# Generalized Alexander quandles of finite groups and their characterizations

Akihiro Higashitani  and  Hirotake Kurihara  Akihiro Higashitani, Department
of Pure and Applied Mathematics, Graduate School of Information Science and
Technology, Osaka University, Yamadaoka 1-5, Suita, Osaka 565-0871, Japan
<EMAIL_ADDRESS>Hirotake Kurihara, Department of Applied
Science, Yamaguchi University, Tokiwadai 2-16-1, Ube 755-8611, Japan
<EMAIL_ADDRESS>
###### Abstract.

The goal of this paper is to characterization generalized Alexander quandles
of finite groups in the language of the underlying groups. Firstly, we prove
that if finite groups $G$ are simple, then the quandle isomorphic classes of
generalized Alexander quandles of $G$ one-to-one correspond to the conjugacy
classes of the automorphism groups of $G$. This correspondence can be also
claimed for the case of symmetric groups. Secondly, we give a characterization
of generalized Alexander quandles of finite groups $G$ under some assumptions
in terms of $G$. As corollaries of this characterization, we obtain several
characterizations in some particular groups, e.g., abelian groups and dihedral
groups. Finally, we perform a characterization of generalized Alexander
quandles arising from groups with their order up to $15$.

2010 Mathematics Subject Classification: Primary; 20N02, Secondary: 20B05,
53C35.
Keywords: generalized Alexander quandles, finite groups, automorphisms.

###### Contents

  1. 1 Introduction
    1. 1.1 Backgrounds and motivations
    2. 1.2 Key notion
    3. 1.3 Main results
    4. 1.4 Structures of this paper
  2. 2 Terminologies and notation
    1. 2.1 Terminologies for quandles
    2. 2.2 Terminologies for groups
    3. 2.3 Quandle triplets and generalized Alexander quandles
  3. 3 Proofs of the main results
    1. 3.1 Normality for $P$ and $P^{2}$
    2. 3.2 Description of $\operatorname{Inn}_{\mathrm{qu}}(Q(G,\psi))$
    3. 3.3 Proofs of Theorem 1.2 and Corollary 1.3
    4. 3.4 More properties for $P$ and $P^{2}$
    5. 3.5 Proof of Theorem 1.4
  4. 4 Application of Theorem 1.4 to dihedral groups
    1. 4.1 Conjugacy classes for $\operatorname{Aut}_{\mathrm{gr}}(D_{n})$
    2. 4.2 Invariants for $Q(D_{n},a,b)$
    3. 4.3 Proof of Corollary 1.6
  5. 5 Relationships among generalized Alexander quandles arising from $C_{2n}$ and $D_{n}$
    1. 5.1 Proof of Corollary 1.5
    2. 5.2 Proof of Corollary 1.7
  6. 6 Generalized Alexander quandles arising from finite groups of small orders
    1. 6.1 The case $|G|=p$
    2. 6.2 The case $|G|=2p$
    3. 6.3 The case $|G|=p^{2}$
    4. 6.4 $n=8$
    5. 6.5 $n=12$
    6. 6.6 $n=15$
    7. 6.7 $n=16$

##  1\. Introduction

###  1.1. Backgrounds and motivations

Let us recall what a quandle is. A set $Q$ equipped with a binary operation
$*$ is said to be a quandle if the following three conditions are satisfied:

  * (Q1)

$x*x=x$ for any $x\in Q$;

  * (Q2)

for each $x,y\in Q$, there is a unique $z\in Q$ such that $z*y=x$;

  * (Q3)

for any $x,y,z\in Q$, we have $(x*y)*z=(x*z)*(y*z)$.

A notion of quandles was originally introduced in [5] in the context of knot
theory. In fact, these axioms correspond to the Reidemeister moves.

