# Quasitriangular Hopf algebras of dimension $pq^{2}$

Kun Zhou and Gongxiang Liu  Department of Mathematics, Nanjing University,
Nanjing 210093, China<EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract.

Let $p$ and $q$ be distinct odd primes and assume $\Bbbk$ is an algebraically
closed field of characteristic zero. We classify all quasitriangular Hopf
algebras of dimension $pq^{2}$ over $\Bbbk$, which are not simple as Hopf
algebras. Moreover, we obtained all quasitriangular structures on these Hopf
algebras.

###### Key words and phrases:

Quasitriangular Hopf algebra, Abelian extension.

###### 2010 Mathematics Subject Classification:

16T05 (primary), 16T25 (secondary)

†Supported by NSFC 11722016.

##  1\. Introduction

Quasitriangular Hopf algebras were introduced by Drinfeld [7] to give
solutions of the quantum Yang-Baxter equations. By definition, a
quasitriangular Hopf algebra is a Hopf algebras whose finite-dimensional
representations form a braided rigid tensor category, which naturally relates
to low dimensional topology (see [15, 16, 13, 32]). Since then, they have been
intensively studied.

The classification of finite-dimensional Hopf algebras has attracted the
attention of many mathematicians in recent years. And the classification of
finite-dimensional quasitriangular Hopf algebras can be regarded as an
important step towards the classification of all finite-dimensional Hopf
algebras: indeed, if $H$ is a finite-dimensional Hopf algebra its _Drinfeld
double_ , $D(H)$, is a quasitriangular Hopf algebra endowed with a Hopf
algebra inclusion $H\hookrightarrow D(H)$.

Let $p$ be a prime number. Hopf algebras of dimension $p$ over an
algebraically closed field $\Bbbk$ of characteristic zero were shown by Zhu
[34] to be isomorphic to the group algebra $\Bbbk[\mathbb{Z}_{p}]$. For a Hopf
algebra $H$ over $\Bbbk$ whose dimension is a product of two primes, there are
also many results: If $H$ is Hopf algebra with dimension $p^{2}$, then $H$ is
isomorphic to group algebra or the Taft algebra of dimension $p^{2}$ (see [21,
28]); Now let $p,q$ be distinct primes. If $p=2$, Ng (see [27]) proved that
$H$ must be semisimple. It is shown by Etingof and Gelaki (see [10]) that
semisimple Hopf algebra of dimension $pq$ is always trivial, that is, they are
isomorphic to group algebras or the duals of group algebras. In general, it is
still an open question whether every Hopf algebra with dimension $pq$ is
semisimple or not although it is widely believed that this should be true. For
the quasitriangular case, S. Natale has showed that all quasitriangular Hopf
algebras of dimension $pq$ with $p,q$ are odd primes, are always semisimple
and isomorphic to group algebras (see [26]).

The classification of a Hopf algebra whose dimension is three product of
primes in general is also open. The study of Hopf algebras of dimension
$p^{3}$ was carried out by García [11] and he proved that the only ribbon Hopf
algebras of dimension $p^{3}$ are group algebras and Frobenius-Lusztig
kernels; The Hopf algebras with dimension $2p^{2}$ ($p$ is an odd prime) has
been researched by several experts: Andruskiewitsch and Natale (see [2])
classified all such non-semisimple pointed Hopf algebras, and Hilgemann-Ng
(see [14]) proved that if a Hopf algebra $H$ of dimension $2p^{2}$ is not
semisimple, then $H$ or $H^{\ast}$ must be pointed. Cheng-Ng (see [4])
considered Hopf algebras with dimension $4p$ and proved that every such non-
semisimple Hopf algebra $H$ is pointed if and only if $H$ satisfies that
$|G(H)|>2$. The non-simple and semisimple Hopf algebras of dimension $pq^{2}$,
where $p,q$ are distinct primes, were classified in [25].

In this paper, we study non-simple quasitriangular Hopf algebras of dimension
$pq^{2}$, where $p,q$ are distinct odd primes. At first, we give a complete
list of them.

######  Theorem 1.1.

Let $H$ be a non-simple Hopf algebra of dimension $pq^{2}$ over $\Bbbk$.
Assume that $H$ admits a quasitriangular structure, then $H$ is semisimple and
is isomorphic to one of the following Hopf algebras:

  * (i)

a group algebra;

  * (ii)

either $\mathscr{A}_{0}$ or $\mathscr{B}_{0},\mathscr{B}_{\lambda_{j}}$ _(_
see Examples 2.9-2.10 for their definitions _)_ for $1\leq j\leq(p-1)/2$.

Then we determine all possible quasitriangular structures on them: All
possible quasitriangular structures on a group algebra of dimension $pq^{2}$
is given by Propositions 4.5 and 4.7; For Hopf algebras $\mathscr{A}_{0}$ and
$\mathscr{B}_{0},\mathscr{B}_{\lambda_{j}}\;(1\leq j\leq(p-1)/2)$, the
corresponding result is given by the next theorem.

######  Theorem 1.2.

With related notations given in Section $2$, we have the following:

  * (i)

All the braiding structures on $\mathscr{A}_{0}$ are given by $\langle
e_{h}g^{i},e_{k}g^{j}\rangle=\delta_{h,b^{j}}\delta_{k,b^{-i}}\lambda^{ij}$,
where $\lambda\in\Bbbk$ such that $\lambda^{q}=1$ and
$h,k\in\mathbb{Z}_{p}\rtimes\mathbb{Z}_{q}$, $0\leq i,j\leq q-1$;

  * (ii)

All the quasitriangular structures on $\mathscr{B}_{\lambda}$,
$\lambda\in\\{0,\lambda_{j}|\;1\leq j\leq(p-1)/2\\}$, are given by

$R=\sum\limits_{\begin{subarray}{l}0\leq i,j\leq q\\\ 0\leq k,l\leq q\\\
\end{subarray}}w(a^{i}b^{j},a^{k}b^{l})e_{a^{i}b^{j}}\otimes e_{a^{k}b^{l}},$

where $w$ is a bicharacter on $\mathbb{Z}_{q}\times\mathbb{Z}_{q}$ and is
determined by the following conditions

$\displaystyle
w(a,a)=w(a^{m},a^{m}),\;w(a,b)=w(a^{m},b^{m^{\lambda}})\eta(a^{m},b^{m^{\lambda}},g),$
$\displaystyle
w(b,a)=w(b^{m^{\lambda}},a^{m})\eta(b^{m^{\lambda}},a^{m},g),\;w(b,b)=w(b^{m^{\lambda}},b^{m^{\lambda}}).$

Note that $\mathscr{A}_{0}$ is self-dual as a Hopf algebra, a braiding
structure on $\mathscr{A}_{0}$ corresponds to a quasitriangular structure on
$\mathscr{A}_{0}$ and vise versa.

The paper is organized as follows. Section 2 is devoted to give some notation
and preliminary results. In Section 3, we will prove that all non-simple
quasitriangular Hopf algebras of dimension $pq^{2}$ are semisimple. Combining
with Natale’s ([25]) result, we get an appropriate range of these Hopf
algebras now. The last section concentrates on the research of possible
quasitriangular structures on them.

######  Convention 1.3.

_Throughout the paper we work over an algebraically closed field $\Bbbk$ of
characteristic 0. All Hopf algebras in this paper are finite dimensional. For
the symbol $\delta$, we mean the classical Kronecker’s symbol. Our references
for the theory of Hopf algebras are [22, 30]. For a Hopf algebra $H$, the
antipode of $H$ will denoted by $S$. We shall also use the notation $S_{H}$
when special emphasis is needed. For a Hopf algebra $H$, the group of group-
like elements in $H$ will be denoted by $G(H)$. A Hopf algebra $H$ is called
simple if it contains no proper normal Hopf subalgebras in the sense of [22,
3.4.1]; $H$ is called semisimple if it is semisimple as an algebra, and
cosemisimple if it is cosemisimple as a coalgebra._

##  2\. Preliminaries

We collect some necessary notions and results in this section.

###  2.1. Quasitriangular Hopf algebra.

Recall that a quasitriangular Hopf algebra is a pair $(H,R)$ where $H$ is a
Hopf algebra over $\Bbbk$ and $R=\sum R^{(1)}\otimes R^{(2)}$ is an invertible
element in $H\otimes H$ such that

$(\Delta\otimes\operatorname{Id})(R)=R_{13}R_{23},\;(\operatorname{Id}\otimes\Delta)(R)=R_{13}R_{12},\;\Delta^{op}(h)R=R\Delta(h),$

for $h\in H$. Here by definition $R_{12}=\sum R^{(1)}\otimes R^{(2)}\otimes 1$
and similarly for $R_{13}$ and $R_{23}$. The element $R$ is called a universal
$\mathcal{R}$-matrix of $H$ or a _quasitriangular structure_ on $H$. Dually,
the definition of coquasitriangular Hopf algebra can be given as follows. A
coquasitriangular Hopf algebra is a pair $(H,\langle,\rangle)$ where $H$ is a
Hopf algebra over $\Bbbk$ and $\langle,\rangle:H\otimes H\rightarrow\Bbbk$ is
invertible in the convolution algebra $\operatorname{Hom}_{\Bbbk}(H\otimes
H,\Bbbk)$ satisfying

$\langle ab,c\rangle=\langle a,c_{(1)}\rangle\langle
b,c_{(2)}\rangle,\;\langle a,bc\rangle=\langle a_{(1)},c\rangle\langle
c_{(2)},b\rangle,$ $\langle a_{(1)},b_{(1)}\rangle a_{(2)}b_{(2)}=\langle
a_{(2)},b_{(2)}\rangle b_{(1)}a_{(1)},$

for $a,b,c\in H$. The linear map $\langle,\rangle$ is called a _braiding
structure_ on $H$.

For a quasitriangular Hopf algebra $(H,R)$, there are Hopf algebra maps
$l_{R}:H^{\ast cop}\rightarrow H$ and $r_{R}:H^{\ast op}\rightarrow H$, given
respectively by

$l_{R}(f):=(f\otimes\operatorname{Id})(R),\;\;r_{R}(f):=(\operatorname{Id}\otimes
f)(R),\;f\in H^{\ast}.$

Let $H_{l}$ and $H_{r}$ denote, respectively, the images of $l_{R}$ and of
$r_{R}$. Then $H_{l}$ and $H_{r}$ are Hopf subalgebras of $H$ of dimension
$n>1$, unless $H$ is cocommutative and $R=1\otimes 1$. By [31, Proposition 2],
we have $H_{l}^{\ast cop}\cong H_{r}$. Since we will discuss the
semisimplicity of Hopf algebras, we recall the following theorem which will be
used tacitly in the remaining discussion, and the proof of it can be seen in
[18, 17].

######  Lemma 2.1.

The following statements on a finite-dimensional Hopf algebra $H$ over an
algebraically closed field of characteristic zero with antipode $S$ are
equivalent:

  * (i)

$H$ is semisimple;

  * (ii)

$H$ is cosemisimple;

  * (iii)

$\emph{Tr}(S^{2})\neq 0$;

  * (iv)

$S^{2}=\operatorname{Id}_{H}$.

Let $H_{R}\subseteq H$ be the minimal quasitriangular Hopf subalgebra of $H$
corresponding to $R$ (see [31]). It is proved in [31, Theorem 2] that $H_{R}$
is a quotient of $D(H_{l})$. In particular, if $H_{l}$ is semisimple, then
$H_{r}$ is semisimple, and therefore $H_{R}$ is semisimple. The following
result is just Theorem 1.3.5 in [12].

