1. Introduction
Let 
$\mathbb{K}$ be a field of characteristic zero. In [Reference Kats and Palyutkin5] (see also [Reference Masuoka6]), Kac and Paljutkin introduced an eight-dimensional semisimple Hopf algebra 
$H_8$ given by generators 
$x,y,z$ subject to the relations
\begin{equation} zx=yz, \quad zy=xz, \quad xy=yx, \quad x^2=y^2=1, \quad z^2=\frac{1}{2}\left(1+x+y-xy\right),\end{equation}such that 
$x$ and 
$y$ are group-like and
\begin{equation}
\Delta(z)=\frac{1}{2}\left(1\otimes 1 + x\otimes 1 + 1\otimes y - x\otimes y\right)(z\otimes z).\end{equation} The antipode of 
$H_8$ is the identity map. The algebra 
$H_8$ can be seen as a quotient of a skew polynomial ring 
$R[z;\sigma]/I$, where the ring of coefficients is the group ring 
$R=\mathbb{K}[\mathbb{Z}_2\times \mathbb{Z}_2]$, with an automorphism 
$\sigma$ that interchanges the generators 
$x$ and 
$y$ of 
$\mathbb{Z}_2\times \mathbb{Z}_2$ and 
$I$ is the ideal generated by the relation of 
$z^2$ in Equation (1). In [Reference Pansera9, Reference Pansera10], Pansera introduced a 
$2n^2$-dimensional semisimple Hopf algebra 
$H_{2n^2}$ as 
$R[z;\sigma]/I$, where 
$R=\mathbb{K}[\mathbb{Z}_n\times \mathbb{Z}_n]$, for a given integer 
$n\geq 2$. The comultiplication of 
$z$ is given by
\begin{equation}
\Delta(z) = \frac{1}{n} \left(\sum_{i,j=0}^{n-1} q^{-ij} x^i\otimes y^j\right)(z\otimes z),
\end{equation}where 
$q$ is a primitive 
$n$th root of unity in 
$\mathbb{K}$. The group algebra 
$R$ is a Hopf subalgebra of 
$H_{2n^2}$ and the antipode 
$S$ of 
$H_{2n^2}$ is extended from 
$R$ by setting 
$S(z)=z$. Motivated by Etingof–Walton’s Theorem, which says that if a semisimple Hopf algebra acts on a commutative domain then its action factors through a group algebra, Pansera showed in [Reference Pansera9] that 
$H_{2n^2}$ admits an action on a quantum plane 
$A$ that does not factor through a group algebra, i.e. 
$H_{2n^2}$ acts inner-faithfully on 
$A$.
The purpose of this note is to extend Pansera’s construction and to provide further examples of non-trivial semisimple Hopf algebras acting inner-faithfully on non-commutative domains. The basic observation is that, in the construction of 
$H_{2n^2}$, the action of the automorphism 
$\sigma$ of order 
$2$ can be regarded as the action of the symmetric group 
$\Sigma_2$ on the tensor product 
$\mathbb{K}[\mathbb{Z}_n] \otimes \mathbb{K}[\mathbb{Z}_n] = \mathbb{K}[\mathbb{Z}_n\times \mathbb{Z}_n]$. As a generalization, we consider 
$\Sigma_m$ acting on the 
$m$-fold tensor 
$R=B^{\otimes m}$, for some bialgebra 
$B$.Footnote 1 The standard generators 
$s_1, \ldots, s_{m-1}$ of 
$\Sigma_m$, i.e. the transpositions 
$s_i=(i, i+1)$, generate a free monoid 
$M=\langle \overline{s}_1, \ldots, \overline{s}_{m-1}\rangle$ that acts on 
$R$ and allows to consider the skew monoid algebra 
$R\# M$, which in the case 
$m=2$ corresponds to the skew polynomial ring 
$R[z;\sigma]$. The comultiplication of 
$\overline{s_i}$ can be defined as 
$\Delta(\overline{s}_i) = J_i (\overline{s}_i \otimes \overline{s}_i )$, for a suitable twist 
$J_i\in R\otimes R$. Under further assumptions on these twists 
$J_i$, we define a Hopf structure on the quotient 
$R\# M / I$, where 
$I$ is the ideal generated by the usual relations of the symmetric group, 
$\overline{s}_i\overline{s}_{i+1}\overline{s}_i = \overline{s}_{i+1}\overline{s}_i\overline{s}_{i+1}$ and 
$\overline{s}_i\overline{s}_j=\overline{s}_j\overline{s}_i$ for 
$|i-j| \gt 1$, but with 
$\overline{s}_i^2 = t_i := \mu_R(J_i)$, where 
$\mu_R$ is the multiplication of 
$R$. The obtained Hopf algebra 
$H$ is an extension of 
$\mathbb{K}\Sigma_m$ by 
$R$ and can be considered a crossed product 
$R\#_{\gamma} \Sigma_m$ for a suitable 
$2$-cocycle 
$\gamma:\Sigma_m\times \Sigma_m \to R^\times$. For 
$B=\mathbb{K}[\mathbb{Z}_n]$ and 
$\mathbb{K}$ containing a primitive 
$n$th root of unity, we provide twists 
$J_i$ that satisfy all our assumptions and yield a family of semisimple Hopf algebras 
$H_{n,m}=\mathbb{K}[\mathbb{Z}_n]^{\otimes m}\#_{\gamma} \Sigma_m$ of dimension 
$n^m m!$. The original Kac–Paljutkin Hopf algebra appears as 
$H_{2,2}$, while Pansera’s algebras appear as 
$H_{n,2}$.
In the last section we examine actions of 
$H_{n,m}$ on a quantum polynomial algebra 
$A$ in 
$m$ generators 
$u_1, \ldots, u_m$. We will show that there exists an 
$mn^m$-dimensional semisimple Hopf subalgebra 
$H$ of 
$H_{n,m}$ that acts inner-faithfully on 
$A$ and whose ring of invariants 
$A^H$ is the subring of cyclic polynomials in 
$u_1^n, \ldots, u_m^n$, in case 
$n$ is even.
2. Constructing Hopf algebras via twists
An invertible element 
$J\in B\otimes B$ of a bialgebra 
$B$ is called a twist for 
$B$ if
\begin{equation}
(\Delta_B \otimes \mathrm{id}_B)(J)(J\otimes 1_B) = (\mathrm{id}_B \otimes \Delta_B)(J)(1_B\otimes J)
\end{equation}and 
$(\epsilon \otimes \mathrm{id} )(J)=1=(\mathrm{id}\otimes\epsilon)(J)$ hold. Writing symbolically 
$J=J^{(1)} \otimes J^{(2)}$, Equation (4) means
\begin{equation}
J_1^{(1)} j^{(1)} \otimes J_2^{(2)} j^{(2)} \otimes J^{(2)}
=
J^{(1)} \otimes J_1^{(2)} j^{(1)} \otimes J_2^{(2)} j^{(2)},
\end{equation}where 
$j^{(1)} \otimes j^{(2)}$ is another copy of 
$J$ and where 
$\Delta_B(J^{(i)})=J_1^{(i)}\otimes J_2^{(i)}$ denotes the comultiplication of one of the legs of 
$J^{(i)}$ of 
$J$. Clearly 
$1_B\otimes 1_B$ is a twist for 
$B$. For 
$B=\mathbb{K}\mathbb{Z}_n$, with generator 
$x$, the element 
$J=\frac{1}{n} \sum_{i,j=0}^{n-1} q^{-ij} x^i \otimes x^j$ is a twist for 
$B$. We will use twists to deform the comultiplication of a group-like element of a bialgebra as they appear naturally in our construction. Suppose 
$R\subseteq S$ is an extension of bialgebras with 
$x\in S\setminus R$ an element such that 
$Rx$ is a free 
$R$-module. Suppose further that 
$J\in R\otimes R$ is an invertible element, such that 
$\Delta_S(x)=J(x\otimes x)$ and 
$\epsilon_S(x)=1$. Then the coassociativity for 
$\Delta_S(x)$ and the freeness of 
$Rx$ imply Equation (4), while the counity will imply 
$(\epsilon_R \otimes \mathrm{id}_R)(J)=(\mathrm{id}_R \otimes \epsilon_R)(J)=1$. Hence 
$J$ is a twist for 
$R$. Since our aim is to extend the bialgebra structure of some algebra 
$R=B^{\otimes m}$ to a skew monoid algebra 
$R\# M$, where 
$M$ is generated by some elements 
$\overline{s}_1, \ldots, \overline{s}_k$, such that 
$\Delta(\overline{s}_i) = J_i (\overline{s}_i \otimes \overline{s}_i)$ holds for some 
$J_i\in R\otimes R$, we will need that 
$J_i$ is a twist for 
$R$.
For two bialgebra-twist pairs 
$(B,J)$ and 
$(B',J')$, the element 
$(\mathrm{id}_B \otimes \tau_{B,B'} \otimes \mathrm{id}_{B'})(J\otimes J')$ is a twist for 
$R=B\otimes B'$, where 
$\tau_{B,B'}:B\otimes B' \to B'\otimes B$ denotes the usual flip map. Hence, given a twist 
$J=J^{(1)}\otimes J^{(2)}$ for 
$B$, we could consider
\begin{equation*}J' = \left(J^{(1)}\otimes 1_B \right)\otimes \left(J^{(2)} \otimes 1_B\right) = (\mathrm{id}_B \otimes \tau_{B,B} \otimes \mathrm{id}_B)(J \otimes 1_{B^{\otimes 2}}),\end{equation*}which is a twist for 
$B\otimes B$. However, we would like to embed 
$J$ differently into 
$(B\otimes B) \otimes (B\otimes B)$. For this, let 
$m\geq 2$ and denote by 
$e_i^m: B \to B^{\otimes m}$, the embedding of 
$B$ into the 
$i$th tensorand of 
$B^{\otimes m}$. We would like that 
$(e_i^m\otimes e_j^m)(J) \in B^{\otimes m} \otimes B^{\otimes m}$ is a twist for 
$B^{\otimes m}$, for any 
$1\leq i \lt j\leq m$. For 
$m=2$, we need in particular that 
$ (e_1^2\otimes e_2^2)(J)$ is a twist for 
$B\otimes B$. For 
$J=J^{(1)} \otimes J^{(2)}$, this condition is equivalent to the condition that
\begin{equation*}J' = (e_1^2\otimes e_2^2)(J) = \left(J^{(1)}\otimes 1_B\right) \otimes \left(1_B\otimes J^{(2)}\right)\end{equation*}satisfies Equation (4) for the ring 
$B\otimes B$, which boils down to
\begin{equation}
J_1^{(1)}j^{(1)} \otimes J_2^{(1)} \otimes j^{(2)} \otimes J^{(2)}
=
J^{(1)} \otimes j^{(1)}\otimes J_1^{(2)} \otimes J_2^{(2)}j^{(2)},
\end{equation}where 
$j^{(1)} \otimes j^{(2)}$ is another copy of 
$J$. Conversely, Equation (6) is actually equivalent to 
$ (e_1^2\otimes e_2^2)(J)$ being a twist for 
$B\otimes B$, because the counit condition is automatically verified. Comparing (6) and (5) shows that the difference lies in the middle tensorands. If 
$J$ is central in 
$B\otimes B$, then Equation (6) implies (5). It remains unclear whether the converse holds.
Lemma 1. Let 
$J$ be a twist for 
$B$. Suppose
\begin{equation} (e_1^2\otimes e_2^2)(J) \ \mbox{is a twist for } B \otimes B.\end{equation} Then 
$(e_i^m\otimes e_{j}^m)(J)$ is a twist for 
$B^{\otimes m}$, for any 
$1\leq i \lt j\leq m$.
