1\}$. In particular, we have $(n-y+x+i+1)\alpha_{x+i}\delta=a+2$ and $(n-y+x+i+2)\notin dom(\alpha_{x+i}\delta)$. Here we have a previous case (for $\alpha_{x+i}\delta$). Having $y+1\notin dom\delta$, we obtain $\delta=\alpha_{x,y+1}\alpha_{x,y+1}\delta$, where $w\alpha_{x,y+1}\delta=w\delta$ for $1\leq w\leq x$ and $(x+1)\alpha_{x,y+1}\delta=a+2$, a previous case for the transformation $\alpha_{x,y+1}\delta$.

Altogether, we have shown that there is $\widehat{\delta}\in\mathcal{FI}_{n}$ with $w\widehat{\delta}=w\delta$ for $w\in dom\delta\cap\{1,\ldots,x-1\}$ and $(x+1)\widehat{\delta}=a+2$ such that $\delta\in\langle J_{n},\widehat{\delta}\rangle$. Consequently, we have shown that $\mathcal{FI}_{n}\subseteq\left\langle J_{n}\right\rangle$. ∎

Now, we construct a generating set of $\mathcal{FI}_{n}$ of minimal size. By Proposition 2.3 and since no element in $J_{n}$ can be generated by elements which not belong to $J_{n}$, we have to find a generating set of $J_{n}$ of minimal size. For this, we define $G_{3}:=\{\gamma_{3},\alpha_{1},\alpha_{2},\beta_{2}^{odd},\beta_{2}^{even}\}$ and $G_{n}:=\{\gamma_{n}\}\cup\{\alpha_{i}:i\in\{1,,\ldots,\frac{n+1}{2}\}$ is odd$\}\cup\{\alpha_{i}:i\in\{2,4,\ldots,n-3\}\}\cup\{\beta_{i}^{odd},\beta_{i}^{even}:i\in\{2,\ldots,\frac{n+1}{2}\}$ is even$\}\cup\{\alpha_{i,j}:i,j\in\overline{n}$ are odd with $4\leq j-i<n-1$, $i\leq n-j+1\}$, whenever $n\geq 5$.

Lemma 2.4**.**

Let $\delta\in J_{n}\cap Par_{n}$. Then there is exactly one $x\in dom\delta$ such that $x$ and $x\delta$ have different parity. In particular, it holds $x\delta\in\{1,n\}$ or $x\in\{1,n\}$.

Proof.

Since $\delta\in Par_{n}$, there is an $x\in dom\delta$ such that $x$ and $x\delta$ have different parity. Assume that $x$ is even and $x\delta\notin\{1,n\}$. Then $x-1,x+1\notin dom\delta$ and $x\delta-1,x\delta+1\notin\text{im}\delta$. Hence, since $rank\delta\geq n-2$, there is only one $y\in dom\delta$ such that $y$ and $y\delta$ have different parity, namely $x$. So, we have $\frac{n+1}{2}-2$ odd elements in the domain of $\delta$, which have to be mapped to $\frac{n+1}{2}-1$ odd elements in the image of $\delta$, a contradiction. By the same argumentation, we obtain that $x$ is unique with $x\in\{1,n\}$, whenever $x\delta$ is even. ∎

Lemma 2.5**.**

Let $\delta\in J_{n}$. Then $\delta\in\left\langle G_{n}\right\rangle$.

Proof.

If $\text{rank}\delta=n$ then $\delta=\gamma_{n}\in G_{n}$ or $\delta=id_{\overline{n}}=\gamma_{n}\gamma_{n}$.

Let now $\text{rank}\delta=n-1$. Then there is $i\in\overline{n}\setminus dom\delta$. We observe that $im\delta=dom\delta$ or $im\delta=(dom\delta)\gamma_{n}$.