######  Lemma 2.2.

Let $(H,R)$ be a quasitriangular Hopf algebra over $\Bbbk$. If $H_{R}$ is
semisimple, then $S_{H}^{4}=\operatorname{Id}$.

Combining above results, we get the following observation.

######  Lemma 2.3.

Let $(H,R)$ be an odd-dimensional quasitriangular Hopf algebra over $\Bbbk$.
Then the following statements are equivalent:

  * (i)

$H$ is semisimple;

  * (ii)

$H_{R}$ is semisimple;

  * (iii)

$H_{l}$ is semisimple;

  * (iv)

$H_{r}$ is semisimple.

###### Proof.

Since a Hopf subalgebra of a semisimple Hopf algebra is semisimple, we know
(i) $\Rightarrow$ (ii) and (ii) $\Rightarrow$ (iii). By Lemma 2.1, we have
(iii) $\Leftrightarrow$ (iv). If $H_{r}$ is semisimple, then $H_{R}$ is
semisimple. Thus one can use Lemma 2.2 to get $S_{H}^{4}=\operatorname{Id}$.
By assumption, the dimension of $H$ is odd, therefore
$\text{Tr}(S_{H}^{2})\neq 0$. Thus, by Lemma 2.1, $H$ is semisimple. ∎

###  2.2. Hopf exact sequence.

######  Definition 2.4.

A short exact sequence of Hopf algebras is a sequence of Hopf algebras and
Hopf algebra maps

(2.1) $\;\;K\xrightarrow{\iota}H\xrightarrow{\pi}\overline{H}$

such that

  * (i)

$\iota$ is injective,

  * (ii)

$\pi$ is surjective,

  * (iii)

$\operatorname{Ker}(\pi)=HK^{+}$, $K^{+}$ is the kernel of the counit of $K$.

Take an exact sequence (2.1), then $K$ is a normal Hopf subalgebra of $H$.
Conversely, if $K$ is a normal Hopf subalgebra of a Hopf algebra $H$, then the
quotient coalgebra $H=H/HK^{+}=H/K^{+}H$ is a quotient Hopf algebra and $H$
fits into an extension (2.1), where $\iota$ and $\pi$ are the canonical maps.
The following lemma is shown in [26, Lemma 3.2].

######  Lemma 2.5.

Let the sequence (2.1) be a short exact sequence of finite-dimensional Hopf
algebras. Let also $L\subseteq H$ be a Hopf subalgebra. If $L$ is simple, then
either $L\subseteq\iota(K)$ or $L\cap\iota(K)=\Bbbk 1$. In the last case, the
restriction $\pi|_{L}$ is injective.

The following two results are [10, Theorem 6] and [26, Theorem 1.1]
respectively which will play crucial role in the proof of Proposition 3.3.

######  Lemma 2.6.

Let $H$ be a semisimple Hopf algebra of dimension $pq$ over $\Bbbk$, where $p$
and $q$ are distinct prime numbers. Then $H$ is trivial.

######  Lemma 2.7.

Let $H$ be a Hopf algebra of dimension $pq$ over $\Bbbk$ where $p$ and $q$ are
odd prime numbers. Assume that $H$ admits a quasitriangular structure. Then
$H$ is semisimple and isomorphic to a group algebra $\Bbbk F$, where $F$ is a
group of order $pq$.

An extension (2.1) above such that $K$ is commutative and $\overline{H}$ is
cocommutative is called _abelian_. In this situation, we know the extension
(2.1) can be written in the following form:

$\;\;\Bbbk^{G}\xrightarrow{\iota}H\xrightarrow{\pi}\Bbbk F,$

where $G,F$ are finite groups. Abelian extensions were classified by Masuoka
(see [20, Proposition 1.5]), and the above $H$ can be expressed as
$\Bbbk^{G}\\#_{\sigma,\tau}\Bbbk F$. To give the description of
$\Bbbk^{G}\\#_{\sigma,\tau}\Bbbk F$, we need the following data

  * (i)

A matched pair of groups, i.e. a quadruple
$(F,G,\triangleleft,\triangleright)$, where
$G\stackrel{{\scriptstyle\triangleleft}}{{\leftarrow}}G\times
F\stackrel{{\scriptstyle\triangleright}}{{\rightarrow}}F$ are action of groups
on sets, satisfying the following conditions

$\displaystyle g\triangleright(ff^{\prime})=(g\triangleright
f)((g\triangleleft f)\triangleright
f^{\prime}),\quad(gg^{\prime})\triangleleft
f=(g\triangleleft(g^{\prime}\triangleright f))(g^{\prime}\triangleleft f),$

for $g,g^{\prime}\in G$ and $f,f^{\prime}\in F$.

  * (ii)

$\sigma:G\times F\times F\rightarrow\Bbbk^{\times}$ is a map such that

$\displaystyle\sigma(g\triangleleft
f,f^{\prime},f^{\prime\prime})\sigma(g,f,f^{\prime}f^{\prime\prime})=\sigma(g,f,f^{\prime})\sigma(g,ff^{\prime},f^{\prime\prime})$

and $\sigma(1,f,f^{\prime})=\sigma(g,1,f^{\prime})=\sigma(g,f,1)=1$, for $g\in
G$ and $f,f^{\prime},f^{\prime\prime}\in F$.

  * (iii)

$\tau:G\times G\times F\rightarrow\Bbbk^{\times}$ is a map satisfying

$\displaystyle\tau(gg^{\prime},g^{\prime\prime},f)\tau(g,g^{\prime},g^{\prime\prime}\triangleright
f)=\tau(g^{\prime},g^{\prime\prime},f)\tau(g,g^{\prime}g^{\prime\prime},f)$

and $\tau(g,g^{\prime},1)=\tau(g,1,f)=\tau(1,g^{\prime},f)$, for
$g,g^{\prime},g^{\prime\prime}\in G$ and $f\in F$. Moreover, the $\sigma,\tau$
satisfy the following compatible condition

$\displaystyle\sigma(gg^{\prime},f,f^{\prime})\tau(g,g^{\prime},ff^{\prime})$
$\displaystyle=\sigma(g,g^{\prime}\triangleright f,(g^{\prime}\triangleleft
f)\triangleright f^{\prime})\sigma(g^{\prime},f,f^{\prime})$
$\displaystyle\tau(g,g^{\prime},f)\tau(g\triangleleft(g^{\prime}\triangleleft
f),g^{\prime}\triangleleft f,f^{\prime}),$

for $g,g^{\prime},g^{\prime\prime}\in G$ and $f,f^{\prime},f^{\prime\prime}\in
F$.

######  Definition 2.8.

[1, Section 2.2] The Hopf algebra $\Bbbk^{G}\\#_{\sigma,\tau}\Bbbk F$ is equal
to $\Bbbk^{G}\otimes\Bbbk F$ as vector space and we write $a\otimes x$ as
$a\\#x$. The product, coproduct are given by

$\displaystyle(e_{g}\\#f).(e_{g^{\prime}}\\#f^{\prime})=\delta_{g\triangleleft
f,g^{\prime}}\;\sigma(g,f,f^{\prime})\;e_{g}\\#(ff^{\prime}),$
$\displaystyle\Delta(e_{g}\\#f)=\sum_{g^{\prime}g^{\prime\prime}=g}\;\tau(g^{\prime},g^{\prime\prime},f)\;e_{g^{\prime}}\\#g^{\prime\prime}\triangleright
f\otimes e_{g^{\prime\prime}}\\#f,$

The unit is $\sum_{g\in G}e_{g}\\#1$ and the counit is
$\epsilon(e_{g}\\#f)=\delta_{g,1}$ and the antipode is

$\displaystyle S(e_{g}\\#f)=\sigma(g^{-1},g\triangleright f,(g\triangleright
f)^{-1})^{-1}\;\tau(g^{-1},g,f)^{-1}\;e_{(g\triangleleft
f)^{-1}}\\#(g\triangleright f)^{-1}.$

We need the following examples of $\Bbbk^{G}\\#_{\sigma,\tau}\Bbbk F$ for
further discussion.

######  Example 2.9.

_Let $p,q$ be two prime numbers such that $p\equiv 1(\text{mod}\;q)$ and let
$\omega$ be a primitive $q$th root of 1 in $\Bbbk$. Assume $t\in\mathbb{N}$
satisfying $t^{q}\equiv 1(\text{mod}\;p)$ and $t\not\equiv 1(\text{mod}\;p)$.
Let $0\leq l\leq(q-1)$, then the Hopf algebra $\mathscr{A}_{l}$ [25, Lemma
1.3.9] belongs to $\Bbbk^{G}\\#_{\sigma,\tau}\Bbbk F$. By definition, the data
$(G,F,\triangleleft,\triangleright,\sigma,\tau)$ of $\mathscr{A}_{l}$ is given
by the following way_

  * (i)

$G=\mathbb{Z}_{p}\rtimes\mathbb{Z}_{q}=\langle
a,b|\;a^{p}=b^{q}=1,bab^{-1}=a^{t}\rangle$, $F=\mathbb{Z}_{q}=\langle
g|\;g^{q}=1\rangle$. The action $\triangleright$ is trivial, and
$a\triangleleft g^{i}=a^{t^{i}},b\triangleleft g^{i}=b$, for $0\leq i\leq
q-1$.

  * (ii)

$\sigma(a^{i}b^{j},g^{m},g^{n})=w^{jlq_{mn}}$, where $q_{mn}$ is the quotient
of $m+n$ in the division by $q$ and $1\leq i\leq p-1$, $0\leq j,m,n\leq q-1$.

  * (iii)

$\tau(g,g^{\prime},f)=1$ for $g,g^{\prime}\in G$ and $f\in F$.

_For ease of use, we define a character $\phi:G\rightarrow\Bbbk$ by
$\phi(a):=1,\phi(b):=\omega$. Then the above $\sigma$ in Example 2.9 can be
expressed by $\sigma(a^{i}b^{j},g^{m},g^{n})=\phi^{lq_{mn}}(a^{i}b^{j})$,
where $i,j,m,n,q_{mn}$ are the same as (ii)._

######  Example 2.10.

_Let $p,q$ be two prime numbers such that $q\equiv 1(\text{mod}\;p)$ and let
$\zeta$ be a primitive $q$th root of 1 in $\Bbbk$. Assume $m\in\mathbb{N}$
satisfying $m^{p}\equiv 1(\text{mod}\;q)$ and $m\not\equiv 1(\text{mod}\;p)$.
Let $0\leq\lambda\leq(p-1)$, then the Hopf algebra $\mathscr{B}_{\lambda}$
[25, Proposition 1.4.7] belongs to $\Bbbk^{G}\\#_{\sigma,\tau}\Bbbk F$. By
definition, the data $(G,F,\triangleleft,\triangleright,\sigma,\tau)$ of
$\mathscr{B}_{\lambda}$ is given by the following way_

  * (i)

$G=\mathbb{Z}_{q}\times\mathbb{Z}_{q}=\langle
a,b|\;a^{q}=b^{q}=1,ab=ba\rangle$, $F=\mathbb{Z}_{p}=\langle
g|\;g^{p}=1\rangle$. The action $\triangleright$ is trivial, and
$a\triangleleft g^{-i}=a^{m^{i}},b\triangleleft g^{-i}=b^{m^{\lambda i}}$, for
$0\leq i\leq p-1$.