Proof. Set 
$R=B^{\otimes m}$. To ease notation, we will drop the superscript 
$m$ from 
$e_i^m$. Let 
$J'=(e_i\otimes e_{j})(J) \otimes R\otimes R$. Note that 
$\Delta_R e_i = (e_i\otimes e_i) \Delta_B$ and 
$1_R = e_i(1_B)$. Then
\begin{align*}
( \Delta_R \otimes id_R)(J')(J' \otimes 1_R)
&=
( \Delta_R e_i \otimes e_j) (J) (e_i\otimes e_j \otimes e_j)(J\otimes 1_B)\\
&=
(e_i\otimes e_i \otimes e_j) ( \Delta_B\otimes \mathrm{id}_B)(J) (e_i\otimes e_j \otimes e_j )(J\otimes 1_B)\\
&=
e_i(J^{(1)}_1j^{(1)}) \otimes e_i(J^{(1)}_2)e_j(j^{(2)}) \otimes e_j(J^{(2)})\\
&=
(e_i\otimes e_{i,j}\otimes e_j)\left(J^{(1)}_1j^{(1)} \otimes J^{(1)}_2 \otimes j^{(2)} \otimes J^{(2)}\right),
\end{align*}where 
$j^{(1)}\otimes j^{(2)}$ is another copy of 
$J$ and 
$e_{i,j}: B^{\otimes 2} \to R$ denotes the embedding of 
$B\otimes B$ into the 
$i$th and 
$j$th tensorands of 
$R$. Similarly one shows 
$( \mathrm{id}_R \otimes \Delta_R )(J')(1_R\otimes J' ) = (e_i\otimes e_{i,j}\otimes e_j)\left(J^{(1)} \otimes j^{(1)}\otimes J_1^{(2)} \otimes J_2^{(2)}j^{(2)}\right).$ Since 
$J$ satisfies Equation (6), 
$J^{(1)}_1j^{(1)} \otimes J^{(1)}_2 \otimes j^{(2)} \otimes J^{(2)} =
j^{(1)}\otimes J_1^{(2)} \otimes J_2^{(2)}j^{(2)}$ and therefore 
$( \Delta_R \otimes id_R)(J')(J' \otimes 1_R) = ( \mathrm{id}_R \otimes \Delta_R )(J')(1_R\otimes J' ).$
The main idea of this note is to use the action of the symmetric group 
$\Sigma_m$ on the tensor product 
$R=B^{\otimes m}$ and to extend the bialgebra structure of 
$R$ to that of a suitable quotient of a skew monoid algebra. The comultiplication of the newly added generators, however, shall be twisted by some twists of 
$R$. A pair 
$(\sigma, J)$ of an algebra homomorphism 
$\sigma:R\to R$ and a twist 
$J$ is called a twisted homomorphism (see [Reference Davydov1]) if 
$\epsilon \sigma = \epsilon$ and 
$\Delta \sigma (h) = J(\sigma \otimes \sigma)\Delta(h)J^{-1}$, for 
$h\in R$. We will be mostly interested in the case where 
$J$ commutes with the elements of the image of 
$\Delta$, in which case 
$(\sigma, J)$ is a twisted homomorphism if and only if 
$\sigma$ is a coalgebra homomorphism. Pansera showed in [Reference Pansera9] that if 
$R$ is a bialgebra with twisted homomorphism 
$(\sigma ,J)$, then the bialgebra structure of 
$R$ can be extended to the skew polynomial ring 
$R[x;\sigma]$, such that 
$\Delta(x)=J(x\otimes x)$ and 
$\epsilon(x)=1$. We will extend this result to 
$R= B^{\otimes m}$ and a twist 
$J$ for 
$B$ that satisfies (7). Here the skew polynomial ring will be replaced by a skew monoid algebra.
Each permutation 
$s \in \Sigma_m$ acts on 
$R=B^{\otimes m}$ by 
$\sigma_{s}: R\to R$, with 
$\sigma_{s} (a_1\otimes \cdots \otimes a_n) = a_{s(1)} \otimes \cdots \otimes a_{s(n)}$. Note that 
$\sigma_{s}$ is a bialgebra homomorphism of 
$R$. Thus 
$\Delta \sigma_{s} = (\sigma_{s} \otimes \sigma_{s})\Delta$ and 
$\epsilon\sigma_{s} = \epsilon$. Given a twist 
$J$ for 
$B$, such that 
$(e_1^2\otimes e_2^2)(J)$ is a twist for 
$B \otimes B$, we define for each transposition 
$s = (i j)\in \Sigma_m$, with 
$i \lt j$, the element 
$J_s = (e_i^m \otimes e_j^m)(J)\in R\otimes R$, which is a twist for 
$R$ by Lemma 1.
Theorem 2. Let 
$B$ be a bialgebra with central twist 
$J$ of 
$B$ that satisfies (7). Let 
$m\geq 2$ and 
$X\subseteq \Sigma_m$ a set of transpositions. Then the free monoid 
$M$, generated by 
$\{\overline{s} : s\in X\}$, acts on 
$R=B^{\otimes m}$ and the bialgebra structure on 
$R$ extends to a bialgebra structure on the skew monoid algebra 
$R\# M$, such that 
$\Delta(\overline{s}) = J_{s} (\overline{s} \otimes \overline{s})$ and 
$\epsilon(\overline{s})=1$, for all 
$s\in X$. Moreover, 
$\pi:R\# M \to \mathbb{K}[M]$ with 
$\pi(a\# \overline{w}) = \epsilon(a)\overline{w}$ is a homomorphism of bialgebras.
Proof. Denote by 
$\mathrm{Aut}(R)$ the group of algebra automorphisms of 
$R$. Since 
$M$ is the free monoid on 
$X$, the map 
$X\to \mathrm{Aut}(R)$ given by 
$s\mapsto \sigma_s$ extends to a map of monoids 
$M\to \mathrm{Aut}(R)$, which allows us to consider the skew monoid ring 
$H=R\# M$, whose basis as a left 
$R$-module is given by the words in 
$\{\overline{s}: s\in X\}$. We identify elements 
$a\in R$ with elements 
$a\#1_M$, where 
$1_M$ is the identity of 
$M$. In particular, for each 
$s\in X$ and 
$a\in R$, we have 
$ \overline{s}a = \sigma_s(a)\overline{s}.$ To define 
$\Delta_H: H\to H\otimes H$, we set
\begin{equation}
\Delta_H(a\overline{w}) := \Delta_R(a) J_{s_1}(\overline{s}_1\otimes \overline{s}_1) \cdots J_{s_k}(\overline{s}_k\otimes \overline{s}_k),
\end{equation}for 
$a\in R$ and 
$w=s_1\ldots s_k$ a word in 
$X$, In particular, 
${\Delta_H}_{\mid_R}=\Delta_R$ and 
$\Delta_H(\overline{s}) = J_s (\overline{s}\otimes \overline{s})$, for 
$s\in X$. Since 
$J$ is central in 
$B^{\otimes 2}$, 
$J_s$ is also central in 
$R\otimes R$, and since 
$\sigma_s$ is a bialgebra homomorphism we have
\begin{equation} \Delta_H(\overline{s})\Delta_R(a)
= J_s(\overline{s}\otimes \overline{s})\Delta_R(a)
= J_s (\sigma_s\otimes \sigma_s)\Delta_R(a) (\overline{s} \otimes \overline{s})
= \Delta_R(\sigma_s(a)) \Delta_H(\overline{s}).\end{equation} Thus, 
$\Delta_H$ is a well-defined algebra homomorphism. The coassociativity follows from the fact that 
$J_s$ is a twist for 
$R$, by Lemma 1:
\begin{align*}
(\Delta_H\otimes \mathrm{id}_H)\Delta_H(\overline{s})
&= (\Delta_R\otimes \mathrm{id}_R)\Delta_R(J_s) \left(\Delta_H(\overline{s})\otimes \overline{s}\right)\\
&= (\Delta_R\otimes \mathrm{id}_R)\Delta_R(J_s) (J_s\otimes 1_R) (\overline{s} \otimes \overline{s}\otimes \overline{s})\\
&= (\mathrm{id}_R\otimes \Delta_R)\Delta_R(J_s) (1_R\otimes J_s) (\overline{s}\otimes \overline{s})\otimes \overline{s})\\
&= (\mathrm{id}_R \otimes \Delta_R)\Delta_R(J_s) \left(\overline{s}\otimes \Delta_H(\overline{s})\right)\\
&=(\mathrm{id}_H\otimes \Delta_H)\Delta_H(\overline{s})
\end{align*} Furthermore, set 
$\epsilon_H(a\overline{w})=\epsilon_R(a)$, for all 
$a\in R$ and 
$w\in M$. Since 
$\epsilon_R\sigma_s=\epsilon_R$, we conclude that 
$\epsilon_H$ is an algebra homomorphism. For any 
$a\in R$ and 
$s\in X$:
\begin{equation*}(\epsilon_H\otimes \mathrm{id}_H)\Delta_H(a\overline{s})
= (\epsilon_H\otimes \mathrm{id}_H) \left(\Delta_R(a) J_s (\overline{s}\otimes \overline{s})\right)
= a (\epsilon_R\otimes \mathrm{id}_R)(J_s) \overline{s} = a\overline{s},\end{equation*}since 
$(\epsilon_R\otimes \mathrm{id}_R)(J_s) = (\epsilon_B\otimes id_B)(J)=1$. Similarly, one checks 
$(id_H\otimes \epsilon_H)$ 
$\Delta_H(a\overline{s}) =1$.
Remark 3. Note that 
$R\# \overline{1}~=\{\gamma \in R\#M \mid (id\otimes \pi)\Delta(\gamma) = \gamma \otimes 1\}.$ Hence 
$R\#M$ is a bialgebra extension of 
$\mathbb{K}[M]$ by 
$R$.
Remark 4. Let 
$s\in \Sigma_m$ be an element of order 
$2$. Then 
$s$ is a product of disjoint transpositions, say 
$s=s_1\cdots s_k$. Since the elements 
$ J_{s_1}, \ldots, J_{s_k}$ commute pairwise, the product 
$J_s := J_{s_1}\cdots J_{s_k}$ is also a twist for 
$R$ and one could also prove a version of Theorem 2 for subsets 
$X\subseteq \Sigma_m$ of permutations of order 
$2$.
Denote by 
$\mu_R$ the multiplication of 
$R$.
Theorem 5. Let 
$B$ be a bialgebra with 
$J$ a central twist for 
$B$ that satisfies (7). Let 
$m\geq 2$, 
$X\subseteq \Sigma_m$ a set of transpositions, 
$M$ the free monoid generated by 
$\{\overline{s} : s\in X\}$ acting on 
$R=B^{\otimes m}$.
(1) Suppose
$J$ satisfies
(10)\begin{equation}
\Delta_{B\otimes B}(J) = (e_1^2 \otimes e_2^2)(J)(e_2^2 \otimes e_1^2)(J)(J\otimes J).
\end{equation}Then the ideal
$I=\langle \overline{s}^2 - t_s \: : \: s\in X\rangle$ is a biideal of
$R\# M$, where
$t_s = \mu_{R}(J_s)$.(2) Suppose moreover that
$B$ is a Hopf algebra with antipode
$S$, such that
$(S\otimes S)(J)=J$ and
$(S\otimes \mathrm{id})(J) = J^{-1}= (\mathrm{id}\otimes S)(J).$ Then
$H=(R\#M)/I$ is a Hopf algebra with
$S(\overline{s}+I) = \overline{s}+I$ in
$H$, for
$s\in X$.
