# Generalization of the basis theorem for the $B$-type Coxeter groups

Sawsan Khaskeia
Department of Mathematics
Ariel University, Israel
<EMAIL_ADDRESS>Robert Shwartz
Department of Mathematics
Ariel University, Israel
<EMAIL_ADDRESS>
###### Abstract

The $OGS$ for non-abelian groups is an interesting generalization of the basis
of finite abelian groups. The definition of $OGS$ states that every element of
a group has a unique presentation as a product of some powers of specific
generators of the group, in a specific given order. In case of the symmetric
groups $S_{n}$ there is a paper of R. Shwartz, which demonstrates a strong
connection between the $OGS$ and the standard Coxeter presentation of the
symmetric group, which is called the standard $OGS$ of $S_{n}$. In this paper
we generalize the standard $OGS$ of $S_{n}$ to the finite classical Coxeter
group $B_{n}$. We describe the exchange laws for the generalized standard
$OGS$ of $B_{n}$, and we connect it to the Coxeter length and the descent set
of $B_{n}$.

##  1 Introduction

The fundamental theorem of finitely generated abelian groups states the
following: Let $A$ be a finitely generated abelian group, then there exists
generators $a_{1},a_{2},\ldots a_{n}$, such that every element $a$ in $A$ has
a unique presentation of a form:

$g=a_{1}^{i_{1}}\cdot a_{2}^{i_{2}}\cdots a_{n}^{i_{n}},$

where, $i_{1},i_{2},\ldots,i_{n}$ are $n$ integers such that for $1\leq k\leq
n$, $0\leq i_{k}<|g_{k}|$, where $a_{k}$ has a finite order of $|a_{k}|$ in
$A$, and $i_{k}\in\mathbb{Z}$, where $a_{k}$ has infinite order in $A$. Where,
the meaning of the theorem is that every abelian group $A$ is direct sum of
finitely many cyclic subgroup $A_{i}$ (where $1\leq i\leq k$), for some
$k\in\mathbb{N}$.

######  Definition 1.0.1.

Let $G$ be a non-abelian group. The ordered sequence of $n$ elements $\langle
g_{1},g_{2},\ldots,g_{n}\rangle$ is called an
$Ordered~{}~{}Generating~{}~{}System$ of the group $G$ or by shortened
notation, $OGS(G)$, if every element $g\in G$ has a unique presentation in the
a form

$g=g_{1}^{i_{1}}\cdot g_{2}^{i_{2}}\cdots g_{n}^{i_{n}},$

where, $i_{1},i_{2},\ldots,i_{n}$ are $n$ integers such that for $1\leq k\leq
n$, $0\leq i_{k}<r_{k}$, where $r_{k}||g_{k}|$ in case the order of $g_{k}$ is
finite in $G$, or $i_{k}\in\mathbb{Z}$, in case $g_{k}$ has infinite order in
$G$. The mentioned canonical form is called $OGS$ canonical form. For every
$q>p$, $1\leq x_{q}<r_{q}$, and $1\leq x_{p}<r_{p}$ the relation

$g_{q}^{x_{q}}\cdot g_{p}^{x_{p}}=g_{1}^{i_{1}}\cdot g_{2}^{i_{2}}\cdots
g_{n}^{i_{n}},$

is called exchange law.

In contrast to finitely generated abelian groups, the existence of an $OGS$ is
generally not true for every finitely generated non-abelian group. Even in
case of two-generated infinite non-abelian groups it is not difficult to find
counter examples. For example, the Baumslag-Solitar groups $BS(m,n)$ [4],
where $m\neq\pm 1$ or $n\neq\pm 1$, or most of the cases of the one-relator
free product of a finite cyclic group generated by $a$, with a finite two-
generated group generated by $b,c$ with the relation $a^{2}\cdot b\cdot a\cdot
c=1$ [10], do not have an $OGS$. Even the question of the existence of an
$OGS$ for a general finite non-abelian group is still open. Moreover, contrary
to the abelian case where the exchange law is just $g_{q}\cdot
g_{p}=g_{p}\cdot g_{q}$, in most of the cases of non-abelian groups with the
existence of an $OGS$, the exchange laws are very complicated. Although there
are some specific non-abelian groups where the exchange laws are very
convenient and have very interesting properties. A very good example of it
this the symmetric group $S_{n}$. In 2001, Adin and Roichman [1] introduced a
presentation of an $OGS$ canonical form for the symmetric group $S_{n}$, for
the hyperoctahedral group $B_{n}$, and for the wreath product
$\mathbb{Z}_{m}\wr S_{n}$. Adin and Roichman proved that for every element of
$S_{n}$ presented in the standard $OGS$ canonical form, the sum of the
exponents of the $OGS$ equals the major-index of the permutation. Moreover, by
using an $OGS$ canonical form, Adin and Roichman generalized the theorem of
MacMahon [6] to the $B$-type Coxeter group, and to the wreath product
$\mathbb{Z}_{m}\wr S_{n}$. A few years later, this $OGS$ canonical form was
generalized for complex reflection groups by Shwartz, Adin and Roichman [11].
Recently, Shwartz [9] significantly extended the results of [1], [11], where
the $OGS$ of $S_{n}$ is strongly connected to the Coxeter length and to the
descent set of the elements. Moreover, in [9], there are described the
exchange laws for the $OGS$ canonical forms of the symmetric group $S_{n}$,
which have very interesting and surprising properties. In the paper we try to
generalize the results of [9] to the finite classical Coxeter group $B_{n}$.
Similarly to the symmetric group $S_{n}$, the group $B_{n}$ can be considered
as permutation group as well. Therefore, we recall the notations of
permutations, the $OGS$ of $S_{n}$ and the corresponding exchange laws, from
[9].

######  Definition 1.0.2.

Let $S_{n}$ be the symmetric group on $n$ elements, then :

  * •

The symmetric group $S_{n}$ is an $n-1$ generated simply-laced finite Coxeter
group of order $n!$, which has the presentation of:

$\langle s_{1},s_{2},\ldots,s_{n-1}|s_{i}^{2}=1,~{}~{}(s_{i}\cdot
s_{i+1})^{3}=1,~{}~{}(s_{i}\cdot s_{j})^{2}=1~{}~{}for~{}~{}|i-j|\geq
2\rangle;$

  * •

The group $S_{n}$ can be considered as the permutation group on $n$ elements.
A permutation $\pi\in S_{n}$ is denoted by

$\pi=[\pi(1);~{}\pi(2);\ldots;~{}\pi(n)]$

(i.e., $\pi=[2;~{}4;~{}1;~{}3]$ is a permutation in $S_{4}$ which satisfies
$\pi(1)=2$, $\pi(2)=4$, $\pi(3)=1$, and $\pi(4)=3$);

  * •

Every permutation $\pi\in S_{n}$ can be presented in a cyclic notation, as a
product of disjoint cycles of the form $(i_{1},~{}i_{2},~{}\ldots,~{}i_{m})$,
which means $\pi(i_{k})=i_{k+1}$, for $1\leq k\leq m-1$, and
$\pi(i_{m})=i_{1}$ (i.e., The cyclic notation of $\pi=[3;~{}4;~{}1;~{}5;~{}2]$
in $S_{5}$, is $(1,~{}3)(2,~{}4,~{}5)$);

  * •

The Coxeter generator $s_{i}$ can be considered the permutation which
exchanges the element $i$ with the element $i+1$, i.e., the transposition
$(i,i+1)$;

  * •

We consider multiplication of permutations in left to right order; i.e., for
every $\pi_{1},\pi_{2}\in S_{n}$, $\pi_{1}\cdot\pi_{2}(i)=\pi_{2}(j)$, where,
$\pi_{1}(i)=j$ (in contrary to the notation in [1] where Adin, Roichman, and
other people have considered right to left multiplication of permutations);

  * •

For every permutation $\pi\in S_{n}$, the Coxeter length $\ell(\pi)$ is the
number of inversions in $\pi$, i.e., the number of different pairs $i,j$, s.
t. $i<j$ and $\pi(i)>\pi(j)$;

  * •

For every permutation $\pi\in S_{n}$, the set of the locations of the descents
is defined to be

$des\left(\pi\right)=\\{1\leq i\leq n-1|\pi(i)>\pi(i+1)\\},$

and

$i\in
des\left(\pi\right)~{}~{}if~{}and~{}only~{}if~{}~{}\ell(s_{i}\cdot\pi)<\ell(\pi)$

(i.e., $i$ is a descent of $\pi$ if and only if multiplying $\pi$ by $s_{i}$
in the left side shortens the Coxeter length of the element.);

  * •

For every permutation $\pi\in S_{n}$, the major-index is defined to be

$maj\left(\pi\right)=\sum_{\pi(i)>\pi(i+1)}i$

(i.e., major-index is the sum of the locations of the descents of $\pi$.).

  * •

By [5] Chapter 3.4, every element $\pi$ of $S_{n}$ can be presented uniquely
in the following normal reduced form, which we denote by $norm(\pi)$:

$norm(\pi)=\prod_{u=1}^{n-1}\prod_{r=0}^{y_{u}-1}s_{u-r}.$

such that $y_{u}$ is a non-negative integer where, $0\leq y_{u}\leq u$ for
every $1\leq u\leq n-1$. By our notation of $norm(\pi)$ the Coxeter length of
an element $\pi$ as follow:

$\ell(\pi)=\sum_{u=1}^{n-1}y_{u}.$

For example: Let $m=8$, $y_{2}=2$, $y_{4}=3$, $y_{5}=1$, $y_{8}=4$, and
$y_{1}=y_{3}=y_{6}=y_{7}=0$, then

$norm(\pi)=(s_{2}\cdot s_{1})\cdot(s_{4}\cdot s_{3}\cdot s_{2})\cdot
s_{5}\cdot(s_{8}\cdot s_{7}\cdot s_{6}\cdot s_{5}).$ $\ell(\pi)=2+3+1+4=10.$

######  Theorem 1.0.3.

Let $S_{n}$ be the symmetric group on $n$ elements. For every $2\leq m\leq n$,
define $t_{m}$ to be the product $\prod_{j=1}^{m-1}s_{j}$. The element $t_{m}$
is the permutation

$t_{m}=[m;~{}1;~{}2;\ldots;~{}m-1;~{}m+1;\ldots;~{}n]$

which is the $m$-cycle $(m,~{}m-1,~{}\ldots,~{}1)$ in the cyclic notation of
the permutation. Then, the elements $t_{n},t_{n-1},\ldots,t_{2}$ generates
$S_{n}$, and every element of $S_{n}$ has a unique presentation in the
following $OGS$ canonical form:

$t_{2}^{i_{2}}\cdot t_{3}^{i_{3}}\cdots
t_{n}^{i_{n}},~{}~{}~{}where~{}~{}~{}0\leq i_{k}<k~{}~{}~{}for~{}~{}~{}2\leq
k\leq n$

######  Proposition 1.0.4.

The following holds:

In order to transform the element $t_{q}^{i_{q}}\cdot t_{p}^{i_{p}}$ ($p<q$)
onto the $OGS$ canonical form
$t_{2}^{i_{2}}\cdot t_{3}^{i_{3}}\cdots t_{n}^{i_{n}}$, i.e., according to the
standard $OGS$, one needs to use the following exchange laws:

$t_{q}^{i_{q}}\cdot t_{p}^{i_{p}}=\begin{cases}t_{i_{q}+i_{p}}^{i_{q}}\cdot
t_{p+i_{q}}^{i_{p}}\cdot t_{q}^{i_{q}}&q-i_{q}\geq p\\\ \\\
t_{i_{q}}^{p+i_{q}-q}\cdot t_{i_{q}+i_{p}}^{q-p}\cdot
t_{q}^{i_{q}+i_{p}}&i_{p}\leq q-i_{q}\leq p\\\ \\\
t_{p+i_{q}-q}^{i_{q}+i_{p}-q}\cdot t_{i_{q}}^{p-i_{p}}\cdot
t_{q}^{i_{q}+i_{p}-p}&q-i_{q}\leq i_{p}.\end{cases}$

######  Remark 1.0.5.

The standard $OGS$ canonical form of $t_{q}^{i_{q}}\cdot t_{p}^{i_{p}}$ is a
product of non-zero powers of two different canonical generators if and only
if $q-i_{q}=p$ or $q-i_{q}=i_{p}$, as follow:

  * •

If $q-i_{q}=p$ then by considering $q-i_{q}\geq p$:

$t_{i_{q}+i_{p}}^{i_{q}}\cdot t_{p+i_{q}}^{i_{p}}\cdot
t_{q}^{i_{q}}=t_{i_{q}+i_{p}}^{i_{q}}\cdot t_{q}^{i_{p}}\cdot t_{q}^{i_{q}}$

and by considering $q-i_{q}\leq p$:

$t_{i_{q}}^{p+i_{q}-q}\cdot t_{i_{q}+i_{p}}^{q-p}\cdot
t_{q}^{i_{q}+i_{p}}=t_{i_{q}}^{0}\cdot t_{i_{q}+i_{p}}^{i_{q}}\cdot
t_{q}^{i_{q}+i_{p}};$

  * •

If $q-i_{q}=i_{p}$ then by considering $q-i_{q}\geq i_{p}$:

$t_{i_{q}}^{p+i_{q}-q}\cdot t_{i_{q}+i_{p}}^{q-p}\cdot
t_{q}^{i_{q}+i_{p}}=t_{i_{q}}^{p-i_{p}}\cdot t_{q}^{q-p}\cdot
t_{q}^{q}=t_{i_{q}}^{p-i_{p}}\cdot t_{q}^{q-p}$

and by considering $q-i_{q}\leq i_{p}$:

$t_{p+i_{q}-q}^{i_{q}+i_{p}-q}\cdot t_{i_{q}}^{p-i_{p}}\cdot
t_{q}^{i_{q}+i_{p}-p}=t_{p+i_{q}-q}^{0}\cdot t_{i_{q}}^{p-i_{p}}\cdot
t_{q}^{q-p}.$

Hence we have

$t_{q}^{i_{q}}\cdot t_{p}^{i_{p}}=\begin{cases}t_{i_{q}+i_{p}}^{i_{q}}\cdot
t_{q}^{i_{q}+i_{p}}&q-i_{q}=p\\\ \\\ t_{i_{q}}^{p-i_{p}}\cdot
t_{q}^{q-p}&q-i_{q}=i_{p}.\end{cases}$

Moreover, the most significant achievement of the paper [9] is the definition
of the standard $OGS$ elementary factorization. By using the standard $OGS$
elementary factorization, it is possible to give a very interesting formula
for the Coxeter length and a complete classification of the descent set of any
arbitrary element of $S_{n}$. In the paper we try to generalize the standard
$OGS$ elementary factorization to the $B$-type Coxeter groups, in order to
find similar properties (Coxeter length and the descent set) for the elements
of the group. Hence, we recall the definition of the standard $OGS$ elementary
element and factorization for the symmetric group $S_{n}$ as it is defined in
[9], and theorems concerning the Coxeter length and the descent set of
elements of $S_{n}$ as it is mentioned and proved in [9] .

######  Definition 1.0.6.

Let $\pi\in S_{n}$, where $\pi=\prod_{j=1}^{m}t_{k_{j}}^{i_{k_{j}}}$ is
presented in the standard $OGS$ canonical form, with $i_{k_{j}}>0$ for every
$1\leq j\leq m$. Then, $\pi$ is called standard $OGS$ elementary element of
$S_{n}$, if

$\sum_{j=1}^{m}i_{k_{j}}\leq k_{1}.$

######  Theorem 1.0.7.

[9] Let $\pi=\prod_{j=1}^{m}t_{k_{j}}^{i_{k_{j}}}$ be a standard $OGS$
elementary element of $S_{n}$, presented in the standard $OGS$ canonical form,
with $i_{k_{j}}>0$ for every $1\leq j\leq m$. Then, the following are
satisfied:

  * •

$\ell(\pi)=\sum_{j=1}^{m}k_{j}\cdot
i_{k_{j}}-(i_{k_{1}}+i_{k_{2}}+\cdots+i_{k_{m}})^{2}=\sum_{j=1}^{m}k_{j}\cdot
i_{k_{j}}-\left(maj\left(\pi\right)\right)^{2};$

  * •

Every subword of $\pi$ is a standard $OGS$ elementary element too. In
particular, for every two subwords $\pi_{1}$ and $\pi_{2}$ of $\pi$, such that
$\pi=\pi_{1}\cdot\pi_{2}$, it is satisfied:

$\ell(\pi)=\ell(\pi_{1}\cdot\pi_{2})<\ell(\pi_{1})+\ell(\pi_{2});$

  * •

$\ell(s_{r}\cdot\pi)=\begin{cases}\ell(\pi)-1&r=\sum_{j=1}^{m}i_{k_{j}}\\\
\ell(\pi)+1&r\neq\sum_{j=1}^{m}i_{k_{j}}\end{cases}.$

i.e., $des\left(\pi\right)$ contains just one element, which means
$des\left(\pi\right)=\\{maj\left(\pi\right)\\}$.

######  Definition 1.0.8.

Let $\pi\in S_{n}$. Let $z(\pi)$ be the minimal number, such that $\pi$ can be
presented as a product of standard $OGS$ elementary elements, with the
following conditions:

  * •

$\pi=\prod_{v=1}^{z(\pi)}\pi^{(v)},~{}~{}~{}~{}where~{}~{}~{}~{}\pi^{(v)}=\prod_{j=1}^{m^{(v)}}t_{h^{(v)}_{j}}^{\imath_{j}^{(v)}},$

by the presentation in the standard $OGS$ canonical form for every $1\leq
v\leq z(\pi)$ and $1\leq j\leq m^{(v)}$ such that:

    * –

$\imath_{j}^{(v)}>0;$

    * –

$\sum_{j=1}^{m^{(1)}}\imath_{j}^{(1)}\leq h^{(1)}_{1}$ i.e.,
$maj\left(\pi^{(1)}\right)\leq h^{(1)}_{1}$;

    * –

$h^{(v-1)}_{m^{(v-1)}}\leq\sum_{j=1}^{m^{(v)}}\imath_{j}^{(v)}\leq
h^{(v)}_{1}$ for $2\leq v\leq z$

i.e., $h^{(v-1)}_{m^{(v-1)}}\leq maj\left(\pi^{(v)}\right)\leq
h^{(v)}_{1}~{}~{}for~{}~{}2\leq v\leq z$.

Then, the mentioned presentation is called Standard $OGS$ elementary
factorization of $\pi$. Since the factors $\pi^{(v)}$ are standard $OGS$
elementary elements, they are called standard $OGS$ elementary factors of
$\pi$.

######  Theorem 1.0.9.

[9] Let $\pi=\prod_{j=1}^{m}t_{k_{j}}^{i_{k_{j}}}$ be an element of $S_{n}$
presented in the standard $OGS$ canonical form, with $i_{k_{j}}>0$ for every
$1\leq j\leq m$. Consider the standard $OGS$ elementary factorization of $\pi$
with all the notations used in Definition 1.0.8. Then, the following
properties hold:

  * •

The standard $OGS$ elementary factorization of $\pi$ is unique, i.e., the
parameters $z(\pi)$, $m^{(v)}$ for $1\leq v\leq z(\pi)$, $h^{(v)}_{j}$, and
$\imath_{j}^{(v)}$ for $1\leq j\leq m^{(v)}$, are uniquely determined by the
standard $OGS$ canonical form of $\pi$, such that:

    * –

For every $h^{(v)}_{j}$ there exists exactly one $k_{j^{\prime}}$ (where,
$1\leq j^{\prime}\leq m$), such that $h^{(v)}_{j}=k_{j^{\prime}}$;

    * –

If $h^{(v)}_{j}=k_{j^{\prime}}$, for some $1\leq v\leq z(\pi)$, $1<j<m^{(v)}$,
and $1\leq j^{\prime}\leq m$, then $\imath_{j}^{(v)}=i_{k_{j^{\prime}}}$;

    * –

If $h^{(v_{1})}_{j_{1}}=h^{(v_{2})}_{j_{2}}$, where $1\leq v_{1}<v_{2}\leq
z(\pi)$, $1\leq j_{1}\leq m^{(v_{1})}$, and
$1\leq j_{2}\leq m^{(v_{2})}$, then necessarily $v_{1}=v_{2}-1$,
$j_{1}=m^{(v_{1})}$, $j_{2}=1$, and

$h^{(v_{2}-1)}_{m^{(v_{2}-1)}}=h^{(v_{2})}_{1}=maj\left(\pi_{(v_{2})}\right)=k_{j^{\prime}},$

for some $j^{\prime}$, such that
$\imath_{m^{(v_{2}-1)}}^{(v_{2}-1)}+\imath_{1}^{(v_{2})}=i_{k_{j^{\prime}}}$;

  * •

$\ell(s_{r}\cdot\pi)=\begin{cases}\ell(\pi)-1&r=\sum_{j=1}^{m^{(v)}}\imath_{j}^{(v)}~{}~{}for~{}~{}1\leq
v\leq z(\pi)\\\ \ell(\pi)+1&otherwise\end{cases}.$

i.e.,

$des\left(\pi\right)=\bigcup_{v=1}^{z(\pi)}des\left(\pi^{(v)}\right)=\\{maj\left(\pi^{(v)}\right)~{}|~{}1\leq
v\leq z(\pi)\\};$

  * •

$\displaystyle\ell(\pi)$
$\displaystyle=\sum_{v=1}^{z(\pi)}\ell(\pi^{(v)})=\sum_{v=1}^{z(\pi)}\sum_{j=1}^{m^{(v)}}h^{(v)}_{j}\cdot\imath_{j}^{(v)}-\sum_{v=1}^{z(\pi)}\left(maj\left(\pi^{(v)}\right)\right)^{2}$
$\displaystyle=\sum_{x=1}^{m}k_{x}\cdot
i_{k_{x}}-\sum_{v=1}^{z(\pi)}\left(maj\left(\pi^{(v)}\right)\right)^{2}$
$\displaystyle=\sum_{x=1}^{m}k_{x}\cdot
i_{k_{x}}-\sum_{v=1}^{z(\pi)}{{\left(c^{(v)}\right)}}^{2},~{}~{}where~{}~{}c^{(v)}\in
des\left(\pi\right).$

The paper is organized as follow: In section 2, we recall some important
definitions and basic properties concerning the group $B_{n}$, Then we
generalize the definition of the standard $OGS$ and we show the arising
exchange laws. In section 3, we focus on the parabolic subgroup of $B_{n}$
which does not contain $s_{0}$ in it’s reduced Coxeter presentation. It is
easy to show that the subgroup is isomorphic to the symmetric group $S_{n}$,
where we denote the subgroup by $\dot{S}_{n}$. We show the presentation of
$\dot{S}_{n}$ by the generalized standard $OGS$, and then we show the
properties of the standard $OGS$ elementary elements and the the standard
$OGS$ elementary factorization of the elements (as it described in [9]) in
$\dot{S}_{n}$. In section 4, we generalize the definition of the standard
$OGS$ elementary factorization for the group $B_{n}$. In section 5, we
introduce a formula for the length function of $B_{n}$ by using the
generalized standard $OGS$ elementary factorization which is defined in the
previous section. In section 6, we characterize the descent set of the
elements in the Coxeter group $B_{n}$, by using the generalized standard $OGS$
presentation for the elements of $B_{n}$.