  * (ii)

$\sigma(g,f,f^{\prime})=1$ for $g\in G$ and $f,f^{\prime}\in F$.

  * (iii)

$\tau(a^{i}b^{j},a^{k}b^{l},g^{n})=\zeta_{n}^{jk}$, where
$\zeta_{n}=\zeta^{c_{n}(m^{\lambda+1})}$, here $c_{n}(r):=1+r+...+r^{n-1}$ for
$r\in\mathbb{Z}$ and $0\leq i,j,k,l\leq q$, $0\leq n\leq p-1$.

Let $0\leq\lambda\leq(p-1)$. For convenience, we will denote the dual of a
Hopf algebra $H$ as $H^{\ast}$. Since we want to express
$\mathscr{B}_{\lambda}^{\ast}$ as the form $\Bbbk^{G}\\#_{\sigma,\tau}\Bbbk
F$, we give the following lemma.

######  Lemma 2.11.

Assume the data of $\Bbbk^{G}\\#_{\sigma,\tau}\Bbbk F$ is
$(G,F,\triangleleft,\triangleright,\sigma,\tau)$. If the action
$\triangleright$ is trivial, i.e $g\triangleright f=f$ for $g\in G,f\in F$,
then $(\Bbbk^{G}\\#_{\sigma,\tau}\Bbbk
F)^{\ast}\cong\Bbbk^{G^{\prime}}\\#_{\sigma^{\prime},\tau^{\prime}}\Bbbk
F^{\prime}$ as a Hopf algebra, where the data
$(G^{\prime},F^{\prime},\triangleleft^{\prime},{{}^{\prime}\triangleright},\sigma^{\prime},\tau^{\prime})$
of $\Bbbk^{G^{\prime}}\\#_{\sigma^{\prime},\tau^{\prime}}\Bbbk F^{\prime}$ is
given as follows

  * (i)

$G^{\prime}=F,\;F^{\prime}=G$. The action $\triangleleft^{\prime}$ is trivial
and $f(^{\prime}\triangleright)g=g\triangleleft f^{-1}$;

  * (ii)

$\sigma^{\prime}:G^{\prime}\times F^{\prime}\times
F^{\prime}\rightarrow\Bbbk^{\times}$ is defined by
$\sigma^{\prime}(f,g,g^{\prime})=\tau(g\triangleleft
f^{-1},g^{\prime}\triangleleft f^{-1},f)$ for $g,g^{\prime}\in G$ and $f\in
F$.

  * (iii)

$\tau^{\prime}:G^{\prime}\times G^{\prime}\times
F^{\prime}\rightarrow\Bbbk^{\times}$ is given by
$\tau^{\prime}(f,f^{\prime},g)=\sigma(g\triangleleft(ff^{\prime})^{-1},f,f^{\prime})$
for $g\in G$ and $f,f^{\prime}\in F$.

###### Proof.

For the Hopf algebra $\Bbbk^{G}\\#_{\sigma,\tau}\Bbbk F$, we denote the dual
basis of $\\{e_{g}\\#f|\;g\in G,f\in F\\}$ by $\\{E_{g;f}|\;g\in G,f\in F\\}$,
that is,
$E_{g;f}(e_{g^{\prime}}\\#f^{\prime})=\delta_{g,g^{\prime}}\delta_{f,f^{\prime}}$
for $g,g^{\prime}\in G$ and $f,f^{\prime}\in F$. Define
$\varphi:(\Bbbk^{G}\\#_{\sigma,\tau}\Bbbk
F)^{\ast}\rightarrow\Bbbk^{G^{\prime}}\\#_{\sigma^{\prime},\tau^{\prime}}\Bbbk
F^{\prime}$ by $\varphi(E_{g;f})=e_{f}\\#g\triangleleft f$ for $g\in G$ and
$f\in F$. Then we claim $\varphi$ is an isomorphism of Hopf algebras.

By definition, we will get
$E_{g;f}E_{g^{\prime};f^{\prime}}=\delta_{f,f^{\prime}}\tau(g,g^{\prime},f)E_{gg^{\prime},f}$
for $g,g^{\prime}\in G$ and $f,f^{\prime}\in F$. Thus we have

(2.2) $\displaystyle\varphi(E_{g;f}E_{g^{\prime};f^{\prime}})$
$\displaystyle=\delta_{f,f^{\prime}}\tau(g,g^{\prime},f)\varphi(E_{gg^{\prime},f})$
$\displaystyle=\delta_{f,f^{\prime}}\tau(g,g^{\prime},f)e_{f}\\#(gg^{\prime})\triangleleft
f.$

Directly, we have

(2.3) $\displaystyle\varphi(E_{g;f})\varphi(E_{g^{\prime};f^{\prime}})$
$\displaystyle=(e_{f}\\#g\triangleleft
f)(e_{f^{\prime}}\\#g^{\prime}\triangleleft f^{\prime})$
$\displaystyle=\delta_{f\triangleleft^{\prime}(g\triangleleft
f),f^{\prime}}\sigma^{\prime}(f,g\triangleleft f,g^{\prime}\triangleleft
f)e_{f}\\#((gg^{\prime})\triangleleft f)$
$\displaystyle=\delta_{f,f^{\prime}}\tau(g,g^{\prime},f)e_{f}\\#(gg^{\prime})\triangleleft
f.$

By the equations (2.2)-(2.3), we know $\varphi$ is an algebra map. More, it is
not hard to see that $\varphi^{\ast}(E_{f;g})=e_{g\triangleleft f^{-1}}\\#f$
for $f\in F$ and $g\in G$, where $\varphi^{\ast}$ is the dual map of
$\varphi$. To show $\varphi$ is coalgebra map, we only need to prove that
$\varphi^{\ast}$ is an algebra map. Similarly, for the Hopf algebra
$\Bbbk^{G^{\prime}}\\#_{\sigma^{\prime},\tau^{\prime}}\Bbbk F^{\prime}$, we
denote the dual basis of $\\{e_{f}\\#g|\;g\in G,f\in F\\}$ by
$\\{E_{f;g}|\;g\in G,f\in F\\}$, that is,
$E_{f;g}(e_{f^{\prime}}\\#g^{\prime})=\delta_{f,f^{\prime}}\delta_{g,g^{\prime}}$
for $g,g^{\prime}\in G$ and $f,f^{\prime}\in F$. Then we know
$E_{f;g}E_{f^{\prime};g^{\prime}}=\delta_{g\triangleleft
f^{\prime},g^{\prime}}\sigma(g^{\prime}\triangleleft(ff^{\prime})^{-1},f,f^{\prime})E_{ff^{\prime};g^{\prime}}$
for $g,g^{\prime}\in G$ and $f,f^{\prime}\in F$. Thus we have

(2.4) $\displaystyle\varphi^{\ast}(E_{f;g}E_{f^{\prime};g^{\prime}})$
$\displaystyle=\delta_{g\triangleleft
f^{\prime},g^{\prime}}\sigma(g^{\prime}\triangleleft(ff^{\prime})^{-1},f,f^{\prime})\varphi^{\ast}(E_{ff^{\prime};g^{\prime}})$
$\displaystyle=\delta_{g\triangleleft
f^{\prime},g^{\prime}}\sigma(g^{\prime}\triangleleft(ff^{\prime})^{-1},f,f^{\prime})e_{g^{\prime}\triangleleft(ff^{\prime})^{-1}}\\#ff^{\prime}$
$\displaystyle=\delta_{g,g^{\prime}\triangleleft(f^{\prime})^{-1}}\sigma(g^{\prime}\triangleleft(ff^{\prime})^{-1},f,f^{\prime})e_{g^{\prime}\triangleleft(ff^{\prime})^{-1}}\\#ff^{\prime}$
$\displaystyle=\delta_{g,g^{\prime}\triangleleft(f^{\prime})^{-1}}\sigma(g\triangleleft
f^{-1},f,f^{\prime})e_{g\triangleleft(f)^{-1}}\\#ff^{\prime}.$

Another way, we get

(2.5)
$\displaystyle\varphi^{\ast}(E_{f;g})\varphi^{\ast}(E_{f^{\prime};g^{\prime}})$
$\displaystyle=(e_{g\triangleleft
f^{-1}}\\#f)(e_{g^{\prime}\triangleleft(f^{\prime})^{-1}}\\#f^{\prime})$
$\displaystyle=\delta_{g,g^{\prime}\triangleleft(f^{\prime})^{-1}}\sigma(g\triangleleft
f^{-1},f,f^{\prime})e_{g\triangleleft f^{-1}}\\#ff^{\prime}.$

Therefore $\varphi$ is coalgebra map by equations (2.4)-(2.5). By definition,
we get $\varphi$ is isomorphism as vector space. Hence we have completed the
proof. ∎

Using this lemma, we get the following characterization of
$\mathscr{B}_{\lambda}^{\ast}$ (see Example 2.10).

######  Example 2.12.

_Let $p,q$ be two prime numbers such that $q\equiv 1(\text{mod}\;p)$ and let
$\zeta$ be a primitive $q$th root of 1 in $\Bbbk$. Assume $m\in\mathbb{N}$
satisfying $m^{p}\equiv 1(\text{mod}\;q)$ and $m\not\equiv 1(\text{mod}\;p)$.
Let $0\leq\lambda\leq(p-1)$, then $\mathscr{B}_{\lambda}^{\ast}$ belongs to
$\Bbbk^{G}\\#_{\sigma,\tau}\Bbbk F$. Due to Lemma 2.11 and the Example 2.10,
the data $(G,F,\triangleleft,\triangleright,\sigma,\tau)$ of
$\mathscr{B}_{\lambda}^{\ast}$ is given by the following way_

  * (i)

$G=\mathbb{Z}_{p}=\langle g|\;g^{p}=1\rangle$,
$F=\mathbb{Z}_{q}\times\mathbb{Z}_{q}=\langle
a,b|\;a^{q}=b^{q}=1,ab=ba\rangle$. The action $\triangleleft$ is trivial, and
$g^{i}\triangleright a=a^{m^{i}},g^{i}\triangleright b=b^{m^{\lambda i}}$, for
$0\leq i\leq p-1$.

  * (ii)

$\sigma(g^{n},a^{i}b^{j},a^{k}b^{l})=\zeta_{n}^{jkm^{(\lambda+1)n}}$, where
$\zeta_{n}=\zeta^{c_{n}(m^{\lambda+1})}$, here $c_{n}(r):=1+r+...+r^{n-1}$ for
$r\in\mathbb{Z}$ and $0\leq i,j,k,l\leq q$, $0\leq n\leq p-1$.

  * (iii)

$\tau(f,f^{\prime},g)=1$ for $g\in G$ and $f,f^{\prime}\in F$.

######  Remark 2.13.

_Recall the definition of $\mathscr{B}_{\lambda}$, we know $\Bbbk
G(\mathscr{B}_{\lambda})=\Bbbk^{G}$, where
$G=\mathbb{Z}_{q}\times\mathbb{Z}_{q}$. By definition, we get
$G(\mathscr{B}_{\lambda}^{\ast})$ is the set of characters of
$\mathscr{B}_{\lambda}$. Through a direct calculation, we get
$|G(\mathscr{B}_{0}^{\ast})|=pq$ and $|G(\mathscr{B}_{\lambda}^{\ast})|=p$ for
$\lambda\neq 0$. Moreover, by Example 2.12, we further obtain $\Bbbk
G(\mathscr{B}_{0}^{\ast})=\langle e_{g^{i}}b^{j}|\;0\leq i\leq p-1,0\leq i\leq
q-1\rangle$ as vector space while $\Bbbk
G(\mathscr{B}_{\lambda}^{\ast})=\langle e_{g^{i}}|\;0\leq i\leq p-1\rangle$ as
vector space for $\lambda\neq 0$._

##  3\. Non-simple quasitriangular Hopf algebras of dimension $pq^{2}$ is
semisimple

In the following content, we always assume that $p,q$ are distinct odd primes.
In this section, we will prove that a non-simple quasitriangular Hopf algebra
of dimension $pq^{2}$ must be semisimple. We start with two basic
observations.