Proof. (1) Let 
$s = (i j) \in X$, with 
$i \lt j$. Note that 
$e_{ij}^m:=\mu_R (e_i^m\otimes e_j^m):B\otimes B\to R$ is an algebra homomorphism that satisfies
\begin{equation}
\Delta_R\circ e_{ij}^m = (e_{ij}^m\otimes e_{ij}^m)\circ \Delta_{B\otimes B},
\end{equation}as well as 
$e_{ij}^me _1^2 = e_i^m$ and 
$e_{ij}^m e_2^2 = e_j^m$. Set 
$J_s := (e_i^m \otimes e_j^m)(J)\in R^{\otimes 2}$ and 
$t_s = \mu_R(J_s)=e_{ij}^m (J)\in R$. Since 
$J$ is central in 
$B\otimes B$, so is 
$t_s$ in 
$R$. Then Equations (10) and (11) imply
\begin{align*}
\Delta_{R}(t_s) = \Delta_R(e_{ij}^m(J))
&= e_{ij}^m\otimes e_{ij}^m \left(\Delta_{B\otimes B}(J)\right) \\
&= e_{ij}^m\otimes e_{ij}^m \left((e_1^2\otimes e_2^2)(J) (e_2^2\otimes e_1^2)(J)(J\otimes J)\right) \\
&=
(e_i^m\otimes e_j^m)(J) (e_j^m\otimes e_i^m)(J) (e_{ij}^m(J) \otimes e_{ij}^m(J)) \\
&= J_s (\sigma_s\otimes \sigma_s)(J_s)(t_s\otimes t_s).
\end{align*} Hence, 
$\Delta(\overline{s}^2 - t_s)
= J_s(\overline{s}\otimes \overline{s})J_s(\overline{s}\otimes \overline{s}) - \Delta(t_s)
= J_s(\sigma_s\otimes \sigma_s)(J)\left(\overline{s}^2\otimes \overline{s}^2 - t_s\otimes t_s\right)
$ is an element of 
$I\otimes (R\# M) + (R\# M)\otimes I$. Since 
$\epsilon_H(t_s)=\epsilon_R(\mu_R(J)))=1=\epsilon_H(\overline{s}^2)$, we conclude that 
$I$ is a coideal of 
$R\# M$.
(2) Set 
$H=(R\# M)/I$ and assume that 
$B$ is a Hopf algebra with antipode 
$S$. Then 
$R$ is a Hopf algebra with antipode 
$S^{\otimes m}$, which we will also denote by 
$S$. Since 
$(S\otimes S)(J)=J$ holds, we have 
$S(t_s) = S(e_{ij}^m(J))=e_{ij}^m(S\otimes S)(J) = t_s.$ Hence, 
$S(\overline{s})=\overline{s}$ modulo 
$I$ is well defined. We also set
\begin{equation}
S(a\overline{s}_1\cdots \overline{s}_k +I)=\overline{s}_k\cdots \overline{s}_1 S(a)+I,
\end{equation}for all 
$a\in R$ and 
$s_1,\ldots, s_k \in X$. Since 
$S\sigma_s = \sigma_s S$, definition (12) leads to a well-defined algebra anti-homomorphism 
$S:H\to H$. Writing 
$J$ again symbolically as 
$J=J^{(1)} \otimes J^{(2)}$, we calculate modulo 
$I$:
\begin{align*} \mu_H(\mathrm{id}_H \otimes S)\Delta_H (\overline{s})
&= \mu_H(\mathrm{id}_H \otimes S) J_s (\overline{s}\otimes \overline{s})\\
&= e_i^m(J^{(1)})\overline{s} S(\overline{s}) S(e_j^m(J^{(2)}))\\
&= e_i^m(J^{(1)})S(e_j^m(J^{(2)}))t_s\\
&= e_{ij}^m\left( (id\otimes S)(J)\right) e_{ij}^m(J)
= e_{ij}^m\left( J^{-1}J\right) = 1
\end{align*} Similarly 
$\mu_H(S \otimes \mathrm{id}_H)\Delta_H (\overline{s}) = 1$ holds modulo 
$I$, because of 
$(S\otimes \mathrm{id})(J) = J^{-1}$. Thus 
$S$ defines an antipode for the bialgebra 
$H$.
Since our aim is to construct semisimple Hopf algebra quotients of 
$R\# M$, they must be necessarily finite dimensional over 
$\mathbb{K}$. Thus we will consider quotients of 
$R\# M$, by using the usual Coxeter presentation of 
$\Sigma_m$ with generators 
$s_i = (i, i+1)$.
Theorem 6
(Extension by the symmetric group). Let 
$B$ be a Hopf algebra with antipode 
$S$ and 
$J$ a central twist for 
$B$ that satisfies (7) and (10), such that 
$(S\otimes S)(J)=J$ and 
$(\mathrm{id}\otimes S)(J)=J^{-1}=(S\otimes \mathrm{id})(J)$. For any 
$m\geq 2$, let 
$X = \{s_1, \ldots, s_{m-1}\}$ be the set of transposition 
$s_i = (i, i+1)$ that generates 
$\Sigma_m$ and let 
$M$ be the free monoid generated by 
$\{\overline{s_i} : 1\leq i \lt m \}$, which acts naturally on 
$R = B^{\otimes m}$. Consider the ideal 
$I$ of 
$R\# M$ generated by
\begin{equation*} \overline{s}_i^2 - t_{s_i}, \qquad \overline{s}_i\overline{s}_{i+1}\overline{s}_i - \overline{s}_{i+1}\overline{s}_i\overline{s}_{i+1}, \qquad \overline{s}_i\overline{s}_{j} - \overline{s}_j\overline{s}_i,\qquad\forall i,j \ \mbox{with } |i-j| \gt 1.\end{equation*} Then 
$H=(R\# M)/I$ is a Hopf algebra of dimension 
$\dim(B)^{m} m!$ with Hopf subalgebra 
$R=B^{\otimes m}$ and Hopf quotient 
$\mathbb{K} \Sigma_m$. Furthermore,
(1)
$\{\overline{w}: w\in \Sigma_m\}$ is an
$R$-basis of
$H$;(2) there exists a
$2$-cocycle
$\gamma:\Sigma_m\times \Sigma_m \to Z(R)^\times$ such that
$H \simeq R \#_\gamma \Sigma_m$ with multiplication given by
(13)for all\begin{equation}
(a\# \overline{w})(b\# \overline{v}) =
a\sigma_w(b) \gamma(w,v)\# \overline{wv}.
\end{equation}
$a,b \in R$ and
$w,v\in \Sigma_m$, where
$\gamma(w,v)$ is a product of elements
$\sigma_u(t_s)$, for some
$u\in \Sigma_m$ and
$s\in X$.(3)
$\epsilon(\gamma(w,v))=1$, for all
$w,v\in \Sigma_m$.(4) if
$B$ is semisimple, then
$H$ is semisimple with integral
$\int \sum_{w\in \Sigma_m} \overline{w}$, where
$\int$ is the integral of
$R$.
Proof. From Theorem 5 we know already that 
$\langle \overline{s}_i^2-t_{s_i}: 1\leq i \lt m\rangle$ is a coideal of 
$R\# M$. Furthermore, for 
$i,j$ such that 
$|i-j| \gt 1$, the transpositions 
$s_i$ and 
$s_j$ are disjoint. Hence 
$J_{s_i}$ and 
$\overline{s}_j\otimes \overline{s}_j$, as well as 
$J_{s_j}$ and 
$\overline{s}_i\otimes \overline{s}_i$ commute in 
$R\# M$. Thus
\begin{equation*}\Delta(\overline{s}_i\overline{s}_j - \overline{s}_j\overline{s}_i)
= J_{s_i}J_{s_j}
\left(\overline{s}_i\overline{s}_j\otimes \overline{s}_i\overline{s}_j - \overline{s}_j\overline{s}_i\otimes \overline{s}_j\overline{s}_i\right) \in (R\#M)\otimes I + I \otimes (R\#M).\end{equation*} For 
$i \lt m-1$, we have 
${\sigma_{s_i}}e_i^m = e_{i+1}^m$ and 
${\sigma_{s_i}}e_{i+1}^m = e_{i}^m$, while 
${\sigma_{s_i}}e_j^m = e_j^m$ if 
$j\not\in\{i, i+1\}$. Hence
\begin{equation}(\sigma_{s_i}\otimes \sigma_{s_i})J_{s_{i+1}}
= (e_{i}^m \otimes e_{i+2}^m)(J)
= (\sigma_{s_{i+1}}\otimes \sigma_{s_{i+1}})J_{s_{i}}
\end{equation} In particular, 
$(\sigma_{s_i}\sigma_{s_{i+1}}\otimes \sigma_{s_i}\sigma_{s_{i+1}})(J_{s_{i}}) = J_{s_{i+1}}$ and 
$
(\sigma_{s_{i+1}}\sigma_{s_{i}}\otimes \sigma_{s_{i+1}}\sigma_{s_{i}})(J_{s_{i+1}}) = J_{s_i}$. Thus,
\begin{equation}
J_{s_i}(\sigma_{s_i}\otimes \sigma_{s_i})(J_{s_{i+1}})(\sigma_{s_i}\sigma_{s_{i+1}}\otimes \sigma_{s_i}\sigma_{s_{i+1}})(J_{s_{i}})
=
J_{s_i}(\sigma_{s_i}\otimes \sigma_{s_i})(J_{s_{i+1}}) J_{s_{i+1}}
\end{equation}
\begin{equation}
J_{s_{i+1}}
(\sigma_{s_{i+1}}\otimes \sigma_{s_{i+1}})(J_{s_{i}})
(\sigma_{s_{i+1}}\sigma_{s_{i}}\otimes \sigma_{s_{i+1}}\sigma_{s_{i}})(J_{s_{i+1}}) =
J_{s_{i+1}}(\sigma_{s_{i}}\otimes \sigma_{s_{i}})(J_{s_{i+1}}) J_{s_i}
\end{equation} Since 
$J$ is central, the values in (15) and (16) are equal; call it 
$\gamma$. Then
\begin{equation*}
\Delta(\overline{s}_{i}\overline{s}_{i+1}\overline{s}_{i})
= J_{s_1}(\overline{s}_{i}\otimes \overline{s}_{i}) J_{s_{i+1}} (\overline{s}_{i+1}\otimes \overline{s}_{i+1})J_{s_1}(\overline{s}_{i}\otimes \overline{s}_{i})
= \gamma(\overline{s}_{i}\overline{s}_{i+1}\overline{s}_{i}
\otimes \overline{s}_{i}\overline{s}_{i+1}\overline{s}_{i})
\end{equation*}and similarly 
$
\Delta(\overline{s}_{i+1}\overline{s}_{i}\overline{s}_{i+1})
= \gamma(\overline{s}_{i+1}\overline{s}_{i}\overline{s}_{i+1}
\otimes \overline{s}_{i+1}\overline{s}_{i}\overline{s}_{i+1}).