##  2 Generalization of the standard OGS for the Coxeter group $B_{n}$

First, we recall some important definitions and basic properties concerning
the group $B_{n}$, for example see [5]

######  Definition 2.0.1.

Let $B_{n}$ be the Coxeter group with $n$ generators, with the following
presentation:

$\begin{array}[]{r}\left\langle
s_{0},s_{1},\ldots,s_{n-1}\right|s_{i}^{2}=1,\left(s_{0}\cdot
s_{1}\right)^{4}=1,\left(s_{i}\cdot s_{i+1}\right)^{3}=1\text{ for }1\leq
i\leq n-1,\left.\left(s_{i}\cdot s_{j}\right)^{2}=1\text{ for }|i-j|\geq
2\right\rangle\end{array}$

Basic properties of $B_{n}$

  * •

The group $B_{n}$ can be presented as a permutation group of the set $[\pm
n]$, Where :

$[\pm n]=\\{i\in\mathbb{Z}|1\leqslant i\leqslant
n\quad\text{or}\quad-n\leqslant i\leqslant-1\\}$

With the following property:

$\pi(-i)=-\pi(i)\quad\text{ for every }\quad i\in[\pm n].$

  * •

$B_{n}$, can be considered as a signed permutation group, where $\pi$ is
uniquely determined by $\pi(i)$ for $1\leq i\leq n$.

  * •

A signed permutation $\pi\in B_{n}$ is denoted by

$[\pi(1);~{}\pi(2);\ldots;~{}\pi(n)]$

(e.g., $\pi=[2;~{}-4;~{}1;~{}3]$ is a permutation in $B_{4}$ which satisfies
$\pi(1)=2$, $\pi(2)=-4$, $\pi(3)=1$, $\pi(4)=3$, and $\pi(-1)=-2$,
$\pi(-2)=4$, $\pi(-3)=-1$, $\pi(-4)=-3$)

  * •

$[|\pi(1)|;~{}|\pi(2)|;\ldots;~{}|\pi(n)|]$ is a permutation of $S_{n}$

  * •

Similarly to $S_{n}$, for $1\leq i\leq n-1$, the Coxeter generators $s_{i}$
can be considered the permutation which exchanges the element $i$ with the
element $i+1$ and additionally exchanges $-i$ with $-(i+1)$, and $s_{0}$ is
the permutation which exchanges the element $1$ with the element $-1$.

  * •

$B_{n}\simeq\mathbb{Z}_{2}^{n}\rtimes S_{n}$.

  * •

$|B_{n}|=2^{n}\cdot n!$.

######  Definition 2.0.2.

We define des($\pi$) to be the left descent set of $\pi$.

$s_{i}\in des(\pi)\quad if\quad\ell(s_{i}\cdot\pi)<\ell(\pi).$

Similarly to $S_{n}$, the following properties holds in $B_{n}$:

$des(\pi)=\\{0\leqslant i\leqslant
n-1\mid\pi(i)>\pi(i+1)\\},\quad(\pi(0)\text{ defined to be }(0))$

Now, we recall the definition of the normal reduced form for $B_{n}$, as it is
defined in [5] Chapter 3.4.

######  Definition 2.0.3.

$norm(\pi)=\prod_{i=0}^{n-1}\prod_{j=0}^{y_{i}-1}s_{|i-j|}$

where, $y_{i}$ is a non-negative integer such that $0\leq y_{i}\leq 2i+1$, and
the Coxeter length of $\pi$ as follow

$\ell(\pi)=\sum_{i=0}^{n-1}y_{i}.$

######  Example 2.0.4.

Consider the normal form of the following $\pi\in B_{5}$

$norm(\pi)=s_{0}\cdot(s_{1}\cdot s_{0}\cdot s_{1})\cdot(s_{3}\cdot s_{2}\cdot
s_{1}\cdot s_{0})\cdot(s_{4}\cdot s_{2}\cdot s_{2}\cdot s_{1}\cdot s_{0}\cdot
s_{1}\cdot s_{2})\cdot(s_{5}\cdot s_{4}\cdot s_{3})$

Then, $y_{0}=1,\ y_{1}=3,\ y_{2}=0,\ y_{3}=4,\ y_{4}=7,\ y_{5}=3$, and then

$\ell(\pi)=1+3+0+4+7+3=18.$

###  2.1 The generalized standard OGS of $B_{n}$

In this subsection we generalize the definition of the standard OGS (defined
in [9]) for the group $B_{n}$, and we show the arising exchange laws.

Now, similar to the definition of $t_{k}$ in $S_{n}$ [9], we define $\tau_{k}$
in $B_{n}$ for $k=1,2,\ldots,n$ as follow:

######  Definition 2.1.1.

For $k=1,2,\ldots,n$ let $\tau_{k}$ be :

$\tau_{k}=\prod_{j=0}^{k-1}s_{j}$

######  Remark 2.1.2.

For every $1\leqslant k\leqslant n$, $\tau_{k}$ satisfies the following
properties:

$\begin{array}[]{l}\tau_{k}(1)=-k\\\ \tau_{k}(j)=j-1,\text{ for }2\leq j\leq
k\\\ \tau_{k}(j)=j,\text{ for }k+1\leq j\leq n\end{array}$

######  Remark 2.1.3.

$\tau_{k}=s_{0}\cdot t_{k}.$

######  Remark 2.1.4.

For $1\leq k\leq n$, let $\tau_{k}$ be an element of $B_{n}$ as it is defined
in Definition 2.1.1, then the following holds:

  * •

For $i_{k}=-k$

$\tau_{k}^{i_{k}}(j)=\left\\{\begin{array}[]{ll}-j&\text{ for }1\leq j\leq
k\\\ \\\ j&\text{ for }k+1\leq j\leq n\end{array}\right.$

  * •

For $0<i_{k}<k$

$\tau_{k}^{i_{k}}(j)=\left\\{\begin{array}[]{ll}-(j-i_{k}+k)&\text{ for }1\leq
j\leq i_{k}\\\ \\\ j-i_{k}&\text{ for }i_{k}+1\leq j\leq k\\\ \\\ j&\text{ for
}k+1\leq j\leq n\end{array}\right.$

  * •

For $-k<i_{k}<0$

$\tau_{k}^{i_{k}}(j)=\left\\{\begin{array}[]{ll}j-i_{k}&\text{ for }1\leq
j\leq i_{k}+k\\\ \\\ -(j-i_{k}-k)&\text{ for }i_{k}+k+1\leq j\leq k\\\ \\\
j&\text{ for }k+1\leq j\leq n\end{array}\right.$

Now, we define the generalized standard $OGS$ for the group $B_{n}$ as follow

######  Theorem 2.1.5.

For $k=1,2,\ldots,n$ let $\tau_{k}$ be the elements of $B_{n}$ as defined in
Definition 2.1.1 then the following holds:

Every element $g\in B_{n}$ has a unique presentation in the following form:

$\tau_{1}^{i_{1}}\cdot\tau_{2}^{i_{2}}\cdots\tau_{n}^{i_{n}}$

such that, $-k\leq i_{k}<k$.

###### Proof.

The proof is by induction on $n$. for $n=1$, it is easy to see that $B_{1}$ is
generated by $\tau_{1}$ since $B_{1}$ is a cyclic group of order $2$. Now,
assume by induction that theorem holds for every $k$ such that $k\leq n-1$.
Denote by $\dot{B}_{n-1}$ the parabolic subgroup of $B_{n}$ generated by
$s_{0},s_{1},\ldots s_{n-2}$. Easy to see that $\dot{B}_{n-1}$ is isomorphic
to $B_{n-1}$ (which satisfy the theorem by the induction hypothesis). Notice,
that in the permutation presentation of $B_{n}$, every element
$x\in\dot{B}_{n-1}$ satisfies $x(n)=n$. Now, consider the right cosets of
$\dot{B}_{n-1}$ in $B_{n}$. There are $2n$ different right cosets, where every
two elements $x$ and $y$ in the same right coset of $\dot{B}_{n-1}$ in $B_{n}$
satisfy $x(n)=y(n)$. Now, notice that the powers of $\tau_{n}$ satisfy the
following properties:

  * •

For $0\leq i_{n}<n$, $\tau_{n}^{i_{n}}(n)=n-i_{n}$;

  * •

For $-n\leq i_{n}<0$, $\tau_{n}^{i_{n}}(n)=i_{n}$.

Hence, for $-n\leq i_{n}\leq n-1$, the elements $\tau_{n}^{i_{n}}$ gives the
$2n$ different images of $n$ in the permutation presentation. Hence, for
$-n\leq i_{n}\leq n-1$, we have the following $2n$ different right cosets of
$\dot{B}_{n-1}$ in $B_{n}$ :

$\dot{B}_{n-1}\tau_{n}^{i_{n}}.$

Then the result of the theorem holds for $k=n$. ∎

######  Example 2.1.6.

Consider the element $\pi=[-2;~{}-1;~{}-4;~{}-3]$ of $B_{4}$. Now, we
construct the $OGS$ presentation of the element, as it is described in Theorem
2.1.5. First, notice that $\pi(4)=-3$. Hence $\tau_{4}^{i_{4}}(i_{4})=-3$,
which implies that $i_{4}=-3$. Thus we conclude:

$\pi\in\dot{B}_{3}~{}\tau_{4}^{-3}.$

Now, consider $\pi(3)=-4$. Notice, that $\tau_{4}^{-3}(-1)=-4$. Hence, we
consider $i_{3}$, such that $\tau_{3}^{i_{3}}(3)=-1$, which implies
$i_{3}=-1$. Thus we conclude:

$\pi\in\dot{B}_{2}~{}\tau_{3}^{-1}\cdot\tau_{4}^{-3}.$

Continuing by the same process, consider $\pi(2)=-1$. Notice, that
$[\tau_{3}^{-1}\cdot\tau_{4}^{-3}](1)=-1$. Hence, we consider $i_{2}$, such
that $\tau_{2}^{i_{2}}(2)=1$, which implies $i_{2}=2-1=1$. Thus we conclude:

$\pi=\dot{B}_{1}~{}\tau_{2}\cdot\tau_{3}^{-1}\cdot\tau_{4}^{-3}.$

Finally, considering $\pi(1)=-2$. Notice, that
$[\tau_{2}\cdot\tau_{3}^{-1}\cdot\tau_{4}^{-3}](-1)=-2$. Hence, we consider
$i_{1}$, such that $\tau_{1}^{i_{1}}(1)=-1$, which implies $i_{1}=-1$. Thus we
conclude:

$\pi=\tau_{1}^{-1}\cdot\tau_{2}\cdot\tau_{3}^{-1}\cdot\tau_{4}^{-3}.$

######  Remark 2.1.7.

We call the presentation of elements of $B_{n}$ which has been shown in
Theorem 2.1.5 the generalized standard $OGS$ presentation of $B_{n}$.

Now, we show some important properties of the generalized standard $OGS$ of
$B_{n}$.
We start with the exchange laws;

######  Proposition 2.1.8.

In order to transform the element $\tau_{q}^{r_{q}}\cdot\tau_{p}^{r_{p}}$
$(p<q)$ into the generalized standard $OGS$ presentation of the form
$\tau_{1}^{i_{1}}\cdot\tau_{2}^{i_{2}}\cdots\tau_{n}^{i_{n}}$ , one need to
use the following exchange laws:

  * •

The case $0<r_{p}<p$:

$\tau_{q}^{r_{q}}\cdot\tau_{p}^{r_{p}}=\left\\{\begin{array}[]{ll}\tau_{r_{q}}^{-r_{q}}\cdot\tau_{r_{q}+r_{p}}^{r_{q}}\cdot\tau_{p+r_{q}}^{r_{p}}\cdot\tau_{q}^{r_{q}}&q-r_{q}\geq
p\\\ \\\
\tau_{r_{q}}^{p-q}\cdot\tau_{r_{q}+r_{p}}^{q-p}\cdot\tau_{q}^{r_{q}+r_{p}}&r_{p}\leq
q-r_{q}\leq p\\\ \\\
\tau_{p+r_{q}-q}^{r_{q}+r_{p}-q}\cdot\tau_{r_{q}}^{p-r_{p}-r_{q}}\cdot\tau_{q}^{r_{q}+r_{p}-p-q}&q-r_{q}\leq
r_{p}\end{array}\right.$

  * •

The case $r_{p}=-p$:

$\tau_{q}^{r_{q}}\cdot\tau_{p}^{r_{p}}=\left\\{\begin{array}[]{ll}\tau_{r_{q}}^{-r_{q}}\cdot\tau_{p+r_{q}}^{-p-r_{q}}\cdot\tau_{q}^{r_{q}}&q-r_{q}\geq
p\\\ \\\
\tau_{p+r_{q}-q}^{-p-r_{q}+q}\cdot\tau_{r_{q}}^{-r_{q}}\cdot\tau_{q}^{r_{q}-q}&q-r_{q}<p\end{array}\right.$

  * •

The case $-p<r_{p}<0$:

$\tau_{q}^{r_{q}}\cdot\tau_{p}^{r_{p}}=\left\\{\begin{array}[]{ll}\tau_{r_{q}+r_{p}+p}^{r_{q}}\cdot\tau_{p+r_{q}}^{r_{p}-r_{q}}\cdot\tau_{q}^{r_{q}}&q-r_{q}\geq
p\\\ \\\
\tau_{p+r_{q}-q}^{-p-r_{q}+q}\cdot\tau_{r_{q}}^{p+r_{q}-q}\cdot\tau_{r_{q}+r_{p}+p}^{q-p}\cdot\tau_{q}^{p+r_{p}-q+r_{q}}&r_{p}+p\leq
q-r_{q}\leq p\\\ \\\
\tau_{p+r_{q}-q}^{r_{p}}\cdot\tau_{r_{q}}^{-r_{p}}\cdot\tau_{q}^{r_{q}+r_{p}}&q-r_{q}\leq
r_{p}+p\end{array}\right.$

###### Proof.

By Remark $2.1.3.$ we have

$\tau_{i}=s_{0}\cdot t_{i}.$

Then,

$\tau_{q}^{r_{q}}\cdot\tau_{p}^{r_{p}}=(s_{0}\cdot
t_{q})^{r_{q}}\cdot(s_{0}\cdot t_{p})^{r_{p}}$

Similar to the exchange laws for the standard $OGS$ of $S_{n}$ as it is
presented in Proposition 1.0.4, we consider several cases, depending on the
values of $p,q,r_{p},r_{q}$. We have the following three main cases:

  1. 1.

$0<r_{p}<p$;

  2. 2.

$r_{p}=-p$;

  3. 3.

$-p<r_{p}<0$.

Let start with the first case.

  1. 1.

The case $0<r_{p}<p$

Similar to three cases of exchange laws which arises from the standard $OGS$
presentation of $S_{n}$, (as it is presented in Proposition 1.0.4), we have
the following three subcases of exchange laws, depending on how the value of
$q-r_{q}$ compares to the values of $p$ and $r_{p}$:

     * •

$q-r_{q}\geq p$;

     * •

$r_{p}\leq q-r_{q}\leq p$;

     * •

$q-r_{q}\leq r_{p}$.

Now, we start with the first subcase:

$\bullet$ $q-r_{q}\geq p$ :

$\tau_{q}^{r_{q}}\cdot\tau_{p}^{r_{p}}=(s_{0}\cdot
t_{q})^{r_{q}}\cdot(s_{0}\cdot t_{p})^{r_{p}}$

Consider the permutation presentation of $(s_{0}\cdot
t_{q})^{r_{q}}\cdot(s_{0}\cdot t_{p})^{r_{p}}$ :

$1\rightarrow-(q-r_{q}+1)\rightarrow-(q-r_{q}+1).$
$r_{q}\rightarrow-q\rightarrow-q.$ $r_{q}+1\rightarrow
1\rightarrow-(p-r_{p}+1).$ $r_{q}+r_{p}\rightarrow r_{p}\rightarrow-(p).$
$r_{q}+r_{p}+1\rightarrow r_{p}+1\rightarrow 1.$ $q\rightarrow
q-r_{q}\rightarrow q-r_{q}\quad\text{or}\quad p-r_{p}.$

By the exchange laws of $S_{n}$, as it is presented in Proposition 1.0.4:

$t_{q}^{r_{q}}\cdot t_{p}^{r_{p}}=t_{r_{q}+r_{p}}^{r_{q}}\cdot
t_{p+r_{q}}^{r_{p}}\cdot t_{q}^{r_{q}}$

Now, we consider the permutation presentation of

$(s_{0}\cdot t_{r_{q}+r_{p}})^{r_{q}}\cdot(s_{0}\cdot
t_{p+r_{q}})^{r_{p}}\cdot(s_{0}\cdot t_{q})^{r_{q}}$

$1\rightarrow-(r_{p}+1)\rightarrow-1\rightarrow q-r_{q}+1.$
$r_{q}\rightarrow-(r_{p}+r_{q})\rightarrow-r_{q}\rightarrow q-r_{q}+r_{q}=q.$
$r_{q}+1\rightarrow
1\rightarrow-1\rightarrow-(p+r_{q}-r_{p}+1)\rightarrow-(p-r_{p}+1).$
$r_{q}+r_{p}\rightarrow r_{p}\rightarrow-(p+r_{q})\rightarrow-p.$
$r_{q}+r_{p}+1\rightarrow r_{q}+r_{p}+1\rightarrow r_{q}+1\rightarrow 1.$
$q\rightarrow q-r_{q}\rightarrow q-r_{q}.$

Then the permutation presentations of $\tau_{q}^{r_{q}}\cdot\tau_{p}^{r_{p}}$
and
$\tau_{r_{q}+r_{p}}^{r_{p}}\cdot\tau_{p+r_{q}}^{r_{p}}\cdot\tau_{q}^{r_{q}}$
satisfy the following properties:

     * •

For $1\leq j\leq r_{q}$:
$\tau_{q}^{r_{q}}\cdot\tau_{p}^{r_{p}}(j)=-\tau_{r_{q}+r_{p}}^{r_{q}}\cdot\tau_{p+r_{q}}^{r_{p}}\cdot\tau_{q}^{r_{q}}(j)$;

     * •

For $r_{q}+1\leq j\leq n$:
$\tau_{q}^{r_{q}}\cdot\tau_{p}^{r_{p}}(j)=\tau_{r_{q}+r_{p}}^{r_{q}}\cdot\tau_{p+r_{q}}^{r_{p}}\cdot\tau_{q}^{r_{q}}(j)$.

Then, we get the following result:

$\tau_{q}^{r_{q}}\cdot\tau_{p}^{r_{p}}=\tau_{r_{q}}^{-r_{q}}\cdot\tau_{r_{q}+r_{p}}^{r_{q}}\cdot\tau_{p+r_{q}}^{r_{p}}\cdot\tau_{q}^{r_{q}},$

in case $q-r_{q}\geq p$ and $0<r_{p}<p$.

Now, we turn to the second subcase of the case $0<r_{p}<p$.

$\bullet$ $r_{p}\leqslant q-r_{q}\leqslant p$

$\tau_{q}^{r_{q}}\cdot\tau_{p}^{r_{p}}=(s_{0}\cdot
t_{q})^{r_{q}}\cdot(s_{0}\cdot t_{p})^{r_{p}},$

By the same way, we can see that the permutation presentations of
$\tau_{q}^{r_{q}}\cdot\tau_{p}^{r_{p}}$ and
$\tau_{r_{q}}^{p-q+r_{q}}\cdot\tau_{r_{q}+r_{p}}^{q-p}\cdot\tau_{q}^{r_{q}+r_{p}}$
satisfy the following properties:

     * •

For $1\leq j\leq r_{q}$:
$\tau_{q}^{r_{q}}\cdot\tau_{p}^{r_{p}}(j)=-\tau_{r_{q}}^{p-q+r_{q}}\cdot\tau_{r_{q}+r_{p}}^{q-p}\cdot\tau_{q}^{r_{q}+r_{p}}(j)$;

     * •

For $r_{q}+1\leq j\leq n$:
$\tau_{q}^{r_{q}}\cdot\tau_{p}^{r_{p}}(j)=\tau_{r_{q}}^{p-q+r_{q}}\cdot\tau_{r_{q}+r_{p}}^{q-p}\cdot\tau_{q}^{r_{q}+r_{p}}(j)$.

Then, we get the following result:

$\tau_{q}^{r_{q}}\cdot\tau_{p}^{r_{p}}=\tau_{r_{q}}^{p-q}\cdot\tau_{r_{q}+r_{p}}^{q-p}\cdot\tau_{q}^{r_{q}+r_{p}},$

in case $r_{p}\leqslant q-r_{q}\leqslant p$ and $0<r_{p}<p$.

Now, we turn to the third subcase of the case $0<r_{p}<p$.