######  Lemma 3.1.

Let $(H,R)$ be a quasitriangular Hopf algebra of dimension $pq^{2}$ over
$\Bbbk$. Assume $H$ is not simple as Hopf algebra and $H$ is not semisimple.
Then $H$ fits into a short exact sequence (2.1) such that
$\emph{dim}(K)\in\\{pq,q^{2}\\}$.

###### Proof.

By assumption, $H$ is not simple. Thus we can assume that $H$ fits into a
short exact sequence (2.1) and $K\neq\Bbbk 1$ is a proper Hopf subalgebra.
Since $(H,R)$ is a quasitriangular Hopf algebra and $\overline{H}$ is a
quotient of $H$, we get $\overline{H}$ is a quasitriangular Hopf algebra. By
Nichols-Zoeller theorem, we have $\text{dim}(K)\in\\{p,q,pq,q^{2}\\}$. To
complete the proof, we only need to show $\text{dim}(K)\not\in\\{p,q\\}$.
Assume $\text{dim}(K)=p$, then $K$ is a group algebra [34] and
$\text{dim}(\overline{H})=q^{2}$. Therefore $K$ is semisimple Hopf algebra.
Due to Lemma 2.7, $\overline{H}$ is also a group algebra. Hence $\overline{H}$
is semisimple Hopf algebra. Since an extension of semisimple Hopf algebras is
also semisimple [3], we get $H$ is semisimple. But this contradicts our
assumption about non-semisimplicity of $H$. Therefore $\text{dim}(K)\neq p$.
Similarly, we have $\text{dim}(K)\neq q$. ∎

######  Lemma 3.2.

Let $(H,R)$ be a quasitriangular Hopf algebra of dimension $pq^{2}$ over
$\Bbbk$. Assume $H$ is not simple as Hopf algebra and $H$ is not semisimple.
Then $H^{\ast cop}\not\cong H$.

###### Proof.

By Lemma 3.1, we can assume that $H$ fits into a short exact sequence (2.1)
such that $\text{dim}(K)\in\\{pq,q^{2}\\}$. Assume $H^{\ast cop}\cong H$.
Since $(H,R)$ is quasitriangular Hopf algebra, we know $H^{\ast cop}$ is also
a quasitriangular Hopf algebra. This implies $H^{\ast}$ is quasitriangular
Hopf algebra. Because $K^{\ast}$ is a quotient of $H^{\ast}$, we obtain
$K^{\ast}$ is quasitriangular Hopf algebra. Moreover,
$\text{dim}(K^{\ast})\in\\{pq,q^{2}\\}$. By Lemma 2.7, $K^{\ast}$ is trivial
and hence $K$ is semisimple. However, we know $\overline{H}$ is also
semisimple due to the dimensional reason. Because an extension of semisimple
Hopf algebras is also semisimple [3], we get that $H$ must be a semisimple
Hopf algebra. This contradicts our assumption about non-semisimplicity of $H$.
Therefore we have $H^{\ast cop}\not\cong H$. ∎

Now we can get what we want.

######  Proposition 3.3.

Let $(H,R)$ be a quasitriangular Hopf algebra of dimension $pq^{2}$ over
$\Bbbk$. Assume $H$ is not simple as Hopf algebra, then $H$ is semisimple.

###### Proof.

Suppose that $H$ is not semisimple. By Lemma 2.3, $H_{R}$ is not semisimple.
Due to Nichols-Zoeller theorem [29] and the classification of Hopf algebras of
prime dimension [34], we must have
$\text{dim}(H_{R})\in\\{pq,q^{2},pq^{2}\\}$. By Lemma 2.7, we must have
$\text{dim}(H_{R})=pq^{2}$. Since $H_{R}$ is a quotient of $D(H_{l})$ [31,
Theorem 2], it follows that $\text{dim}(H_{R})|\text{dim}(H_{l})^{2}$. Thus
$\text{dim}(H_{l})\in\\{pq,pq^{2}\\}$.

Assume $\text{dim}(H_{l})=pq$. By Lemma 2.3, we get $H_{l}$ is not semisimple.
Since the dimensional reason, we obtain $H_{l}$ must be simple as Hopf
algebra. Due to Lemma 3.1, we can assume that $H$ fits into a short exact
sequence (2.1) such that $\text{dim}(K)\in\\{pq,q^{2}\\}$. Assume
$\text{dim}(K)=pq$. Since $H_{l}$ is simple as Hopf algebra, we can use Lemma
2.5 to get $H_{l}\subseteq K$ or $H_{l}\cap K=\Bbbk 1$. Since
$\text{dim}(H_{l})=pq$ and $\text{dim}(\overline{H})=q$, we know
$\pi|_{H_{l}}$ is not injective. By Lemma 2.5, we get $H_{l}\cap K\neq\Bbbk
1$. Thus $H_{l}\subseteq K$. Because $\text{dim}(H_{l})=\text{dim}(K)=pq$ in
this case, we obtain $H_{l}=K$. Similarly, we have $H_{r}=K$. Therefore
$H_{R}=K$. Since $\text{dim}(K)=pq$, we have $\text{dim}(H_{R})=pq$. By Lemma
2.7, we get $H_{R}$ is semisimple. But this contradicts with the previous
conclusion about $H_{R}$, so $\text{dim}(K)\neq pq$. Suppose
$\text{dim}(K)=q^{2}$. By Lemma 2.5, $H_{l}\subseteq K$ or $H_{l}\cap K=\Bbbk
1$. Since $\text{dim}(H_{l})=pq$ and $\text{dim}(K)=q^{2}$, we get $H_{l}\neq
K$. Hence $H_{l}\cap K=\Bbbk 1$. This implies that $\pi|_{H_{l}}$ is injective
by Lemma 2.5 again. Since $\text{dim}(\overline{H})=p$ and
$\text{dim}(H_{l})=pq$, we get $\pi|_{H_{l}}$ is not injective. Thus
$\text{dim}(K)\neq q^{2}$. Then we get a conclusion that
$\text{dim}(H_{l})\neq pq$.

Assume $\text{dim}(H_{l})=pq^{2}$. Then we have $H_{l}=H_{r}=H$ in this case.
Thus $H^{\ast cop}\cong H$. But this can’t happen by Lemma 3.2. Hence the non-
semisimple hypothesis for $H$ is not true. ∎

##  4\. Determination of Quasitriangular Structures

In this section, we will settle the proof of Theorems 1.1 and 1.2.

###  4.1. An range of Hopf algebras.

The classification of non-trivial semisimple Hopf algebras of dimension
$pq^{2}$ over $\Bbbk$ which are not simple as Hopf algebras has been given in
[25]. To state this result, we need a set of numbers. Since $p$ is an odd
prime, we can find a subset $\\{\lambda_{i}|\;1\leq i\leq(p-1)/2\\}$ from the
set $\\{1,2,...,p-2\\}$ such that $\lambda_{i}\lambda_{j}\neq
1(\text{mod}\;p)$ if $i\neq j$. Next, we always assume that we have chosen a
subset $\\{\lambda_{i}|\;1\leq i\leq(p-1)/2\\}$ so that
$\lambda_{i}\lambda_{j}\neq 1(\text{mod}\;p)$ if $i\neq j$. For example, if
$p=5$ then we can choose $\\{1,2\\}$ to be a needed subset. The following
result is [25, Theorem 3.12.4].

######  Lemma 4.1.

Suppose $A$ is a non-trivial semisimple Hopf algebras of dimension $pq^{2}$
over $\Bbbk$ which are not simple as Hopf algebra. Then either

  * (i)

$p\equiv 1(\emph{mod}\;q)$, and $A$ is isomorphic to either of the Hopf
algebras $\mathscr{A}_{l}$, $0\leq l\leq q-1$ in Example 2.9, or

  * (ii)

$q\equiv 1(\emph{mod}\;p)$, and $A$ is isomorphic to either of the Hopf
algebras $\mathscr{B}_{0}$ or $\mathscr{B}_{\lambda_{j}}$, $1\leq
j\leq(p-1)/2$ and their duals. See Example 2.10.

Combining Proposition 3.3 and this lemma, we have

######  Corollary 4.2.

With notions above and let $H$ be a non-simple quasitriangular Hopf algebra of
dimension $pq^{2}$, then it is isomorphic to one of the following Hopf
algebras:

  * 1)

A group algebra;

  * 2)

$\mathscr{A}_{l}$ for $0\leq l\leq q-1$;

  * 3)

$\mathscr{B}_{0}$, $\mathscr{B}_{\lambda_{j}}$, $\mathscr{B}_{0}^{\ast}$,
$\mathscr{B}_{\lambda_{j}}^{\ast}$ for $1\leq j\leq(p-1)/2$.

It is well-known that a group algebra is cocommutative and thus it always has
a trivial quasitriangular structure. Therefore, to complete the classification
of non-simple quasitriangular Hopf algebras of dimension $pq^{2}$, we need
consider possible quasitriangular structures on Hopf algebras stated in 2) and
3) of above corollary.

###  4.2. Quasitriangular structures on a group algebra of dimension $pq^{2}$.

Let $G$ be a finite group. If $G$ is happened to be an abelian group, then it
is well-known that quasitriangular structures are just bicharacters and so we
only consider the question for nonabelian groups in this subsection. For the
purpose, we first need the following lemma.

######  Lemma 4.3.

Let $G$ be a group and $K\subseteq G$ a normal and commutative subgroup.
Assume that $\\{e_{k}\\}_{k\in K}$ are the orthogonal idempotents of
$\Bbbk[K]$ and $R=\sum_{k,k^{\prime}\in K}w(k,k^{\prime})e_{k}\otimes
e_{k^{\prime}}$, where $w$ is a bicharacter on $K$. For $g\in G$, we denote
$ge_{k}g^{-1}=\varphi_{g}(k)$, $k\in K$. Then $R$ is a quasitriangular
structure on $\Bbbk[G]$ if and only if
$w(k,k^{\prime})=w(\varphi_{g}(k),\varphi_{g}(k^{\prime}))$ for
$k,k^{\prime}\in K$ and $g\in G$.

###### Proof.

By a direct computation. ∎

The following lemma is the result in [5, Section 4]. And we try to keep the
notation in [5, Section 4] as much as possible.