$ Hence 
$\Delta(\overline{s}_{i}\overline{s}_{i+1}\overline{s}_{i}-\overline{s}_{i+1}\overline{s}_{i}\overline{s}_{i+1})$ is an element of 
$I\otimes (R\#M) + (R\#M) \otimes I$. Defining 
$S(\overline{s}_i)=\overline{s}_i$, for all 
$i$, we obtain that 
$H=(R\# M)/I$ is a Hopf algebra. Note that 
$R\# M$ is graded by the length of words in 
$M$. Since elements of the form 
$\overline{s}_i\overline{s}_{j}-\overline{s}_j\overline{s}_{i}$ and 
$\overline{s}_{i}\overline{s}_{i+1}\overline{s}_{i}-\overline{s}_{i+1}\overline{s}_{i}\overline{s}_{i+1}$ are homogeneous of degree 
$2$ and 
$3$ respectively, the only possible 
$a\in R\cap I$ must be of the form 
$a=\sum_{i=1}^{m-1} a_i(\overline{s}_i^2-t_i)$. Since words in 
$M$ form an 
$R$-basis of 
$R\# M$, we must have 
$a_i =0$, for all 
$i$; hence 
$R\cap I=0$ and 
$R$ embeds into 
$H=(R\# M)/I$ as a Hopf subalgebra of 
$H$. Actually, 
$R$ is a normal Hopf subalgebra and 
$HR^+=R^+H$ is a Hopf ideal of 
$H$, where 
$R^+ = R\cap \mathrm{Ker}(\epsilon)$.
(1) Since 
$s_1,\ldots, s_{n-1}$ generate 
$\Sigma_m$, we choose for each 
$w\in \Sigma_m$ a representation 
$w = s_{i_1}\cdots s_{i_{k}}$, with 
$k$ minimal. We will write 
$\overline{w}= \overline{s}_{i_1}\cdots \overline{s}_{i_{k}}$, for the corresponding element in 
$R\# M$. Since the 
$s_i$ satisfy 
$s_is_{i+1}s_i = s_{i+1}s_is_{i+1}$ and 
$s_is_j = s_js_i$, for 
$|i-j| \gt 1$, any word in 
$\overline{s}_1, \ldots, \overline{s}_{m-1}$ reduces to an element 
$\overline{w}$ modulo 
$I$. Thus 
$\{\overline{w}+ I : w\in \Sigma_m\}$ generates 
$H$ as a left 
$R$-module. On the other hand, an element of the form 
$\sum_{w\in \Sigma_m} a_w \overline{w} $, with 
$a_w\in R$ belongs to 
$I$ if and only if 
$a_w=0$, for any 
$w\in \Sigma_m$. Thus 
$H$ is free over 
$R$ with basis 
$\{\overline{w} : w\in \Sigma_m\}$ and 
$\dim(H)=\dim(R)^m m!$ follows.
(2) Let 
$w,v \in \Sigma_m$ then there exists a unique element 
$\gamma(w,v)\in R$ such that 
$\overline{w}\:\overline{v} = \gamma(w,v)\overline{wv}$. More precisely, let 
$w= s_{i_1}\cdots s_{i_k}$ and 
$v= s_{j_1}\cdots s_{j_l}$ be the chosen minimal presentations of 
$w$ and 
$v$. If 
$wv$ has length 
$k+l$, then set 
$\gamma(w,v)=1$. If the length of 
$wv$ is less than 
$k+l$, then some reduction occurred, which means that some relations of the type 
$s_i^2=\mathrm{id}$ must have reduced the length of 
$wv$. In 
$H$, the reduction takes the form of replacing 
$\overline{s}_i^2$ by 
$t_{s_i}\in R^\times$. Depending on where in 
$wv$ the replacement takes place, some automorphism of the form 
$\sigma_u$ is applied to shift 
$t_{s_i}$ to the ‘left’ of 
$\overline{wv}$. Hence 
$\overline{w}\:\overline{v} = \gamma(w,v) \overline{wv}$, where 
$\gamma(w,v)$ is a product of elements of the form 
$\sigma_u(t_{s_i})$, for some 
$u\in \Sigma_m$ and some 
$i$.
(3) Since 
$J_{s_i}$ is central in 
$R\otimes R$, also 
$t_{s_i} = \mu_R(J_{s_i})$ is central in 
$R$. Furthermore, as 
$(\epsilon_B\otimes \mathrm{id}_B)(J)=1$, also 
$\epsilon(t_{s_i})=1$ and therefore 
$\epsilon(\sigma_u(t_{s_i}))=1$. Thus 
$\epsilon(\gamma(w,v))=1$, for any 
$w,v\in \Sigma_m$.
(4) If 
$B$ is semisimple, then 
$R$ is semisimple and the antipode of 
$R$ is an involution. Since 
$S^2(\overline{w})=\overline{w}$, for 
$w\in \Sigma_m$, the antipode of 
$H$ is an involution and 
$H$ is semisimple. Let 
$\int_B$ be the integral of 
$B$, then 
$\int = \int_B\otimes \cdots \otimes \int_B\in R=B^{\otimes m}$ is an integral for 
$R$. Note that 
$\int$ is invariant under the action of 
$\Sigma_m$. Hence, for 
$v\in \Sigma_m$, we have
\begin{equation}
\overline{v}\int\sum_{w\in \Sigma_m}\overline{w} = \sigma_v(\int)\sum_{w\in \Sigma_m} \gamma(v,w)\overline{vw} = \sum_{w\in \Sigma_m}\int\epsilon(\gamma(v,w)) \overline{vw} = \int \sum_{w\in \Sigma_m} \overline{w}.
\end{equation}3. Generalized Kac–Paljutkin algebras
The group algebra of a finite cyclic group provides a twist that satisfies all the requirements of the last Theorem. Let 
$n,m\geq 2$, 
$q$ a primitive 
$n$th root of unity and 
$\mathbb{Z}_n=\langle x : x^n=1\rangle$ the cyclic group of order 
$n$. For each 
$0\leq k \lt n$ set
\begin{equation*} e_k = \frac{1}{n} \sum_{i=0}^{n-1} q^{-ik} x^i.\end{equation*} Then 
$\{e_0, \ldots, e_{n-1}\}$ is a complete set of orthogonal idempotents of 
$B=\mathbb{K} \mathbb{Z}_n$.
Lemma 7. Let 
$B=\mathbb{K}\mathbb{Z}_n$. Then 
$J=\sum_{k=0}^{n-1} e_k\otimes x^k \in B\otimes B$ is a twist for 
$B$ satisfying (7) and (10), such that 
$(S\otimes S)(J)=J$ and 
$(\mathrm{id}\otimes S)(J)=J^{-1} = (S\otimes \mathrm{id})(J)$ hold.
Proof. It is well-known that 
$J$ is a twist for 
$B$. We calculate 
$(S\otimes S)(J) = \frac{1}{n}\sum_{i,k=0}^{n-1} q^{-ik} x^{-i} \otimes x^{-k} = J$. Moreover,
\begin{equation}
(\mathrm{id}\otimes S)(J) J = \sum_{k,l=0}^{n-1} e_ke_l \otimes x^{l-k} = \sum_{k=0}^{n-1} e_k \otimes 1_B = 1_B\otimes 1_B,
\end{equation}shows 
$(\mathrm{id} \otimes S)(J)=J^{-1}$. Since 
$S^2=\mathrm{id}$ and 
$(S\otimes S)(J^{-1})=J^{-1}$, also 
$(S\otimes \mathrm{id})(J)=J^{-1}$ holds. Next we show that
\begin{equation*}J' =(e_1^2\otimes e_2^2)(J) = \sum_{k=0}^{n-1} e_k \otimes 1_B \otimes 1_B \otimes x^k =
\sum_{k=0}^{n-1} x^k \otimes 1_B \otimes 1_B \otimes e_k\end{equation*}is a twist for 
$B\otimes B$:
\begin{align*}
&(\Delta_{B\otimes B} \otimes \mathrm{id}_{B\otimes B})(J')(J'\otimes 1_{B\otimes B})\\
&\quad =
\sum_{k,l} (\Delta_{B\otimes B}(x^k\otimes 1)\otimes 1\otimes e_k) (e_l\otimes 1\otimes 1 \otimes x^l \otimes 1\otimes 1)\\
&\quad = \sum_{k,l} x^ke_l \otimes 1 \otimes x^k \otimes x^l \otimes 1 \otimes e_k\\
&\quad =
\sum_{k,l} e_l \otimes 1 \otimes x^k \otimes x^l \otimes 1 \otimes x^le_k\\
&\quad =
\sum_{k,l} \left(e_l \otimes 1 \otimes \Delta_{B\otimes B}(1\otimes x^l)\right) (1\otimes 1 \otimes x^k \otimes 1 \otimes 1\otimes e_k)\\
&\quad = (\mathrm{id}_{B\otimes B} \otimes \Delta_{B\otimes B}(J')(1_{B\otimes B} \otimes J')
\end{align*}where we use 
$x^ke_l =q^{kl}e_l$ and 
$q^{kl}e_k = x^le_k$. Lastly, we will verify Equation (10). For this consider
\begin{align*}
(e_1\otimes e_2)(J) (e_2\otimes e_1)(J) (J\otimes J)
&=
\sum_{k,l,s,t} (e_k\otimes x^l \otimes e_l \otimes x^k)(e_s\otimes x^s\otimes e_t \otimes x^t)\\
&=\sum_{k,l} e_k\otimes x^{k+l} \otimes e_l \otimes x^{k+l}\\
&= \sum_{k,j} e_k \otimes x^{j} \otimes e_{j-k} \otimes x^{j}
= \Delta_{B\otimes B}(J),
\end{align*}because 
$ \Delta_B(e_j)=\sum_{k} e_k \otimes e_{j-k}$ holds.
From Theorem 6 and Lemma 7, we deduce that there exists a non-trivial semisimple Hopf algebra 
$H_{n,m} = \left(\mathbb{K}\mathbb{Z}_n\right)^{\otimes m} \#_\gamma \Sigma_m$ of dimension 
$n^m m!$, for any integers 
$n,m\geq 2$ (provided 
$\mathbb{K}$ contains a primitive 
$n$th root of unity).
Theorem 8
(Generalized Kac–Paljutkin algebras). For any 
$n,m \geq 2$, 
$\mathbb{K}$ a field of characteristic zero with primitive 
$n$th root of unity 
$q$, there exists a semisimple Hopf algebra 
$H_{n,m} = \mathbb{K}\mathbb{Z}_n^{\otimes m} \#_\gamma \Sigma_m$ of dimension 
$n^m m!$ that is an extension of 
$\mathbb{K} \Sigma_m$, by 
$\mathbb{K}\mathbb{Z}_n^{\otimes m}$, which is generated by elements 
$x_1,\ldots, x_m, z_1,\ldots, z_{m-1}$ such that
\begin{equation*}x_i^n=1, \qquad x_ix_j=x_ix_j, \qquad z_k x_i = x_{s_k(i)} z_k, \qquad \forall \:i,j,k,\end{equation*}where 
$s_k$ is the transposition 
$s_k=(k, k+1)$. Furthermore, for all 
$k,l$ with 
$|k-l| \gt 1$:
\begin{equation}
z_kz_l=z_lz_k, \qquad z_kz_{k+1}z_k = z_{k+1}z_kz_{k+1}, \qquad z_k^2=\frac{1}{n} \sum_{i,j=0}^{n-1} q^{-ij} x_{k}^i x_{k+1}^j.
\end{equation} The elements 
$x_i$ are group-like, while
\begin{equation}
\Delta(z_k) = \left(\frac{1}{n} \sum_{i,j=0}^{n-1} q^{-ij} x_{k}^i \otimes x_{k+1}^j \right) (z_k\otimes z_k), \quad \epsilon(z_k)=1, \quad S(z_k)=z_k.