If $i\leq n-3$ is even then $\delta=\alpha_{i}\in G_{n}$ or $\delta=\alpha_{i}^{2}$ or $\delta=\alpha_{i}\gamma_{n}$ or $\delta=\alpha_{i}^{2}\gamma_{n}$ or

\delta=\left(\begin{array}[]{ccccccc}1&\cdots&i-1&i&i+1&\cdots&n\\ i-1&\cdots&1&&i+1&\cdots&n\end{array}\right)=\gamma_{n}\alpha_{n-i+1}\gamma_{n} or $\delta=\gamma_{n}\alpha_{n-i+1}$ or

\delta=\left(\begin{array}[]{ccccccc}1&\cdots&i-1&i&i+1&\cdots&n\\ i-1&\cdots&1&&n&\cdots&i+1\end{array}\right)=\gamma_{n}\alpha_{n-i+1}\gamma_{n}\alpha_{i} or $\delta=\gamma_{n}\alpha_{n-i+1}\gamma_{n}\alpha_{i}\gamma_{n}$.

If $i=n-1$  then $\delta=\gamma_{n}\alpha_{{}_{2}}$ or $\delta=\gamma_{n}\alpha_{2}\gamma_{n}$ or $\delta=\gamma_{n}\alpha_{2}^{2}$ or $\delta=\gamma_{n}\alpha_{2}^{2}\gamma_{n}$.

If $i$ is odd and $i\leq\frac{n+1}{2}$ then $\delta=\alpha_{i}\in G_{n}$ or $\delta=\alpha_{i}\gamma_{n}$ and if $i>\frac{n+1}{2}$ then $\delta=\gamma_{n}\alpha_{n-i+1}$ or $\delta=\gamma_{n}\alpha_{n-i+1}\gamma_{n}$, where $-i<-\frac{n+1}{2}$, i.e. $n-i+1<\frac{n+1}{2}$.

Now, we consider the case $\text{rank}\delta=n-2$ and $\delta\notin Par_{n}$. Then there are $i,j\in\overline{n}\setminus dom\delta$ with $i<j$. Let both $i$ and $j$ be odd. If $4\leq j-i<n-1$ and $i\leq n-j+1$ then $\delta=$ $\alpha_{i,j}\in G_{n}$ or $\delta=\alpha_{i,j}\gamma_{n}$. Suppose now that $j-i=n-1$, i.e. $i=1$ and $j=n$. Here, we obtain $\delta=\alpha_{1,n}=\gamma_{n}\alpha_{1}\alpha_{n}$ or $\delta=\alpha_{1,n}^{2}=\alpha_{1}\alpha_{n}$. If $2=j-i<n-1$, i.e. $i+2=j$ then \delta=\left(\begin{array}[]{ccccccccc}1&\cdots&i-1&i&i+1&i+2&i+3&\cdots&n\\ 1&\cdots&i-1&&i+1&&i+3&\cdots&n\end{array}\right)=\alpha_{i}\alpha_{i+2} or \delta=\left(\begin{array}[]{ccccccccc}1&\cdots&i-1&i&i+1&i+2&i+3&\cdots&n\\ n&\cdots&n-i+2&&n-i&&n-i-2&\cdots&1\end{array}\right)=\alpha_{i}\alpha_{i+2}\gamma_{n}. Suppose now $i>n-j+1$. Then we put $r:=n-j+1$ and $s:=n-i+1$. We observe that $r=n-j+1<i=n-(n-i+1)+1=n-s+1$ and $\delta=\gamma_{n}\alpha_{r,s\text{.}}\gamma_{n}$. Admit now that $i$ is even or $j$ is even. The following is not hard to verify. If $i$ is even and $j$ is odd then $\delta\in\{\alpha_{i},\alpha_{i}^{2},\gamma_{n}\alpha_{n-i+1}\gamma_{n},\alpha_{i}\gamma_{n},\alpha_{i}^{2}\gamma_{n},\gamma_{n}\alpha_{n-i+1},\gamma_{n}\alpha_{n-i+1}\gamma_{n}\alpha_{i},\gamma_{n}\alpha_{n-i+1}\gamma_{n}\alpha_{i}\gamma_{n}\}\{\alpha_{j},\alpha_{j}\gamma_{n}\}$. If $i$ is odd and $j$ is even then we obtain $\delta\in\left\langle G_{n}\right\rangle$, dually. If both $i$ and $j$ are even then we have $\alpha_{i,j}=\alpha_{i}\alpha_{j-i}\alpha_{i}$ and $\delta\in\Gamma_{1}\Gamma_{2}\{\gamma_{n},id_{\overline{n}}\}$, where $\Gamma_{1}=\{id_{\overline{n}},\alpha_{j}\}\{id_{\overline{n}},\alpha_{i,j}\}\{id_{\overline{n}},\gamma_{n}\alpha_{n-i+1}\gamma_{n}\}$ and $\Gamma_{2}=\{\alpha_{i}^{2}\alpha_{j}^{2},\alpha_{i}\alpha_{j}^{2},\alpha_{i}^{2},\gamma_{n}\alpha_{n-j+1}\gamma_{n}\}$.