######  Lemma 4.4.

Let $G$ be a group of order $pq^{2}$_(_ $q>p$_)_. Then $G$ is isomorphic to
the following cases

  * (1)

$\beta_{1}=\mathbb{Z}_{pq^{2}}$;

  * (2)

$\beta_{2}=\mathbb{Z}_{p}\bigoplus\mathbb{Z}_{q}\bigoplus\mathbb{Z}_{q}$;

  * (3)

$p\mid(q^{2}-1)$ and $\beta_{3}=\langle
s,t|\;s^{q^{2}}=t^{p}=1,tst^{-1}=s^{m}\rangle$, $m\in\mathbb{N}$ satisfying
$m^{p}\equiv 1(\emph{mod}\;q^{2})$ and $m\not\equiv 1(\emph{mod}\;q^{2})$;

  * (4)

$p\mid(q-1)$ and $\beta_{4}=\langle
s,t,u|\;s^{q}=t^{q}=u^{p}=1,tst^{-1}=s,usu^{-1}=s,utu^{-1}=t^{m}\rangle$,
$m\in\mathbb{N}$ satisfying $m^{p}\equiv 1(\emph{mod}\;q)$ and $m\not\equiv
1(\emph{mod}\;q)$;

  * (5)

$p\mid(q-1)$ and $\beta_{5}=\langle
s,t,u|\;s^{q}=t^{q}=u^{p}=1,tst^{-1}=s,usu^{-1}=s^{m},utu^{-1}=t^{m}\rangle$,
$m\in\mathbb{N}$ satisfying $m^{p}\equiv 1(\emph{mod}\;q)$ and $m\not\equiv
1(\emph{mod}\;q)$;

  * (6)

$p\mid(q-1)$ and $\beta_{6}=\langle
s,t,u|\;s^{q}=t^{q}=u^{p}=1,tst^{-1}=s,usu^{-1}=s^{m},utu^{-1}=t^{n}\rangle$,
$m\in\mathbb{N}$ satisfying $m^{p}\equiv 1(\emph{mod}\;q),n^{p}\equiv
1(\emph{mod}\;q)$ and $m\not\equiv 1(\emph{mod}\;q),n\not\equiv
1(\emph{mod}\;q)$ and $m\not\equiv n(\emph{mod}\;q)$;

  * (7)

$p\mid(q+1)$ and $\beta_{7}=\langle
s,t,u|\;s^{q}=t^{q}=u^{p}=1,tst^{-1}=s,usu^{-1}=t,utu^{-1}=s^{m}t^{n},u^{q}su^{-q}=s\rangle$,
$m,n\in\mathbb{N}$.

Moreover, if $G$ is not abelian, i.e. $G\in\\{\beta_{i}|\;3\leq i\leq 7\\}$,
then $G$ has a largest non-trivial commutative normal subgroup, that is, it
contains all other non-trivial commutative normal subgroups _(_ see next
proposition for the construction of this largest subgroup _)_.

######  Proposition 4.5.

We have

  * (1)

All the quasitriangular structures on $\beta_{3}$ are given by $R=\sum_{g,h\in
K}w(g,h)e_{g}\otimes e_{h},$ where $w$ is a bicharacter on $K=\langle
s\rangle$ such that $w(s^{m},s^{m})=w(s,s)$;

  * (2)

All the quasitriangular structures on $\beta_{4}$ are given by $R=\sum_{g,h\in
K}w(g,h)e_{g}\otimes e_{h},$ where $w$ is a bicharacter on $K=\langle
s,t\rangle$ such that

$w(s,t^{m})=w(s,t),\;w(t^{m},s)=w(t,s),\;w(t^{m},t^{m})=w(t,t);$

  * (3)

All the quasitriangular structures on $\beta_{5}$ are given by $R=\sum_{g,h\in
K}w(g,h)e_{g}\otimes e_{h},$ where $w$ is a bicharacter on $K=\langle
s,t\rangle$ such that

$w(s^{m},s^{m})=w(s,s),\;w(s^{m},t^{m})=w(s,t),$
$\;w(t^{m},s^{m})=w(t,s),\;w(t^{m},t^{m})=w(t,t);$

  * (4)

All the quasitriangular structures on $\beta_{6}$ are given by $R=\sum_{g,h\in
K}w(g,h)e_{g}\otimes e_{h},$ where $w$ is a bicharacter on $K=\langle
s,t\rangle$ such that

$w(s^{m},s^{m})=w(s,s),\;w(s^{m},t^{n})=w(s,t),$
$\;w(t^{n},s^{m})=w(t,s),\;w(t^{n},t^{n})=w(t,t);$

  * (5)

All the quasitriangular structures on $\beta_{7}$ are given by $R=\sum_{g,h\in
K}w(g,h)e_{g}\otimes e_{h},$ where $w$ is a bicharacter on $K=\langle
s,t\rangle$ such that

$w(t,t)=w(s,s),\;w(t,s^{m}t^{n})=w(s,t),$
$\;w(s^{m}t^{n},t)=w(t,s),\;w(s^{m}t^{n},s^{m}t^{n})=w(t,t);$

###### Proof.

By Lemma 4.4, we can assume $K$ is the largest non-trivial commutative normal
subgroup of $G$. Assume $R$ is a quasitriangular structures on $\Bbbk G$, then
$H_{R}\in\Bbbk[K]\otimes\Bbbk[K]$ by [6, Theorem 3]. Due to Lemma 4.3, we get
what we want. ∎

Assume $q<p$, then the following lemma is the result in [5, Section 5]. And we
also keep the notation in [5, Section 5] as much as possible.

######  Lemma 4.6.

Let $G$ be a group of order $pq^{2}$($q<p$). Then $G$ is isomorphic to the
following cases

  * (1)

$\gamma_{1}=\mathbb{Z}_{pq^{2}}$;

  * (2)

$\gamma_{2}=\mathbb{Z}_{p}\bigoplus\mathbb{Z}_{q}\bigoplus\mathbb{Z}_{q}$;

  * (3)

$q\mid(p-1)$ and $\gamma_{3}=\langle
s,t|\;s^{p}=t^{q^{2}}=1,tst^{-1}=s^{m}\rangle$, $m\in\mathbb{N}$ satisfying
$m^{q}\equiv 1(\emph{mod}\;p)$ and $m\not\equiv 1(\emph{mod}\;p)$;

  * (4)

$q^{2}\mid(p-1)$ and $\gamma_{4}=\langle
s,t|\;s^{p}=t^{q^{2}}=1,tst^{-1}=s^{m}\rangle$, $m\in\mathbb{N}$ satisfying
$m^{q^{2}}\equiv 1(\emph{mod}\;p)$ and $m^{q}\not\equiv 1(\emph{mod}\;p)$;

  * (5)

$q\mid(p-1)$ and $\gamma_{5}=\langle
s,t,u|\;s^{p}=t^{q}=u^{q}=1,tut^{-1}=u,tst^{-1}=s,usu^{-1}=s^{m}\rangle$,
$m\in\mathbb{N}$ satisfying $m^{q}\equiv 1(\emph{mod}\;p)$ and $m\not\equiv
1(\emph{mod}\;p)$;

  * (6)

$q\mid(p-1)$ and $\gamma_{6}=\langle
s,t,u|\;s^{p}=t^{q}=u^{q}=1,tut^{-1}=u,tst^{-1}=s^{m},usu^{-1}=s^{n}\rangle$,
$m,n\in\mathbb{N}$ satisfying $m^{q}\equiv n^{q}\equiv 1(\emph{mod}\;p)$ and
$m\not\equiv 1(\emph{mod}\;p),n\not\equiv 1(\emph{mod}\;p)$ and $m\not\equiv
n(\emph{mod}\;p)$;

Moreover, if $G$ is not abelian, i.e. $G\in\\{\gamma_{i}|\;1\leq i\leq 6,i\neq
1,2\\}$, then $G$ has a largest non-trivial commutative and normal subgroup.

######  Proposition 4.7.

We have

  * (1)

All the quasitriangular structures on $\gamma_{3}$ are given by
$R=\sum_{g,h\in K}w(g,h)e_{g}\otimes e_{h},$ where $w$ is a bicharacter on
$K=\langle s,t^{q}\rangle$ such that

$w(s^{m},s^{m})=w(s,s),\;w(s^{m},t^{q})=w(s,t^{q}),\;w(t^{q},s^{m})=w(t^{q},s);$

  * (2)

All the quasitriangular structures on $\gamma_{4}$ are given by
$R=\sum_{g,h\in K}w(g,h)e_{g}\otimes e_{h},$ where $w$ is a bicharacter on
$K=\langle s\rangle$ such that $w(s^{m},s^{m})=w(s,s)$;

  * (3)

All the quasitriangular structures on $\gamma_{5}$ are given by
$R=\sum_{g,h\in K}w(g,h)e_{g}\otimes e_{h},$ where $w$ is a bicharacter on
$K=\langle s,t\rangle$ such that

$w(s^{m},s^{m})=w(s,s),\;w(s^{m},t)=w(s,t),\;w(t,s^{m})=w(t,s);$

  * (4)

All the quasitriangular structures on $\gamma_{6}$ are given by
$R=\sum_{g,h\in K}w(g,h)e_{g}\otimes e_{h},$ where $w$ is a bicharacter on
$K=\langle s\rangle$ such that

$w(s^{m},s^{m})=w(s,s),\;w(s^{n},s^{n})=w(s,s).$

###### Proof.

Similar to the proof of Proposition 4.5. ∎

###  4.3. Quasitriangular structures on $\mathscr{A}_{l}$.

At first, Natale (see [25, Lemma 1.3.9]) already proved that $\mathscr{A}_{l}$
are self dual for $0\leq l\leq q-1$. Therefore, it is enough to consider the
braiding structures on $\mathscr{A}_{l}$ instead of their quasitriangular
structures. Secondly, we need point out that Gelaki already constructed the
Hopf algebra $\mathscr{A}_{1}$ in [12, Theorem 3.11] and showed
$\mathscr{A}_{1}$ admits no quasitriangular structure. As the first step to
our aim, we will show a more general result than Gelaki’s, that is, we will
show that all $\mathscr{A}_{l}$ for $1\leq l\leq q-1$ admit no quasitriangular
structures. To do this, we start with the following two lemmas.

######  Lemma 4.8.

With notions given in Example 2.9, the only braiding structure on $\Bbbk^{G}$
is trivial, i.e. $\langle e_{g},e_{h}\rangle=\epsilon(e_{g})\epsilon(e_{h})$,
where $g,h\in G$.

###### Proof.

Assume $R$ is a quasitriangular structure on $\Bbbk G$. We will show
$R=1\otimes 1$ and thus we complete the proof. By [6, Theorem 3], there are
commutative and normal subgroups $H,K\subseteq G$ such that $R\in\Bbbk
H\otimes\Bbbk K$. Since $G=\langle a,b\;|a^{p}=b^{q}=1,bab^{-1}=a^{t}\rangle$,
the only non-trivial commutative and normal subgroup of $G$ is
$\mathbb{Z}_{p}=\langle a\rangle$. Thus
$R\in\Bbbk\mathbb{Z}_{p}\otimes\Bbbk\mathbb{Z}_{p}$.