\end{equation} As before, we write 
$t_k:=\frac{1}{n} \sum_{i,j=0}^{n-1} q^{-ij} x_{k}^i x_{k+1}^j$ and 
$J_k := \frac{1}{n} \sum_{i,j=0}^{n-1} q^{-ij} x_{k}^i \otimes x_{k+1}^j $. Since 
$(\mathrm{id}\otimes S)(J_k) = J_k^{-1}$, we also have
\begin{equation*}t_k^{-1} = \frac{1}{n} \sum_{i,j=0}^{n-1} q^{-ij} x_{k}^i x_{k+1}^{-j}.\end{equation*}Remark 9. Let 
$G:=\mathbb{Z}_n^m$. Then 
$\mathbb{K} G \simeq \mathbb{K}\mathbb{Z}_n^{\otimes m}$. By construction, 
$\mathbb{K} G$ is a Hopf subalgebra of 
$H:=H_{n,m}$ and 
$\mathbb{K} \Sigma_m$ is a Hopf quotient algebra of 
$H$. Since the ring of 
$\mathbb{K} \Sigma_m$-coninvariants of 
$H$ is 
$\mathbb{K} G$ and since 
$\mathbb{K} G \simeq \mathbb{K}^G$ as Hopf algebras,Footnote 2 as 
$G$ is finite Abelian, the Hopf algebra 
$H$ is an extension of 
$\mathbb{K} \Sigma_m$ by 
$\mathbb{K}^G$ and hence equivalent to an element of 
$\operatorname{Opext}(\mathbb{K} \Sigma_m, \mathbb{K}^G)$, the set of equivalence classes of all extensions of 
$\mathbb{K}\Sigma_m$ by 
$\mathbb{K}^G$, associated with a matched pair 
$(\Sigma_m, G)$. Here the matched pair is given by 
$\Sigma_m$ acting on 
$G$ by permutations of its components and 
$G$ acting trivially on 
$\Sigma$. The set 
$\operatorname{Opext}(\mathbb{K} \Sigma_m, \mathbb{K}^G)$ forms a group with identity element being the group ring 
$\mathbb{K}[G\rtimes \Sigma_m]$ of the semidirect product of 
$G$ and 
$\Sigma_m$. Since 
$H_{n,m}$ is not cocommutative, 
$H_{n,m}$ is equivalent to a non-trivial element of 
$\operatorname{Opext}(\mathbb{K}\Sigma_m, \mathbb{K}^G)$. By [Reference Masuoka8, Proposition 1.5], elements of 
$\operatorname{Opext}(\mathbb{K}\Sigma_m, \mathbb{K}^G)$ are equivalent to a bicrossed products 
$\mathbb{K}^G {\#^\tau_\gamma}\: \mathbb{K}\Sigma_m$, for cocycles 
$\gamma: \Sigma_m \times \Sigma_m \to \mathbb{K}^G$ and 
$\tau: G\times G\to \mathbb{K}^{\Sigma_m}$. From the multiplication of 
$H_{n,m}$, we see that 
$\tau$ is constant 
$1_{K^{\Sigma_m}}$, while 
$\gamma$ depends on the chosen twist 
$J$.
For 
$m=2$, the elements of 
$\operatorname{Opext}(\mathbb{K}\Sigma_2, \mathbb{K}^{\mathbb{Z}_n\times \mathbb{Z}_n })$ have been classified in [Reference Masuoka7, Theorem 2.1] and are in bijective correspondence with all 
$n$th roots of unity in 
$\mathbb{K}$. Thus, the Hopf algebras 
$H_{n,2}$ had already been discovered in [Reference Masuoka7], prior to [Reference Pansera10]. In particular, for each root of unity 
$q$, there exists a twist 
$J$ (as defined above) that determines the comultiplication of such an extension via 
$\Delta(z)=J(z\otimes z)$. The cocycle 
$\gamma $ is determined by the element 
$t=\mu(J)$, where 
$\gamma(s,s)=t$ for the generator 
$s$ of 
$\Sigma_2=\mathbb{Z}_2$.
For 
$m\geq 3$, it would be interesting to classify 
$\operatorname{Opext}(\mathbb{K}\Sigma_m, \mathbb{K}^G)$ in general, whereas 
$H_{n,m}$ provides a non-trivial example of it. Again, for each 
$n$-th root of unity 
$q$, we use the same twist 
$J$ of 
$B=\mathbb{K}\mathbb{Z}_n$ as above. For each Coxeter generator 
$s_{i}$ of 
$\Sigma _{m}$, we deform the comultiplication of 
$z_i = \overline{s_i}$ by defining 
$\Delta(z_i)=J_i(z_i\otimes z_i)$, where 
$J_i = (e_i^m\otimes e_{i+1}^m)(J) \in R\otimes R$ for 
$R=B^{\otimes m}=\mathbb{K} G$ means that 
$J$ is placed into the 
$i$-th and 
$(i+1)$-st components, respectively. As Example 11 illustrates, the cocycle 
$\gamma $ is determined by 
$t=\mu(J)$, which is embedded suitably into 
$R$ such that 
$\gamma(s_i,s_i)=t_i:=\mu(J_i)$ and 
$\gamma(s_i,s_j)=1$ for 
$i\neq j$. Hence, the non-triviality of 
$H_{n,m}$ as an element of 
$\operatorname{Opext}(\mathbb{K}\Sigma_m, \mathbb{K}^G)$ stems essentially from the non-triviality of the Kac–Paljutkin algebra 
$H_{n,2}$ and has just been placed into an 
$m$-fold tensor product of 
$\mathbb{K}\mathbb{Z}_n$.
Note that it is shown in [Reference Halbig and Lomp4] that 
$H_{n,m}$ is isomorphic, as an algebra, to the group ring 
$\mathbb{K}[\mathbb{Z}_n \wr \Sigma_m]$ of the wreath product of 
$\mathbb{Z}_n$ and 
$\Sigma_m$. Since the wreath product 
$\mathbb{Z}_n\wr \Sigma_m$ is the semidirect 
$\mathbb{Z}_n^m \rtimes \Sigma_m$, the Hopf algebra 
$H_{n,m}$ is isomorphic as an algebra to the smash product 
$\mathbb{K}^G \#\: \mathbb{K}\Sigma_m$ (without cocycle). However, due to the fact that 
$H_{n,m}$ is not cocommutative, this isomorphism is not an isomorphism of Hopf algebras.
Example 10 (Case
$m=2$)
Write 
$x$ for the generator of 
$B=\mathbb{K}\mathbb{Z}_n$ and write the elements of 
$R=B^{\otimes 2} = \mathbb{K} \mathbb{Z}_n \otimes \mathbb{K}\mathbb{Z}_n$ as linear combinations of elements 
$x^iy^j$, where 
$x$ stands for 
$x\otimes 1$ and 
$y$ for 
$1\otimes x$. Since 
$m=2$, we have only one transposition, which is 
$s=(12)$. Hence 
$\sigma_s(x)=y$, 
$\sigma_s(y)=x$ and the twist 
$J=\sum_{j=0}^{n-1} e_j\otimes x^j$ for 
$B$ becomes the twist
\begin{equation*}J_s = \sum_{j=0}^{n-1} e_j \otimes y^j = \frac{1}{n} \sum_{i,j=0}^{n-1} q^{-ij} x^i \otimes y^j \end{equation*}for 
$R$. Instead of 
$\overline{s}$ write 
$z$ for the generator of the monoid 
$M=\langle \overline{s}\rangle$. Set 
$t:= \mu(J_s) = \frac{1}{n}\sum_{i,j=0}^{n-1} q^{-ij} x^iy^j$. The Hopf algebra 
$H_{n,2}=(R\#M)/\langle z^2-t\rangle$ has therefore generators 
$x,y,z$ subject to
\begin{equation*} xy=yx, \quad x^n=1=y^n, \quad zx=yz, \quad zy=xz, \quad z^2= t.\end{equation*} While 
$x$ and 
$y$ are group-like,
\begin{equation*}\Delta(z) = \left( \frac{1}{n} \sum_{i,j=0}^{n-1} q^{-ij} x^i \otimes y^j \right)(z\otimes z).\end{equation*} A basis for 
$H_{n,2}$ are the monomials 
$\{x^iy^jz^k : 0\leq i,j \lt n, k\in\{0,1\}\}$. Hence 
$H_{n,2}$ is precisely the generalized Kac–Paljutkin algebra 
$H_{2n^2}$ considered by Pansera. In particular, 
$H_{2,2}$ is the Kac–Paljutkin Hopf algebra itself. Note that 
$H_{n,2} = \mathbb{K}[\mathbb{Z}_n\times \mathbb{Z}_n] \#_\gamma \Sigma_2$, for 
$\gamma:\Sigma_2 \times \Sigma_2 \to R^\times$ given by
\begin{equation*}\gamma(\mathrm{id},\mathrm{id})=\gamma(\mathrm{id}, (12)) = \gamma((12),\mathrm{id}) = 1, \qquad \gamma((12),(12)) = t.\end{equation*}Example 11 (Case
$m=3$)
Let 
$q$ be a primitive third root of unity. Write 
$x$ for the generator of 
$B=\mathbb{K}\mathbb{Z}_n$ and 
$R=B^{\otimes 3} = \mathbb{K} \mathbb{Z}_n \otimes \mathbb{K}\mathbb{Z}_n\otimes \mathbb{K}\mathbb{Z}_n$. Elements of 
$R$ are linear combinations of monomials 
$x_1^ix_ 2^jx_3^k$, where 
$x_1$ stands for 
$x\otimes 1\otimes 1$, 
$x_2$ stands for 
$1\otimes x\otimes 1$ and 
$x_3$ stands for 
$1\otimes 1\otimes x$. The transpositions 
$s_1=(12)$ and 
$s_2=(23)$ are generators of 
$\Sigma_3$. Write 
$\sigma_i = \sigma_{s_i}$ for the corresponding automorphism of 
$R$. The twist 
$J=\sum_{j=0}^{n-1} e_j\otimes x^j$ for 
$\mathbb{K} \mathbb{Z}_n$ yields two twists
\begin{equation*}J_{s_1} = \frac{1}{n} \sum_{i,j=0}^{n-1} q^{-ij} x_1^i \otimes x_2^j \qquad \mbox{ and } \qquad
J_{s_2} = \frac{1}{n} \sum_{i,j=0}^{n-1} q^{-ij} x_2^i \otimes x_3^j\end{equation*}for 
$R$. Set 
$z_1:= \overline{s_1}$ and 
$z_2:= \overline{s_2}$ for the generator of the free monoid 
$M=\langle \overline{s_1}, \overline{s_2}\rangle$. The Hopf algebra
\begin{equation}
H_{n,3}=(R\#M)/\langle z_1^2-t_{1}, z_2^2-t_{2}, z_1z_2z_1-z_2z_1z_2\rangle = \left(\mathbb{K} \mathbb{Z}_n\right)^{\otimes 3} \#_\gamma \Sigma_3\end{equation}has therefore generators 
$x_1,x_2,x_3,z_1,z_2$ subject to 
$x_ix_j=x_jx_i$ and 
$x_i^n=1$, for all 
$i,j$ and
\begin{equation*} z_1x_1=x_2z_1, \quad z_1x_2=x_1z_1, \quad z_1x_3=x_3z_1\end{equation*}
\begin{equation*} z_2x_1=x_1z_2, \quad z_2x_2=x_3z_2, \quad z_2x_3=x_2z_2\end{equation*}
\begin{equation*}z_1z_2z_1 = z_2z_1z_2, \qquad z_1^2=t_1:=\frac{1}{n} \sum_{i,j=0}^{n-1} q^{-ij} x_1^ix_2^j, \qquad z_2^2=t_2:=\frac{1}{n}\sum_{i,j=0}^{n-1} q^{-ij} x_2^ix_3^j.\end{equation*} While the 
$x_i$ are group-like, we have