Now, we consider the case $\text{rank}\delta=n-2$ and $\delta\in Par_{n}$. By Lemma 2.4, there is exactly one $x\in dom\delta$ such that $x$ and $x\delta$ have different parity and either $x$ is even and $x\delta\in\{1,n\}$ or $x\in\{1,n\}$. Hence, there is an even $a\in\overline{n}$ such that $\delta=\beta_{a}^{odd}$ or \delta=\left(\begin{array}[]{ccccccccc}1&2&3&\cdots&n-a+1&n-a+2&n-a+3&\cdots&n\\ a&&n&\cdots&a+2&&a-2&\cdots&1\end{array}\right)=\alpha_{2}\beta_{a}^{odd} or $\delta=\beta_{a}^{even}$ or \delta=\left(\begin{array}[]{ccccccccc}1&\cdots&a-2&a-1&a&a+1&a+2&\cdots&n\\ n&&n-a+3&&1&&n-a+1&\cdots&3\end{array}\right)=\beta_{a}^{even}\alpha_{2}. Suppose that $a>\frac{n+1}{2}$. Then we have $\beta_{a}^{odd}=\alpha_{2}\beta_{n-a+1}^{odd}\gamma_{n}$ and $\beta_{a}^{even}=\gamma_{n}\beta_{n-a+1}^{even}\alpha_{2}$, where $n-a+1<n-(\frac{n+1}{2})+1=\frac{n+1}{2}$. ∎

Lemma 2.5 shows $\left\langle J_{n}\right\rangle=\left\langle G_{n}\right\rangle$. Thus, Proposition 2.3 provides that $G_{n}$ is a generating set for $\mathcal{FI}_{n}$.

Corollary 2.6**.**

$\mathcal{FI}_{n}=\left\langle G_{n}\right\rangle$.

It remains to show that $G_{n}$ is a generating set of minimal size. It is useful to classify the partial injections in $\mathcal{FI}_{n}$ with rank $n-1$. For $1\leq i\leq\frac{n+1}{2}$, let $R_{i}:=\{\alpha\in\mathcal{FI}_{n}:dom\alpha=\overline{n}\setminus\{i\}$ or $dom\alpha=\overline{n}\setminus\{n-i+1\}\}$. Clearly, $\bigcup\{R_{i}:1\leq i\leq\frac{n+1}{2}\}=\{\alpha\in\mathcal{FI}_{n}:\text{rank}\alpha=n-1\}$. Moreover, any generating set of $\mathcal{FI}_{n}$ contains elements form each $R_{i}$, $1\leq i\leq\frac{n+1}{2}$, as the following lemma will show.

Lemma 2.7**.**

Let $G\subseteq\mathcal{FI}_{n}$ with $\mathcal{FI}_{n}=\left\langle G\right\rangle$. Then $G\cap R_{i}\neq\emptyset$ for all $i\in\{1,2,\ldots,\frac{n+1}{2}\}$.

Proof.