In this paper, we employ the following definition of quandles using “point
symmetry”, which is a map globally defined at each point. We say that a set
$Q$ equipped with a point symmetry $s_{x}:Q\rightarrow Q$ for each $x\in Q$ is
a quandle if the following three conditions are satisfied:

  * (Q1’)

$s_{x}(x)=x$ for each $x\in Q$;

  * (Q2’)

$s_{x}$ is a bijection on $Q$ for each $x\in Q$;

  * (Q3’)

$s_{x}\circ s_{y}=s_{s_{x(y)}}\circ x_{x}$ holds for each $x,y\in Q$.

We can directly verify that those definitions are equivalent by letting
$s_{x}(\cdot):Q\rightarrow Q$ be $(\cdot)*x$. To be precise, we should denote
a quandle $Q$ by $(Q,*)$ or $(Q,s)$, i.e., a pair of a set $Q$ and a binary
operation $*$ or a point symmetry $s_{x}(\cdot)(x\in Q)$, but we often write
it for $Q$ if there is no confusion, while we also write, e.g., $(Q,s)$ or
$(Q,s^{\prime})$ if we want to distinguish the equipped operations.

The above equivalent definition derives from the theory of symmetric spaces
since quandles can be regarded as symmetric spaces “without differential
structure”. In particular, due to the similarity of the theory of homogeneous
spaces, homogeneous quandles have been intensively studies (see, e.g., [1, 4,
6, 15, 16, 17, 18]). Here, a quandle $Q$ is said to be homogeneous if the
automorphism group of $Q$ acts transitively on $Q$. Moreover, it is proved
that any homogeneous quandles can be obtained from a triple of a group $G$,
its subgroup $K$ and its automorphism $\psi$ satisfying a certain condition,
called a quandle triplet. (See [4, Section 3] and [5, Section 7].) In
particular, in the case where $K=\\{e\\}$, we can always construct a certain
homogeneous quandle, which we will denote $Q(G,\psi)$. This homogeneous
quandle is called a generalized Alexander quandle (also known as a principal
quandle) of $G$ and $\psi$. See Subsection 2.3 for more details. Note that
Alexander quandles correspond to the case where $G$ is abelian.

The main object of this paper is generalized Alexander quandles $Q(G,\psi)$
for finite groups $G$. The symbols that appear in the following sections will
be explained in Section 2.

Given a finite group $G$, let ${\mathcal{Q}}(G)$ be the set of quandle
isomorphic classes of $Q(G,\psi)$’s, i.e.,

${\mathcal{Q}}(G):=\\{Q(G,\psi):\psi\in\operatorname{Aut}_{\mathrm{gr}}(G)\\}/\cong_{\mathrm{qu}}.$

The following problem naturally arises.

######  Problem 1.1 ([7, Problem 3.5]).

Determine ${\mathcal{Q}}(G)$ for a given group $G$.

Nelson completely solves this problem in the case of finite abelian groups in
[11, Theorem 2.1]. The goal of this paper is to give a more general statement
for this problem.

###  1.2. Key notion

Let $G$ be a finite group with its identity $e$, let $\psi$ be an automorphism
of $G$ and let $Q=Q(G,\psi)$ be a generalized Alexander quandle. In this
paper, we introduce an invariant of generalized Alexander quandle:

(1) $\begin{split}P=P(Q)&:=\operatorname{Inn}_{\mathrm{qu}}(Q)e\\\ &=\\{x\in
G:\exists a_{1},\ldots,\exists a_{r}\in G\text{ such that
}x=s_{a_{1}}\circ\cdots\circ s_{a_{r}}(e)\\}.\end{split}$

Namely, $P$ is the orbit of $e$ by the action of
$\operatorname{Inn}_{\mathrm{qu}}(Q(G,\psi))$. Note that the same notion has
already appeared in [1]. See Remark 3.5.

This $P$ plays a crucial role for the solution of Problem 1.1. In fact, $P$
satisfies several important properties on quandle isomorphism and will play
key roles for the proofs of our main results.

###  1.3. Main results

The first main result of this paper is the following.