\begin{equation*}\Delta(z_1) = \left( \frac{1}{n} \sum_{i,j=0}^{n-1} q^{-ij} x_1^i \otimes x_2^j \right) (z_1\otimes z_1) \qquad
\Delta(z_2) = \left(\frac{1}{n} \sum_{i,j=0}^{n-1} q^{-ij} x_2^i \otimes x_3^j \right) (z_2\otimes z_2).\end{equation*} 
$H_{n,3}$ has dimension 
$6n^3$ and basis 
$ \left\{x_1^i \ x_2^j \ x_3^k \ \overline{w} : 0\leq i,j,l \lt n, \:\overline{w} \in \{1, z_1, z_2, z_1z_2, z_2z_1, \right.$ 
$\left. z_1z_2z_1\} \right\}$. The 
$2$-cocycle 
$\gamma$ is determined by the multiplication in 
$H_{n,3}$ and given by 
$\gamma(\mathrm{id}, w)=\gamma(w,\mathrm{id})=1$, for 
$w\in \Sigma_3$, and the following table. Note that 
$t_2=\sigma_1\sigma_2(t_1)=t^{\sigma_2\sigma_1}$ and 
$\sigma_1(t_2)=\sigma_2(t_1) = \frac{1}{n}\sum_{i,j=0}^{n-1} q^{-ij} x_1^ix_3^j =t^{\sigma_2}$. The values of the 
$2$-cocycle 
$\gamma$ can be calculated and are as follows:
\begin{equation*}\begin{array}{r|ccccc}
\gamma & s_1 & s_2 & s_1s_2 & s_2s_1 & s_1s_2s_1\\\hline
s_1 & t_1 & 1 & t_1 & 1 & t_1 \\
s_2 & 1 & t_2 & 1 & t_2 & t_2 \\
s_1s_2 & 1 & t_3 & t_1 & t_1t_3 & t_1t_3 \\
s_2s_1 & t_3 & 1 & t_2t_3 & t_2 & t_2t_3 \\
s_1s_2s_1 & t_2 & t_1 & t_2t_3 & t_1t_3 & t_1t_2t_3
\end{array}\end{equation*}Remark 12. We see that 
$H_{n,2}$ embeds into 
$H_{n,3}$ by sending 
$x,y, z$ of 
$H_{n,2}$ to the corresponding elements 
$x_1,x_2,z_1$ in 
$H_{n,3}$. More generally, for 
$n,m\geq 2$, the algebra 
$R=\mathbb{K}\mathbb{Z}_n^{\otimes m}$ embeds into 
$R' = \mathbb{K}\mathbb{Z}_n^{\otimes m+1}$ by sending 
$a_1\otimes \cdots \otimes a_m$ into 
$a_1\otimes \cdots \otimes a_m \otimes 1_B$. Similarly, 
$\Sigma_m$ can be considered a subgroup of 
$\Sigma_{m+1}$, where we identify the generators 
$s_1,\ldots, s_{m-1}$ with the first 
$m-1$ generators of 
$\Sigma_{m+1}$. Hence, if 
$x_1,\ldots, x_m$ denotes the basis of 
$R=\mathbb{K}\mathbb{Z}_n^{\otimes m+1}$ and 
$z_1,\ldots, z_{m-1}$ the remaining algebra generators of 
$H_{n,m}$ and if 
$x_1', \ldots, x_{m+1}'$ denote the basis of 
$R'$ and 
$z_1',\ldots , z_{m+1}'$ the remaining generators of 
$H_{n,m+1}$, then mapping 
$x_i$ to 
$x_i'$ and 
$z_i$ to 
$z_i'$ yields an algebra embedding of 
$H_{n,m}$ into 
$H_{n,m+1}$, which is also an embedding of Hopf algebras.
Remark 13. Let 
$H=H_{n,m}$ and 
$w\in \Sigma_m$. We claim that the comultiplication of 
$\overline{w}$ in 
$H$ looks like 
$\Delta(\overline{w}) = J(w) (\overline{w}\otimes \overline{w})$ for an invertible element 
$J(w)\in R\otimes R$ with 
$\epsilon(\mu_R(J(w)))=1$. We will prove this by induction on the length of the chosen representation 
$w=s_{i_1}\ldots s_{i_k}$ in the generators 
$s_1,\ldots s_{m-1}$. For 
$k=1$, we have 
$w=s_i$, for some 
$i$ and by definition 
$\Delta(z_i)=J_i(z_i\otimes z_i)$ for 
$z_i=\overline{s_i}$ and 
$J_i = J_{s_i} = (e_i^m\otimes e_{i+1}^m)(J)$, which is invertible, since 
$J$ is invertible. Moreover, 
$\mu_R(J_i)=e_{i, i+1}^m(J) = t_i$ and 
$\epsilon(t_i)=1$. Now suppose that 
$w=s_iv$ has length greater than 
$1$, where 
$v\in \Sigma_m$ and there exists 
$J(v)\in R\otimes R$ such that 
$\Delta(\overline{v})=J(v)(\overline{v}\otimes \overline{v})$. Then
\begin{equation}
\Delta(\overline{w}) = J_i (z_i\otimes z_i) J(v) (\overline{v}\otimes \overline{v}) = J_i J(v)^{\sigma_i} (\overline{w} \otimes \overline{w}),
\end{equation}where we denote by 
$J(w)^{\sigma_i} = (\sigma_i\otimes \sigma_i)(J(w))$. Setting 
$J(w)=J(s_iv)=J_iJ(v)^{\sigma_i} $ proves our claim, since also 
$\mu_R(J(w)) = t_i \sigma_i( \mu_R(J(v)))$ has counit 
$1$.
Thus, given a subgroup 
$N$ of 
$\Sigma_m$, we can consider the Hopf subalgebra of 
$H$ generated by 
$R$ and 
$\{\overline{w} : w\in N\}$, which we shall denote by 
$R\#_\gamma N$. Since for any 
$w\in N$, 
$\Delta(\overline{w})=J(w)(\overline{w}\otimes \overline{w}) \in (R\#_\gamma N) \otimes (R\#_\gamma N)$ we conclude that the subalgebra 
$R\#_\gamma N$ is a semisimple Hopf subalgebra of 
$H$.
Example 14. Consider 
$H_{n,m}=R\#_\gamma \Sigma_m$ with generators as above and consider 
$\theta$, the product of all 
$z_i's$, i.e. 
$\theta =z_1 \cdots z_{m-1}$. Let 
$H$ be the subalgebra of 
$H_{n,m}$ generated by 
$R$ and 
$\theta$. We claim that 
$H$ is equal to 
$R\#_{\gamma} \langle s\rangle$, where 
$s=(12\cdots m)$ is the cycle of length 
$m$. Clearly, 
$\theta = \overline{s}$ and we claim that 
$\theta^k = \left(\prod_{i=1}^{k-1} \gamma(s^i,s)\right) \overline{s^k}$, for 
$2\leq k\leq m$, since inductively, if 
$\theta^k = c_k \overline{s^k}$, for some 
$c_k\in R^\times$, then 
$\theta^{k+1} = c_k \overline{s^k} \: \overline{s} = c_k\gamma(s^k,s) \overline{s^{k+1}}$. Thus 
$H=R\#_{\gamma} \langle s\rangle$, as 
$\left(\prod_{i=1}^{k-1} \gamma(s^i,s)\right) \in R^\times$. Note that 
$\theta^m = \left(\prod_{i=1}^{m-1} \gamma(s^i,s)\right) =:t \in R^\times$. Hence 
$H\simeq R[\theta;\sigma]/\langle \theta^m-t\rangle$, where 
$\sigma = \sigma_s$. In particular, 
$H$ is generated by 
$x_1,\ldots, x_m$ and 
$\theta$ subject to 
$\theta x_i = x_{s(i)} \theta$ and 
$\theta^m=t$. While 
$\Delta(\theta) = J(s)(\theta\otimes \theta)$, with 
$J(s)$ as in Remark 13. Note that 
$ \theta^{m-1} = \tilde{t}\: \overline{s}^{m-1}$, for 
$\tilde{t} = t \gamma(s^{m-1},s)^{-1}$. Since 
$s^{-1} = s^{m-1}$ we obtain
\begin{equation}
\theta^{m-1} = \tilde{t} \:\overline{s} ^{ -1} =\tilde{t} \:z_{m-1} \cdots z_1 = \tilde{t} S(\theta).
\end{equation} Hence 
$S(\theta) = \tilde{t}^{-1} \theta^{m-1} = t^{-1} \gamma(s^{m-1},s) \theta^{m-1}$. The dimension of 
$H$ is 
$mn^m$. For 
$m=2$, we obtain 
$H'=H_{n,2}$.
4. Actions of the generalized Kac–Paljutkin algebra
As before, let 
$H=H_{n,m} = R\#_\gamma \Sigma_m$, with 
$R=B^{\otimes m}$ and 
$B=\mathbb{K}\mathbb{Z}_n$, for some 
$n,m\geq 2$.
The aim of this section is to define an action of 
$H$ on a non-commutative algebra 
$A$ such that the action does not factor through a group algebra – in other words, the action is inner-faithful. To achieve this, we first examine 
$m$-dimensional simple 
$H$-modules 
$V$ (Lemma 15) in order to define a quantum polynomial algebra 
$A$ generated by the basis 
$u_1,\ldots, u_m$ of 
$V$, such that 
$V$ is an 
$H$-submodule of 
$A$. After introducing commutation relations between the generators 
$u_i$, we obtain a Noetherian domain with an inner-faithful 
$H$-action (Theorem 20).
Let 
$q$ be the primitive 
$n$th root of unity used in the definition of the twist 
$J$ of 
$B$. Since 
$\mathbb{K} \Sigma_m$ is a factor ring of 
$H$, any irreducible representation over 
$\Sigma_m$ is a simple module over 
$H$. As above, let 
$x_1, \ldots, x_m$ and 
$z_1,\ldots, z_{m-1}$ be the generators of 
$H$. For each transposition 
$s_i = (i, i+1)$ we have 
$z_i=\overline{s}_i$ and set 
$\sigma_i := \sigma_{s_i}$.
Assume that there exists a square root 
$p$ of 
$q$. Given 
$0\leq a,b \lt n$ we define a left 
$H$-module structure on an 
$m$-dimensional vector space 
$V_{a,b}$ with basis 
$u_1,\ldots, u_m$ by letting 
$x_i$ and 
$z_k$ respectively act as 
$m\times m$-matrices:
\begin{equation}
X_i = \left[\begin{array}{cccccc}
q^b & & & & & \\
& \ddots & & & &\\
& & q^a & & & \\
& & & &\ddots & \\
& & & & & q^b
\end{array}\right]
\qquad
Z_k = \left[\begin{array}{cccccc}
p^{b^2} & & & & & \\
& \ddots & & & &\\
& & 0 & 1 & & \\
& & q^{ab} & 0 & &\\
& & & &\ddots & \\
& & & & & p^{b^2}
\end{array}\right]
\end{equation}where 
$X_i$ has on its diagonal the value 
$q^b$, except on the 
$i$th row, where it has 
$q^a$. The matrix 
$Z_k$ has on its diagonal the value 
$p^{b^2}$ except for the 
$k$th and 
$k+1$st rows (resp. columns), where one has a 
$2\times 2$ block of the form 
$\left[\begin{array}{cc} 0 & 1 \\ q^{ab} & 0\end{array}\right]$. It is clear, that if 
$1\leq k \lt l \lt m$ such that 
$|k-l| \gt 1$, then 
$Z_k$ and 
$Z_l$ commute, since their 
$2\times 2$-block is in different rows and columns. Hence 
$z_kz_l\cdot u_j=z_lz_k\cdot u_j$ holds, for 
$|k-l| \gt 1$.