Let $\omega$ be a primitive $p$th root of 1. If we define
$e_{a^{i}}=p^{-1}\sum_{j=1}^{p}\omega^{ij}a^{j}$, $0\leq i\leq p-1$, then
$\\{e_{a^{i}}\\}_{i=0}^{p-1}$ is a basis of orthogonal idempotents of
$\Bbbk\mathbb{Z}_{p}$. Moreover, we have
$be_{a^{i}}b^{-1}=e_{a^{it^{\prime}}}$ for $0\leq i\leq(p-1)$, where
$t^{\prime}:=t^{q-1}$. Since
$R\in\Bbbk\mathbb{Z}_{p}\otimes\Bbbk\mathbb{Z}_{p}$, we can write
$R=\sum_{i,j=0}^{p-1}w(a^{i},a^{j})e_{a^{i}}\otimes e_{a^{j}}$, where $w$ is a
bicharacter on $\mathbb{Z}_{p}$. Because $\Delta^{cop}(b)R=R\Delta(b)$, we get
$w(a^{i},a^{j})=w(a^{it},a^{jt})$, $0\leq i,j\leq p-1$. In particular,
$w(a,a)=w(a^{t},a^{t})$. By assumption, $t^{q}\equiv 1(\text{mod}\;p)$ and
$t\not\equiv 1(\text{mod}\;p)$, $q>2$, thus $t^{2}\not\equiv
1(\text{mod}\;p)$. And this implies $w(a,a)=1$. Since $w$ is bicharacter on
$\mathbb{Z}_{p}$, we have $R=1\otimes 1$. ∎

######  Lemma 4.9.

With notions given in Example 2.9 and assume $\langle,\rangle$ is a braiding
structure on $\mathscr{A}_{l}$ for $0\leq l\leq q-1$. Then there are
$g_{0},g_{1}\in G$ such that

  * (i)

$\langle e_{h},g^{i}\rangle=\delta_{h,g_{0}^{i}}$ and $\langle
g^{i},e_{h}\rangle=\delta_{h,g_{1}^{i}}$, $0\leq i\leq q$ and $h\in G$;

  * (ii)

$\phi^{l}(g_{0})=\phi^{l}(g_{1})$;

  * (iii)

$h\triangleleft g=g_{0}hg_{0}^{-1}$ and $h\triangleleft g=g_{1}^{-1}hg_{1}$,
$h\in G$.

###### Proof.

Since $\langle,\rangle$ is a braiding structure and $g$ is a grouplike element
in $\mathscr{A}_{l}$, we know
$\langle,g\rangle:\mathscr{A}_{l}\rightarrow\Bbbk$ is an algebra map. Thus the
restriction $\langle,g\rangle|_{\Bbbk^{G}}$ is an algebra map, i.e a character
of $\Bbbk^{G}$. Let $g_{0}\in G$ such that $\langle
e_{h},g\rangle=\delta_{h,g_{0}}$. By definition, $gg^{i}=\sum_{k\in
G}\sigma(k,g,g^{i})e_{k}g^{i+1}$. Thus $\langle e_{h},gg^{i}\rangle=\sum_{k\in
G}\sigma(k,g,g^{i})\langle e_{h},e_{k}g^{i+1}\rangle$ for $i\geq 0$. Since
$\langle,\rangle$ is a braiding structure, we have $\langle
e_{h},gg^{i}\rangle=\sum_{rs=h}\langle e_{r},g^{i}\rangle\langle
e_{s},g\rangle=\langle e_{hg_{0}^{-1}},g^{i}\rangle$. Similarly, we have
$\langle e_{h},e_{k}g^{i+1}\rangle=\sum_{rs=h}\langle
e_{r},g^{i+1}\rangle\langle e_{s},e_{k}\rangle$. By Lemma 4.8, the restriction
$\langle,\rangle|_{\Bbbk^{G}}$ is trivial. Thus $\langle
e_{h},e_{k}g^{i+1}\rangle=\delta_{k,1}\langle e_{h},g^{i+1}\rangle$. This
implies $\sum_{k\in G}\sigma(k,g,g^{i})\langle
e_{h},e_{k}g^{i+1}\rangle=\langle e_{h},g^{i+1}\rangle$. Therefore $\langle
e_{hg_{0}^{-1}},g^{i}\rangle=\langle e_{h},g^{i+1}\rangle$. By induction way,
we have $\langle e_{h},g^{i}\rangle=\delta_{h,g_{0}^{i}}$, $0\leq i\leq q$.
Similarly, we can find $g_{1}\in G$ such that $\langle
g^{i},e_{h}\rangle=\delta_{h,g_{1}^{i}}$, $0\leq i\leq q$. Thus we have (i).

By definition, we have $gg^{i}=\sum_{k\in G}\sigma(k,g,g^{i})e_{k}g^{i+1}$ for
$i\geq 0$. Thus we get $\langle gg^{i},g\rangle=\sum_{k\in
G}\sigma(k,g,g^{i})\langle e_{k}g^{i+1},g\rangle$. Let $\langle
g,g\rangle:=\lambda$. Because $\langle,g\rangle$ is an algebra map, $\langle
gg^{i},g\rangle=\lambda\langle g^{i},g\rangle$ and $\langle
e_{k}g^{i+1},g\rangle=\delta_{k,g_{0}}\langle g^{i+1},g\rangle$. Thus $\langle
g^{i+1},g\rangle=\lambda\sigma(g_{0},g,g^{i})^{-1}\langle g^{i},g\rangle$. By
induction way, we have $\langle
g^{q},g\rangle=\lambda^{q}\prod_{i=1}^{q-1}\sigma(g_{0},g,g^{i})^{-1}$. By
definition, we have $\sigma(g_{0},g,g^{i})=1$ for $1\leq i\leq q-2$ and
$\sigma(g_{0},g,g^{q-1})=\phi^{l}(g_{0})$. Thus $\langle
g^{q},g\rangle=\lambda^{q}\phi^{l}(g_{0})^{-1}$. By definition, $g^{q}=1$.
Therefore $\lambda^{q}=\phi^{l}(g_{0})$. Similarly, we have
$\lambda^{q}=\phi^{l}(g_{1})$. And hence we have (ii).

Because $\langle,\rangle$ is a braiding structure, we get $\sum_{rs=h}\langle
e_{r},g\rangle e_{s}g=\sum_{rs=h}ge_{r}\langle e_{s},g\rangle$. Since $\langle
e_{r},g\rangle=\delta_{r,g_{0}}$ and $\langle
e_{s},g\rangle=\delta_{s,g_{0}}$, we have $e_{g_{0}^{-1}h}g=ge_{hg_{0}^{-1}}$.
By definition, $e_{g_{0}^{-1}h}g=ge_{(g_{0}^{-1}h)\triangleleft g}$. Thus
$ge_{(g_{0}^{-1}h)\triangleleft g}=ge_{hg_{0}^{-1}}$. Because
$gg^{q-1}=\sum_{k\in G}\sigma(k,g,g^{q-1})e_{k}$, we know $g$ is invertible.
Hence $(g_{0}^{-1}h)\triangleleft g=hg_{0}^{-1}$. In particular,
$g_{0}^{-1}\triangleleft g=g_{0}^{-1}$ by letting $h=1$. And hence
$h\triangleleft g=g_{0}hg_{0}^{-1}$. Similarly, we have $\sum_{rs=h}\langle
g,e_{r}\rangle ge_{s}=\sum_{rs=h}e_{r}g\langle g,e_{s}\rangle$. Since $\langle
g,e_{r}\rangle=\delta_{r,g_{1}}$ and $\langle
g,e_{s}\rangle=\delta_{s,g_{1}}$, we have $ge_{g_{1}^{-1}h}=e_{hg_{1}^{-1}}g$.
Because $e_{hg_{1}^{-1}}g=ge_{(hg_{1}^{-1})\triangleleft g}$ and $g$ is
invertible, we have $(hg_{1}^{-1})\triangleleft g=g_{1}^{-1}h$. Thus
$g_{1}^{-1}\triangleleft g=g_{1}^{-1}$ by letting $h=1$. And hence we get
$h\triangleleft g=g_{1}^{-1}hg_{1}$. ∎

Now we can use Lemmas 4.8-4.9 to get the following promised result.

######  Proposition 4.10.

The Hopf algebras $\mathscr{A}_{l}$, $1\leq l\leq q-1$ admit no
quasitriangular structure.

###### Proof.

Let $0\leq l\leq q-1$. Assume $\langle,\rangle$ is a braiding structure on
$\mathscr{A}_{l}$. Let $S:=\\{h\in G|\;h\triangleleft g=h\\}$. Directly, we
know $S$ is proper subgroup of $G$. Since $b^{i}\in S$ for $1\leq i\leq q$, we
have $S=\\{b^{i}|\;1\leq i\leq q\\}$. By (iii) of Lemma 4.9, we have $g_{0}\in
S$. Let $g_{0}=b^{i}$, where $1\leq i\leq q$. Because $h\triangleleft
g=g_{0}hg_{0}^{-1}$, we get $a^{t}=b^{i}ab^{-i}$ by letting $h=a$. By
definition, $b^{i}ab^{-i}=a^{t^{i}}$. Thus $a^{t}=a^{t^{i}}$. This implies
$t^{i-1}\equiv 1(\text{mod}\;p)$. By assumption, we have $0\leq i-1\leq q-1$.
Thus we have $i=1$. And hence $g_{0}=b$. Similarly, we get $g_{1}=b^{-1}$. By
(ii) of Lemma 4.9, we have $\phi^{l}(b)=\phi^{l}(b^{-1})$. Therefore we get
$\omega^{2l}=1$. If $1\leq l\leq q-1$, then $\omega^{2l}\neq 1$. But this
contradicts the previous conclusion, so $\mathscr{A}_{l}$ admits no
quasitriangular structure. ∎

The last task of this subsection is to find all possible quasitriangular
structures on $\mathscr{A}_{0}$.

######  Proposition 4.11.

All the braiding structures on $\mathscr{A}_{0}$ are given by $\langle
e_{h}g^{i},e_{k}g^{j}\rangle=\delta_{h,b^{j}}\delta_{k,b^{-i}}\lambda^{ij}$,
where $\lambda\in\Bbbk$ such that $\lambda^{q}=1$ and
$h,k\in\mathbb{Z}_{p}\rtimes\mathbb{Z}_{q}$, $0\leq i,j\leq q-1$.

###### Proof.

Assume $\langle,\rangle$ is a braiding structure on $\mathscr{A}_{0}$. By
Lemma 4.8, we have $\langle e_{g},e_{h}\rangle=\epsilon(e_{g})\epsilon(e_{h})$
for $g,h\in G$. Similar to the proof of Proposition 4.10, we get $g_{0}=b$ and
$g_{1}=b^{-1}$. Let $\langle g,g\rangle=\lambda$. Note that $g$ is a grouplike
element, and by induction way, we have $\langle
g^{i},g^{j}\rangle=\lambda^{ij}$. This implies $\langle
g^{q},g\rangle=\lambda^{q}$. Because $g^{q}=1$, we get $\lambda^{q}=1$. Since
$\langle,\rangle$ is a braiding structure, $\langle
e_{h}g^{i},e_{k}g^{j}\rangle=\sum_{rs=k}\langle e_{h},e_{r}g^{j}\rangle\langle
g^{i},e_{s}g^{j}\rangle$. Directly, we have $\langle
e_{h},e_{r}g^{j}\rangle=\sum_{uv=h}\langle e_{u},g^{j}\rangle\langle
e_{v},e_{r}\rangle=\delta_{r,1}\delta_{h,b^{j}}$ and $\langle
g^{i},e_{s}g^{j}\rangle=\langle g^{i},e_{s}\rangle\langle
g^{i},e_{s}g^{j}\rangle=\delta_{s,b^{-i}}\lambda^{ij}$. Thus $\langle
e_{h}g^{i},e_{k}g^{j}\rangle=\delta_{h,b^{j}}\delta_{k,b^{-i}}\lambda^{ij}$.