$\bullet$ $q-r_{q}\leqslant r_{p}$

$\tau_{q}^{r_{q}}\cdot\tau_{p}^{r_{p}}=(s_{0}\cdot
t_{q})^{r_{q}}\cdot(s_{0}\cdot t_{p})^{r_{p}}.$

By the same way, we can see that the permutation presentations of
$\tau_{q}^{r_{q}}\cdot\tau_{p}^{r_{p}}$ and
$\tau_{p+r_{q}-q}^{r_{q}+r_{p}-q}\cdot\tau_{r_{q}}^{p-r_{p}}\cdot\tau_{q}^{r_{q}+r_{p}-q}$
satisfy the following properties:

     * •

For $1\leq j\leq r_{q}$:
$\tau_{q}^{r_{q}}\cdot\tau_{p}^{r_{p}}(j)=\tau_{p+r_{q}-q}^{r_{q}+r_{p}-q}\cdot\tau_{r_{q}}^{p-r_{p}}\cdot\tau_{q}^{r_{q}+r_{p}-q}(j)$;

     * •

For $r_{q}+1\leq j\leq n$:
$\tau_{q}^{r_{q}}\cdot\tau_{p}^{r_{p}}(j)=-\tau_{p+r_{q}-q}^{r_{q}+r_{p}-q}\cdot\tau_{r_{q}}^{p-r_{p}}\cdot\tau_{q}^{r_{q}+r_{p}-q}(j)$.

Then, we get the result:

$\tau_{q}^{r_{q}}\cdot\tau_{p}^{r_{p}}=\tau_{p+r_{q}-q}^{r_{q}+r_{p}-q}\cdot\tau_{r_{q}}^{p-r_{p}-r_{q}}\cdot\tau_{q}^{r_{q}+r_{p}-p-q},$

in case $q-r_{q}\leqslant r_{p}$ and $0<r_{p}<p$.

  2. 2.

The case $r_{p}=-p$.

The case is divided into the following two subcases, depending on the value of
$q-r_{q}$ compares to the values of $p$:

     * •

$q-r_{q}\geq p$;

     * •

$q-r_{q}<p$.

Now, we start with the first subcase:

$\bullet$ $q-r_{q}\geq p.$

Consider the permutation of $\tau_{q}^{r_{q}}\cdot\tau_{p}^{-p}$:

$\tau_{q}^{r_{q}}=[-(q-r_{q}+1);~{}-(q-r_{q});\ldots;~{}-q;~{}1;~{}2;\ldots;~{}q-r_{q}]$

$\tau_{p}^{-p}=[-1;~{}-2;\ldots;~{}-p;~{}p+1;~{}p+2;\ldots;~{}q].$

Therefore the permutation presentation of
$\tau_{q}^{r_{q}}\cdot\tau_{p}^{-p}$:

$1\rightarrow-(q-r_{q}+1)\rightarrow-(q-r_{q}+1).$
$r_{q}\rightarrow-q\rightarrow-q.$ $r_{q}+1\rightarrow 1\rightarrow-1.$
$r_{q}+p\rightarrow p\rightarrow-p.$

Now, consider the permutation presentation of
$\tau_{r_{q}}^{-r_{q}}\cdot\tau_{r_{q}+p}^{-(r_{q}+p)}$:

$\tau_{r_{q}}^{-r_{q}}\cdot\tau_{r_{q}+p}^{-(r_{q}+p)}=[1;~{}2;\ldots;~{}r_{q};~{}-(r_{q}+1);\ldots-(r_{q}+p);~{}r_{q}+p+1;\ldots;~{}q].$

Hence we get

$\tau_{q}^{r_{q}}\cdot\tau_{p}^{-p}=\tau_{r_{q}}^{-r_{q}}\cdot\tau_{r_{q}+p}^{-(r_{q}+p)}\cdot\tau_{q}^{r_{q}},$

in case $q-r_{q}\geq p$ and $r_{p}=-p$.

$\bullet$ Now we turn to the subcase $q-r_{q}<p.$

Since $q-r_{q}<p$ is the permutation presentation of
$\tau_{q}^{r_{q}}\cdot\tau_{p}^{-p}$ as follow:

$1\rightarrow-(q-r_{q}+1)\rightarrow q-r_{q}+1.$
$p+r_{q}-q\rightarrow-p\rightarrow p.$
$p+r_{q}-q+1\rightarrow-(p+1)\rightarrow-(p+1).$
$r_{q}\rightarrow-q\rightarrow-q.$ $r_{q}+1\rightarrow 1\rightarrow-1.$
$q\rightarrow q-r_{q}\rightarrow-(q-r_{q}).$

Now, considering the permutation presentation of
$\tau_{r_{q}+p-q}^{-(r_{q}+p-q)}\cdot\tau_{r_{q}}^{-r_{q}}\cdot\tau_{q}^{-q}$:

$\tau_{r_{q}+p-q}^{-(r_{q}+p-q)}\cdot\tau_{r_{q}}^{-r_{q}}\cdot\tau_{q}^{-q}=[-1;~{}-2;\ldots;~{}-(r_{q}+p-q);~{}r_{q}+p-q+1;\ldots~{}r_{q};~{}-(r_{q}+1);\ldots;~{}-q].$

Hence we get

$\tau_{q}^{r_{q}}\cdot\tau_{p}^{-p}=\tau_{r_{q}+p}^{-(r_{q}+p)}\cdot\tau_{r_{q}}^{-r_{q}}\cdot\tau_{q}^{r_{q}-q},$

in case $q-r_{q}<p$ and $r_{p}=-p$.

  3. 3.

The case $-p<r_{p}<0$

In this case, we have the following three subcases, depending on how the value
of $q-r_{q}$ compares to the values of $p$ and $p+r_{p}$:

     * •

$q-r_{q}\geq p$;

     * •

$r_{p}+p\leq q-r_{q}\leq p$;

     * •

$q-r_{q}\leq r_{p}+p$.

Now, we start with the first subcase:

$\bullet$ $q-r_{q}\geq p.$

$\tau_{q}^{r_{q}}\cdot\tau_{p}^{r_{p}}=\tau_{q}^{r_{q}}\cdot\tau_{p}^{p+r_{p}}\cdot\tau_{p}^{-p}$

Consider the exchange law for $\tau_{q}^{r_{q}}\cdot\tau_{p}^{p+r_{p}}$ where
$0<p+r_{p}<p$ and $q-r_{q}\geq p$.

$\tau_{q}^{r_{q}}\cdot\tau_{p}^{p+r_{p}}=\tau_{r_{q}}^{-r_{q}}\cdot\tau_{r_{q}+r_{p}+p}^{r_{q}}\cdot\tau_{p+r_{q}}^{p+r_{p}}\cdot\tau_{q}^{r_{q}}.$

Then we get:

$\tau_{q}^{r_{q}}\cdot\tau_{p}^{p+r_{p}}\cdot\tau_{p}^{-p}=\tau_{r_{q}}^{-r_{q}}\cdot\tau_{r_{q}+r_{p}+p}^{r_{q}}\cdot\tau_{p+r_{q}}^{p+r_{p}}\cdot\tau_{q}^{r_{q}}\cdot\tau_{p}^{-p}$

Then, by using the exchange law for $\tau_{q}^{r_{q}}\cdot\tau_{p}^{-p}$ where
$q-r_{q}\geq p$ we get:

$\tau_{q}^{r_{q}}\cdot\tau_{p}^{r_{p}}=\tau_{r_{q}}^{-r_{q}}\cdot\tau_{r_{q}+r_{p}+p}^{r_{q}}\cdot\tau_{p+r_{q}}^{p+r_{p}}\cdot\tau_{r_{q}}^{-r_{q}}\cdot\tau_{p+r_{q}}^{-p-r_{q}}\cdot\tau_{q}^{r_{q}}.$

$=\tau_{r_{q}}^{-r_{q}}\cdot\tau_{r_{q}+r_{p}+p}^{r_{q}}\cdot\tau_{p+r_{p}}^{-(p+r_{p})}\cdot\tau_{p+r_{q}+r_{p}}^{-(p+r_{q}+r_{p})}\cdot\tau_{p+r_{q}}^{p+r_{p}}\cdot\tau_{p+r_{q}}^{-(p+r_{q})}\cdot\tau_{q}^{r_{q}}.$

$=\tau_{r_{q}}^{-r_{q}}\cdot\tau_{r_{q}}^{-r_{q}}\cdot\tau_{p+r_{p}+r_{q}}^{-(p+r_{p}+r_{q})}\cdot\tau_{p+r_{p}+r_{q}}^{r_{q}}\cdot\tau_{p+r_{q}+r_{p}}^{-(p+r_{q}+r_{p})}\cdot\tau_{p+r_{q}}^{p+r_{p}}\cdot\tau_{p+r_{q}}^{-(p+r_{q})}\cdot\tau_{q}^{r_{q}}.$

Then, we get the result:

$\tau_{q}^{r_{q}}\cdot\tau_{p}^{r_{p}}=\tau_{r_{q}+r_{p}+p}^{r_{q}}\cdot\tau_{p+r_{q}}^{r_{p}-r_{q}}\cdot\tau_{q}^{r_{q}},$

in case $q-r_{q}\geq p$ and $-p<r_{p}<0$.

Now, we turn to the second subcase of the case $-p<r_{p}<0$.

$\bullet$ $r_{p}+p\leqslant q-r_{q}\leqslant p.$

$\tau_{q}^{r_{q}}=\tau_{q}^{r_{q}}\cdot\tau_{p}^{p+r_{p}}\cdot\tau_{p}^{-p}.$

$\tau_{q}^{r_{q}}\cdot\tau_{p}^{p+r_{p}}\cdot\tau_{p}^{-p}=\tau_{r_{q}}^{p-q}\cdot\tau_{r_{q}+p+r_{p}}^{q-p}\cdot\tau_{q}^{r_{q}+p+r_{p}}\cdot\tau_{p}^{-p}.$

Consider the exchange law for $\tau_{q}^{r_{q}+p+r_{p}}\cdot\tau_{p}^{-p}.$
where $q-r_{q}<p$ .

$=\tau_{r_{q}}^{p-q}\cdot\tau_{r_{q}+p+r_{p}}^{q-p}\cdot\tau_{p+r_{q}+p+r_{p}-q}^{-p-r_{q}-(p+r_{p}+q)}\cdot\tau_{r_{q}+p+r_{p}}^{-(r_{q}+p+r_{p})}\cdot\tau_{q}^{r_{q}+p+r_{p}-q}$

Consider the exchange law for
$\tau_{r_{q}+p+r_{p}}^{q-p}\cdot\tau_{p+r_{q}+p+r_{p}-q}^{-p-r_{q}-(p+r_{p}+q)}$
where $q-r_{q}=p$ .

$=\tau_{r_{q}}^{p-q}\cdot\tau_{q-p}^{-(q-p)}\cdot\tau_{r_{q}+p+r_{p}-q}^{-(r_{q}+p+r_{p})}\cdot\tau_{r_{q}+p+r_{p}}^{q-p}\cdot\tau_{r_{q}+p+r_{p}}^{-(r_{q}+p+r_{p})}\cdot\tau_{q}^{r_{q}+p+r_{p}-q}.$

$\tau_{r_{q}}^{p-q}\cdot\tau_{q-p}^{-(q-p)}=\tau_{r_{q}}^{r_{q}-(q-p)}\cdot\tau_{r_{q}}^{-r_{q}}\cdot\tau_{q-p}^{-(q-p)}$

Consider the exchange law for
$\tau_{r_{q}}^{r_{q}-(q-p)}\cdot\tau_{q-p}^{-(q-p)}$, where $q-r_{q}=p$.

$=\tau_{r_{q}-(q-p)}^{-(r_{q}-(q-p))}\cdot\tau_{r_{q}}^{-r_{q}}\cdot\tau_{r_{q}}^{r_{q}-(q-p)}\cdot\tau_{r_{q}}^{-r_{q}}\cdot\tau_{r_{q}+p+r_{p}}^{q-p}\cdot\tau_{q}^{r_{q}+p+r_{p}-q}.$

Then, we get the result:

$\tau_{q}^{r_{q}}\cdot\tau_{p}^{r_{p}}=\tau_{r_{q}-q+p}^{-r_{q}+q-p}\cdot\tau_{r_{q}}^{r_{q}+p-q}\cdot\tau_{r_{q}+p+r_{p}}^{q-p}\cdot\tau_{q}^{r_{q}+p+r_{p}-q},$

in case $r_{p}+p\leqslant q-r_{q}\leqslant p$ and $-p<r_{p}<0$.

Now, we turn to the third subcase of the case $0<r_{p}<p$.

$\bullet$ $q-r_{q}\leqslant r_{p}+p.$

$\tau_{q}^{r_{q}}\cdot\tau_{p}^{r_{p}}=\tau_{q}^{r_{q}}\cdot\tau_{p}^{p+r_{p}}\cdot\tau_{p}^{-p}$

By the exchange law of the case $0\leqslant r_{p}\leqslant p$ where
$q-r_{q}\leqslant r_{p}$ we get:

$\tau_{q}^{r_{q}}\cdot\tau_{p}^{p+r_{p}}\cdot\tau_{p}^{-p}=\tau_{p+r_{q}-q}^{r_{q}+p+r_{p}-q}\cdot\tau_{r_{q}}^{p-(p+r_{p})-r_{q}}\cdot\tau_{q}^{r_{q}+p+r_{p}-p-q}\cdot\tau_{p}^{-p}.$

$=\tau_{p+r_{q}-q}^{r_{q}+p+r_{p}-q}\cdot\tau_{r_{q}}^{-r_{p}-r_{q}}\cdot\tau_{q}^{r_{q}+r_{p}-q}\cdot\tau_{p}^{-p}.$

$\tau_{q}^{r_{q}+r_{p}-q}\cdot\tau_{p}^{-p}=\tau_{q}^{r_{q}+r_{p}}\cdot\tau_{q}^{-q}\cdot\tau_{p}^{-p}.$

Consider the exchange law for $\tau_{q}^{r_{q}+r_{p}}\cdot\tau_{p}^{-p}$
where, $q-r_{q}<p$

$=\tau_{p+r_{q}-q}^{r_{q}+p+r_{p}-q}\cdot\tau_{r_{q}}^{-r_{p}-r_{q}}\cdot\tau_{p+r_{q}+r_{p}-q}^{-(p+r_{q}+r_{p}-q)}\cdot\tau_{r_{q}+r_{p}}^{-(r_{q}+r_{p})}\cdot\tau_{q}^{r_{q}+r_{p}}\cdot\tau_{q}^{-q}\cdot\tau_{q}^{-q}$

$\tau_{r_{q}}^{-r_{p}-r_{q}}\cdot\tau_{p+r_{q}+r_{p}-q}^{-(p+r_{q}+r_{p}-q)}=\tau_{r_{q}}^{-r_{p}}\cdot\tau_{r_{q}}^{-r_{q}}\cdot\tau_{p+r_{q}+r_{p}-q}^{-(p+r_{q}+r_{p}-q)}.$

Consider the exchange law for
$\tau_{r_{q}}^{-r_{p}}\cdot\tau_{p+r_{q}+r_{p}-q}^{-(p+r_{q}+r_{p}-q)}$ where
$q-r_{q}>p$

$=\tau_{p+r_{q}-q}^{r_{q}+p+r_{p}-q}\cdot\tau_{-r_{p}}^{r_{p}}\cdot\tau_{p+r_{q}+r_{p}-q-r_{p}}^{-(p+r_{q}-q)}\cdot\tau_{r_{q}}^{-r_{p}}\cdot\tau_{r_{q}}^{-r_{q}}\cdot\tau_{r_{q}+r_{p}}^{-(r_{q}+r_{p})}\cdot\tau_{q}^{r_{q}+r_{p}}.$

$=\tau_{p+r_{q}-q}^{r_{q}+p+r_{p}-q}\cdot\tau_{-r_{p}}^{r_{p}}\cdot\tau_{p+r_{q}-q}^{-(p+r_{q}-q)}\cdot\tau_{r_{q}}^{-r_{p}}\cdot\tau_{r_{q}}^{-r_{q}}\cdot\tau_{r_{q}+r_{p}}^{-(r_{q}+r_{p})}\cdot\tau_{q}^{r_{q}+r_{p}}.$

Consider the exchange law
$\tau_{r_{q}}^{-r_{p}}\cdot\tau_{r_{q}+r_{p}}^{-(r_{q}+r_{p})}$ where
$q-r_{q}=p$

$=\tau_{p+r_{q}-q}^{r_{q}+p+r_{p}-q}\cdot\tau_{-r_{p}}^{r_{p}}\cdot\tau_{p+r_{q}-q}^{-(p+r_{q}-q)}\cdot\tau_{-r_{p}}^{r_{p}}\cdot\tau_{r_{q}+r_{p}-r_{p}}^{-r_{q}}\cdot\tau_{r_{q}}^{-r_{p}}\cdot\tau_{r_{q}}^{-r_{q}}\cdot\tau_{q}^{r_{q}+r_{p}}.$

Consider the exchange law for
$\tau_{p+r_{q}-q}^{r_{q}+p+r_{p}-q}\cdot\tau_{-r_{p}}^{r_{p}}$ where
$q-r_{q}=p$

$=\tau_{r_{q}+p+r_{p}-q}^{-(r_{q}+p+r_{p}-q)}\cdot\tau_{-r_{p}+r_{q}+p+r_{p}-q}^{-(r_{q}+p-q)}\cdot\tau_{p+r_{q}-q}^{r_{q}+p+r_{p}-q}\cdot\tau_{p+r_{q}-q}^{-(p+r_{q}-q)}\cdot\tau_{-r_{p}}^{r_{p}}\cdot\tau_{r_{q}}^{-r_{q}}\cdot\tau_{r_{q}}^{-r_{p}}\cdot\tau_{r_{q}}^{-r_{q}}\cdot\tau_{q}^{r_{q}+r_{p}}.$

$=\tau_{r_{q}+p+r_{p}-q}^{-(r_{q}+p+r_{p}-q)}\cdot\tau_{p+r_{q}-q}^{r_{q}+p+r_{p}-q}\cdot\tau_{-r_{p}}^{r_{p}}\cdot\tau_{r_{q}}^{-r_{p}}\cdot\tau_{q}^{r_{q}+r_{p}}.$

Consider the exchange law for
$\tau_{p+r_{q}-q}^{r_{q}+p+r_{p}-q}\cdot\tau_{-r_{p}}^{r_{p}}$ where
$q-r_{q}=p.$

$=\tau_{r_{q}+p+r_{p}-q}^{-(r_{q}+p+r_{p}-q)}\cdot\tau_{r_{q}+p+r_{p}-q}^{-(r_{q}+p+r_{p}-q)}\cdot\tau_{-r_{p}+r_{q}+r_{p}+p-q}^{-(r_{q}+p-q)}\cdot\tau_{p+r_{q}-q}^{r_{q}+p+r_{p}-q}\cdot\tau_{r_{q}}^{-r_{p}}\cdot\tau_{q}^{r_{q}+r_{p}}.$

$=\tau_{r_{q}+p-q}^{-(r_{q}+p-q)}\cdot\tau_{p+r_{q}-q}^{r_{q}+p+r_{p}-q}\cdot\tau_{r_{q}}^{-r_{p}}\cdot\tau_{q}^{r_{q}+r_{p}}.$

Then, we get the result:

$\tau_{q}^{r_{q}}\cdot\tau_{p}^{r_{p}}=\tau_{r_{q}+p-q}^{r_{p}}\cdot\tau_{r_{q}}^{-r_{p}}\cdot\tau_{q}^{r_{q}+r_{p}},$

in case $q-r_{q}\leqslant r_{p}+p$ and $-p<r_{p}<0$.

∎

##  3 The presentation of the parabolic subgroup $\dot{S}_{n}$ of $B_{n}$ by
the generalized standard OGS

This section deals with the properties of the parabolic subgroup of $B_{n}$
which does not contain $s_{0}$ in its reduced Coxeter presentation.

######  Definition 3.0.1.

Denote by $\dot{S}_{n}$ the parabolic subgroup of $B_{n}$ which is generated
by $s_{1},s_{2},\ldots,s_{n-1}$ (i.e, The elements of $B_{n}$ which can be
written without any occurrence of $s_{0}$).

######  Remark 3.0.2.

By the Coxeter diagram of $B_{n}$ one can easily see that the subgroup
$\dot{S}_{n}$ which is defined in Definition 3.0.1 is isomorphic to the
symmetric group $S_{n}$.

Now, we show the generalized standard $OGS$ decomposition of the elements of
the subgroup $\dot{S}_{n}$ of $B_{n}$ by using the following two lemmas

######  Lemma 3.0.3.

Consider the elements of $B_{n}$ are presented by the generalized standard OGS
presentation, as it is described in Theorem 2.1.5, then the following holds:

$\bullet$ $\tau_{k}^{-1}\cdot\tau_{k+r}$ is the element
$\prod_{j=k}^{k+r-1}s_{j}$ of $\dot{S}_{n}$ for every $1\leqslant i\leqslant
k$.

###### Proof.

We start with the case of $r=1$:

$\tau_{k}=s_{0}\cdot s_{1}\cdots s_{k-1}$

Then,

$\tau_{k}^{-1}\cdot\tau_{k+1}=s_{k-1}\cdots s_{1}\cdot s_{0}\cdot s_{0}\cdot
s_{1}\cdots s_{k-1}\cdot s_{k}$ $\tau_{k}^{-1}\cdot\tau_{k+1}=s_{k}$

Now, we turn to the case: $\tau_{k}^{-1}\cdot\tau_{k+r}$.

The normal form of the element $\tau_{k}^{-1}\cdot\tau_{k+r}$ as follow:

$\tau_{k}^{-1}\cdot\tau_{k+r}=(\tau_{k}^{-1}\cdot\tau_{k+1})\cdot(\tau_{k+1}^{-1}\cdot\tau_{k+2})\cdots(\tau_{k+r-1}^{-1}\cdot\tau_{k+r})$

Then we conclude :

$\tau_{k}^{-1}\cdot\tau_{k+r}=s_{k}\cdot s_{k+1}\cdots s_{k+r-1}$

∎

######  Lemma 3.0.4.

Consider the elements of $B_{n}$ are presented by the generalized the standard
OGS presentation, as it is described in Theorem 2.1.5, then the following
holds:

$\bullet$ $\tau_{k_{1}}^{-i}\cdot\tau_{k_{2}}^{i}$ is the element
$\prod_{j_{1}=k_{1}}^{k_{2}-1}\prod_{j_{2}=0}^{j_{1}+i-1}s_{j_{1}-j_{2}}$ of
$\dot{S}_{n}$ for every $1\leqslant i\leqslant k$.