Assume there is an $i\in\{1,2,\ldots,\frac{n+1}{2}\}$ with $G\cap R_{i}=\emptyset$. Then $\alpha_{i}\notin G$ and there are $g_{1},\ldots,g_{s}\in G\setminus\{id_{\bar{n}}\}$ such that $\alpha_{i}=g_{1}\cdots g_{s}$, where $g_{k}g_{k+1}\neq id_{\bar{n}}$ for $1\leq k<s$. This implies $\text{rank}g_{1}=n$, i.e. $g_{1}=\gamma_{n}$, or $g_{1}\in R_{i}$. The latter one is not possible. But $g_{1}=\gamma_{n}$ provides $\text{rank}g_{2}=n$, i.e. $g_{2}=\gamma_{n}$, and thus $g_{1}g_{2}=id_{\bar{n}}$, or $g_{2}\in R_{i}$. Both are not possible. ∎

Now, we are able to state the minimal size of a generating set of $\mathcal{FI}_{n}$. It will coincide with the size of $G_{n}$, which gives us the rank of $\mathcal{FI}_{n}$.

Proposition 2.8**.**

Let $G$ be a generating set for $\mathcal{FI}_{n}$. Then $\left|G\right|\geq 5$ if $n=3$ and $\left|G\right|\geq\frac{n-5}{2}+\left\lfloor\frac{n+6}{4}\right\rfloor\left\lfloor\frac{n+7}{4}\right\rfloor$ if $n\geq 5$.

Proof.

Since $\gamma_{n}$ and $id_{n}=\gamma_{n}\gamma_{n}$ are the only transformations in $\mathcal{FI}_{n}$ with rank $n$, we can conclude that we have at least one transformation with rank $n$ in $G$, namely $\gamma_{n}$.

Lemma 2.7 provides that there are at least $\left\lceil\frac{n}{4}\right\rceil$ transformations in $G\cap\bigcup\{R_{i}:1\leq i$ $\leq\frac{n+1}{2}$, $i$ is odd$\}$, where $\left\lceil\frac{n}{4}\right\rceil$ $=\left|\{m\in\{1,\ldots,\frac{n+1}{2}\}:m\text{ is odd}\}\right|$. Moreover, there is at least one element in $G\cap R_{2}$, by Lemma 2.7. If $n=3$ then we have at least $\left\lceil\frac{n}{4}\right\rceil+1=2$ elements in $G$ with rank $n-1$. Suppose that $n\geq 5$. If $\frac{n+1}{2}$ is even then there is at least one element in $G\cap R_{\frac{n+1}{2}}$, by Lemma 2.7, too. If $n=5$ then we have at least $\left\lceil\frac{n}{4}\right\rceil+1=\left\lceil\frac{n}{4}\right\rceil+\frac{n-3}{2}$ elements in $G$ with rank $n-1$. If $n=7$ then we have at least $\left\lceil\frac{n}{4}\right\rceil+1+1=\left\lceil\frac{n}{4}\right\rceil+\frac{n-3}{2}$ elements in $G$ with rank $n-1$. Let now $n\geq 9$. We consider the case that $i\in\{4,\ldots,\frac{n-1}{2}\}$ is even. Assume that $\left|G\cap R_{i}\right|\leq 1$. It is easy to verify that $\left|R_{i}\right|=16$, but on the other hand, for any $\alpha\in R_{i}$, we have $\left|\left\langle\alpha,\mathcal{FI}_{n}\setminus R_{i}\right\rangle\cap R_{i}\right|\leq 8$, because $\left\langle\alpha,\mathcal{FI}_{n}\setminus R_{i}\right\rangle\cap R_{i}\subseteq\{\alpha_{i},\alpha_{i}^{2},\alpha_{i}\gamma_{n},\alpha_{i}^{2}\gamma_{n},\gamma_{n}\alpha_{i},\gamma_{n}\alpha_{i}^{2},\gamma_{n}\alpha_{i}\gamma_{n},\gamma_{n}\alpha_{i}^{2}\gamma_{n}\}$. This implies that $R_{i}\nsubseteqq\left\langle G\right\rangle$, a contradiction. If $\frac{n+1}{2}$ is even then $\frac{n-7}{4}=\left|\{m\in\{4,\ldots,\frac{n-1}{2}\}:m\text{ is even}\}\right|$, i.e. we have at least $1+1+2\left(\frac{n-7}{4}\right)=\frac{n-3}{2}$ elements in $G\cap\bigcup\{R_{i}:1\leq i$ $\leq\frac{n+1}{2}$, $i$ is even$\}$. If $\frac{n+1}{2}$ is odd then $\frac{n-5}{4}=\left|\{m\in\{4,\ldots,\frac{n-1}{2}\}:m\text{ is even}\}\right|$, i.e. we have at least $1+2\left(\frac{n-5}{4}\right)=\frac{n-3}{2}$ elements in $G\cap\bigcup\{R_{i}:1\leq i$ $\leq\frac{n+1}{2}$, $i$ is even$\}$. Altogether, we have at least $\left\lceil\frac{n}{4}\right\rceil+\frac{n-3}{2}$ elements in $G$ with rank $n-1$, whenever $n\geq 5$.