######  Theorem 1.2.

Let $G$ be a finite simple group. Then ${\mathcal{Q}}(G)$ one-to-one
corresponds to the set of conjugacy classes of
$\operatorname{Aut}_{\mathrm{gr}}(G)$.

Although it is well known that the symmetric group on $\\{1,\ldots,n\\}$ is
not simple for $n\geq 3$, we immediate obtain the following as a corollary of
this theorem.

######  Corollary 1.3.

Let $\mathfrak{S}_{n}$ be the symmetric group on $\\{1,\ldots,n\\}$. Then
${\mathcal{Q}}(\mathfrak{S}_{n})$ one-to-one corresponds to the set of
conjugacy classes of $\operatorname{Aut}_{\mathrm{gr}}(\mathfrak{S}_{n})$.

See Subsection 3.3 for the proofs of Theorem 1.2 and Corollary 1.3. Note that
this corollary implies [7, Theorem 5.14], which claims the same result in the
case $n\in\\{3,4,\ldots,30\\}\setminus\\{15\\}$, and gives an affirmative
answer for [7, Question 5.15].

Next, we consider generalized Alexander quandles satisfying (P1) and (P2)
(definitions of (P1) and (P2) will appear in Section 3.4). We can give an
equivalent condition of the quandle isomorphic property among generalized
Alexander quandles satisfying (P1) and (P2). This equivalent condition is the
second main result of this paper and the detail is the following.

######  Theorem 1.4.

Let $G$ and $G^{\prime}$ be finite groups, let
$\psi\in\operatorname{Aut}_{\mathrm{gr}}(G)$ and
$\psi^{\prime}\in\operatorname{Aut}_{\mathrm{gr}}(G^{\prime})$, and let
$Q=Q(G,\psi)$ and $Q^{\prime}=Q(G^{\prime},\psi^{\prime})$, respectively.
Assume that $Q$ and $Q^{\prime}$ satisfy (P1) and (P2). Then
$Q\cong_{\mathrm{qu}}Q^{\prime}$ holds if and only if the following conditions
are satisfied:

  * (A)

$|G|=|G^{\prime}|$;

  * (B)

$|\operatorname{Fix}(\psi,G)|=|\operatorname{Fix}(\psi^{\prime},G^{\prime})|$;

  * (C)

there is a group isomorphism $h$ from $P=P(Q)$ to $P^{\prime}=P(Q^{\prime})$
(i.e., $P\cong_{\mathrm{gr}}P^{\prime}$) satisfying the following conditions:

    * (C-1)

$h\circ\psi|_{P}=\psi^{\prime}|_{P^{\prime}}\circ h$ and

    * (C-2)

for any $a\in G$ there is $a^{\prime}\in G^{\prime}$ such that
$h(s_{a}(e))=s_{a^{\prime}}^{\prime}(e^{\prime})$ holds.

This theorem induces many interesting results. In fact, there are many
generalized Alexander quandles satisfying the conditions (P1) and (P2). The
following three corollaries follow from Theorem 1.4.

######  Corollary 1.5 ([11, Theorem 2.1]).

Let $G$ and $G^{\prime}$ be finite abelian groups and let
$\psi\in\operatorname{Aut}_{\mathrm{gr}}(G)$ and
$\psi^{\prime}\in\operatorname{Aut}_{\mathrm{gr}}(G^{\prime})$, respectively.
Let $\rho=\operatorname{id}_{G}-\psi$ and let
$\rho^{\prime}=\operatorname{id}_{G^{\prime}}-\psi^{\prime}$. Then
$Q(G,\psi)\cong_{\mathrm{qu}}Q(G^{\prime},\psi^{\prime})$ holds if and only if
$|G|=|G^{\prime}|$ holds and there is a group isomorphism
$h:\operatorname{Im}\rho\rightarrow\operatorname{Im}\rho^{\prime}$ satisfying
$h\circ\psi|_{\operatorname{Im}\rho}=\psi^{\prime}|_{\operatorname{Im}\rho^{\prime}}\circ
h$.