###### Proof.

The proof is by induction on the value of $i$.

The case of $i=1$ has been proved in Lemma 3.0.3

Assume by induction that the lemma holds for $i=i_{0}$, i.e.,

$\tau_{k_{1}}^{-i_{0}}\cdot\tau_{k_{2}}^{i_{0}}=\prod_{j_{1}=k_{1}}^{k_{2}-1}\prod_{j_{2}=0}^{j_{1}+i_{0}-1}s_{j_{1}-j_{2}}.$

Now we prove the lemma for $i=i_{0}+1$.

First, notice:

$\tau_{k}^{-(i_{0}+1)}\cdot\tau_{k+1}^{i_{0}+1}=\tau_{k}^{-1}\cdot(\tau_{k}^{-i_{0}}\cdot\tau_{k+1}^{i_{0}})\cdot\tau_{k+1}$

Then by using the induction assumption and using the property $s_{p}\cdot
s_{q}=s_{q}\cdot s_{p}$ for $|p-q|\geq 2$ we have,

$\tau_{k}^{-1}\cdot(\tau_{k}^{-i_{0}}\cdot\tau_{k+1}^{i_{0}})\cdot\tau_{k+1}=(s_{k-1}\cdots
s_{1}\cdot s_{0})\cdot(s_{k}\cdot s_{k-1}\cdots
s_{k-(i_{0}-1)})\cdot(s_{0}\cdot s_{1}\cdots s_{k-1}\cdot s_{k}).$
$=(s_{k-1}\cdots s_{k-(i_{0}-2)}\cdot s_{k-(i_{0}-1)}\cdot
s_{k-i_{0}})\cdot(s_{k-(i_{0}+1)}\cdots s_{0})\cdot(s_{k}\cdot s_{k-1}\cdots$
$s_{k-(i_{0}-1)})\cdot(s_{0}\cdot s_{1}\cdots
s_{k-(i_{0}+1)})\cdot(s_{k-i_{0}}\cdot s_{k-(i_{0}-1)}\cdots s_{k}).$
$=(s_{k-1}\cdots s_{k-(i_{0}-2)}\cdot s_{k-(i_{0}-1)}\cdot
s_{k-i_{0}})\cdot(s_{k}\cdot s_{k-1}\cdots s_{k-(i_{0}-1)}\cdot
s_{k-i_{0}}\cdot s_{k-(i_{0}-1)}\cdot s_{k-(i_{0}-1)+1}\cdots s_{k}).$

Since

$s_{k}\cdot s_{k-1}\cdots s_{k-(i_{0}-1)}\cdot s_{k-i_{0}}\cdot
s_{k-(i_{0}-1)}\cdots s_{k}=s_{k-i_{0}}\cdot s_{k-(i_{0}-1)}\cdots s_{k}\cdot
s_{k-1}\cdots s_{k-(i_{0}-1)}\cdot s_{k-i_{0}}$

We get:

$\tau_{k}^{-1}\cdot(\tau_{k}^{-i_{0}}\cdot\tau_{k+1}^{i_{0}})\cdot\tau_{k+1}=s_{k-1}\cdot
s_{k-2}\cdots s_{k-i_{0}}\cdot s_{k-i_{0}}\cdot s_{k-(i_{0}-1)}\cdots
s_{k-1}\cdot s_{k}\cdot s_{k-1}\cdots s_{k-i_{0}}.$ $=s_{k}\cdot s_{k-1}\cdots
s_{k-i_{0}}.$

Hence, the lemma holds for $i=i_{0}+1$, and therefore it holds for every value
of $i$. ∎

######  Theorem 3.0.5.

The presentation of every element $\pi\in\dot{S}_{n}$ by the generalized
standard $OGS$ (as it is described in Theorem 2.1.5) has the following form:

  1. 1.

$\tau_{k_{1}}^{i_{k_{1}}}\cdot\tau_{k_{2}}^{i_{k_{2}}}\cdots\tau_{k_{m}}^{i_{k_{m}}}$
where $-k_{j}\leqslant i_{k_{j}}\leqslant k_{j}-1$, for every $1\leqslant
j\leqslant m$, $i_{k_{1}}<0$.

  2. 2.

$\sum_{j=1}^{m}i_{k_{j}}=0$.

  3. 3.

$0\leqslant\sum_{j=r}^{m}i_{k_{j}}\leqslant k_{r-1}$ for $2\leqslant
r\leqslant m$.

Where, 3. is equivalent to

$-k_{r}\leqslant\sum_{j=1}^{r}i_{k_{j}}\leqslant 0$ for $1\leqslant r\leqslant
m-1$.

###### Proof.

Consider $\pi\in\dot{S}_{n}$. Then the presentation of $\pi$ by the normal
form as it is defined in [5], Chapter 3.4, is as follow:

$\pi=(s_{r_{1}}\cdot s_{r_{1}-1}\cdots s_{r_{1}-v_{1}})\cdot(s_{r_{2}}\cdot
s_{r_{2}-1}\cdots s_{r_{2}-v_{2}})\cdots(s_{r_{z}}\cdot s_{r_{z}-1}\cdots
s_{r_{z}-v_{z}}),$

where $z$ is the positive integer, $r_{1}<r_{2}<\cdots<r_{z}$, and $0\leqslant
v_{j}\leqslant r_{j}-1$, for every $1\leqslant j\leqslant z$. By Lemma 3.0.4
we have,

$s_{r_{j}}\cdot s_{r_{j}-1}\cdots
s_{r_{j}-v_{j}}=\tau_{r_{j}}^{-(v_{j}+1)}\cdot\tau_{r_{j}+1}^{v_{j}+1}$

for every $1\leqslant j\leqslant z$. Therefore,

$\pi=\tau_{r_{1}}^{-(v_{1}+1)}\cdot\tau_{r_{1}+1}^{v_{1}+1}\cdot\tau_{r_{2}}^{-(v_{2}+1)}\cdot\tau_{r_{2}+1}^{v_{2}+1}\cdots\tau_{r_{z}}^{-(v_{z}+1)}\cdot\tau_{r_{z}+1}^{v_{z}+1}.$
(1)

Now, notice for the case where $r_{j}+1=r_{j+1}$ and $v_{j}=v_{j+1}$ for some
$1\leq j\leq z-1$. Then,

$\tau_{r_{j}+1}^{v_{j}+1}\cdot\tau_{r_{j+1}}^{-(v_{j+1}+1)}=\tau_{r_{j}+1}^{v_{j}+1}\cdot\tau_{r_{j}+1}^{-(v_{j}+1)}=1.$
(2)

Hence, we get that Equation 1 is equivalent to

$\pi=\tau_{r_{1}}^{-(v_{1}+1)}\cdots\tau_{r_{j}}^{-(v_{j}+1)}\cdot\tau_{r_{j+1}+1}^{v_{j+1}+1}\cdots\tau_{r_{z}+1}^{v_{z}+1}.$
(3)

Hence, by substituting the identity which arises from Equation 2 in Equation
1, for every $1\leq j\leq z-1$, such that $r_{j}+1=r_{j+1}$ and
$v_{j}=v_{j+1}$, Equation 1 becomes of the following form:

$\pi=\tau_{r_{1}}^{-(v_{1}+1)}\cdot\tau_{r_{q_{1}}+1}^{v_{1}+1}\cdot\tau_{r_{q_{1}+1}}^{-(v_{q_{1}}+1)}\cdot\tau_{r_{q_{2}}+1}^{v_{q_{1}}+1}\cdots\tau_{r_{q_{z^{\prime}}+1}}^{-(v_{q_{z^{\prime}}}+1)}\cdot\tau_{r_{z}+1}^{v_{q_{z^{\prime}}}+1}.$
(4)

such that the following holds:

  * •

$1\leq q_{1}$;

  * •

$q_{j}+1\leq q_{j+1}$ for every $1\leq j\leq z^{\prime}-1$;

  * •

$q_{z^{\prime}}=z$.

Now, consider the generalized standard $OGS$ presentation of $\pi$ as it is
described in Theorem 2.1.5.

$\pi=\tau_{k_{1}}^{i_{k_{1}}}\cdot\tau_{k_{2}}^{i_{k_{2}}}\cdots\tau_{k_{m}}^{i_{k_{m}}}$
(5)

such that

$-k_{j}\leqslant i_{k_{j}}\leqslant k_{j}-1$ , for every $1\leqslant
j\leqslant m$.

Then, by concerning the uniqueness of presentation of an element $\pi\in
B_{n}$ by the generalized standard $OGS$, and using the presentation of $\pi$
as it is presented in Equation 4, the following properties are satisfied.

  * •

$\sum_{j=1}^{m}i_{k_{j}}=-(v_{1}+1)+(v_{1}+1)+\sum_{j=1}^{z^{\prime}}(-(v_{q_{j}}+1)+(v_{q_{j}}+1))=0.$
(6)

Hence, part 2 of the theorem holds.

Now, we show part 3 of the theorem, where we use all notations which were used
from the beginning of the proof up to this point. We consider the generalized
standard $OGS$ presentation of $\pi$ with the notations of Equation 5. Then,
consider the following equation.

$-k_{r}\leqslant\sum_{j=1}^{r}i_{k_{j}}\leqslant 0\quad for\quad 1\leqslant
r\leqslant m.$ (7)

We prove Equation 7 by induction on $r$. Then, we show that Equation 7 is
equivalent to part 3 of the theorem.

  * •

Since $k_{1}=r_{1}$, we have $-k_{1}\leq i_{k_{1}}=-(v_{1}+1)<0$. Hence,
Equation 7 holds for $r=1$. Now, we show Equation 7 for $r=2$, and then we
prove it by induction for every $r\leq m$.

  * •

Now, we consider $k_{2}$ and $i_{k_{2}}$. Notice, $k_{2}=r_{q_{1}}+1$ and by
Equation 4, either

$r_{q_{1}}+1<r_{{q_{1}}+1}$

or

$r_{q_{1}}+1=r_{{q_{1}}+1}.$

First, assume

$r_{q_{1}}+1<r_{{q_{1}}+1}.$

Then, $i_{k_{2}}=v_{1}+1$, and then

$i_{k_{1}}+i_{k_{2}}-(v_{1}+1)+(v_{1}+1)=0.$ (8)

Now, assume

$r_{q_{1}}+1=r_{{q_{1}}+1}.$

Then,

$\tau_{r_{q_{1}}+1}^{v_{1}+1}\cdot\tau_{r_{q_{1}+1}}^{-(v_{q_{1}}+1)}=\tau_{r_{q_{1}}+1}^{v_{1}-v_{q_{1}}}.$

Hence, we get

$i_{k_{2}}={v_{1}-v_{q_{1}}}.$

Therefore,

$i_{k_{1}}+i_{k_{2}}=-(v_{1}+1)+(v_{1}-v_{q_{1}})=-(v_{q_{1}}+1)<0,$

and since

$-(v_{q_{1}}+1)\geq-\tau_{r_{q_{1}+1}}=-\tau_{r_{q_{1}}+1}=-k_{2},$

we get

$-k_{2}\leqslant i_{k_{1}}+i_{k_{2}}<0.$ (9)

Hence, by considering both Equations 8 and 9 we get

$-k_{2}\leqslant i_{k_{1}}+i_{k_{2}}\leqslant 0,$

for every $\pi\in\dot{S}_{n}.$

  * •

Now, assume by induction on $p$, for every $1\leq p\leq p_{0}$,

$-k_{p}\leqslant\sum_{j=1}^{p}i_{k_{j}}\leqslant 0.$

Consider $k_{{p_{0}}+1}$ and $i_{k_{{p_{0}}+1}}$. Notice, either
$k_{{p_{0}}+1}=r_{q_{y}}+1$ or $k_{{p_{0}}+1}=r_{q_{y}+1}$ for some $1\leq
y\leq z^{\prime}$.

Assume $k_{{p_{0}}+1}=r_{q_{y}}+1$. Then by Equation 4, either

$r_{q_{y}}+1<r_{{q_{y}}+1}$

or

$r_{q_{y}}+1=r_{{q_{y}}+1},$

for some $1\leq y\leq z^{\prime}$. First, assume

$r_{q_{y}}+1<r_{{q_{y}}+1}.$

Then, $i_{k_{p_{0}+1}}=v_{q_{y-1}}+1$, and then by using Equation 4,

$\sum_{j=1}^{p_{0}+1}i_{k_{j}}=-(v_{1}+1)+(v_{1}+1)+\sum_{j=1}^{y-1}(-(v_{q_{j}}+1)+(v_{q_{j}}+1))=0.$
(10)

Now, assume

$r_{q_{y}}+1=r_{{q_{y}}+1}.$

Then,

$\tau_{r_{q_{y}}+1}^{v_{q_{y-1}}+1}\cdot\tau_{r_{q_{y}+1}}^{-(v_{q_{y}}+1)}=\tau_{r_{q_{1}}+1}^{v_{q_{y-1}}-v_{q_{y}}}\quad\text{which
implies}\quad i_{k_{p_{0}+1}}=v_{q_{y-1}}-v_{q_{y}}$ (11)

By induction

$-k_{p_{0}}\leq\sum_{j=1}^{p_{0}}i_{k_{j}}\leq 0,$ (12)

By the uniqueness of the presentation by the generalized standard $OGS$ (as it
is described in Theorem 2.1.5), and by using Equation 4, the following holds:

$\sum_{j=1}^{p_{0}}i_{k_{j}}=-(v_{1}+1)+(v_{1}+1)+\sum_{j=1}^{y-2}(-(v_{q_{j}}+1)+(v_{q_{j}}+1))-(v_{q_{y-1}}+1)=-(v_{q_{y-1}}+1).$
(13)

If $k_{{p_{0}}+1}=r_{q_{y}}+1$, then $k_{p_{0}}=r_{q_{y-1}+1}$. Then, by
Equation 4,
$v_{q_{y-1}}+1\leq r_{q_{y-1}+1}$. Hence,

$v_{q_{y-1}}+1\leq k_{p_{0}}.$

Therefore,

$-k_{p_{0}+1}=-(r_{{q_{y}}+1})\leq-(v_{q_{y}}+1)=-(v_{q_{y-1}}+1)+(v_{q_{y-1}}-v_{q_{y}})\leq-
k_{p_{0}}+(v_{q_{y-1}}-v_{q_{y}}).$

By Equation 11, $i_{k_{p_{0}+1}}=v_{q_{y-1}}-v_{q_{y}}$, and then by using
Equation 12, we get

$-k_{p_{0}}+(v_{q_{y-1}}-v_{q_{y}})\leq\sum_{j=1}^{p_{0}}i_{k_{j}}+i_{k_{p_{0}+1}}=\sum_{j=1}^{p_{0}+1}i_{k_{j}}$

Then By using Equations 13, 11

$\sum_{j=1}^{p_{0}+1}i_{k_{j}}=\sum_{j=1}^{p_{0}}i_{k_{j}}+i_{k_{p_{0}+1}}-(v_{q_{y-1}}+1)+v_{q_{y-1}}-v_{q_{y}}=-1-v_{q_{y}}\leq
0.$

Now, assume $k_{{p_{0}}+1}=r_{q_{y}+1}$. Then by Equation 4, either

$r_{q_{y}}+1<r_{{q_{y}}+1}\quad\text{or}\quad r_{q_{y}}+1=r_{{q_{y}}+1},$

for some $1\leq y\leq z^{\prime}$. In case $r_{q_{y}}+1=r_{{q_{y}}+1}$, we
have already proved $-k_{p_{0}+1}\leq\sum_{j=1}^{p_{0}+1}i_{k_{j}}\leq 0.$
Hence, assume

$r_{q_{y}}+1<r_{{q_{y}}+1}.$

Then, $i_{k_{p_{0}+1}}=-(v_{q_{y}}+1)$, and then by using Equation 4,

$\sum_{j=1}^{p_{0}+1}i_{k_{j}}=-(v_{1}+1)+(v_{1}+1)+\sum_{j=1}^{y-1}(-(v_{q_{j}}+1)+(v_{q_{j}}+1))-(v_{q_{y}}+1)=-(v_{q_{y}}+1)=i_{k_{p_{0}+1}}.$

Therefore,

$-k_{p_{0}+1}\leq
i_{k_{p_{0}+1}}=\sum_{j=1}^{p_{0}+1}i_{k_{j}}=-(v_{q_{y}}+1)<0$

Hence, we get

$-k_{p}\leqslant\sum_{j=1}^{p}i_{k_{j}}\leqslant 0\quad for\quad 1\leqslant
p\leqslant m-1.$ (14)

  * •

By Equation 14, $-k_{p-1}\leqslant\sum_{j=1}^{p-1}i_{k_{j}}\leqslant 0$, and
by Equation 6, $\sum_{j=1}^{m}i_{k_{j}}=0$.

Therefore,

$0\leqslant\sum_{j=p}^{m}i_{k_{j}}=\sum_{j=1}^{m}i_{k_{j}}-\sum_{j=1}^{p-1}i_{k_{j}}\leqslant
k_{p-1}.$

Hence, part 3 of the theorem holds.

∎

######  Example 3.0.6.

Let

$\pi=\tau_{9}^{-8}\cdot\tau_{10}\cdot\tau_{11}^{-3}\cdot\tau_{13}^{10}=(\tau_{9}^{-8}\cdot\tau_{10}^{8})\cdot(\tau_{10}^{-7}\cdot\tau_{11}^{7})\cdot(\tau_{11}^{-10}\cdot\tau_{13}^{10})$

The permutation presentation of $\pi$

$[9;~{}-1;~{}-2;~{}-3;~{}-4;~{}-5;~{}-6;~{}-7;~{}-8;~{}10;~{}11;~{}12;~{}13]$
$\cdot[-10;~{}1;~{}2;~{}3;~{}4;~{}5;~{}6;~{}7;~{}8;~{}9;~{}11;~{}12;~{}13]$
$\cdot[4;~{}5;~{}6;~{}7;~{}8;~{}9;~{}10;~{}11;~{}-1;~{}-2;~{}-3;~{}12;~{}13]$
$\cdot[-4;~{}-5;~{}-6;~{}-7;~{}-8;~{}-9;~{}-10;~{}-11;~{}-12;~{}-13;~{}1;~{}2;~{}3]$
$=[1;~{}5;~{}7;~{}8;~{}9;~{}10;~{}11;~{}12;~{}13;~{}4;~{}6;~{}2;~{}3].$

The following holds:

  * •

$k_{1}=9,\quad k_{2}=10,\quad k_{3}=11,\quad k_{4}=13$;

  * •

$i_{k_{1}}=-8,\quad i_{k_{2}}=1,\quad i_{k_{3}}=-3,\quad i_{k_{4}}=10;$

  * •

$i_{k_{4}}=10>0,\quad i_{k_{3}}+i_{k_{4}}=7\geq 0,\quad
i_{k_{2}}+i_{k_{3}}+i_{k_{4}}=8\geq 0;$

  * •

$i_{k_{1}}+i_{k_{2}}+i_{k_{3}}+i_{k_{4}}=0$.

######  Lemma 3.0.7.

Let $\pi$ be an element of $\dot{S}_{n}$. By Theorem 3.0.5, the generalized
standard $OGS$ presentation of $\pi$ as an element of $B_{n}$ has the
following form

$\pi=\tau_{k_{1}}^{i_{k_{1}}}\cdot\tau_{k_{2}}^{i_{k_{2}}}\cdots\tau_{k_{m}}^{i_{k_{m}}}$

such that

  * •

$-k_{j}\leqslant i_{k_{j}}\leqslant k_{j}-1$, for every $1\leqslant j\leqslant
m$.

  * •

$\sum_{j=1}^{m}i_{k_{j}}=0$.

  * •

$0\leqslant\sum_{j=r}^{m}i_{k_{j}}\leqslant k_{r-1}$, for every $2\leq r\leq
m$.

Then, by considering $\pi$ as an element of $S_{n}$, the standard $OGS$
presentation of $\pi$ as it is presented in Theorem 1.0.3, has the following
form:

$t_{k_{1}}^{i_{k_{1}}^{\prime}}\cdot t_{k_{2}}^{i_{k_{2}}^{\prime}}\cdots
t_{k_{m}}^{i_{k_{m}}^{\prime}}$

where,

  * •

$i_{k_{j}}^{\prime}=i_{k_{j}}\quad\text{if}\quad 0\leqslant i_{k_{j}}<k_{j}.$

  * •

$i_{k_{j}}^{\prime}=k_{j}+i_{k_{j}}\quad\text{if}\quad-k_{j}\leqslant
i_{k_{j}}<0.$

###### Proof.

Consider the generalized standard $OGS$ presentation of the following element
$\pi\in\dot{S}_{n}$, as it is described in Theorem 3.0.5:

$\pi=\tau_{k_{1}}^{i_{k_{1}}}\cdot\tau_{k_{2}}^{i_{k_{2}}}\cdots\tau_{k_{m}}^{i_{k_{m}}}$

such that

  * •

$-k_{j}\leqslant i_{k_{j}}\leqslant k_{j}-1$, for every $1\leqslant j\leqslant
m$.

  * •

$\sum_{j=1}^{m}i_{k_{j}}=0$.

  * •

$0\leqslant\sum_{j=r}^{m}i_{k_{j}}\leqslant k_{r-1}$, for every $2\leq r\leq
m$.

Then

$\pi=(\tau_{k_{1}}^{-(\sum_{j=2}^{m}i_{k_{j}})}\cdot\tau_{k_{2}}^{(\sum_{j=2}^{m}i_{k_{j}})})\cdots(\tau_{k_{m-2}}^{-(i_{k_{m-1}}+i_{k_{m}})}\cdot\tau_{k_{m-1}}^{(i_{k_{m-1}}+i_{k_{m}})})\cdot(\tau_{k_{m-1}}^{-i_{k_{m}}}\cdot\tau_{k_{m}}^{i_{k_{m}}})$

Then , we conclude the following identity, where in left hand side of Equation
15 using Lemma 3.0.4 and in right hand side of Equation 15 using the standard
$OGS$ presentation of $\pi$ as it is described in Theorem 1.0.3 by considering
$\pi$ as an element of $S_{n}$.