Assume that $\left|G\cap Par_{n}\right|<2\left\lfloor\frac{n+1}{4}\right\rfloor$, where $\left\lfloor\frac{n+1}{4}\right\rfloor=\left|\left\{m\in\{2,\ldots,\frac{n+1}{2}\}:m\text{ is even}\right\}\right|$. Note, there exists exactly one $x\in dom\alpha$ with $x$ and x$\alpha$ have different parity, whenever $\alpha\in J_{n}\cap Par_{n}$, by Lemma 2.4. This provides that there is an even $j\in\{2,\ldots,\frac{n+1}{2}\}$ such that $j\alpha,(n-j+1)\alpha\notin\{1,n\}$ for all $\alpha\in G$ or $1\alpha,n\alpha\notin\{j,n-j+1\}$ for all $\alpha\in G$.

Suppose that $1\alpha,n\alpha\notin\{j,n-j+1\}$ for all $\alpha\in G$. In particular, this implies that $\beta_{j}^{odd}\notin G$. Hence, there are $g_{1},\ldots,g_{s}\in G\setminus\{id_{\bar{n}}\}$ such that $\beta_{j}^{odd}=g_{1}\cdots g_{s}$. Since $1\beta_{j}^{odd}$ and $1$ have different parity, we conclude that there is $k\in\{1,\ldots,s\}$ such that $g_{k}\in Par_{n}$. Without loss of generality, we can assume that $g_{r}\notin Par_{n}$ for $k<r\leq s$. By Lemma 2.4, then there is an even $m\in\overline{n}$ such that $1g_{k}=m$ or $ng_{k}=m$. Since $1g_{k},ng_{k}\notin\{j,n-j+1\}$, we conclude that $m\notin\{j,n-j+1\}$. Note that $\text{im}\beta_{j}^{odd}=\overline{n}\setminus\{j-1,j+1\}$. If $k=s$ then $\text{im}\beta_{j}^{odd}=\text{im}(g_{1}\cdots g_{s})=\text{im}g_{k}=\overline{n}\setminus\{m-1,m+1\}\neq\overline{n}\setminus\{j-1,j+1\}$, a contradiction. If $k<s$ then we put

[TABLE]