Conversely, let $\lambda\in\Bbbk$ such that $\lambda^{q}=1$ and we can define
$\langle
e_{h}g^{i},e_{k}g^{j}\rangle=\delta_{h,b^{j}}\delta_{k,b^{-i}}\lambda^{ij}$,
where $h,k\in\mathbb{Z}_{p}\rtimes\mathbb{Z}_{q}$, $0\leq i,j\leq q-1$. Due to
$\lambda^{q}=1$, we have $\langle
e_{h}g^{i},e_{k}g^{j}\rangle=\delta_{h,b^{j}}\delta_{k,b^{-i}}\lambda^{ij}$
for $i,j\in\mathbb{N}$. Next we will show $\langle,\rangle$ is a braiding
structure. Firstly, we show
$\langle(e_{h}g^{i})(e_{k}g^{j}),e_{r}g^{m}\rangle=\langle
e_{h}g^{i},(e_{r}g^{m})_{(1)}\rangle\langle
e_{k}g^{j},(e_{r}g^{m})_{(2)}\rangle$. Since

$\displaystyle\langle(e_{h}g^{i})(e_{k}g^{j}),e_{r}g^{m}\rangle$
$\displaystyle=\langle\delta_{h\triangleleft
g^{i},k}e_{h}g^{i+j},e_{r}g^{m}\rangle$ $\displaystyle=\delta_{h\triangleleft
g^{i},k}\delta_{h,b^{m}}\delta_{r,b^{-(i+j)}}\lambda^{(i+j)m}$
$\displaystyle=\delta_{h,k}\delta_{h,b^{m}}\delta_{r,b^{-(i+j)}}\lambda^{(i+j)m}$

and

$\displaystyle\langle e_{h}g^{i},(e_{r}g^{m})_{(1)}\rangle\langle
e_{k}g^{j},(e_{r}g^{m})_{(2)}\rangle$ $\displaystyle=\sum_{cd=r}\langle
e_{h}g^{i},e_{c}g^{m}\rangle\langle e_{k}g^{j},e_{d}g^{m}\rangle$
$\displaystyle=\sum_{cd=r}\delta_{h,b^{m}}\delta_{c,b^{-i}}\lambda^{im}\langle
e_{k}g^{j},e_{d}g^{m}\rangle$
$\displaystyle=\sum_{cd=r}\delta_{h,b^{m}}\delta_{c,b^{-i}}\lambda^{im}\delta_{k,b^{m}}\delta_{d,b^{-j}}\lambda^{jm}$
$\displaystyle=\delta_{h,k}\delta_{h,b^{m}}\delta_{r,b^{-(i+j)}}\lambda^{(i+j)m},$

we have $\langle(e_{h}g^{i})(e_{k}g^{j}),e_{r}g^{m}\rangle=\langle
e_{h}g^{i},(e_{r}g^{m})_{(1)}\rangle\langle
e_{k}g^{j},(e_{r}g^{m})_{(2)}\rangle$. Secondly, we will show $\langle
e_{h}g^{i},e_{k}g^{j}e_{r}g^{m}\rangle=\langle(e_{h}g^{i})_{(1)},e_{r}g^{m}\rangle\langle(e_{h}g^{i})_{(2)},e_{k}g^{j}\rangle$.
Since

$\displaystyle\langle e_{h}g^{i},e_{k}g^{j}e_{r}g^{m}\rangle$
$\displaystyle=\langle e_{h}g^{i},\delta_{k\triangleleft
g^{j},r}e_{k}g^{j+m}\rangle$ $\displaystyle=\delta_{k\triangleleft
g^{j},r}\delta_{h,b^{m+j}}\delta_{k,b^{-i}}\lambda^{i(j+m)}$
$\displaystyle=\delta_{k,r}\delta_{h,b^{m+j}}\delta_{k,b^{-i}}\lambda^{i(j+m)}$

and

$\displaystyle\langle(e_{h}g^{i})_{(1)},e_{r}g^{m}\rangle\langle(e_{h}g^{i})_{(2)},e_{k}g^{j}\rangle$
$\displaystyle=\sum_{cd=h}\langle e_{c}g^{i},e_{r}g^{m}\rangle\langle
e_{d}g^{i},e_{k}g^{j}\rangle$
$\displaystyle=\sum_{cd=h}\delta_{c,b^{m}}\delta_{r,b^{-i}}\lambda^{im}\langle
e_{d}g^{i},e_{k}g^{j}\rangle$
$\displaystyle=\sum_{cd=h}\delta_{c,b^{m}}\delta_{r,b^{-i}}\lambda^{im}\delta_{d,b^{j}}\delta_{k,b^{-i}}\lambda^{ij}$
$\displaystyle=\delta_{k,r}\delta_{h,b^{m+j}}\delta_{k,b^{-i}}\lambda^{i(j+m)},$

we have $\langle
e_{h}g^{i},e_{k}g^{j}e_{r}g^{m}\rangle=\langle(e_{h}g^{i})_{(1)},e_{r}g^{m}\rangle\langle(e_{h}g^{i})_{(2)},e_{k}g^{j}\rangle$.
Finally, we will show the following equation

(4.1)
$\displaystyle\langle(e_{h}g^{i})_{(1)},(e_{k}g^{j})_{(1)}\rangle(e_{h}g^{i})_{(2)}(e_{k}g^{j})_{(2)}=(e_{k}g^{j})_{(1)}(e_{h}g^{i})_{(1)}\langle(e_{h}g^{i})_{(2)},(e_{k}g^{j})_{(2)}\rangle.$

Since

$\displaystyle\langle(e_{h}g^{i})_{(1)},(e_{k}g^{j})_{(1)}\rangle(e_{h}g^{i})_{(2)}(e_{k}g^{j})_{(2)}$
$\displaystyle=\sum_{rs=h}\sum_{cd=k}\langle
e_{r}g^{i},e_{c}g^{j}\rangle(e_{s}g^{i})(e_{d}g^{j})$
$\displaystyle=\sum_{rs=h}\sum_{cd=k}\delta_{r,b^{j}}\delta_{c,b^{-i}}\lambda^{ij}(e_{s}g^{i})(e_{d}g^{j})$
$\displaystyle=\lambda^{ij}(e_{b^{-j}h}g^{i})(e_{b^{i}k}g^{j})$
$\displaystyle=\lambda^{ij}\delta_{b^{-j}h,kb^{i}}e_{b^{-j}h}g^{i+j}$

and

$\displaystyle(e_{k}g^{j})_{(1)}(e_{h}g^{i})_{(1)}\langle(e_{h}g^{i})_{(2)},(e_{k}g^{j})_{(2)}\rangle$
$\displaystyle=\sum_{rs=h}\sum_{cd=k}(e_{c}g^{j})(e_{r}g^{i})\langle
e_{s}g^{i},e_{d}g^{j}\rangle$
$\displaystyle=\sum_{rs=h}\sum_{cd=k}(e_{c}g^{j})(e_{r}g^{i})\delta_{s,b^{j}}\delta_{d,b^{-i}}\lambda^{ij}$
$\displaystyle=\lambda^{ij}(e_{kb^{i}}g^{j})(e_{hb^{-j}}g^{i})$
$\displaystyle=\lambda^{ij}\delta_{kb^{i},b^{-j}h}e_{kb^{i}}g^{i+j}$
$\displaystyle=\lambda^{ij}\delta_{b^{-j}h,kb^{i}}e_{b^{-j}h}g^{i+j},$

we have the equation (4.1). ∎

###  4.4. Quasitriangular structures on $\mathscr{B}_{\lambda}$ and
$\mathscr{B}_{\lambda}^{\ast}$.

Let $0\leq\lambda<p-1$. We will consider the quasitriangular structures on
$\mathscr{B}_{\lambda}$ and $\mathscr{B}_{\lambda}^{\ast}$ in this subsection.
The following result seems has its own interesting.

######  Lemma 4.12.

Let $(H,R)$ be a quasitriangular Hopf algebra over $\Bbbk$. Assume $G(H)$ is
an abelian group and all proper Hopf subalgebras are trivial. Then $H_{R}=H$
or $H_{R}\subseteq\Bbbk G(H)$. Moreover, we have $H^{\ast cop}\cong H$ if
$H_{R}=H$.

###### Proof.

Assume $H_{R}\neq H$, then we will show $H_{R}\subseteq\Bbbk G(H)$. Since
$H_{l}\subseteq H_{R}$, we have $H_{l}\neq H$. By assumption, we know $H_{l}$
is trivial, i.e it is group algebra or the dual of group algebra. Since
$H_{l}^{\ast cop}\cong H_{r}$, we obtain $H_{l}$ or $H_{r}$ is group algebra.
Assume $H_{l}$ is group algebra. By definition, we have $H_{l}\subseteq\Bbbk
G(H)$. Because $G(H)$ is an abelian group, we know $H_{l}$ is commutative.
This implies $H_{r}$ is also group algebra. Thus $H_{r}\subseteq\Bbbk G(H)$.
Therefore, $H_{R}\subseteq\Bbbk G(H)$. Similarly, if $H_{r}$ is group algebra,
then we also have $H_{R}\subseteq\Bbbk G(H)$. So we have shown
$H_{R}\subseteq\Bbbk G(H)$ when $H_{R}\neq H$.

Assume $H_{R}=H$. We will prove $H^{\ast cop}\cong H$ in this case. If
$H_{l}\neq H$, then we can repeat the previous process and get
$H_{R}\subseteq\Bbbk G(H)$. This implies $\Bbbk G(H)=H$. Since $G(H)$ is an
abelian group, we know $H^{\ast cop}\cong H$. If $H_{l}=H$, then we have
$H_{r}=H$. Because $H_{l}^{\ast cop}\cong H_{r}$, we also obtain $H^{\ast
cop}\cong H$. ∎

With the help of this lemma, we can discuss the quasitriangular structures on
$\mathscr{B}_{\lambda}$ (resp. $\mathscr{B}_{\lambda}^{\ast}$), which are
defined in Example 2.10 (resp. Example 2.12). At first, we find that

######  Proposition 4.13.

The Hopf algebras $\mathscr{B}_{\lambda}^{\ast}$ admit no quasitriangular
structure, where $0\leq\lambda<p-1$.

###### Proof.

Let $0\leq\lambda<p-1$. Assume $R$ is a quasitriangular structure on
$\mathscr{B}_{\lambda}^{\ast}$. We first show that
$(\mathscr{B}_{\lambda}^{\ast},R)$ satisfies the conditions of Lemma 4.12. By
Remark 2.13, we get $G(\mathscr{B}_{\lambda}^{\ast})$ is an abelian group. Due
to the dimension reason and Lemma 2.6, we know all proper Hopf subalgebras of
$\mathscr{B}_{\lambda}^{\ast}$ are trivial. Thus
$\mathscr{B}_{\lambda}^{\ast}$ satisfies the conditions of Lemma 4.12. By
Remark 2.13 again, we have $G(\mathscr{B}_{\lambda}^{cop})\not\cong
G(\mathscr{B}_{\lambda}^{\ast})$ and hence
$\mathscr{B}_{\lambda}^{cop}\not\cong\mathscr{B}_{\lambda}^{\ast}$. Thus
$R\in\Bbbk G(\mathscr{B}_{\lambda}^{\ast})\otimes\Bbbk
G(\mathscr{B}_{\lambda}^{\ast})$.