$\tau_{k_{1}}^{-j}\cdot\tau_{k_{2}}^{j}=t_{k_{1}}^{k_{1}-j}\cdot
t_{k_{2}}^{j}$ (15)

Hence,

$\pi=(t_{k_{1}}^{k_{1}-(\sum_{j=2}^{m}i_{k_{j}})}\cdot
t_{k_{2}}^{\sum_{j=2}^{m}i_{k_{j}}})\cdots(t_{k_{m-2}}^{k_{m-2}-(\sum_{j=m-1}^{m}i_{k_{j}})}\cdot
t_{k_{m-1}}^{\sum_{j=m-1}^{m}i_{k_{j}}})\cdot(t_{k_{m-1}}^{k_{m-1}-i_{k_{m}}}\cdot
t_{k_{m}}^{i_{k_{m}}})=$

$=t_{k_{1}}^{k_{1}-(\sum_{j=2}^{m}i_{k_{j}})}\cdot
t_{k_{2}}^{k_{2}+i_{k_{2}}}\cdots t_{k_{m}}^{k_{m}+i_{k_{m}}}$

Since $t_{k_{j}}^{k_{j}}=1$ and $k_{1}\geq|i_{k_{1}}|=\sum_{j=2}^{m}i_{k_{j}}$
we get the following:

$t_{k_{1}}^{k_{1}-(\sum_{j=2}^{m}i_{k_{j}})}\cdot
t_{k_{2}}^{k_{2}+i_{k_{2}}}\cdots
t_{k_{m}}^{k_{m}+i_{k_{m}}}=t_{k_{1}}^{i^{\prime}_{k_{1}}}\cdots
t_{k_{m}}^{i^{\prime}_{k_{m}}}.$

∎

Theorem 3.0.8 shows the characterization of the standard $OGS$ elementary
elements of $S_{n}$ (defined in Definition 1.0.6) by the generalized standard
$OGS$ of $\dot{S}_{n}$ according to Theorem 3.0.5.

######  Theorem 3.0.8.

Let $\pi$ be an element of $\dot{S}_{n}$ in $B_{n}$. Then, by considering
$\pi$ as an element of $S_{n}$, $\pi$ is a standard $OGS$ elementary element
as it is defined in Definition 1.0.6 if and only if the generalized standard
$OGS$ presentation of $\pi$ as it is described in Theorem 2.1.5, has the
following form:

$\pi=\tau_{k_{1}}^{i_{k_{1}}}\cdot\tau_{k_{2}}^{i_{k_{2}}}\cdots\tau_{k_{m}}^{i_{k_{m}}},$

where,

$-k_{1}\leqslant i_{k_{1}}<0\quad,0<i_{k_{j}}<k_{j}\quad\text{for}\quad j\geq
2,\quad\sum_{j=1}^{m}i_{k_{j}}=0.$

###### Proof.

Consider:

$\pi=\tau_{k_{1}}^{i_{k_{1}}}\cdot\tau_{k_{2}}^{i_{k_{2}}}\cdots\tau_{k_{m}}^{i_{k_{m}}}$

and assume

$-k_{1}\leqslant i_{k_{1}}<0\quad,0<i_{k_{j}}<k_{j}\quad\text{for}\quad j\geq
2,\quad\sum_{j=1}^{m}i_{k_{j}}=0.$

First, we prove that $\pi\in\dot{S}_{n}$ and then by considering $\pi$ as an
element of $S_{n}$, we prove that $\pi$ is a standard $OGS$ elementary element
(as it is defined in Definition 1.0.6).

Consider $\sum_{j=r}^{m}i_{k_{j}}$, for $2\leq r\leq m$. Since $i_{k_{r}}>0$
for every $r\geq 2$, we have
$\sum_{j=r}^{m}i_{k_{j}}\leq\sum_{j=2}^{m}i_{k_{j}}$. Since
$\sum_{j=1}^{m}i_{k_{j}}=0$, we have $\sum_{j=2}^{m}i_{k_{j}}=-i_{k_{1}}$
Therefore, $-k_{1}\leq i_{k_{1}}<0$ implies that
$0<\sum_{j=2}^{m}i_{k_{j}}\leq k_{1}$. Hence, for every $2\leq r\leq m$:

$0<\sum_{j=r}^{m}i_{k_{j}}\leq\sum_{j=2}^{m}i_{k_{j}}\leq k_{1}<k_{r}.$

Hence, by Theorem 3.0.5, $\pi\in\dot{S}_{n}$. Now consider the presentation of
$\pi$ as an element of $S_{n}$ according to the standard $OGS$ presentation by
using Lemma 3.0.7.

$\pi=t_{k_{1}}^{i^{\prime}_{k_{1}}}\cdot t_{k_{2}}^{i^{\prime}_{k_{2}}}\cdots
t_{k_{m}}^{i^{\prime}_{k_{m}}}$

such that the following holds:

  * •

$i^{\prime}_{k_{r}}=i_{k_{r}}$, since $i_{k_{r}}>0$ for $2\leq r\leq m$.

  * •

$i^{\prime}_{k_{1}}=k_{1}+i_{k_{1}}$, since $i_{k_{1}}<0$.

  * •

$\sum_{j=1}^{m}i^{\prime}_{k_{j}}=k_{1}+\sum_{j=1}^{m}i_{k_{j}}=k_{1}$, since
$\sum_{j=1}^{m}i_{k_{j}}=0$.

Hence, concerning the value of $i_{k_{1}}$, $\pi$ has one of the two following
presentation as an element of $S_{n}$:

  * •

In the case $-k_{1}<i_{k_{1}}<0$,we have
$0<i^{\prime}_{k_{1}}=k_{1}+i_{k_{1}}<k_{1}$, and then the standard $OGS$
presentation (as it is presented in Theorem 1.0.3) of $\pi$ as an element of
$S_{n}$ as follow:

$\pi=t_{k_{1}}^{i^{\prime}_{k_{1}}}\cdot t_{k_{2}}^{i^{\prime}_{k_{2}}}\cdots
t_{k_{m}}^{i^{\prime}_{k_{m}}},$

such that $1\leq i^{\prime}_{k_{j}}<k_{j}$ for every $1\leq j\leq m$, and
$\sum_{j=1}^{m}i^{\prime}_{k_{j}}=k_{1}$. Hence, by Definition 1.0.6, the
element $t_{k_{1}}^{k_{1}+i_{k_{1}}}\cdot t_{k_{2}}^{i_{k_{2}}}\cdots
t_{k_{m}}^{i_{k_{m}}}$ is a standard $OGS$ elementary element;

  * •

In the case $i_{k_{1}}=-k_{1}$, we have
$i^{\prime}_{k_{1}}=k_{1}+i_{k_{1}}=0$, and then the standard $OGS$
presentation (as it is presented in Theorem 1.0.3) of $\pi$ as an element of
$S_{n}$ as follow:

$\pi=t_{k_{1}}^{k_{1}+i_{k_{1}}}\cdot t_{k_{2}}^{i_{k_{2}}}\cdots
t_{k_{m}}^{i_{k_{m}}}=t_{k_{2}}^{i_{k_{2}}}\cdots t_{k_{m}}^{i_{k_{m}}},$

such that
$\sum_{j=2}^{m}i_{k_{j}}=k_{1}+i_{k_{1}}+\sum_{j=2}^{m}i_{k_{j}}=k_{1}<k_{2}$.
Hence, by Definition 1.0.6 the element

$\pi=t_{k_{2}}^{i_{k_{2}}}\cdots t_{k_{m}}^{i_{k_{m}}}$

is a standard $OGS$ elementary element.

Now, we turn to the proof second direction of the theorem.

Assume $\pi\in\dot{S}_{n}$, and by considering $\pi$ as an element of $S_{n}$,
$\pi$ is a standard $OGS$ elementary element. Then, by Definition 1.0.6,

$\pi=t_{k_{1}}^{i_{k_{1}}}\cdot t_{k_{2}}^{i_{k_{2}}}\cdots
t_{k_{m}}^{i_{k_{m}}},$

where, $1\leq i_{k_{j}}<k_{j}$ and $\sum_{j=1}^{m}i_{k_{j}}\leq k_{1}$.Then,
by [9] Theorem 28

$norm(\pi)=\prod_{u=\rho_{1}}^{k_{1}-1}\prod_{r=0}^{\rho_{1}-1}s_{u-r}\cdot\prod_{u=k_{1}}^{k_{2}-1}\prod_{r=0}^{\rho_{2}-1}s_{u-r}\cdot\prod_{u=k_{2}}^{k_{3}-1}\prod_{r=0}^{\rho_{3}-1}s_{u-r}\cdots\prod_{u=k_{m-1}}^{k_{m}-1}\prod_{r=0}^{\rho_{m}-1}s_{u-r},$

where, $\rho_{j}=\sum_{x=j}^{m}i_{k_{x}}$ for $1\leq j\leq m$.

Now, consider $\pi$ presented in terms of generators $\tau_{k_{j}}$ (as it
defined in Definition 2.1.1), for $1\leq j\leq m$. By Lemma 3.0.4,

$\pi=(\tau_{maj(\pi)}^{-\rho_{1}}\cdot\tau_{k_{1}}^{\rho_{1}})\cdot(\tau_{k_{1}}^{-\rho_{2}}\cdot\tau_{k_{2}}^{\rho_{2}})\cdot(\tau_{k_{2}}^{-\rho_{3}}\cdot\tau_{k_{3}}^{\rho_{3}})\cdots(\tau_{k_{m-1}}^{-\rho_{m}}\cdot\tau_{k_{m}}^{\rho_{m}})$
$=\tau_{maj(\pi)}^{-\rho_{1}}\cdot\tau_{k_{1}}^{\rho_{1}-\rho_{2}}\cdot\tau_{k_{2}}^{\rho_{2}-\rho_{3}}\cdots\tau_{k_{m-1}}^{\rho_{m-1}-\rho_{m}}\cdot\tau_{k_{m}}^{\rho_{m}}.$

Since $\rho_{j}=\sum_{x=j}^{m}i_{k_{x}}$, the following holds:

  * •

$\rho_{1}=maj(\pi)$;

  * •

$\rho_{j}-\rho_{j+1}=i_{k_{j}}$ for $1\leq j\leq m-1$;

  * •

$\rho_{m}=i_{k_{m}}$.

Now, notice that $\pi$ is a standard $OGS$ elementary element, therefore by
Definition 1.0.6, $maj(\pi)\leq k_{1}$, i.e., either $maj(\pi)<k_{1}$ or
$maj(\pi)=k_{1}$.

Hence, in case $maj(\pi)<k_{1}$, the following holds:

$\pi=\tau_{maj(\pi)}^{-\rho_{1}}\cdot\tau_{k_{1}}^{\rho_{1}-\rho_{2}}\cdot\tau_{k_{2}}^{\rho_{2}-\rho_{3}}\cdots\tau_{k_{m-1}}^{\rho_{m-1}-\rho_{m}}\cdot\tau_{k_{m}}^{\rho_{m}}$
$=\tau_{maj(\pi)}^{-maj(\pi)}\cdot\tau_{k_{1}}^{i_{k_{1}}}\cdot\tau_{k_{2}}^{i_{k_{2}}}\cdots\tau_{k_{m-1}}^{i_{k_{m-1}}}\cdot\tau_{k_{m}}^{i_{k_{m}}}.$

such that $0<i_{k_{j}}<k_{j}$ for $1\leq j\leq m$, and $-maj(\pi)<0$. Hence,
the Theorem holds in case $maj(\pi)<k_{1}$.

Now, assume $maj(\pi)=k_{1}$, then

$\rho_{2}=\sum_{j=2}^{m}i_{k_{j}}=\sum_{j=1}^{m}i_{k_{j}}-i_{k_{1}}=maj(\pi)-i_{k_{1}}=k_{1}-i_{k_{1}},$

and then,

$\tau_{maj(\pi)}^{-\rho_{1}}\cdot\tau_{k_{1}}^{\rho_{1}-\rho_{2}}=\tau_{k_{1}}^{-\rho_{2}}=\tau_{k_{1}}^{i_{k_{1}}-k_{1}}$

Hence, the following holds:

$\pi=\tau_{maj(\pi)}^{-\rho_{1}}\cdot\tau_{k_{1}}^{\rho_{1}-\rho_{2}}\cdot\tau_{k_{2}}^{\rho_{2}-\rho_{3}}\cdots\tau_{k_{m-1}}^{\rho_{m-1}-\rho_{m}}\cdot\tau_{k_{m}}^{\rho_{m}}$
$=\tau_{k_{1}}^{i_{k_{1}}-k_{1}}\cdot\tau_{k_{2}}^{i_{k_{2}}}\cdots\tau_{k_{m-1}}^{i_{k_{m-1}}}\cdot\tau_{k_{m}}^{i_{k_{m}}}.$

such that $0<i_{k_{j}}<k_{j}$ for $2\leq j\leq m$, and $i_{k_{1}}-k_{1}<0$.
Hence, the Theorem holds in case $maj(\pi)=k_{1}$ as well.

∎

######  Example 3.0.9.

Let

$\pi=\tau_{6}^{-5}\cdot\tau_{8}^{2}\cdot\tau_{9}^{2}\cdot\tau_{10}$

Notice,

$i_{k_{1}}=-5<0,\quad i_{k_{2}}=2>0,\quad i_{k_{3}}=2>0,\quad i_{k_{4}}=1>0,$
$i_{k_{1}}+i_{k_{2}}+i_{k_{3}}+i_{k_{4}}=-5+2+2+1=0.$

Hence, the generalized standard $OGS$ presentation of $\pi$ satisfies the
conditions of Theorem 3.0.8. By Lemma 3.0.7, the presentation of $\pi$ in the
standard $OGS$ presentation, by considering $\pi$ as an element of $S_{n}$ as
follow:

$\pi=t_{6}^{6-5}\cdot t_{8}^{2}\cdot t_{9}^{2}\cdot t_{10}=t_{6}\cdot
t_{8}^{2}\cdot t_{9}^{2}\cdot t_{10}.$

Notice, $maj(\pi)=1+2+2+1=6=k_{1}$. Hence, by Defintion 1.0.6, $\pi$ is a
standard $OGS$ elementary element by considering $\pi$ as an elemnt of
$S_{n}$.

Now, consider the permutation presentation of $\pi$.

$\pi=[6;~{}-1;~{}-2;~{}-3;~{}-4;~{}-5;~{}7;~{}8;~{}9;~{}10]\cdot[-7;~{}-8;~{}1;~{}2;~{}3;~{}4;~{}5;~{}6;~{}9;~{}10]$
$\cdot[-8;~{}-9;~{}1;~{}2;~{}3;~{}4;~{}5;~{}6;~{}7;~{}10]\cdot[-10;~{}1;~{}2;~{}3;~{}4;~{}5;~{}6;~{}7;~{}8;~{}9]$
$=[1;~{}4;~{}5;~{}7;~{}8;~{}10;~{}2;~{}3;~{}6;~{}9].$

Notice, that the following holds:

$\pi(1)<\pi(2)<\pi(3)<\pi(4)<\pi(5)<\pi(6),\quad\pi(6)>\pi(7),$
$\pi(7)<\pi(8)<\pi(9)<\pi(10).$

Therefore, $des(\pi)=\\{6\\}$. By [9] Theorem 28, $\pi\in S_{n}$ is a standard
$OGS$ elementary element if and only if $des(\pi)$ contains a single element.
Hence, $\pi$ is a standard $OGS$ elementary element by considering $\pi$ as an
element of $S_{n}$.

Theorem 3.0.10 shows a connection between the generalized standard $OGS$ of an
element $\pi\in\dot{S}_{n}$ as it is presented in Theorem 3.0.5, and the
standard $OGS$ factorization of $\pi$ as it is defined in Definition 1.0.8 by
considering $\pi$ as an element of $S_{n}$.

######  Theorem 3.0.10.

Let $\pi$ be an element of $\dot{S}_{n}$, such that the presentation of $\pi$
by the generalized standard $OGS$ , as it is presented in Theorem 3.0.5, as
follow:

$\pi=\tau_{k_{1}}^{i_{k_{1}}}\cdot\tau_{k_{2}}^{i_{k_{2}}}\cdots\tau_{k_{m}}^{i_{k_{m}}}$

such that

  * •

$-k_{j}\leqslant i_{k_{j}}\leqslant k_{j}-1$, for every $1\leqslant j\leqslant
m$.

  * •

$\sum_{j=1}^{m}i_{k_{j}}=0$.

  * •

$0\leqslant\sum_{j=r}^{m}i_{k_{j}}\leqslant k_{r-1}$, for every $2\leq r\leq
m$.

Then, consider the standard $OGS$ elementary factorization of $\pi$ as it is
defined in Definition 1.0.8, where we consider $\pi$ as an element in $S_{n}$:

$\pi=\prod_{v=1}^{z(\pi)}\pi^{(v)}$

The value of $z(\pi)$ satisfies the following property:

$z(\pi)=|\\{1\leqslant j\leqslant n\quad|\quad i_{k_{j}}<0.\\}|$

###### Proof.

Let $\pi\in\dot{S}_{n}$ is presented by the generalized standard $OGS$
presentation as follow:

$\pi=\tau_{k_{1}}^{i_{k_{1}}}\cdot\tau_{k_{2}}^{i_{k_{2}}}\cdots\tau_{k_{m}}^{i_{k_{m}}}.$

Denote by $q$ the number of $j$ such that $1\leq j\leq m$ and $i_{k_{j}}<0$,
and by $r_{1},r_{2}\ldots,r_{q}$, all the indices of $k$, such that
$r_{1}<r_{2}<r_{3},...<r_{q}$, and $i_{k_{r_{j}}}<0$ for $1\leq j\leq q$.
Since, $\pi\in\dot{S}_{n}$, by Theorem 3.0.5, $i_{k_{1}}<0$. Hence, $r_{1}=1$.

Now, for every $1\leq\alpha\leq q$, let define $i^{\prime}_{\alpha}$ by the
following recursive algorithm:

  * •

Let define $i^{\prime}_{q}$ to be

$i^{\prime}_{q}=i_{k_{m}};$

  * •

For $\alpha=q-1$, let define $i^{\prime}_{q-1}$ to be

$i^{\prime}_{q-1}=i_{r_{q}}+\sum_{j=r_{q}+1}^{m-1}i_{k_{j}}+i^{\prime}_{q};$

  * •

Then for every $1\leq\alpha\leq q-2$, let define $i^{\prime}_{\alpha}$ to be

$i^{\prime}_{\alpha}=i_{r_{\alpha+1}}+\sum_{j=r_{\alpha+1}+1}^{r_{\alpha+2}-1}i_{k_{j}}+i^{\prime}_{\alpha+1}.$

Then, consider the following presentation of $\pi$

$\pi=\tau_{k_{1}}^{i_{k_{1}}}\cdot\tau_{k_{2}}^{i_{k_{2}}}\cdots\tau_{k_{m}}^{i_{k_{m}}}$
$=(\tau_{k_{r_{1}}}^{i_{k_{r_{1}}}}\cdot\tau_{k_{2}}^{i_{k_{2}}}\cdots\tau_{k_{r_{2}-1}}^{i_{k_{r_{2}-1}}}\cdot\tau_{k_{r_{2}}}^{i^{\prime}_{1}})\cdot(\tau_{k_{r_{2}}}^{-\sum_{j=r_{2}+1}^{r_{3}-1}i_{k_{j}}-i^{\prime}_{2}}\cdot\tau_{k_{r_{2}+1}}^{i_{k_{r_{2}+1}}}\cdots\tau_{k_{r_{3}-1}}^{i_{k_{r_{3}-1}}}\cdot\tau_{k_{r_{3}}}^{i^{\prime}_{2}})\cdots$
$\cdots(\tau_{k_{r_{q}}}^{-\sum_{j=r_{q}+1}^{m-1}i_{k_{j}}-i^{\prime}_{q}}\cdot\tau_{k_{r_{q}+1}}^{i_{k_{r_{q}+1}}}\cdots\tau_{k_{m-1}}^{i_{k_{m-1}}}\cdot\tau_{k_{m}}^{i^{\prime}_{q}}).$

Since $\pi\in\dot{S}_{n}$, by Theorem 3.0.5, $\sum_{j=1}^{m}i_{k_{j}}=0$.
Therefore,

$i_{k_{1}}=i_{k_{r_{1}}}=-\sum_{j=2}^{r_{2}-1}i_{k_{j}}-i^{\prime}_{1}.$

Hence, $\pi$ can be presented as follow:

$\pi=(\tau_{k_{r_{1}}}^{-\sum_{j=2}^{r_{2}-1}i_{k_{j}}-i^{\prime}_{1}}\cdot\tau_{k_{2}}^{i_{k_{2}}}\cdots\tau_{k_{r_{2}-1}}^{i_{k_{r_{2}-1}}}\cdot\tau_{k_{r_{2}}}^{i^{\prime}_{1}})\cdot(\tau_{k_{r_{2}}}^{-\sum_{j=r_{2}+1}^{r_{3}-1}i_{k_{j}}-i^{\prime}_{2}}\cdot\tau_{k_{r_{2}+1}}^{i_{k_{r_{2}+1}}}\cdots\tau_{k_{r_{3}-1}}^{i_{k_{r_{3}-1}}}\cdot\tau_{k_{r_{3}}}^{i^{\prime}_{2}})\cdots$
$\cdots(\tau_{k_{r_{q}}}^{-\sum_{j=r_{q}+1}^{m-1}i_{k_{j}}-i^{\prime}_{q}}\cdot\tau_{k_{r_{q}+1}}^{i_{k_{r_{q}+1}}}\cdots\tau_{k_{m-1}}^{i_{k_{m-1}}}\tau_{k_{m}}^{i^{\prime}_{q}}).$

Now, notice, the following properties holds:

  * •

$i_{k_{j}}>0$ for every $1\leq j\leq m$ such that $j\neq r_{p}$ for $1\leq
p\leq q$.

  * •

Notice, $i^{\prime}_{q}=i_{k_{m}}$. Hence, by Theorem 3.0.5,
$i^{\prime}_{q}>0$.