If $g=id_{n}$ then we get a contradiction by the previous arguments. If $g=\gamma_{n}$ then $\text{im}\beta_{j}^{odd}=\text{im}(g_{1}\cdots g_{k}g)=\{n-m,n-m+2\}\neq\overline{n}\setminus\{j-1,j+1\}$ since $m\neq n-j+1$, a contradiction. If $\text{rank}g=n-1$ then $\text{im}g=\overline{n}\setminus\{j-1\}$ or $\text{im}g=\overline{n}\setminus\{j+1\}$. First, we consider the case $\text{im}g=\overline{n}\setminus\{j-1\}$. Since $j-1$ is odd, we can conclude that $g\in\{\alpha_{j-1},\gamma_{n}\alpha_{j-1}\}$. Suppose that $g=\alpha_{j-1}$. Since $\text{rank}g_{1}\cdots g_{k}g=\text{rank}g_{k}=n-2$, we have $\text{im}g_{k}\subseteq domg$. This implies $j-1\in\{m-1,m+1\}$. Because of $j\neq m$, we have $j-1=m+1$. This provides $m-1=j-3$, i.e. $j-3\notin\text{im}g_{k}=\text{im}(g_{1}\cdots g_{k})=\text{im}(g_{1}\cdots g_{k}\alpha_{j-1})=\text{im}(g_{1}\cdots g_{k}g)$, a contradiction. Suppose that $g=\gamma_{n}\alpha_{j-1}$. Clearly, $domg=\overline{n}\setminus\{n-j+2\}$. Then $\text{im}g_{k}\subseteq domg$ provides $n-j+2\in\{m-1,m+1\}$. But $n-j+1\neq m$ gives $n-j+2=m-1$. Hence, $m+1=n-j+4$ and we obtain $j-3=(n-j+4)\gamma_{n}\alpha_{j-1}$. Thus $j-3\notin\text{im}(g_{1}\cdots g_{k}g)$, a contradiction. Dually, we can treat the case $\text{im}g=\overline{n}\setminus\{j+1\}$. If $\text{rank}g=n-2$ then $\text{im}g=\overline{n}\setminus\{j-1,j+1\}$. Since both $j-1$ and $j+1$ are odd, we can conclude that $domg=\overline{n}\setminus\{j-1,j+1\}$ or $domg=\overline{n}\setminus\{n-j,n-j+2\}$, since $g=\alpha_{j-1}\alpha_{j+1}$ or $g=\gamma_{n}\alpha_{j-1}\alpha_{j+1}$. But because of $m\neq n-j+1$ and $m\neq j$, we have $m-1\neq n-j$ and $m-1\neq j-1$, respectively. This implies $img_{k}=\{m-1,m+1\}\neq domg$, a contradiction.

Suppose that $j\alpha,(n-j+1)\alpha\notin\{1,n\}$ for all $\alpha\in G$. Then we conclude dually that there are $\delta_{1},\ldots,\delta_{t}\in G\setminus\{id_{\bar{n}}\}$ such that $\beta_{j}^{even}=\delta_{1}\cdots\delta_{t}$ but $dom\delta_{1}\cdots\delta_{t}\neq dom$ $\beta_{j}^{even}$, i.e. we obtain a contradiction, too.

Suppose that $n\geq 5$. By straightforward combinatorial calculations, one obtains that there are exactly $\left\lfloor\frac{n}{4}\right\rfloor\left\lfloor\frac{n+2}{4}\right\rfloor-1$ pairs $(i,j)$ of odd numbers $i,j\in\overline{n}$ with $4\leq j-i<n-1$ and $i\leq n-j+1$. Assume that there are less than $\left\lfloor\frac{n}{4}\right\rfloor\left\lfloor\frac{n+2}{4}\right\rfloor-1$ elements in $G\setminus Par_{n}$ with rank $n-2$. Note that for odd numbers $i,j\in\overline{n}$ with $4\leq j-i<n-1$ and $i\leq n-j+1$, it holds $4\leq(n-i+1)-(n-j+1)<n-1$ but $(n-j+1)>n-(n-i+1)+1$, whenever $i\neq n-j+1$. In particular, $i=n-j+1$ implies $j=n-i+1$. This justifies the existence of a pair $(i,j)$ of odd numbers $i,j\in\overline{n}$ with $4\leq j-i<n-1$ and $i\leq n-j+1$ such that $dom\alpha\neq\overline{n}\setminus\{i,j\}$ and $dom\alpha\neq\overline{n}\setminus\{n-i+1,n-j+1\}$ for all $\alpha\in G$. In particular, $\alpha_{i,j}\notin G$ and there are $h_{1},\ldots,h_{u}\in G\setminus\{id_{\bar{n}}\}$ such that $\alpha_{i,j}=h_{1}\cdots h_{u}$. Note that each $\alpha\in\mathcal{FI}_{n}$ with $dom\alpha=\overline{n}\setminus\{\upsilon\}$ for some odd $\upsilon\in\overline{n}$ is either order-preserving or order-reversing. Assume that $rankh_{q}\geq n-1$ for all $q\in\{1,\ldots,u\}$. Since $\alpha_{i,j}$ is neither order-preserving nor order-reversing, there is an even $v\in\overline{n}$ with $v\notin dom(h_{1}\cdots h_{u})$, a contradiction. This implies the existence of a $k\in\{1,\ldots,u\}$ with $\text{rank}h_{k}=n-2$. Without loss of generality, we can assume that $rankh_{q}\geq n-1$ for $1\leq q<k$. In particular, an odd number is missing in $domh_{q}$, for $q\in\{1,\ldots,k\}$. Let $domh_{k}=\{s,t\}$. Since $h_{k}\in G$, we have $\{s,t\}\neq\{i,j\}$ as well as $\{s,t\}\neq\{n-i+1,n-j+1\}$, i.e. $\{n-s+1,n-t+1\}\neq\{i,j\}$. Clearly, $domh_{q}\subseteq\text{im}h_{q-1}$ for $1\leq q\leq k$. Therefore and since $h_{q}$ is order-preserving or order-reversing for $1\leq q<k$, there is $h\in\{id_{n},\gamma_{n}\}$ such that $h_{1}\cdots h_{k-1}h_{k}=hh_{k}$. Hence, $dom(h_{1}\cdots h_{k})=\{s,t\}$ or $dom(h_{1}\cdots h_{k})=\{n-s+1,n-t+1\}$. Since $\text{rank}(h_{1}\cdots h_{k})=2=\text{rank}(h_{1}\cdots h_{u})$, we can conclude that $dom\alpha_{i,j}=dom(h_{1}\cdots h_{u})=dom(h_{1}\cdots h_{k})\neq\{i,j\}$, a contradiction.