For the braid relation 
$z_kz_{k+1}z_k = z_{k+1}z_kz_{k+1}$ it is enough to look at the 
$3\times 3$-submatrix of the rows and columns of index 
$k,k+1$ and 
$k+2$ of the product 
$Z_kZ_{k+1}Z_k$, i.e.
\begin{equation}
\left[\begin{array}{ccc}
0 & 1 & 0\\
q^{ab} & 0 & 0\\
0 & 0 & p^{b^2}
\end{array}\right]
\left[\begin{array}{ccc}
p^{b^2} & 0 & 0\\
0 & 0 & 1\\
0 & q^{ab} & 0
\end{array}\right]
\left[\begin{array}{ccc}
0 & 1 & 0\\
q^{ab} & 0 & 0\\
0 & 0 & p^{b^2}
\end{array}\right]
= p^{b^2}
\left[\begin{array}{ccc}
0 & 0 & 1\\
0& q^{ab} & 0\\
q^{2ab}&0 & 0
\end{array}\right]
\end{equation} Similarly, for 
$Z_{k+1}Z_kZ_{k+1}$ one obtains:
\begin{equation}
\left[\begin{array}{ccc}
p^{b^2} & 0 & 0\\
0 & 0 & 1\\
0 & q^{ab} & 0
\end{array}\right]
\left[\begin{array}{ccc}
0 & 1 & 0\\
q^{ab} & 0 & 0\\
0 & 0 & p^{b^2}
\end{array}\right]
\left[\begin{array}{ccc}
p^{b^2} & 0 & 0\\
0 & 0 & 1\\
0 & q^{ab} & 0
\end{array}\right]
= p^{b^2}
\left[\begin{array}{ccc}
0 & 0 & 1\\
0& q^{ab} & 0\\
q^{2ab}&0 & 0
\end{array}\right]
\end{equation} It is not difficult to check that 
$Z_kX_i = X_{s_k(i)}Z_k$ holds for all 
$i$ and 
$k$. Moreover, using 
$p^2=q$,
\begin{equation}
Z_k^2 = \left[\begin{array}{llllll}
q^{b^2} & & & & & \\
& \ddots & & & &\\
& & q^{ab} & & & \\
& & & q^{ab} & &\\
& & & &\ddots & \\
& & & & & q^{b^2}
\end{array}\right] = \frac{1}{n}\sum q^{-st} X_k^s X_{k+1}^t,
\end{equation}since the latter matrix is a diagonal matrix with 
$\frac{1}{n}\sum_{s,t} q^{-st+bs+bt} = q^{b^2}$ on its 
$i$th row, for 
$i\notin\{k,k+1\}$, and 
$\frac{1}{n}\sum_{s,t} q^{-st+as+bt} = q^{ab}$ on its 
$k$th and 
$k+1$st row. Thus, 
$V_{a,b}$ is a left 
$H$-module.
Lemma 15. 
$V_{a,b}$ is a simple left 
$H_{n,m}$-module, if 
$a\neq b \:(\mathrm{mod}\: n)$.
Proof. Let 
$H=H_{n,m}$ and assume 
$a\neq b\:(\mathrm{mod}\: n)$. Let 
$v=\sum_{i=1}^m \lambda_i u_i$ be a non-zero element of 
$v$ with 
$\lambda_k \neq 0$. Then 
$\lambda_k^{-1}(x_k-q^b)\cdot v = (q^a-q^b)u_k \neq 0$. Thus, 
$u_k\in H\cdot v$. However, since 
$\Sigma_m$ acts transitively on 
$\{1,\ldots, m\}$, any basis element 
$u_j$ of 
$V_{a,b}$ can be obtained by multiplying 
$u_k$ by a suitable product of 
$Z_i's$. Hence 
$V_{a,b}$ is simple.
Lemma 16. Let 
$m\geq 3$ and 
$a\neq b \:(\mathrm{mod}\: n)$ and 
$a'\neq b' \:(\mathrm{mod}\: n)$. Then 
$V_{a,b} \simeq V_{a',b'}$ as left 
$H_{n,m}$-module if and only if 
$(a,b) = (a',b')$ in 
$\mathbb{Z}_n^2$.
Proof. Suppose 
$f: V_{a,b} \to V_{a',b'}$ is an isomorphism of 
$H_{n,m}$-modules and 
$(\lambda_{ij})_{1\leq i,j\leq m}$ is a scalar matrix such that 
$f(u_i)=\sum_{j=1}^m \lambda_{ij} u_j'$, where 
$\{u_1,\ldots, u_m\}$ respectively 
$\{u_1', \ldots, u_m'\}$ is a basis for 
$V_{a,b}$ respectively 
$V_{a',b'}$. We will show that 
$(\lambda_{ij})$ is a diagonal matrix. Suppose there exists 
$i\neq j$ with 
$\lambda_{ij}\neq 0$. Let 
$l\neq i,j$ (using 
$m\geq 3$) and consider the following equations:
\begin{equation}
\sum_{r=1}^m \lambda_{ir}q^b u_r' = q^b f(u_i) = x_j\cdot f(u_i) = \lambda_{ij} q^{a'} u_j' + \sum_{r\neq j}^m \lambda_{ir}q^{b'} u_r'
\end{equation}
\begin{equation}
\sum_{r=1}^m \lambda_{ir}q^b u_r' = q^b f(u_i) = x_l\cdot f(u_i) = \lambda_{il} q^{a'} u_l' + \sum_{r\neq l}^m \lambda_{ir}q^{b'} u_r'.
\end{equation} Comparing the coefficients of 
$u_j'$ on both sides of each equation, we conclude 
$b=a' \:(\mathrm{mod}\: n)$ and 
$b=b' \:(\mathrm{mod}\: n)$, which contradicts the assumption 
$a'\neq b' \:(\mathrm{mod}\: n)$. Hence 
$f(u_i)=\lambda_{ii} u_i'$ and 
$\lambda_{ii}\neq 0$, for all 
$i$. Thus 
$\lambda_{ii}q^au_i' = q^af(u_i)=x_i \cdot f(u_i)=\lambda_{ii}q^{a'}u_i'$ and 
$ \lambda_{ii}q^bu_i' = q^bf(u_i)=x_j \cdot f(u_i)=\lambda_{ii}q^{b'}u_i'$, for any 
$j\neq i$ shows 
$a=a'$ and 
$b=b'$ modulo 
$n$. The converse statement holds, since the map 
$f:V_{a,b}\to V_{a',b'}$ with 
$f(u_i)=u_i'$ is an isomorphism of 
$H_{n,m}$-modules, in case 
$(a,b)=(a',b')$ in 
$\mathbb{Z}_n^2$.
Denote by 
$M_{m,a,b}\in M_{m}(\mathbb{Z})$ the 
$m\times m$-matrix, such that the diagonal entries are all equal to 
$a$ and all other entries are equal to 
$b$. For example, for 
$m=2$ and 
$m=3$, we have the matrices
\begin{equation}
M_{2,a,b} = \left[\begin{array}{cc} a & b \\ b & a \end{array}\right], \quad
M_{3,a,b} = \left[\begin{array}{ccc} a & b& b \\ b & a &b \\ b & b& a \end{array}\right]. \quad
\end{equation} The determinant of such matrices are 
$\det(M_{2,a,b}) = a^2-b^2$ and 
$\det(M_{3,a,b}) = a^3+2b^3-3ab^2$. More generally, if 
$e_i$ denotes the canonical basis of 
$\mathbb{K}^m$, then 
$v_1=\sum_{i=1}^m e_i$ is an eigenvector of 
$M_{m,a,b}$ with eigenvalue 
$a+(m-1)b$ and the vectors 
$v_i := e_{i-1}-e_{i}$, for 
$2\leq i \leq n$ are eigenvectors with eigenvalue 
$a-b$. Hence,
\begin{equation}
\det(M_{m,a,b}) = (a+(m-1)b)(a-b)^{m-1}.
\end{equation} An 
$H$-module 
$V$ is inner-faithful if the only Hopf ideal that annihilates 
$V$ is the zero ideal.
Lemma 17. 
$V_{a,b}$ is an inner-faithful 
$R$-module if 
$\gcd((a+(m-1)b)(a-b)^{m-1}, n)=1$.
Proof. Let 
$I$ be a Hopf ideal of 
$R$. Since 
$R$ is a group algebra, there exists a subgroup 
$N\subseteq G=\mathbb{Z}_n^m$ such that 
$I = \mathbb{K} G (\mathbb{K} N)^+$ (see [Reference Farnsteiner2, Lemma 4] or [Reference Passman and Quinn11, p. 329]). Suppose 
$I\cdot V_{a,b}=0$. Then 
$\omega= 1-x_1^{\alpha_1} \cdots x_m^{\alpha_m}\ \in I$, for some 
$x_1^{\alpha_1} \cdots x_m^{\alpha_m}\in N$. For any 
$1\leq j \leq m$ we obtain an equation 
$\omega\cdot u_j = 0$, i.e. 
$1 = q^{\alpha_j a + \sum_{l\neq j} \alpha_l b}$. Thus, 
$\alpha_j a + \sum_{l\neq j} \alpha_l b = 0 \: (\mathrm{mod}\: n),$ or in matrix notation 
$\alpha M_{m,a,b} = 0 \: (\mathrm{mod}\: n)$. If 
$\gcd(\det(M_{m,a,b}), n)=1$, then 
$M_{m,a,b}$ is invertible modulo 
$n$. Thus 
$\alpha_i=0$, for all 
$i$, which means that 
$N$ is trivial and so is 
$I$.
Next we show that 
$H_{n,m}$ acts on a quantum polynomial algebra 
$A$ in 
$m$ variables, such that 
$V_{a,b}$ is its degree one space.
Theorem 18. Let 
$a\neq b \:(\mathrm{mod}\: n)$, 
$q$ a primitive 
$n$th root of unity in 
$\mathbb{K}$ and a square root 
$p$ of 
$q$. Set 
$\lambda = q^{b^2-ab}$ and 
$\mu = p^{b^2-a^2}$ and let 
$(r_{ij})$ be the 
$m\times m$-matrix such that 
$r_{ii}=1$, 
$ r_{ij} = \lambda^{j-i-1} \mu$ for 
$i \lt j$ and 
$r_{ij}=r_{ji}^{-1}$ for 
$j \lt i$. Consider the quantum polynomial algebra in 
$m$ variables 
$u_1,\ldots, u_m$ subject to 
$u_i u_j = r_{ij} u_ju_i$, for all 
$i,j$.
\begin{equation}
A_{a,b} = \frac{\mathbb{K}\langle u_1,\ldots, u_m\rangle}{\langle u_iu_j - r_{ij}u_ju_i :i,j\rangle}.