  * •

Since
$\sum_{j=r_{q}+1}^{m-1}i_{k_{j}}+i^{\prime}_{q}=\sum_{j=r_{q}+1}^{m}i_{k_{j}}$,
by Theorem 3.0.5, $\sum_{j=r_{q}+1}^{m-1}i_{k_{j}}+i^{\prime}_{q}\leq
k_{r_{q}}$. Hence,
$-k_{r_{q}}\leq-\sum_{j=r_{q}+1}^{m-1}i_{k_{j}}-i^{\prime}_{q}<0$.

  * •

Notice,
$i^{\prime}_{q-1}=i_{r_{q}}+\sum_{j=r_{q}+1}^{m-1}i_{k_{j}}+i^{\prime}_{q}=\sum_{j=r_{q}}^{m}i_{k_{j}}$.
By Theorem 3.0.5, $\sum_{j=r_{q}}^{m}i_{k_{j}}\geq 0$. Hence,
$i^{\prime}_{q-1}\geq 0$.

  * •

Now, by considering
$\sum_{j=r_{\alpha+1}+1}^{r_{\alpha+2}-1}i_{k_{j}}+i^{\prime}_{\alpha+1}$ and
$i^{\prime}_{\alpha}$ for $1\leq\alpha\leq q-2$, the following properties
holds (which can be proved by the same argument as it have been proved for
$\alpha=q-1$):

    * –

$-k_{r_{\alpha+1}}\leq-\sum_{j=r_{\alpha+1}+1}^{r_{\alpha+2}-1}i_{k_{j}}-i^{\prime}_{\alpha+1}<0$;

    * –

$i^{\prime}_{\alpha}\geq 0$.

Hence, by Theorem 3.0.8, every component $\pi_{\alpha}$ of $\pi$ (where
$1\leq\alpha\leq q$) of the following form

$\pi_{\alpha}=\tau_{k_{r_{\alpha}}}^{-\sum_{j=r_{\alpha}+1}^{r_{\alpha+1}-1}i_{k_{j}}-i^{\prime}_{\alpha}}\cdot\tau_{k_{r_{\alpha}+1}}^{i_{k_{r_{\alpha}+1}}}\cdots\tau_{k_{r_{\alpha+1}-1}}^{i_{k_{r_{\alpha+1}-1}}}\cdot\tau_{k_{r_{\alpha+1}}}^{i^{\prime}_{\alpha}}$

is a standard $OGS$ elementary element by considering $\pi_{\alpha}$ as an
element of $S_{n}$.

Notice the following observations:

  * •

Since $i^{\prime}_{\alpha}\geq 0$ for $1\leq\alpha\leq q-1$, we have that
either $i^{\prime}_{\alpha}>0$ or $i^{\prime}_{\alpha}=0$ for $1\leq\alpha\leq
q-1$. ( $i^{\prime}_{q}=i_{k_{m}}>0$ for every $\pi\in\dot{S}_{n}$).

  * •

Since $\sum_{j=r_{\alpha}+1}^{r_{\alpha+1}-1}i_{k_{j}}+i^{\prime}_{\alpha}\leq
k_{r_{\alpha}}$, we have that either
$k_{r_{\alpha}}-\sum_{j=r_{\alpha}+1}^{r_{\alpha+1}-1}i_{k_{j}}-i^{\prime}_{\alpha}>0$
or
$k_{r_{\alpha}}-\sum_{j=r_{\alpha}+1}^{r_{\alpha+1}-1}i_{k_{j}}-i^{\prime}_{\alpha}=0$
for $1\leq\alpha\leq q$.

Then, by Lemma 3.0.7, the presentation of the component $\pi_{\alpha}$ of
$\pi$ by the standard $OGS$ of $S_{n}$ is one of the four following form:

  * •

$\pi_{\alpha}=t_{k_{r_{\alpha}}}^{k_{r_{\alpha}}-\sum_{j=r_{\alpha}+1}^{r_{\alpha+1}-1}i_{k_{j}}-i^{\prime}_{\alpha}}\cdot
t_{k_{r_{\alpha}+1}}^{i_{k_{r_{\alpha}+1}}}\cdots
t_{k_{r_{\alpha+1}-1}}^{i_{k_{r_{\alpha+1}-1}}}\cdot
t_{k_{r_{\alpha+1}}}^{i^{\prime}_{\alpha}},$

in case $i^{\prime}_{\alpha}>0$ and
$\sum_{j=r_{\alpha}+1}^{r_{\alpha+1}-1}i_{k_{j}}+i^{\prime}_{\alpha}<k_{r_{\alpha}}$
(i.e.,
$k_{r_{\alpha}}-\sum_{j=r_{\alpha}+1}^{r_{\alpha+1}-1}i_{k_{j}}-i^{\prime}_{\alpha}>0$);

  * •

$\pi_{\alpha}=t_{k_{r_{\alpha}}}^{k_{r_{\alpha}}-\sum_{j=r_{\alpha}+1}^{r_{\alpha+1}-1}i_{k_{j}}-i^{\prime}_{\alpha}}\cdot
t_{k_{r_{\alpha}+1}}^{i_{k_{r_{\alpha}+1}}}\cdots
t_{k_{r_{\alpha+1}-1}}^{i_{k_{r_{\alpha+1}-1}}},$

in case $i^{\prime}_{\alpha}=0$ and
$\sum_{j=r_{\alpha}+1}^{r_{\alpha+1}-1}i_{k_{j}}+i^{\prime}_{\alpha}<k_{r_{\alpha}}$
(i.e.,
$k_{r_{\alpha}}-\sum_{j=r_{\alpha}+1}^{r_{\alpha+1}-1}i_{k_{j}}-i^{\prime}_{\alpha}>0$);

  * •

$\pi_{\alpha}=t_{k_{r_{\alpha}+1}}^{i_{k_{r_{\alpha}+1}}}\cdots
t_{k_{r_{\alpha+1}-1}}^{i_{k_{r_{\alpha+1}-1}}}\cdot
t_{k_{r_{\alpha+1}}}^{i^{\prime}_{\alpha}},$

in case $i^{\prime}_{\alpha}>0$ and
$\sum_{j=r_{\alpha}+1}^{r_{\alpha+1}-1}i_{k_{j}}+i^{\prime}_{\alpha}=k_{r_{\alpha}}$
(i.e.,
$k_{r_{\alpha}}-\sum_{j=r_{\alpha}+1}^{r_{\alpha+1}-1}i_{k_{j}}-i^{\prime}_{\alpha}=0$);

  * •

$\pi_{\alpha}=t_{k_{r_{\alpha}+1}}^{i_{k_{r_{\alpha}+1}}}\cdots
t_{k_{r_{\alpha+1}-1}}^{i_{k_{r_{\alpha+1}-1}}},$

in case $i^{\prime}_{\alpha}=0$ and
$\sum_{j=r_{\alpha}+1}^{r_{\alpha+1}-1}i_{k_{j}}+i^{\prime}_{\alpha}=k_{r_{\alpha}}$
(i.e.,
$k_{r_{\alpha}}-\sum_{j=r_{\alpha}+1}^{r_{\alpha+1}-1}i_{k_{j}}-i^{\prime}_{\alpha}=0$)

Now, by considering the standard $OGS$ presentation of $\pi_{\alpha}$ for
$1\leq\alpha\leq q$ as an element in $S_{n}$ the following holds:

  * •

In all the possibly cases of $\pi_{\alpha}$,

$maj(\pi_{\alpha})=k_{r_{\alpha}}-\sum_{j=r_{\alpha}+1}^{r_{\alpha+1}-1}i_{k_{j}}-i^{\prime}_{\alpha}+\sum_{j=r_{\alpha}+1}^{r_{\alpha+1}-1}i_{k_{j}}+i^{\prime}_{\alpha}=k_{r_{\alpha}}.$

  * •

The smallest $p$ such that a non-zero power of $t_{p}$ is a subword of the
standard $OGS$ presentation of $\pi_{\alpha}$ is
$p=k_{r_{\alpha}}=maj(\pi_{\alpha})$ or
$p=k_{r_{\alpha}+1}>maj(\pi_{\alpha})$.

  * •

The greatest $p$ such that a non-zero power of $t_{p}$ is a subword of the
standard $OGS$ presentation of $\pi_{\alpha-1}$ (for $\alpha\geq 2$) is
$p=k_{r_{\alpha}}=maj(\pi_{\alpha})$ or
$p=k_{r_{\alpha}-1}<maj(\pi_{\alpha})$.

Hence, by Definition 1.0.8,

$\pi=\pi_{1}\cdot\pi_{2}\cdots\pi_{q}$

is a standard $OGS$ elementary factorization of $\pi$, such that every element
$\pi_{\alpha}$ for $1\leq\alpha\leq q$ is the elementary factor
$\pi^{(\alpha)}$ of $\pi$, where $z(\pi)=q$.

∎

######  Example 3.0.11.

Let
$\pi=\tau_{5}^{-3}\cdot\tau_{7}^{2}\cdot\tau_{8}^{-4}\cdot\tau_{9}^{4}\cdot\tau_{11}^{-3}\cdot\tau_{12}^{4}$.

Then by Theorem 3.0.5 $\pi\in\dot{S}_{12}$.

Moreover,

  * •

$i_{k_{1}}=-3<0,\quad i_{k_{3}}=-4<0,\quad i_{k_{5}}=-3<0$;

  * •

$i_{k_{2}}=2>0,\quad i_{k_{4}}=4>0,\quad i_{k_{6}}=4>0$.

Hence, by Theorem 3.0.10 $z(\pi)=3$ and the standard $OGS$ elementary
factorization of $\pi$ as follow:

$\pi^{(1)}=\tau_{5}^{-3}\cdot\tau_{7}^{2}\cdot\tau_{8},\quad\pi^{(2)}=\tau_{8}^{-5}\cdot\tau_{9}^{4}\cdot\tau_{11},\quad\pi^{(3)}=\tau_{11}^{-4}\cdot\tau_{12}^{4}.$

Indeed,

$\pi^{(1)}\cdot\pi^{(2)}\cdot\pi^{(3)}=(\tau_{5}^{-3}\cdot\tau_{7}^{2}\cdot\tau_{8})\cdot(\tau_{8}^{-5}\cdot\tau_{9}^{4}\cdot\tau_{11})\cdot(\tau_{11}^{-4}\cdot\tau_{12}^{4})$
$=\tau_{5}^{-3}\cdot\tau_{7}^{2}\cdot\tau_{8}^{-4}\cdot\tau_{9}^{4}\cdot\tau_{11}^{-3}\cdot\tau_{12}^{4}=\pi.$

##  4 Generalization of the standard OGS factorization for $B_{n}$

There is defined In [9] a standard $OGS$ elementary factorization for the
elements of the symmetric group $S_{n}$. This definition eases the calculation
of the Coxeter length and the descent sets of the elements of $S_{n}$. In this
section, we generalize the definition of the standard OGS elementary
factorization for the group $B_{n}$.

######  Theorem 4.0.1.

Let $\pi\in B_{n}$, then $\pi$ has a unique presentation in the following
form: $u_{1}\cdot v_{1}\cdot u_{2}\cdot v_{2}\cdots u_{r}$ for some $r$, such
that the following holds:

  * •

$u_{j}\in\dot{S}_{n}$, for every $1\leqslant j\leqslant r$.

  * •

$v_{j}=\tau_{p_{j}}^{p_{j}}$, where for $j>j^{\prime}$,
$p_{j}>p_{j^{\prime}}$.

  * •

For every $1\leqslant j\leqslant r$, either $u_{j}=1$ or
$u_{j}=\tau_{p_{j_{1}}}^{i_{p_{j_{1}}}}\cdots\tau_{p_{j_{z_{j}}}}^{i_{p_{j_{z_{j}}}}}$
by the generalized standard $OGS$ presentation (as it is described in Theorem
2.1.5), where $p_{j_{1}}\geq p_{j-1}$, for every $2\leqslant j\leqslant r$,and
$p_{j_{z_{j}}}\leqslant p_{j}$ , for every $1\leqslant j\leqslant r-1$.

###### Proof.

Let
$\pi=\tau_{k_{1}}^{i_{k_{1}}}\cdot\tau_{k_{2}}^{i_{k_{2}}}\cdots\tau_{k_{m}}^{i_{k_{m}}}$
be expressed by the generalized standard $OGS$ presentation.
\- First, We find $u_{r}$ and $v_{r-1}$, where we consider two cases which
depend on the possibly values of $i_{k_{m}}$ as follow:

  1. 1.

$1\leq i_{k_{m}}<k_{m}$.

  2. 2.

$-k_{m}\leq i_{k_{m}}<0$.

\- Consider case 1 (i.e., $1\leq i_{k_{m}}<k_{m}$).

  * •

By Theorem 3.0.5, $\tau_{k_{m}}^{i_{k_{m}}}$ can be considered as a subword of
an element in $\dot{S}_{n}$. Hence, we consider $\tau_{k_{m}}^{i_{k_{m}}}$ as
a subword of $u_{r}$ and $p_{r_{z_{r}}}=k_{m}$.

\- Now, We consider ${k_{m-1}}$ and $i_{k_{m-1}}$ in case 1 ($1\leq
i_{k_{m}}<k_{m}$), where we divide the proof into the following two subcases:

    * –

subcase 1: $k_{m-1}<i_{k_{m}}$.

    * –

subcase 2: $k_{m-1}\geq i_{k_{m}}$.

  * •

subcase 1: If $k_{m-1}<i_{k_{m}}$, then put $u_{r}$ to be
$u_{r}=\tau_{i_{k_{m}}}^{-i_{k_{m}}}\cdot\tau_{k_{m}}^{i_{k_{m}}}$, and put
$v_{r-1}$ to be $v_{r-1}=\tau_{i_{k_{m}}}^{i_{k_{m}}}$.
Then,

$\pi=\tau_{k_{1}}^{i_{k_{1}}}\cdot\tau_{k_{2}}^{i_{k_{2}}}\cdots\tau_{k_{m-1}}^{i_{k_{m-1}}}\cdot\tau_{i_{k_{m}}}^{i_{k_{m}}}\cdot\tau_{i_{k_{m}}}^{-i_{k_{m}}}\cdot\tau_{k_{m}}^{i_{k_{m}}}=\tau_{k_{1}}^{i_{k_{1}}}\cdot\tau_{k_{2}}^{i_{k_{2}}}\cdots\tau_{k_{m-1}}^{i_{k_{m-1}}}\cdot
v_{r-1}\cdot u_{r}.$

Then, we look at the subword

$\tilde{\pi}=\tau_{k_{1}}^{i_{k_{1}}}\cdot\tau_{k_{2}}^{i_{k_{2}}}\cdots\tau_{k_{m-1}}^{i_{k_{m-1}}}$

of $\pi$ , and we consider $u_{r-1}$ and $v_{r-2}$ by applying the same
algorithm as we have done for finding $u_{r}$ and $v_{r-1}$, by considering
$\tilde{\pi}$ instead of $\pi$.

  * •

subcase 2: If $k_{m-1}\geq i_{k_{m}}$, then there are the following two
possibilities:

    * –

Possibility 1: $i_{k_{m}}+i_{k_{m-1}}\geq k_{m-1}$.

    * –

Possibility 2: $i_{k_{m}}+i_{k_{m-1}}<k_{m-1}$.

  * •

Assume possibility 1. Then, $i_{k_{m}}+i_{k_{m-1}}\geq k_{m-1}$. Then we put
$u_{r}$ to be $u_{r}=\tau_{k_{m-1}}^{-i_{k_{m}}}\cdot\tau_{k_{m}}^{i_{k_{m}}}$
and we put $v_{r-1}$ to be $v_{r-1}=\tau_{k_{m-1}}^{k_{m-1}}$. Then,

$\pi=\tau_{k_{1}}^{i_{k_{1}}}\cdot\tau_{k_{2}}^{i_{k_{2}}}\cdots\tau_{k_{m-2}}^{i_{k_{m-2}}}\cdot\tau_{k_{m-1}}^{{i_{k_{m-1}}+i_{k_{m}}-k_{m-1}}}\cdot\tau_{k_{m-1}}^{k_{m-1}}\cdot(\tau_{k_{m-1}}^{-i_{k_{m}}}\cdot\tau_{k_{m}}^{i_{k_{m}}})$
$=\tau_{k_{1}}^{i_{k_{1}}}\cdot\tau_{k_{2}}^{i_{k_{2}}}\cdots\tau_{k_{m-2}}^{i_{k_{m-2}}}\cdot\tau_{k_{m-1}}^{{i_{k_{m-1}}+i_{k_{m}}-k_{m-1}}}\cdot
v_{r-1}\cdot u_{r}.$

Then, we look at the subword

$\tilde{\pi}=\tau_{k_{1}}^{i_{k_{1}}}\cdot\tau_{k_{2}}^{i_{k_{2}}}\cdots\tau_{k_{m-2}}^{i_{k_{m-2}}}\cdot\tau_{k_{m-1}}^{{i_{k_{m-1}}+i_{k_{m}}-k_{m-1}}}$

of $\pi$ , and we consider $u_{r-1}$ and $v_{r-2}$ by applying the same
algorithm as we have done for finding $u_{r}$ and $v_{r-1}$, by considering
$\tilde{\pi}$ instead of $\pi$.

  * •

Assume possibility 2. Then, $i_{k_{m}}+i_{k_{m-1}}<k_{m-1}$. Then,
$\tau_{k_{m-1}}^{i_{k_{m-1}}}\cdot\tau_{k_{m}}^{i_{k_{m}}}$ is a subword of
$u_{r}$ and $p_{r_{z_{r-1}}}=k_{m-1}$. Then, we consider $\tau_{k_{m-2}}$ and
$i_{k_{m-2}}$. Then, We analyze $\tau_{k_{m-2}}$ and $i_{k_{m-2}}$ by the same
argument as we have analysed $\tau_{k_{m-1}}$ and $i_{k_{m-1}}$.
First, dividing the proof to two main subcases

    * –

Subcase 1: $k_{m-2}<i_{k_{m}}+i_{k_{m-1}}$.

    * –

Subcase 2: $k_{m-2}\geq i_{k_{m}}+i_{k_{m-1}}$.

  * •

We analyze subcase 1 by the same algorithm as we have done in the subcase
$k_{m-1}<i_{k_{m}}$. i.e., put $u_{r}$ to be
$u_{r}=\tau_{i_{k_{m}}+i_{k_{m-1}}}^{-i_{k_{m}}-i_{k_{m-1}}}\cdot\tau_{k_{m-1}}^{i_{k_{m-1}}}\cdot\tau_{k_{m}}^{i_{k_{m}}}$
and put $v_{r-1}$ to be
$v_{r-1}=\tau_{i_{k_{m}}+i_{k_{m-1}}}^{i_{k_{m}}+i_{k_{m-1}}}.$

  * •

We analyze subcase 2 by the same algorithm as we have done in the subcase
$k_{m-1}\geq i_{k_{m}}$ . i.e., considering two possibilities.

    * –

Possibility 1: $i_{k_{m}}+i_{k_{m-1}}+i_{k_{m-2}}\geq k_{m-2}$.

    * –

Possibility 2: $i_{k_{m}}+i_{k_{m-1}}+i_{k_{m-2}}<k_{m-2}$.

  * •

Then, in case of possibility 1, we put $u_{r}$ and $v_{r-1}$ by the same way
as we have done for the possibility $i_{k_{m}}+i_{k_{m-1}}\geq k_{m-1}$.

  * •

In case of possibility 2 ($i_{k_{m}}+i_{k_{m-1}}+i_{k_{m-2}}<k_{m-2}$), we
conclude that
$\tau_{k_{m-2}}^{i_{k_{m-2}}}\cdot\tau_{k_{m-1}}^{i_{k_{m-1}}}\cdot\tau_{k_{m}}^{i_{k_{m}}}$
is a subword of $u_{r}$ and $p_{r_{z_{r-2}}}=k_{m-2}$. Then, considering
$\tau_{k_{m-3}}$ and $i_{k_{m-3}}$ by continuing the same process as we have
done in the possibility $i_{k_{m}}+i_{k_{m-1}}<k_{m-1}$ where we have
considered $\tau_{k_{m-2}}$ and $i_{k_{m-2}}$.

  * •

Then we continue the algorithm by the same method.

  * •

Finally, we get the following conclusion from the described algorithm
concerning $u_{r}$ and $v_{r-1}$ in case 1 ($i_{k_{m}}>0$). For every $1\leq
x\leq m$, Let define $\rho_{x}$ to be $\rho_{x}=\sum_{j=x}^{m}i_{k_{j}}$.
Then, we consider the largest $x$ such that $1\leq x\leq m$ such that one of
the following two conditions holds

    * –

$\rho_{x+1}\leq k_{x}\leq\rho_{x}$.

    * –

$k_{x-1}<\rho_{x}<k_{x}$.

Then $u_{r}$ and $v_{r-1}$ have the following presentation by the generalized
standard $OGS$.

    * –

If $\rho_{x+1}\leq k_{x}\leq\rho_{x}$, then

$u_{r}=\tau_{k_{x}}^{-\rho_{x+1}}\cdot\tau_{k_{x+1}}^{i_{k_{x+1}}}\cdot\tau_{k_{x+2}}^{i_{k_{x+2}}}\cdots\tau_{k_{m}}^{i_{k_{m}}}\quad
v_{r-1}=\tau_{k_{x}}^{k_{x}}.$
$\pi=\tau_{k_{1}}^{i_{k_{1}}}\cdot\tau_{k_{2}}^{i_{k_{2}}}\cdots\tau_{k_{x-1}}^{i_{k_{x-1}}}\cdot\tau_{k_{x}}^{\rho_{x}-k_{x}}\cdot
v_{r-1}\cdot u_{r}.$

Hence, we find $u_{r-1}$ and $v_{r-2}$ by considering

$\tilde{\pi}=\tau_{k_{1}}^{i_{k_{1}}}\cdot\tau_{k_{2}}^{i_{k_{2}}}\cdots\tau_{k_{x-1}}^{i_{k_{x-1}}}\cdot\tau_{k_{x}}^{\rho_{x}-k_{x}},$

where we use the same method which is used to find $u_{r}$ and $v_{r-1}$.