Altogether, we have shown that $\left|G\right|\geq 1+2+2\left\lfloor\frac{3+1}{4}\right\rfloor=5$ if $n=3$ and $\left|G\right|\geq 1+\left\lceil\frac{n}{4}\right\rceil+\frac{n-3}{2}+2\left\lfloor\frac{n+1}{4}\right\rfloor+\left\lfloor\frac{n}{4}\right\rfloor\left\lfloor\frac{n+2}{4}\right\rfloor-1$, whenever $n\geq 5$. It is easy to verify that $\left\lceil\frac{n}{4}\right\rceil+2\left\lfloor\frac{n+1}{4}\right\rfloor=n-\left\lfloor\frac{n}{4}\right\rfloor$ and $n-\left\lfloor\frac{n}{4}\right\rfloor+\left\lfloor\frac{n}{4}\right\rfloor\left\lfloor\frac{n+2}{4}\right\rfloor=\left\lfloor\frac{n+6}{4}\right\rfloor\left\lfloor\frac{n+7}{4}\right\rfloor-1$. Hence, $\left|G\right|\geq\left\lfloor\frac{n+6}{4}\right\rfloor\left\lfloor\frac{n+7}{4}\right\rfloor-1+\frac{n-3}{2}=\frac{n-5}{2}+\left\lfloor\frac{n+6}{4}\right\rfloor\left\lfloor\frac{n+7}{4}\right\rfloor$, whenever $n\geq 5$. ∎

It is easy to calculate that $\left|G_{n}\right|=1+\left\lceil\frac{n}{4}\right\rceil+\frac{n-3}{2}+2\left\lfloor\frac{n+1}{4}\right\rfloor+\left\lfloor\frac{n}{4}\right\rfloor\left\lfloor\frac{n+2}{4}\right\rfloor-1=\frac{n-5}{2}+\left\lfloor\frac{n+6}{4}\right\rfloor\left\lfloor\frac{n+7}{4}\right\rfloor$, whenever $n\geq 5$, and $\left|G_{3}\right|=5$. So, we get the following corollary.

Corollary 2.9**.**

$\text{rank}\mathcal{FI}_{3}=5$* and $\text{rank}\mathcal{FI}_{n}=\frac{n-5}{2}+\left\lfloor\frac{n+6}{4}\right\rfloor\left\lfloor\frac{n+7}{4}\right\rfloor$, whenever $n\geq 5$.*

Acknowledgement

The second author was supported by the Thailand Research Fund, Grant No. MRG6180296 and Research Fund of Faculty of Science, Silpakorn University, Grant No. SRF-JRG-2561-08 through a visiting researcher fellowship.