\end{equation} Then 
$A_{a,b}$ is a left 
$H$-module algebra for the semisimple Hopf algebra 
$H=\mathbb{K}\mathbb{Z}_n^{\otimes m}\#_\gamma \Sigma_m$, where 
$V_{a,b}$ determines the 
$H$-action on the generators 
$u_i$, i.e.
\begin{equation*}x_i\cdot u_i = q^a u_i, \qquad x_j\cdot u_i = q^b u_i, \quad \mbox{ for } i\neq j\end{equation*}
\begin{equation*}z_i\cdot u_i = q^{ab} u_{i+1}, \qquad z_i\cdot u_{i+1} = u_i, \qquad z_i \cdot u_j = p^{b^2} u_j, \qquad \mbox{ for } j \neq \{i,i+1\}.\end{equation*}Proof. We claim that 
$A$ is a left 
$H$-module algebra. As before, 
$R=\mathbb{K}\mathbb{Z}_n^{\otimes m}$. Since the group-likes 
$x_i$ act as scalars on 
$u_j$, we conclude that 
$A$ is an 
$R$-module algebra. Recall that
\begin{equation}
\Delta(z_i) = J_i (z_i \otimes z_i) = \frac{1}{n} \sum_{s,t} q^{-st} x_i^sz_i \otimes x_{i+1}^tz_i.
\end{equation} Hence for 
$1\leq i \lt m$:
\begin{align*}
z_i\cdot (u_iu_{i+1})
&=
\frac{1}{n} \sum_{s,t} q^{-st} \left(x_i^sz_i\cdot u_i \right)\left( x_{i+1}^tz_i\cdot u_{i+1}\right)
= q^{ab} q^{b^2} u_{i+1}u_i \\
z_i\cdot (u_{i+1}u_i) &=
\frac{1}{n} \sum_{s,t} q^{-st} \left(x_i^sz_i\cdot u_{i+1} \right)\left( x_{i+1}^tz_i\cdot u_{i}\right)
= q^{ab} q^{a^2} u_{i}u_{i+1} = q^{ab} q^{a^2} \mu u_{i+1}u_{i},
\end{align*}where 
$r_{i,i+1} = \mu=p^{b^2-a^2}$. Thus 
$z_i\cdot(u_iu_{i+1}) = q^{ab} q^{b^2}u_{i+1}u_i = q^{ab}\mu^2 q^{a^2} u_{i+1}u_i = z_i\cdot ( r_{i,i+1} u_{i+1}u_i).$
A similar calculation for 
$j \gt i+1$ shows
\begin{align*}
z_i\cdot (u_iu_j) &= p^{b^2} q^{ab+b^2} u_{i+1}u_j\\
z_i\cdot (u_ju_i) &= p^{b^2} q^{2ab} u_{j}u_{i+1} = p^{b^2}q^{2ab} r_{j,i+1} u_{i+1}u_j
\end{align*} Since 
$r_{ij}r_{j,i+1} = r_{ij}r_{i+1,j}^{-1} = \lambda = q^{b^2-ab}$ we obtain
\begin{equation}
z_i\cdot(u_iu_j)
= p^{b^2}q^{ab+b^2} u_{i+1}u_j = p^{b^2} r_{ij}r_{j,i+1}q^{2ab}u_{i+1}u_j =
z_i\cdot(r_{ij} u_{j}u_i).
\end{equation} Similar calculations show 
$z_i \cdot (u_lu_k) = r_{lk}z_i\cdot(u_ku_l)$, for all 
$i,k,l$.
Proposition 19. Let 
$H_{n,m} = R\#_\gamma \Sigma_m$ and 
$A_{a,b}$ as above, for some 
$a\neq b \:(\mathrm{mod}\: n)$. If 
$\gcd(\det(M_{m,a,b}), n)=1$ and 
$n$ is even, then 
$A_{a,b}^{H_{n,m}}$ is contained in the commutative polynomial ring 
$A_{a,b}^{R}=\mathbb{K}[u_1^n, \ldots, u_m^n]$.
Proof. Set 
$A=A_{a,b}$. Clearly, 
$A^H\subseteq A^R$. Let 
$u_1, \ldots, u_m$ be the algebra generators of 
$A$. The monomials 
$u^\alpha := u_1^{\alpha_1}\cdots u_m^{\alpha_m}$, with 
$\alpha = (\alpha_1,\ldots, \alpha_m)\in {\mathbb{N}}^m$, form a 
$\mathbb{K}$-basis for 
$A$. Let 
$w = \sum a_{\alpha} u^\alpha \in A^R$, with only finitely many non-zero 
$a_{\alpha}\in \mathbb{K}$. Denote by 
$\sup(w) = \{\alpha : a_\alpha\neq 0\}$ the support of 
$w$. For any 
$1\leq i \leq m$ we have
\begin{equation*}w = x_i \cdot w = \sum a_{\alpha} q^{\alpha_i a + \left(\sum_{j\neq i}\alpha_j\right)b} u^\alpha.\end{equation*} Hence 
$\alpha_i a + \left(\sum_{j\neq i}\alpha_j\right)b = 0 \:(\mathrm{mod}\: n)$, for any 
$\alpha \in \sup(w)$ and for any 
$i$, which means that 
$\alpha M_{m,a,b} = 0 \:(\mathrm{mod}\: n)$. Since 
$\gcd(\det(M_{m,a,b}),n)=1$, 
$M_{m,a,b}$ is invertible modulo 
$n$ and 
$\alpha = (0,\ldots, 0) \:(\mathrm{mod}\: n)$, i.e. 
$n\mid \alpha_i$, for all 
$i$. Note that if 
$n$ is even, then 
$r_{ij}^{n^2} = q^{n^2(b^2-ab)} p^{n^2(b^2-a^2)} = (-1)^{n(b^2-a^2)}=1$. Hence 
$u_i^n u_j^n=u_j^nu_i^n$ and 
$A^R \subseteq \mathbb{K}[u_1^n, \ldots, u_m^n]$. The converse inclusion is clear.
The condition 
$\gcd(\det(M_{m,a,b}), n)=1$ is obviously satisfied when 
$a=1$ and 
$b=0$. Then 
$\lambda = q^{b^2-ab}=1$ and 
$\mu = p^{b^2-a^2}=p^{-1}$. Thus the matrix 
$(r_{ij})$ satisfies 
$r_{ji}= p$ for 
$i \lt j$.
Theorem 20. Let 
$n,m\geq 2$, 
$H=R\#_\gamma \Sigma_m$ and 
$R=\mathbb{K}\mathbb{Z}_n^{\otimes m}$. Let 
$q$ be the primitive 
$n$th root of unity used in the definition of the twist 
$J$ and assume there exists a square root 
$p$ of 
$q$. Then
\begin{equation*}A = \frac{\mathbb{K}\langle u_1,\ldots, u_m\rangle}{\langle u_ju_i - pu_iu_j : i \lt j\rangle}\end{equation*}is a left 
$H$-module algebra with action of the generators 
$x_i$ and 
$z_i$ of 
$H$ given by
\begin{equation*}x_iu_i=qu_i, \quad x_iu_j = u_j \: \mbox{ for}\ j\neq i, \quad z_i u_i = u_{i+1}, \quad z_i u_{i+1}=u_i, \quad z_iu_j=u_j \mbox{ for } j\neq i, i+1.\end{equation*} Moreover, 
$A$ is an inner-faithful 
$R$-module and if 
$n$ is even then 
$A^H\subseteq \mathbb{K}[u_1^n, \ldots, u_m^n]$.
In [Reference Ferraro, Ellen Kirkman and Won3], Ferraro et al. showed that Pansera’s Hopf algebras 
$H_{n,2}$ act inner-faithfully on an Artin-Schelter regular algebra 
$A$, such that its ring of invariants is again Artin-Schelter regular – an analogous property to the Chevalley–Shephard–Todd Theorem for group actions on polynomial algebras. The authors of [Reference Ferraro, Ellen Kirkman and Won3] termed such Hopf algebras ‘reflection Hopf algebra’ for 
$A$.
Example 21. Like in Example 14, consider the Hopf subalgebra 
$H=R\#_\gamma \langle s\rangle$ of 
$H_{n,m}$ generated by 
$R$ and 
$\theta=z_1z_2\cdots z_{m-1}$, where 
$s=(12\cdots m)$. Then 
$H$ has dimension 
$mn^m$ and is generated as algebra by 
$x_1,\ldots, x_m$ and 
$\theta$. By Theorem 20, 
$H_{n,m}$ and hence 
$H$ acts on
\begin{equation*}A =A_{1,0}= \frac{\mathbb{K}\langle u_1,\ldots, u_m\rangle}{\langle u_ju_i - pu_iu_j : i \lt j\rangle},\end{equation*}where 
$\theta$ acts cyclically on the generators of 
$A$, i.e. 
$\theta \cdot u_j =u_{s(j)}$, for all 
$j$. We claim that 
$A$ is an inner-faithful 
$H$-module algebra. Suppose 
$I$ is a Hopf ideal of 
$H$ with 
$I\cdot A = 0$ and let 
$\omega = \sum_{i=0}^{m-1} a_i \theta^i \in I$, with 
$a_i \in R$. Then 
$\omega \cdot u_j = \sum_{i=0}^{m-1} a_i \cdot u_{s(j)} = 0$, for all 
$1\leq j \leq m$. Since the spaces 
$\mathbb{K} u_j$ are stable under the action of 
$R$, we get 
$a_i \cdot u_j = 0$, for all 
$i$ and 
$j$. Thus 
$a_i \in I\cap R$, for all 
$i$. As 
$I$ is also a Hopf ideal of 
$R$, we conclude by Lemma 17, 
$I\cap R=0$. Hence 
$\omega = 0$, which means that 
$H$ acts inner faithfully on 
$A$.
If 
$n$ is even, then by Theorem 20, 
$A^H\subseteq \mathbb{K}[u_1^n, \ldots, u_m^n]$. Since 
$\theta$ acts cyclically on the generators 
$u_i$, it also acts cyclically on the elements 
$u_i^n$. Hence 
$A^H$ consists of the cyclic polynomials in 
$u_i^n$, i.e. 
$A^H = \mathbb{K}[u_1^n, \ldots, u_m^n]^{\langle s \rangle}$.
Note that the quantum polynomial algebra 
$A=A_{1,0}$ is an Artin-Schelter regular ring. In case 
$m=2$, 
$H=H_{n,2}$ and 
$A^H=\mathbb{K}[u_1^n, u_2^n]^{\langle (12) \rangle} = \mathbb{K}[u_1^n+u_2^n, u_1^nu_2^n]$ is a polynomial algebra, hence Artin-Schelter regular and 
$H$ is a reflection Hopf algebra for 
$A$. In case 
$m\geq 3$, the ring of invariants 
$A^H = \mathbb{K}[u_1^n, \ldots, u_m^n]^{\langle s \rangle}$ is not a polynomial ring and 
$H$ is not a reflection Hopf algebra for 
$A$ in the sense of [Reference Ferraro, Ellen Kirkman and Won3].
Acknowledgments
I dedicate this paper to the cherished memory of my father. Parts of this research were conducted in 2024 during a visit to the Technical University of Dresden, supported by a DAAD grant. I am especially grateful to Uli Krähmer and his team – Zbigniew Wojciechowski, Ivan Bartulović and Tony Zorman – for their warm hospitality and for making my stay in Dresden both productive and enjoyable. My thanks go also to the anonymous referee for her/his valuable comments. Furthermore, I was partially supported by CMUP, member of LASI, which is financed by national funds through FCT – Fundação para a Ciência e a Tecnologia, I.P., under the projects with reference UIDB/00144/2020 and UIDP/00144/2020.
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