    * –

If $k_{x-1}<\rho_{x}<k_{x}$, then

$u_{r}=\tau_{\rho_{x}}^{-\rho_{x}}\cdot\tau_{k_{x}}^{i_{k_{x}}}\cdot\tau_{k_{x+1}}^{i_{k_{x+1}}}\cdots\tau_{k_{m}}^{i_{k_{m}}}\quad
v_{r-1}=\tau_{\rho_{x}}^{\rho_{x}}$
$\pi=\tau_{k_{1}}^{i_{k_{1}}}\cdot\tau_{k_{2}}^{i_{k_{2}}}\cdots\tau_{k_{x-1}}^{i_{k_{x-1}}}\cdot
v_{r-1}\cdot u_{r}.$

Hence, we find $u_{r-1}$ and $v_{r-2}$ by considering

$\tilde{\pi}=\tau_{k_{1}}^{i_{k_{1}}}\cdot\tau_{k_{2}}^{i_{k_{2}}}\cdots\tau_{k_{x-1}}^{i_{k_{x-1}}},$

where we use the same method which is used to find $u_{r}$ and $v_{r-1}$.

\- Now, We turn to case 2 (i.e., $-k_{m}\leq i_{k_{m}}<0$ ).

If $i_{k_{m}}<0$ then, $u_{r}$ is defined to be $1$, and $v_{r-1}$ is defined
to be $\tau_{k_{m}}^{-k_{m}}=\tau_{k_{m}}^{k_{m}}$. Then,

$\tau_{k_{m}}^{i_{k_{m}}}=\tau_{k_{m}}^{-k_{m}+(i_{k_{m}}+k_{m})}=\tau_{k_{m}}^{i_{k_{m}}+k_{m}}\cdot\tau_{k_{m}}^{-k_{m}}$

Since, $-k_{m}\leq i_{k_{m}}<0$, we get that $0\leq k_{m}+i_{k_{m}}<k_{m}$.
Then,

$\pi=\tau_{k_{1}}^{i_{k_{1}}}\cdot\tau_{k_{2}}^{i_{k_{2}}}\cdots\tau_{k_{m}}^{i_{k_{m}+k_{m}}}\cdot
v_{r-1}\cdot u_{r},$

where $0\leqslant i_{k_{m}}+k_{m}<k_{m}$ . Now, we consider the following two
subcases:

  * •

Subcase 1: $i_{k_{m}}+k_{m}>0$.

  * •

Subcase 2: If $i_{k_{m}}+k_{m}=0$.

Consider subcase 1. Then, $i_{k_{m}}+k_{m}>0$, and then we find $u_{r-1}$ by
the same algorithm as we have done for finding $u_{r}$ in case 1 (The case of
$1\leq i_{k_{m}}<k_{m}$).

Now, consider subcase 2. Then, $i_{k_{m}}+k_{m}=0$. Hence,

$\pi=\tau_{k_{1}}^{i_{k_{1}}}\cdot\tau_{k_{2}}^{i_{k_{2}}}\cdots\tau_{k_{m-1}}^{i_{k_{m-1}}}\cdot\tau_{k_{m}}^{i_{k_{m}+k_{m}}}\cdot
v_{r-1}\cdot
u_{r}=\tau_{k_{1}}^{i_{k_{1}}}\cdot\tau_{k_{2}}^{i_{k_{2}}}\cdots\tau_{k_{m-1}}^{i_{k_{m-1}}}\cdot
v_{r-1}\cdot u_{r}.$

Then, we look at the subword

$\tilde{\pi}=\tau_{k_{1}}^{i_{k_{1}}}\cdot\tau_{k_{2}}^{i_{k_{2}}}\cdots\tau_{k_{m-1}}^{i_{k_{m-1}}}$

of $\pi$ , and we consider $u_{r-1}$ and $v_{r-2}$ by applying the same
algorithm as we have done for finding $u_{r}$ and $v_{r-1}$, by considering
$\tilde{\pi}$ instead of $\pi$.

∎

######  Example 4.0.2.

Let
$\pi=\tau_{3}^{2}\cdot\tau_{4}^{3}\cdot\tau_{5}^{-2}\cdot\tau_{7}^{4}\cdot\tau_{8}^{2}\cdot\tau_{9}^{4}$.

Then, applying the algorithm as it is described in Theorem 4.0.1.

First notice,

$i_{k_{6}}=4>0.$

Hence

$\tau_{k_{6}}^{i_{k_{6}}}=\tau_{9}^{4}$

is a subword of $u_{r}$. Now notice,

$i_{k_{6}}+i_{k_{5}}=6,\quad i_{k_{6}}+i_{k_{5}}+i_{k_{4}}=11,\quad k_{4}=7,$

Hence, we have:

$i_{k_{6}}+i_{k_{5}}<k_{4}<i_{k_{6}}+i_{k_{5}}+i_{k_{4}}$

Therefore, we put $u_{r}$ and $v_{r-1}$ as follow:

$u_{r}=\tau_{k_{4}}^{-i_{k_{5}}-i_{k_{6}}}\cdot\tau_{k_{5}}^{i_{k_{5}}}\cdot\tau_{k_{6}}^{i_{k_{6}}}=\tau_{7}^{-6}\cdot\tau_{8}^{2}\cdot\tau_{9}^{4},\quad
v_{r-1}=\tau_{k_{4}}^{k_{4}}=\tau_{7}^{7}.$

Hence, we get

$\pi=\tau_{k_{1}}^{i_{k_{1}}}\cdot\tau_{k_{2}}^{i_{k_{2}}}\cdot\tau_{k_{3}}^{i_{k_{3}}}\cdot\tau_{k_{4}}^{i_{k_{4}}+k_{4}-k_{5}-k_{6}}\cdot\tau_{k_{4}}^{k_{4}}\cdot(\tau_{k_{4}}^{-i_{k_{5}}-i_{k_{6}}}\cdot\tau_{k_{5}}^{i_{k_{5}}}\cdot\tau_{k_{6}}^{i_{k_{6}}})=$
$\tau_{3}^{2}\cdot\tau_{4}^{3}\cdot\tau_{5}^{-2}\cdot\tau_{7}^{3}\cdot\tau_{7}^{7}\cdot(\tau_{7}^{-6}\cdot\tau_{8}^{2}\cdot\tau_{9}^{4}).$

Now, look at the following subword of $\pi$:

$\tau_{k_{1}}^{i_{k_{1}}}\cdot\tau_{k_{2}}^{i_{k_{2}}}\cdot\tau_{k_{3}}^{i_{k_{3}}}\cdot\tau_{k_{4}}^{i^{\prime}_{k_{4}}}=\tau_{3}^{2}\cdot\tau_{4}^{3}\cdot\tau_{5}^{-2}\cdot\tau_{7}^{3}.$

Where,

$i^{\prime}_{k_{4}}=i_{k_{4}}+k_{4}-i_{k_{5}}-i_{k_{6}}.$

Since,

$i^{\prime}_{k_{4}}=3>0,$

$\tau_{k_{4}}^{i^{\prime}_{k_{4}}}=\tau_{7}^{3}$

is a subword of $u_{r-1}$.

Now, the following is satisfied:

$i^{\prime}_{k_{4}}+i_{k_{3}}=3-2=1,\quad k_{2}=4,\quad
i^{\prime}_{k_{4}}+i_{k_{3}}+i_{k_{2}}=4$

Hence,

$i^{\prime}_{k_{4}}+i_{k_{3}}<k_{2}\leq
i^{\prime}_{k_{4}}+i_{k_{3}}+i_{k_{2}}$

Therefore, we put $u_{r-1}$ and $v_{r-1}$ to be as follow:

$u_{r-1}=\tau_{k_{2}}^{-i_{k_{3}}-i^{\prime}_{k_{4}}}\cdot\tau_{k_{3}}^{i_{k_{3}}}\cdot\tau_{k_{4}}^{i^{\prime}_{k_{4}}}=\tau_{4}^{-1}\cdot\tau_{5}^{-2}\cdot\tau_{7}^{3},\quad
v_{r-2}=\tau_{k_{2}}^{k_{2}}=\tau_{4}^{4}.$

Then we get:

$\pi=\tau_{k_{1}}^{i_{k_{1}}}\cdot\tau_{k_{2}}^{k_{2}}\cdot(\tau_{k_{2}}^{-i_{k_{3}}-i^{\prime}_{k_{4}}}\cdot\tau_{k_{3}}^{i_{k_{3}}}\cdot\tau_{k_{4}}^{i^{\prime}_{k_{4}}})\cdot\tau_{k_{4}}^{k_{4}}\cdot(\tau_{k_{4}}^{-i_{k_{5}}-i_{k_{6}}}\cdot\tau_{k_{5}}^{i_{k_{5}}}\cdot\tau_{k_{6}}^{i_{k_{6}}})=$
$\tau_{3}^{2}\cdot\tau_{4}^{4}\cdot(\tau_{4}^{-1}\cdot\tau_{5}^{-2}\cdot\tau_{7}^{3})\cdot\tau_{7}^{7}\cdot(\tau_{7}^{-6}\cdot\tau_{8}^{2}\cdot\tau_{9}^{4}).$

Now, consider the following subword of $\pi$:

$\tau_{k_{1}}^{i_{k_{1}}}=\tau_{3}^{2}.$

Since

$k_{1}=2>0,$ $\tau_{k_{1}}^{i_{k_{1}}}=\tau_{3}^{2}$

is a subword of $u_{r-2}$. Now, notice

$i_{k_{1}}=2,\quad k_{1}=3.$

Hence,

$i_{k_{1}}<k_{1}.$

Therefore, we get $r-3=1$, and we put $u_{r-2}=u_{2}$, $v_{r-3}=v_{1}$ and
$u_{r-3}=u_{1}$ as follow:

$u_{r-2}=u_{2}=\tau_{i_{k_{1}}}^{-i_{k_{1}}}\cdot\tau_{k_{1}}^{i_{k_{1}}}=\tau_{2}^{-2}\cdot\tau_{3}^{3},\quad
v_{1}=\tau_{i_{k_{1}}}^{i_{k_{1}}}=\tau_{2}^{2},\quad u_{1}=1.$

Hence,

$\pi=u_{1}\cdot v_{1}\cdot u_{2}\cdot v_{2}\cdot u_{3}\cdot v_{3}\cdot
u_{4}=\tau_{2}^{2}\cdot(\tau_{2}^{-2}\cdot\tau_{3}^{2})\cdot\tau_{4}^{4}\cdot(\tau_{4}^{-1}\cdot\tau_{5}^{-2}\cdot\tau_{7}^{3})\cdot\tau_{7}^{7}\cdot(\tau_{7}^{-6}\cdot\tau_{8}^{2}\cdot\tau_{9}^{4}).$

Where,

$u_{1}=1\quad\quad v_{1}=\tau_{2}^{2}=\tau_{2}^{-2}\quad\quad
u_{2}=\tau_{2}^{-2}\cdot\tau_{3}^{2}\quad\quad
v_{2}=\tau_{4}^{4}=\tau_{4}^{-4}$
$u_{3}=\tau_{4}^{-1}\cdot\tau_{5}^{-2}\cdot\tau_{7}^{3}\quad\quad
v_{3}=\tau_{7}^{7}=\tau_{7}^{-7}\quad\quad
u_{4}=\tau_{7}^{-6}\cdot\tau_{8}^{2}\cdot\tau_{9}^{4}.$

##  5 The Coxeter length function in $B_{n}$ by applying the generalized
standard $OGS$

In this section We give an explicit formula for $\ell(\pi)$ considering the
generalized standard $OGS$ presentation, as it is described in Theorem 2.1.5,
and the presentation that is described in Theorem 4.0.1. We start with a lemma
about the length of elements of the form $\tau_{k}^{-k}$. Then, we find the
length of elements in the subgroup $\dot{S}_{n}$. Finally we give a formula
for the length of a general element of $B_{n}$.

######  Lemma 5.0.1.

For $1\leq k\leq n$, let $\tau_{k}$ be the element of $B_{n}$ as it is defined
in Definition 2.1.1. Then, the following holds:

$\ell(\tau_{k}^{-k})=\sum_{i=0}^{k-1}2i-1=k^{2}$

###### Proof.

By considering the normal form of $\tau_{k}^{-k}\in B_{n}$ as it is defined in
Definition 2.0.3:

$\tau_{k}^{-k}=\prod_{i=0}^{k-1}\prod_{j=0}^{2i}s_{|i-j|}=s_{0}\cdot(s_{1}\cdot
s_{0}\cdot s_{1})\cdot(s_{2}\cdot s_{1}\cdot s_{0}\cdot s_{1}\cdot
s_{2})\cdots(s_{k-1}\cdots s_{1}\cdot s_{0}\cdot s_{1}\cdots s_{k-1}).$

Hence

$\ell(\tau_{k}^{-k})=\sum_{i=0}^{k-1}2i-1=k^{2}$

∎

######  Theorem 5.0.2.

Let $\pi$ be an element in the subgroup $\dot{S}_{n}$ of $B_{n}$, where by the
generalized standard $OGS$ presentation of $\pi$ (as it is described in
Theorem 3.0.5):

$\pi=\tau_{k_{1}}^{i_{k_{1}}}\cdot\tau_{k_{2}}^{i_{k_{2}}}\cdots\tau_{k_{m}}^{i_{k_{m}}}$

such that

  * •

$-k_{j}\leqslant i_{k_{j}}\leqslant k_{j}-1$, for every $1\leqslant j\leqslant
m$.

  * •

$\sum_{j=1}^{m}i_{k_{j}}=0$.

  * •

$0\leqslant\sum_{j=r}^{m}i_{k_{j}}\leqslant k_{r-1}$, for every $2\leq r\leq
m$.

Then,

$\ell(\pi)=\sum_{j=1}^{m}k_{j}\cdot i_{k_{j}}.$

###### Proof.

Let $\pi$ be an element of $\dot{S}_{n}$, presented by the generalized
standard $OGS$ (as it is described in Theorem 3.0.5) as follow:

$\pi=\tau_{k_{1}}^{i_{k_{1}}}\cdot\tau_{k_{2}}^{i_{k_{2}}}\cdots\tau_{k_{m}}^{i_{k_{m}}}$

such that

  * •

$-k_{j}\leqslant i_{k_{j}}\leqslant k_{j}-1$, for every $1\leqslant j\leqslant
m$.

  * •

$\sum_{j=1}^{m}i_{k_{j}}=0$.

  * •

$0\leqslant\sum_{j=r}^{m}i_{k_{j}}\leqslant k_{r-1}$, for every $2\leq r\leq
m$.

The presentation of $\pi$ by the normal form ([5] Chapter 3.4) as follow:

$\pi=(s_{r_{1}}\cdot s_{r_{1}-1}\cdots s_{r_{1}-\mu_{1}})\cdot(s_{r_{2}}\cdot
s_{r_{2}-1}\cdots s_{r_{2}-\mu_{2}})\cdots(s_{r_{z}}\cdot s_{r_{z}-1}\cdots
s_{r_{z}-\mu_{z}}),$

where $r_{j}<r_{j+1}$ for every $1\leq j\leq z-1$. Therefore,
$\ell(\pi)=\sum_{j=1}^{z}\mu_{j}+z$.
Then, By Lemma 3.0.4,

$\pi=\tau_{r_{1}}^{-(\mu_{1}+1)}\cdot\tau_{r_{1}+1}^{\mu_{1}+1}\cdot\tau_{r_{2}}^{-(\mu_{2}+1)}\cdot\tau_{r_{2}+1}^{\mu_{2}+1}\cdots\tau_{r_{z}}^{-(\mu_{z}+1)}\cdot\tau_{r_{z}+1}^{\mu_{z}+1}.$

Thus,

$\ell(\pi)=\sum_{j=1}^{z}(\mu_{j}+1)=\sum_{j=1}^{z}{-(\mu_{j}+1)\cdot
r_{j}+(\mu_{j}+1)\cdot(r_{j}+1)}.$

Since $r_{j}<r_{j+1}$, we conclude $r_{j}+1\leq r_{j+1}$. Hence, we have
either $r_{j}+1=r_{j+1}$ or $r_{j}+1<r_{j+1}$ for some $1\leq j\leq z-1$.
Therefore, in case $r_{j}+1=r_{j+1}$ for some $1\leq j\leq z-1$,

$\pi=\tau_{r_{1}}^{-(\mu_{1}+1)}\cdot\tau_{r_{1}+1}^{\mu_{1}+1}\cdot\tau_{r_{2}}^{-(\mu_{2}+1)}\cdot\tau_{r_{2}+1}^{\mu_{2}+1}\cdots\tau_{r_{z}}^{-(\mu_{z}+1)}\cdot\tau_{r_{z}+1}^{\mu_{z}+1}=$
$=\tau_{r_{1}}^{-(\mu_{1}+1)}\cdots\tau_{r_{j}+1}^{(\mu_{j}+1)-(\mu_{j+1}+1)}\cdots\tau_{r_{z}+1}^{\mu_{z}+1}=$
$=\tau_{k_{1}}^{i_{k_{1}}}\cdots\tau_{k_{m}}^{i_{k_{m}}},$

Then, by the uniqueness of presentation by the generalized standard $OGS$
(Theorem 2.1.5), we conclude:

$\ell(\pi)=\sum_{j=1}^{z}(\mu_{j}+1)\cdot
r_{j}+(\mu_{j}+1)\cdot(r_{j}+1)=\sum_{j=1}^{m}k_{j}\cdot i_{k_{j}}.$

∎

The next theorem demonstrates the connection of the generalized standard $OGS$
presentation of an element $\pi$ of $B_{n}$ (as it is described in Theorem
2.1.5) to the Coxeter length of $\pi$.

######  Theorem 5.0.3.

Let $\pi$ be an element of $B_{n}$, Consider the presentation of $\pi$ in the
form $u_{1}\cdot v_{1}\cdot u_{2}\cdot v_{2}\cdots u_{r}$ for some $r$, as it
is described in Theorem 4.0.1.
Then,

$\ell(\pi)=\ell(u_{r})+\ell(v_{r-1})-\ell(u_{r-1})-\ell(v_{r-1})+\ell(u_{r-2})+\ell(v_{r-2})-\cdots+(-1)^{r-1}\cdot\ell(u_{1}).$

###### Proof.

First, consider the case:

$\pi=u\cdot v=\prod_{j=1}^{m}\tau_{k_{j}}^{i_{k_{j}}}\cdot\tau_{p}^{p}.$

where $p\geq k_{m}$. For $1\leq j<n$, denote by $\dot{B}_{j}$ the parabolic
subgroup of $B_{n}$ which is generated by $s_{0},s_{1},\ldots s_{j-1}$.
(Notice $\dot{B}_{j}$ is isomorphic to $B_{j}$.) By Theorem 2.1.5, $u$ is in
the parabolic subgroup $\dot{B}_{k_{m}}$ of $B_{n}$, and by [5] Chapter 3.4,
the normal form of $v=\tau_{p}^{p}$ as follow

$v=\tau_{p}^{p}=s_{0}\cdot(s_{1}\cdot s_{0}\cdot s_{1})\cdots(s_{p-1}\cdot
s_{p-2}\cdots s_{0}\cdot s_{1}\cdots s_{p-1})$

Since, $p\geq k_{m}$, both elements $u$ and $v$ can be considered as elements
of $\dot{B}_{p}$.