If $\lambda\neq 0$, then we have $\Bbbk
G(\mathscr{B}_{\lambda}^{\ast})=\\{e_{g^{i}}|\;0\leq i\leq p-1\\}$ by Remark
2.13. Since $R\in\Bbbk G(\mathscr{B}_{\lambda}^{\ast})\otimes\Bbbk
G(\mathscr{B}_{\lambda}^{\ast})$, we can assume that $R=\sum_{r,s\in
G}w(r,s)e_{r}\otimes e_{s}$, where $G=G(\mathscr{B}_{\lambda}^{\ast})$ and $w$
is a bicharacter on $G$. Because $R$ is a quasitriangular structure, we have
$\Delta^{cop}(a)R=R\Delta(a)$. Recall the Example 2.12, we get
$\Delta(a)=\sum_{r,s\in G}e_{r}(s\triangleright a)\otimes e_{s}a$. Thus we
have

(4.2) $\displaystyle\Delta^{cop}(a)R$ $\displaystyle=[\sum_{r,s\in
G}e_{s}a\otimes e_{r}(s\triangleright a)]R$ (4.3) $\displaystyle=[\sum_{r,s\in
G}e_{s}a\otimes e_{r}(s\triangleright a)][\sum_{r,s\in G}w(r,s)e_{r}\otimes
e_{s}]$ (4.4) $\displaystyle=\sum_{r,s\in G}w(r,s)e_{r}a\otimes
e_{s}(r\triangleright a)$

and $R\Delta(a)=\sum_{r,s\in G}w(r,s)e_{r}(s\triangleright a)\otimes e_{s}a$.
Since $e_{g}a\otimes e_{g}(g\triangleright a)$ appear in $\Delta^{cop}(a)R$
while not in $R\Delta(a)$, we know $\Delta^{cop}(a)R\neq R\Delta(a)$. But this
contradicts with the previous conclusion, so $\mathscr{B}_{\lambda}^{\ast}$
admits no quasitriangular structure.

If $\lambda=0$. By Remark 2.13, we get $\\{e_{g^{i}}b^{j}|\;0\leq i\leq
p-1,0\leq j\leq q-1\\}$ is a linear basis of $\Bbbk
G(\mathscr{B}_{0}^{\ast})$. Because $R\in\Bbbk
G(\mathscr{B}_{0}^{\ast})\otimes\Bbbk G(\mathscr{B}_{0}^{\ast})$, we can
assume that $R=\sum_{0\leq i,j\leq p-1,0\leq k,l\leq
q-1}\lambda_{i,j}^{k,l}e_{g^{i}}b^{k}\otimes e_{g^{j}}b^{l}$, where
$\lambda_{i,j}^{k,l}\in\Bbbk$. Because $R$ is a quasitriangular structure, we
have $\Delta^{cop}(a)R=R\Delta(a)$. By Example 2.12, we get
$\Delta(a)=\sum_{0\leq i,j\leq p-1}e_{g^{i}}a^{m^{j}}\otimes e_{g^{j}}a$. If
we mimic the previous process for $\mathscr{B}_{\lambda}^{\ast}$, $\lambda\neq
0$, then we get

(4.5) $\displaystyle\Delta^{cop}(a)R=\sum\limits_{\begin{subarray}{l}0\leq
i,j\leq p-1\\\ 0\leq k,l\leq q-1\\\
\end{subarray}}\lambda_{i,j}^{k,l}e_{g^{i}}ab^{k}\otimes
e_{g^{j}}a^{m^{i}}b^{l}.$

Similarly, we have

(4.6) $\displaystyle R\Delta(a)=\sum\limits_{\begin{subarray}{l}0\leq i,j\leq
p-1\\\ 0\leq k,l\leq q-1\\\
\end{subarray}}\lambda_{i,j}^{k,l}e_{g^{i}}b^{k}a^{m^{j}}\otimes
e_{g^{j}}b^{l}a.$

By definition, we get $ba=\sum_{0\leq i\leq
p-1}\zeta_{i}^{m^{(\lambda+1)i}}e_{g^{i}}(ab)$. By induction way, we have
$e_{g^{i}}b^{k}a^{m^{j}}=\zeta_{i}^{km^{(\lambda+1)i}m^{j}}e_{g^{i}}a^{m^{j}}b^{k}$
and $e_{g^{j}}b^{l}a=\zeta_{j}^{m^{(\lambda+1)j}l}e_{g^{j}}ab^{l}$. By
combining these equalities with the equation (4.6), we get

(4.7) $\displaystyle R\Delta(a)=\sum\limits_{\begin{subarray}{l}0\leq i,j\leq
p-1\\\ 0\leq k,l\leq q-1\\\
\end{subarray}}\lambda_{i,j}^{k,l}\zeta_{i}^{km^{(\lambda+1)i}m^{j}}\zeta_{j}^{m^{(\lambda+1)j}l}e_{g^{i}}a^{m^{j}}b^{k}\otimes
e_{g^{j}}ab^{l}.$

If $1\leq i\leq p-1$ and $0\leq j\leq p-1$, $0\leq k,l\leq q-1$, then
$e_{g^{i}}ab^{k}\otimes e_{g^{j}}a^{m^{i}}b^{l}$ will appear in (4.5) while
not in (4.7). Since $\\{e_{g^{i}}a^{k}b^{l}|\;0\leq i\leq p-1,0\leq k,l\leq
q-1\\}$ is a linear basis of $\mathscr{B}_{0}$ and
$\Delta^{cop}(a)R=R\Delta(a)$, we get the coefficient of
$e_{g^{i}}ab^{k}\otimes e_{g^{j}}a^{m^{i}}b^{l}$ in (4.6) is $0$, i.e we have
$\lambda_{i,j}^{k,l}=0$ for $i\neq 0$. Thus $R=\sum_{0\leq j\leq p-1,0\leq
k,l\leq q-1}\lambda_{0,j}^{k,l}e_{1}b^{k}\otimes e_{g^{j}}b^{l}$. Directly we
have $(e_{a}\otimes e_{1})R=0$, and hence $R$ is not invertible. This implies
$R$ is not quasitriangular structure, so $\mathscr{B}_{0}^{\ast}$ admits no
quasitriangular structure. ∎

Secondly, we can describe all possible quasitriangular structures on
$\mathscr{B}_{\lambda}$. Before this, we define
$\eta(h,k,g^{i}):=\tau(h,k,g^{i})\tau(k,h,g^{i})^{-1}$ for
$h,k\in\mathbb{Z}_{q}\times\mathbb{Z}_{q},i\geq 0$. Then we have

######  Proposition 4.14.

All the quasitriangular structures on $\mathscr{B}_{\lambda}$ are given by

$R=\sum_{r,s\in G}w(r,s)e_{r}\otimes e_{s},$

where $w$ is a bicharacter on $\mathbb{Z}_{q}\times\mathbb{Z}_{q}$ such that
the following conditions

$\displaystyle
w(a,a)=w(a^{m},a^{m}),\;w(a,b)=w(a^{m},b^{m^{\lambda}})\eta(a^{m},b^{m^{\lambda}},g),$
$\displaystyle
w(b,a)=w(b^{m^{\lambda}},a^{m})\eta(b^{m^{\lambda}},a^{m},g),\;w(b,b)=w(b^{m^{\lambda}},b^{m^{\lambda}}).$

###### Proof.

Assume $R$ is a quasitriangular structure on $\mathscr{B}_{\lambda}$. By
Remark 2.13, we get $G(\mathscr{B}_{\lambda}^{\ast cop})\not\cong
G(\mathscr{B}_{\lambda})$ and hence $\mathscr{B}_{\lambda}^{\ast
cop}\not\cong\mathscr{B}_{\lambda}$. Due to the dimension reason and Theorem
2.6, we know all proper Hopf subalgebras of $\mathscr{B}_{\lambda}$ are
trivial. By Lemma 4.12, we can suppose that $R=\sum_{r,s\in
G}w(r,s)e_{r}\otimes e_{s}$, where $w$ is a bicharacter on
$\mathbb{Z}_{q}\times\mathbb{Z}_{q}$. Since $R$ is a quasitriangular
structure, we have $\Delta^{cop}(g)R=R\Delta(g)$. By definition, we get
$\Delta(g)=\sum_{r,s\in G}\tau(r,s,g)e_{r}g\otimes e_{s}g$. Thus we have

$\displaystyle\Delta^{cop}(g)R$ $\displaystyle=(\sum_{r,s\in
G}\tau(s,r,g)e_{r}g\otimes e_{s}g)R$ $\displaystyle=(\sum_{r,s\in
G}\tau(s,r,g)e_{r}g\otimes e_{s}g)(\sum_{c,d\in G}w(c,d)e_{c}\otimes e_{d})$
$\displaystyle=\sum_{r,s,c,d\in G}\tau(s,r,g)w(c,d)e_{r}e_{c\triangleleft
g^{-1}}g\otimes e_{s}e_{d\triangleleft g^{-1}}g$ $\displaystyle=\sum_{r,s\in
G}\tau(s,r,g)w(r\triangleleft g,s\triangleleft g)e_{r}g\otimes e_{s}g$

and $R\Delta(g)=\sum_{r,s\in G}\tau(r,s,g)w(r,s)e_{r}g\otimes e_{s}g$.
Therefore $w(r\triangleleft g,s\triangleleft g)=w(r,s)\eta(r,s,g)$ for $r,s\in
G$, where $\eta(r,s,g)=\tau(r,s,g)\tau(s,r,g)^{-1}$. By definition, we know
$\tau(-,-,g)$ is 2-cocycle. Moreover, because $G$ is an abelian group, we get
$\eta(-,-,g)$ is bicharacter on $G$. Thus $w(r\triangleleft g,s\triangleleft
g)=w(r,s)\eta(r,s,g)$ for $r,s\in G$ if and only if they hold for generators
of $G$, i.e they hold for $r,s\in\\{a,b\\}$. For convenience, we use the
equivalent form of the equations $w(r\triangleleft g,s\triangleleft
g)=w(r,s)\eta(r,s,g)$ for $r,s\in\\{a,b\\}$, which is $w(r,s)=w(r\triangleleft
g^{-1},s\triangleleft g^{-1})\eta(r\triangleleft g^{-1},s\triangleleft
g^{-1},g)$ for $r,s\in\\{a,b\\}$. And hence we have

(4.8) $\displaystyle
w(a,a)=w(a^{m},a^{m}),\;w(a,b)=w(a^{m},b^{m^{\lambda}})\eta(a^{m},b^{m^{\lambda}},g),$
(4.9) $\displaystyle
w(b,a)=w(b^{m^{\lambda}},a^{m})\eta(b^{m^{\lambda}},a^{m},g),\;w(b,b)=w(b^{m^{\lambda}},b^{m^{\lambda}}).$

Conversely, assume $w$ be a bicharacter on $G$ such that the equations
(4.8)-(4.9), then we can define $R=\sum_{r,s\in G}w(r,s)e_{r}\otimes e_{s}$.
Due to the equations (4.8)-(4.9), we get $\Delta^{cop}(g)R=R\Delta(g)$.
Because $R$ is invertible and $\\{e_{h},g|\;h\in G\\}$ generates
$\mathscr{B}_{\lambda}$ as algebra, we know $\Delta^{cop}(x)R=R\Delta(x)$ for
$x\in\mathscr{B}_{\lambda}$. Therefore $R$ is a quasitriangular structure on
$\mathscr{B}_{\lambda}$. ∎

Proofs of Theorems 1.1 and 1.2. Due to Corollary 4.2 and Propositions
4.10,4.11 and 4.13,4.14, we know Theorems 1.1 and 1.2 hold.∎

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