By [5] Chapter 3.4 $v=\tau_{p}^{p}$ is the longest element in the parabolic
subgroup $\dot{B}_{p}$ of $B_{n}$. Therefore, by [5] Chapter 2.3,

$\ell(\pi)=\ell(v)-\ell(u).$

Now, assume the following by induction on $m$:

If

$\pi=u_{1}\cdot v_{1}\cdots u_{m-1}\cdot v_{m-1}\cdot u_{m}\cdot v_{m},$

then

$\ell(\pi)=\sum_{i=1}^{m}(-1)^{m-i}\cdot(\ell(v_{i})-\ell(u_{i})).$

Now, consider :

$\pi_{1}=\pi\cdot u_{m+1}.$

where $\pi=u_{1}\cdot v_{1}\cdots u_{m-1}\cdot v_{m-1}\cdot u_{m}\cdot v_{m}$.
Then, $v_{m}=\tau_{q}^{q}$ for some $q<n$, and by Theorem 4.0.1 $\pi$ is in
the parabolic subgroup $\dot{B}_{q}$ of $B_{n}$. Notice, that the generalized
standard $OGS$ presentation of $u_{m+1}$ is
$u_{m+1}=\prod_{j=1}^{m}\tau_{k_{j}}^{i_{k_{j}}}$, where by Theorem 4.0.1,
$k_{1}\geq q$, and then by [5] Chapter 3.4, the normal form of $u_{m+1}$ as
follow:

$u_{m+1}=(s_{r_{1}}\cdot s_{r_{1}-1}\cdots
s_{r_{1}-\mu_{1}})\cdot(s_{r_{2}}\cdot s_{r_{2}-1}\cdots
s_{r_{2}-\mu_{2}})\cdots(s_{r_{z}}\cdot s_{r_{z}-1}\cdots s_{r_{z}-\mu_{z}})$

where $r_{j}<r_{j+1}$ for every $1\leq j\leq z-1$. By the proof of Theorem
5.0.2, $r_{1}=k_{1}$, and since $k_{1}\geq q$, we have that $r_{1}\geq q$.
Since, $\pi\in\dot{B}_{q}$, the normal form of $\pi\cdot u_{m+1}$ by [5]
Chapter 3.4 as follow:

$norm(\pi\cdot u_{m+1})=$ $=norm(\pi)\cdot(s_{r_{1}}\cdot s_{r_{1}-1}\cdots
s_{r_{1}-\mu_{1}})\cdot(s_{r_{2}}\cdot s_{r_{2}-1}\cdots
s_{r_{2}-\mu_{2}})\cdots(s_{r_{z}}\cdot s_{r_{z}-1}\cdots s_{r_{z}-\mu_{z}})=$
$=norm(\pi)\cdot norm(u_{m+1}).$

Hence,

$\ell(\pi\cdot u_{m+1})=\ell(\pi)+\ell(u_{m+1}).$

Now, consider:

$\pi_{2}=\pi_{1}\cdot v_{m+1}=\pi\cdot(u_{m+1}\cdot v_{m+1}).$

By the same argument as in the calculation of the length of $(u\cdot v)$, we
get that $v_{m+1}$ is the longest element in a parabolic subgroup of $B_{n}$,
which contains the elements $u_{1},v_{1},\ldots,u_{m},v_{m},u_{m+1}$. Hence,
by [5] Chapter 2.3,

$\ell(\pi_{2})=\ell(v_{m+1})-\ell(\pi_{1})=\ell(v_{m+1})-[\ell(u_{m+1})+\ell(\pi)]=(\sum_{i=1}^{m+1}(-1)^{m+1-i}\cdot(\ell(v_{i})-\ell(u_{i})).$

Hence,

$\ell(\pi)=\sum_{i=1}^{m+1}(-1)^{m+1-i}(\ell(v_{i})-\ell(u_{i})).$

∎

######  Example 5.0.4.

Let $\pi$ be:

$\pi=\tau_{2}\cdot\tau_{3}\cdot\tau_{4}^{3}\cdot\tau_{5}^{2}$
$\pi=\tau_{1}\cdot(\tau_{1}^{-1}\cdot\tau_{2})\cdot\tau_{2}^{2}\cdot(\tau_{2}^{-2}\cdot\tau_{3}\cdot\tau_{4})\cdot\tau_{4}^{4}\cdot(\tau_{4}^{-2}\cdot\tau_{5}^{2}).$

Hence,

$u_{1}=1\quad\quad v_{1}=\tau_{1}=\tau_{1}^{-1}\quad\quad
u_{2}=\tau_{1}^{-1}\cdot\tau_{2}\quad\quad v_{2}=\tau_{2}^{2}=\tau_{2}^{-2}$
$u_{3}=\tau_{2}^{-2}\cdot\tau_{3}\cdot\tau_{4}\quad\quad
v_{3}=\tau_{4}^{4}=\tau_{4}^{-4}\quad\quad
u_{4}=\tau_{4}^{-2}\cdot\tau_{5}^{2}.$

Then,

$\ell(\pi)=\ell(\tau_{4}^{-2}\cdot\tau_{5}^{2})+\ell(\tau_{4}^{-4})-\ell(\tau_{2}^{-2}\cdot\tau_{3}\cdot\tau_{4})-\ell(\tau_{2}^{-2})+\ell(\tau_{1}^{-1}\cdot\tau_{2})+\ell(\tau_{1}^{-1}).$
$=4\cdot(-2)+5\cdot 2+4\cdot 4-2\cdot(-2)-3-4-2\cdot 2+1\cdot(-1)+2+1=13.$

Indeed, the presentation of $\pi$ in the normal form contains $13$ Coxeter
generators as follow

$\text{norm}(\pi)=s_{0}\cdot s_{1}\cdot(s_{2}\cdot s_{1}\cdot
s_{0})\cdot(s_{3}\cdot s_{2}\cdot s_{1}\cdot s_{0}\cdot s_{1}\cdot
s_{2})\cdot(s_{4}\cdot s_{3}).$

##  6 The descent set of elements of $B_{n}$.

In this subsection, we characterize the descent set of the elements in the
Coxeter group $B_{n}$ by using the generalized standard $OGS$ presentation for
the elements of $B_{n}$

######  Proposition 6.0.1.

Let $\pi$ be an element of $\dot{S}_{n}$ in $B_{n}$ which is expressed by the
generalized standard OGS presentation (i.e.,
$\pi=\tau_{k_{1}}^{i_{k_{1}}}\cdot\tau_{k_{2}}^{i_{k_{2}}}\cdots\tau_{k_{m}}^{i_{k_{m}}}$,
where by Theorem 3.0.5, $\sum_{j=1}^{m}i_{k_{j}}=0$ , and
$k_{j}\leqslant\sum_{j=1}^{r}i_{k_{j}}\leqslant 0$ , $r<m.$)

Then $des(\pi)$, the set of the locations of the descents of $\pi$ satisfies
the following property:

$s_{k_{j}}\in des(\pi)\Longleftrightarrow i_{k_{j}}<0$

###### Proof.

Consider

$\pi=\tau_{k_{1}}^{i_{k_{1}}}\cdot\tau_{k_{2}}^{i_{k_{2}}}\cdots\tau_{k_{m}}^{i_{k_{m}}},\quad\text{where}\quad\sum_{j=1}^{m}i_{k_{j}}=0.$

Denote by $r_{1},\ldots,r_{q}$, all the indices $j$ such that $i_{k_{j}}<0$.
Since by Theorem 3.0.5, $i_{k_{1}}<0$ for every element of $\dot{S}_{n}$, we
have $r_{1}=1$.

Now, for every $1\leq\alpha\leq q$, let define $i^{\prime}_{\alpha}$ as it was
defined in Theorem 3.0.10, by the following recursive method:

  * •

Let define $i^{\prime}_{q}$ to be

$i^{\prime}_{q}=i_{k_{m}};$

  * •

starting with $\alpha=q-1$, let define $i^{\prime}_{q-1}$ to be

$i^{\prime}_{q-1}=i_{r_{q}}+\sum_{j=r_{q}+1}^{m-1}i_{k_{j}}+i^{\prime}_{q};$

  * •

Then for every $1\leq\alpha\leq q-2$, let define $i^{\prime}_{\alpha}$ to be

$i^{\prime}_{\alpha}=i_{r_{\alpha+1}}+\sum_{j=r_{\alpha+1}+1}^{r_{\alpha+2}-1}i_{k_{j}}+i^{\prime}_{\alpha+1}.$

Then by Theorem 3.0.10,

$\pi=(\tau_{k_{r_{1}}}^{-\sum_{j=2}^{r_{2}-1}i_{k_{j}}-i^{\prime}_{1}}\cdot\tau_{k_{2}}^{i_{k_{2}}}\cdots\tau_{k_{r_{2}-1}}^{i_{k_{r_{2}-1}}}\cdot\tau_{k_{r_{2}}}^{i^{\prime}_{1}})\cdot(\tau_{k_{r_{2}}}^{-\sum_{j=r_{2}+1}^{r_{3}-1}i_{k_{j}}-i^{\prime}_{2}}\cdot\tau_{k_{r_{2}+1}}^{i_{k_{r_{2}+1}}}\cdots\tau_{k_{r_{3}-1}}^{i_{k_{r_{3}-1}}}\cdot\tau_{k_{r_{3}}}^{i^{\prime}_{2}})\cdots$
$\cdots(\tau_{k_{r_{q}}}^{-\sum_{j=r_{q}+1}^{m-1}i_{k_{j}}-i^{\prime}_{q}}\cdot\tau_{k_{r_{q}+1}}^{i_{k_{r_{q}+1}}}\cdots\tau_{k_{m-1}}^{i_{k_{m-1}}}\tau_{k_{m}}^{i^{\prime}_{q}}).$

Then, by Theorem 3.0.10 the following properties hold:

  * •

By considering $\pi$ as an element of $S_{n}$, the subword

$\pi_{\alpha}=\tau_{k_{r_{\alpha}}}^{-\sum_{j=r_{\alpha}+1}^{r_{\alpha+1}-1}i_{k_{j}}-i^{\prime}_{\alpha}}\cdot\tau_{k_{r_{\alpha}+1}}^{i_{k_{r_{\alpha}+1}}}\cdots\tau_{k_{r_{\alpha+1}-1}}^{i_{k_{r_{\alpha+1}-1}}}\cdot\tau_{k_{r_{\alpha+1}}}^{i^{\prime}_{\alpha}}$

is the standard $OGS$ elementary factor $\pi^{(\alpha)}$ of $\pi$ for every
$\alpha$ such that $1\leq\alpha\leq q$, where
$maj(\pi^{(\alpha)})=k_{r_{\alpha}}$, and $z(\pi)=q$;

  * •

$i_{k_{j}}>0$ for every $2\leq j\leq m$ such that $j\neq r_{\alpha}$ for
$1\leq\alpha\leq q$.

Hence, by Theorem 1.0.9,

$des\left(\pi\right)=\bigcup_{\alpha=1}^{z(\pi)}des\left(\pi^{(\alpha)}\right)=\\{maj\left(\pi^{(\alpha)}\right)~{}|~{}1\leq\alpha\leq
z(\pi)\\};$

Hence, for $1\leq j\leq k_{m}$, $s_{j}\in des(\pi)$ if and only if
$j=k_{r_{\alpha}}$ for $1\leq\alpha\leq q$. Since by definition of
$r_{\alpha}$, $j=k_{r_{\alpha}}$ if and only if $i_{j}<0$, we get that
$s_{j}\in des(\pi)$ if and only if $i_{j}<0$. ∎

######  Example 6.0.2.

Let $\pi=\tau_{3}^{-2}\cdot\tau_{4}^{-1}\cdot\tau_{5}^{3}$ .

Then, the following holds:

  * •

$k_{1}=3,\quad k_{2}=4,\quad k_{3}=5$;

  * •

$i_{k_{1}}=-2,\quad i_{k_{2}}=-1,\quad i_{k_{3}}=3$;

  * •

$i_{k_{1}}=-2<0,\quad i_{k_{1}}+i_{k_{2}}=-1\leq 0,\quad
i_{k_{1}}+i_{k_{2}}+i_{k_{3}}=0$.

Hence, by Theorem 3.0.5 $\pi\in\dot{S}_{5}$.

The permutation presentation of $\pi$ is as follows:

$\pi=\tau_{3}^{-2}\cdot\tau_{4}^{-1}\cdot\tau_{5}^{3}$
$=[3;~{}-1;~{}-2;~{}4;~{}5]\cdot[2;~{}3;~{}4;~{}-1;~{}5]\cdot[-3;~{}-4;~{}-5;~{}1;~{}2]$
$=[1;~{}4;~{}5;~{}3;~{}2]$

Then we have,

$\pi(1)<\pi(2)<\pi(3),\quad\quad\pi(3)>\pi(4)>\pi(5).$

Hence,

$\text{des}(\pi)=\\{3,4\\}.$

######  Proposition 6.0.3.

Let $\pi$ be the element $\tau_{k}^{-k}$ , Then:

$des(\tau_{k}^{-k})=\\{0,1,2,\cdots,k-1\\}$

###### Proof.

Consider the permutation presentation of $\pi$:

$\pi=[-1;~{}-2;\ldots;~{}-k;~{}k+1;~{}k+2;\ldots;~{}n]$

Then,by definition $2.0.1.$

$des(\tau_{k}^{-k})=\\{0,1,2,\cdots,k-1\\}$

∎

######  Theorem 6.0.4.

Let $\pi$ be the element $v\cdot
u=\tau_{i_{1}}^{-i_{1}}\cdot(\tau_{i_{2}}^{k_{2}}\cdots\tau_{i_{m}}^{k_{m}}).$
where $v$ is the longest element of the parabolic subgroup which is generated
by the Coxeter elements $\\{s_{0},s_{1},\cdots,s_{i_{1}-1}\\}$, and $u$ is an
element of $\dot{S}_{n}$, presented by the generalized standard $OGS$, where
$i_{2}\geq i_{1}$. Then the descent set of $\pi$ satisfies the following:

  * •

$j\in des(v\cdot u)\quad,0\leqslant j<i_{1}\Longleftrightarrow j\notin
des(u).$

  * •

$j\in des(v\cdot u)\quad,j>i_{1}\Longleftrightarrow j\in des(u).$

  * •

$i_{1}\notin des(v\cdot u)$.

###### Proof.

Consider

$\pi=\tau_{i_{1}}^{-i_{1}}\cdot(\tau_{i_{2}}^{k_{2}}\cdots\tau_{i_{m}}^{k_{m}}).$
$v=\tau_{i_{1}}^{-i_{1}}.$
$u=\tau_{i_{2}}^{k_{2}}\cdots\tau_{i_{m}}^{k_{m}}\quad,u\in S_{i_{m}}.$

Then by considering the permutation presentation of the element $vu$:

  * •

$[v\cdot u](j)=-u(j)\quad 1\leqslant j\leqslant i_{1}$

  * •

$[v\cdot u](j)=u(j)\quad i_{1}+1\leqslant j\leqslant i_{m}.$

Hence the following holds:

  * •

For $1\leq j\leq i_{1}-1$:

$j\in des(v\cdot u)\Longleftrightarrow[v\cdot u](j)>[v\cdot
u](j+1)\Longleftrightarrow u(j)<u(j+1)\quad(i.e.,\quad j\notin des(u)).$

  * •

For $i_{1}+1\leq j\leq n-1$:

$j\in des(v\cdot u)\Longleftrightarrow[v\cdot u](j)>[v\cdot
u](j+1)\Longleftrightarrow u(j)>u(j+1)\quad(i.e.,\quad j\in des(u)).$

  * •

For $j=i_{1}$:

$[v\cdot u](i_{1})=-u(i_{1}),\quad[v\cdot
u](i_{1}+1)=u(i_{1}+1)\Longrightarrow[v\cdot u](i_{1})<[v\cdot
u](i_{1}+1)\Longrightarrow i_{1}\notin des(v\cdot u).$

∎

######  Theorem 6.0.5.

Let $\pi$ be the element $u\cdot
v=(\tau_{i_{1}}^{k_{1}}\cdot\tau_{i_{2}}^{k_{2}}\cdots\tau_{i_{m}}^{k_{m}})\cdot\tau_{i_{m+1}}^{i_{m+1}}$
where $v$ is the longest element of the parabolic subgroup generated by the
Coxeter elements $\\{s_{0},s_{1},\cdots,s_{i_{m+1}-1}\\}$ ,and $u$ is element
of $\dot{S}_{n}$ presented by the generalized standard $OGS$, and $i_{m}\leq
i_{m+1}$, then the descent set of $\pi$ satisfies the following property:

  * •

$j\in des(u\cdot v)\quad,0\leqslant j<i_{m}\Longleftrightarrow j\notin
des(u).$

  * •

$j\in des(u\cdot v)\quad,i_{m}\leqslant j<i_{m+1}.$

###### Proof.

Consider

$\pi=(\tau_{i_{1}}^{k_{1}}\cdot\tau_{i_{2}}^{k_{2}}\cdots\tau_{i_{m}}^{k_{m}})\cdot\tau_{i_{m+1}}^{i_{m+1}}\quad,i_{m}\leq
i_{m+1}.$

$u=\tau_{i_{1}}^{k_{1}}\cdot\tau_{i_{2}}^{k_{2}}\cdots\tau_{i_{m}}^{k_{m}}\quad
u\in S_{i_{m}}.$ $des(u)=i_{j}\quad,k_{j}<0.$

$v=\tau_{i_{m+1}}^{i_{m+1}}.$

$des(v)=\\{0,1,\cdots,i_{m+1-1}\\}.$

Then by looking at the permutation presentation of the element $uv$:

$[u\cdot v](j)=-u(j)\quad 1\leqslant j\leqslant i_{m},\quad\quad[u\cdot
v](j)=-j\quad i_{m}<j<i_{m+1}.$

Hence the following holds:

  * •

For $1\leq j\leq i_{m}$:

$j\in des(u\cdot v)\Longleftrightarrow[u\cdot v](j)>[u\cdot
v](j+1)\Longleftrightarrow u(j)<u(j+1)\quad(i.e.,\quad j\notin des(u)).$

  * •

For $i_{m}<j\leq i_{m}+1$:

$[u\cdot v](j)>[u\cdot v](j+1)\Longrightarrow j\in des(u\cdot v)).$

∎

######  Example 6.0.6.

The case $\pi=v\cdot u$:

$\bullet$ Consider

$\pi=\tau_{9}^{4}\cdot\tau_{10}^{4}=\tau_{8}^{-8}\cdot(\tau_{8}^{-8}\cdot\tau_{9}^{4}\cdot\tau_{10}^{4}).$

The permutation presentation of $\pi$ as follow:

$v=\tau_{8}^{-8}=[-1;~{}-2;~{}-3;~{}-4;~{}-5;~{}-6;~{}-7;~{}-8;~{}9;~{}10]$
$u=\tau_{8}^{-8}\cdot\tau_{9}^{4}\cdot\tau_{10}^{4}$
$=[-1;~{}-2;~{}-3;~{}-4;~{}-5;~{}-6;~{}-7;~{}-8;~{}9;~{}10]\cdot[-6;~{}-7;~{}-8;~{}-9;~{}1;~{}2;~{}3;~{}4;~{}5;~{}10]\cdot$
$[~{}-7;~{}-8;~{}-9;~{}-10;~{}1;~{}2;~{}3;~{}4;~{}5;~{}6]$
$=[2;~{}3;~{}4;~{}5;~{}7;~{}8;~{}9;~{}10;~{}1;~{}6].$

Hence,

$\pi=v\cdot
u=\tau_{8}^{-8}\cdot(\tau_{8}^{-8}\cdot\tau_{9}^{4}\cdot\tau_{10}^{4})=\tau_{9}^{4}\cdot\tau_{10}^{4}$
$=[-1;~{}-2;~{}-3;~{}-4;~{}-5;~{}-6;~{}-7;~{}-8;~{}9;~{}10]\cdot[2;~{}3;~{}4;~{}5;~{}7;~{}8;~{}9;~{}10;~{}1;~{}6]$
$=[-2;~{}-3;~{}-4;~{}-5;~{}-7;~{}-8;~{}-9;~{}-10;~{}1;~{}6]$

Then :

$des(v)=des(\tau_{8}^{-8})=\\{0,1,2,...,7\\},\quad
des(u)=des(\tau_{8}^{-8}\cdot\tau_{9}^{4}\cdot\tau_{10}^{4})=\\{8\\}.$

Hence, we conclude that in the permutation presentation of $u$, we get:

$u(1)<u(2)<...<u(8)\quad\quad u(8)>u(9)\quad\quad u(9)<u(10).$

And in the permutation presentation of $v$ ,

$v(i)=-i\quad\text{for}\quad 1<i<8.$

Hence, in the permutation presentation of $v\cdot u$ :

  * •

For every $1<i<8\Longleftrightarrow[v\cdot u](i)=-u(i).$

  * •

For $i=9,10\Longleftrightarrow[v\cdot u](i)=u(i).$

  * •

Therefore $u(8)<0\quad and\quad u(9)>0$, which applies $[v\cdot u](8)<[v\cdot
u](9).$

  * •

Furthermore the following holds:

$0>[v\cdot u](1)>[v\cdot u](2)>[v\cdot u](3)>\cdots>[v\cdot u](8),\quad[v\cdot
u](9)<[v\cdot u](10).$

Hence, the descent set of $\pi$ as follow:

$des(\pi)=\\{0,1,2,3,4,5,6,7\\}.$

######  Example 6.0.7.

The case $\pi=u\cdot v$ :

$\bullet$Let $\pi$ be

$(\tau_{3}^{-1}\cdot\tau_{4}^{-2}\cdot\tau_{5}^{3})\cdot\tau_{5}^{-5}.$

$u=\tau_{3}^{-1}\cdot\tau_{4}^{-2}\cdot\tau_{5}^{3}\in\dot{S}_{5}$ and
$v=\tau_{5}^{-5}$.

The permutation presentation of $\pi$ as follow:

$u=\tau_{3}^{-1}\cdot\tau_{4}^{-2}\cdot\tau_{5}^{3}$
$=[2;~{}3;~{}-1;~{}4;~{}5]\cdot[3;~{}4;~{}-1;~{}-2;~{}5]\cdot[-3;~{}-4;~{}-5;~{}1;~{}2]$
$=[1;~{}3;~{}5;~{}4;~{}2].$

$v=\tau_{5}^{-5}=[-1;~{}-2;~{}-3;~{}-4;~{}-5].$

Hence,

$\pi=u\cdot
v=(\tau_{3}^{-1}\cdot\tau_{4}^{-2}\cdot\tau_{5}^{3})\cdot\tau_{5}^{-5}$
$[-1;~{}-3;~{}-5;~{}-4;~{}-2].$

Then by Proposition 6.0.1:

$des(u)=\\{3,4\\}$

Hence we conclude:

$u(1)<u(2)<u(3)\quad\quad u(3)>u(4)>u(5).$

Now, we turn to the element $v$. We know that:

$v(i)=-i\quad\text{for every}\quad 1<i<5.$

Then, Then by Proposition 6.0.1:

$des(v)=\\{0,1,2,3,4\\}.$

Therefore, in the permutation presentation of $u\cdot v$, We have:

$[uv](i)=-u(i)\quad 1<i<5.$

Hence, we have in the permutation presentation of $u\cdot v$,

$(0>[u\cdot v](1)>[u\cdot v](2)>[u\cdot v](3)),\quad([u\cdot v](3)<[u\cdot
v](4)<[u\cdot v](5)).$

Then the descent set of $\pi=u\cdot v$ is as follow:

$des(u\cdot v)=\\{0,1,2\\}\quad\text{where}\quad des(u\cdot
v)=\\{0,1,2,3,...,k-1\\}-des(u).$

##  7 Conclusions and future plans

This paper is a natural continuation of the paper [9], where there was
introduced a quite interesting generalization of the fundamental theorem for
abelian groups to two important and very elementary families of non-abelian
Coxeter groups, the $I$-type (dihedral groups), and the $A$-type (symmetric
groups). There were introduced canonical forms, with very interesting exchange
laws, and quite interesting properties concerning the Coxeter lengths of the
elements. In this paper we generalized the results of [9] for the $B$-type
Coxeter groups, namely $B_{n}$. The results of the paper motivate us for
further generalizations of the $OGS$ and the arising properties from it for
more families of Coxeter and generalized Coxeter groups, which have an
importance in the classification of Lie algebras and the Lie-type simple
groups, and in other fields of mathematics, such as algebraic geometry for
classification of fundamental groups of Galois covers of surfaces [2]. The
next natural step of generalization is defining a generalized standard OGS
canonical form for $D$-type Coxeter group, which similarly to the $B$-type
Coxeter groups, have presentations as signed permutations. Furthermore, it is
interesting to find generalization of the $OGS$ to the affine classical
families $\tilde{A}_{n}$, $\tilde{B}_{n}$, $\tilde{C}_{n}$, and
$\tilde{D}_{n}$, and also to other generalizations of the mentioned Coxeter
groups, as the complex reflection groups $G(r,p,n)$ [8] or the generalized
affine classical groups, the definition of which is described in [7], [3].

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