Corollary 1.5 gives a group-theoretic interpretation of [11, Theorem 2.1]. See
Section 5 for more details.

######  Corollary 1.6.

Let $D_{n}$ be a dihedral group of order $2n$, let $a,a^{\prime}\in
C_{n}^{\times}$ and let $b,b^{\prime}\in C_{n}$. Put $d=\gcd(n,1-a,b)$ and
$d^{\prime}=\gcd(n,1-a^{\prime},b^{\prime})$. Then
$Q(D_{n},\varphi_{a,b})\cong_{\mathrm{qu}}Q(D_{n},\varphi_{a^{\prime},b^{\prime}})$
holds if and only if the following three conditions are satisfied:

  * •

$|\operatorname{Fix}(\varphi_{a,b},D_{n})|=|\operatorname{Fix}(\varphi_{a^{\prime},b^{\prime}},D_{n})|$;

  * •

$d=d^{\prime}$;

  * •

$a\equiv a^{\prime}\pmod{\frac{n}{d}}$.

See Section 4 for more details.

######  Corollary 1.7.

Let $n$ be a positive integer and fix $a\in C_{2n}^{\times}$. Let $a=2k+1$ for
some nonnegative integer $k$, let $g=\gcd(k,n)$ and let
$\tilde{a}=a-\lfloor\frac{a}{n}\rfloor n$. Then we have a quandle isomorphism
$Q(C_{2n},a)\cong_{\mathrm{qu}}Q(D_{n},\varphi_{\tilde{a},g})$.

This corollary means that any linear Alexander quandle of cardinality $2n$ can
be realized as a generalized Alexander quandle $Q(D_{n},\varphi_{c,d})$ for
some certain $c\in C_{n}^{\times}$ and $d\in C_{n}$. See Section 5 for more
details.

The third main result of this paper is a complete classification of
generalized Alexander quandles up to quandle isomorphism for finite groups
with small orders. Namely, we consider the problem of the determination of
${\mathcal{Q}}_{\operatorname{GAQ}}(n)$ for each $n$’s, where

$\displaystyle{\mathcal{Q}}_{\operatorname{GAQ}}(n):=$
$\displaystyle\\{Q(G,\psi):\text{$G$ is a group of order
$n$},\;\psi\in\operatorname{Aut}_{\mathrm{gr}}(G)\\}/\cong_{\mathrm{qu}}$
$\displaystyle=$ $\displaystyle\bigcup_{\text{$G$ is a group of order
$n$}}{\mathcal{Q}}(G).$

######  Theorem 1.8.

The complete list of $|{\mathcal{Q}}_{\operatorname{GAQ}}(n)|$ for $n\leq 15$
is given in Table 1.

$n$ | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15
---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---
$|{\mathcal{Q}}_{\operatorname{GAQ}}(n)|$ | 1 | 1 | 2 | 3 | 4 | 3 | 6 | 9 | 11 | 5 | 10 | 11 | 12 | 7 | 8
Table 1. List of $|{\mathcal{Q}}_{\operatorname{GAQ}}(n)|$

The results in the cases $n=2,3,5,7,11,13$ come from Subsection 6.1, the
results in the cases $n=6,10,14$ come from Subsection 6.2 and the results in
the cases $n=4,9$ come from Subsection 6.3. The cases $n=8$, $n=12$ and $n=15$
are discussed in Subsections 6.4, 6.5 and 6.6, respectively. Finally, we
mention in Subsection 6.7 that Theorem 1.4 is not available in the case
$n=16$.

###  1.4. Structures of this paper

An organization of this paper is as follows. In Section 2, we collect the
notions and the notation and recall some previous results, which will us
throughout the paper. In Section 3, we give our proofs of Theorem 1.2,
Corollary 1.3 and Theorem 1.4. As applications of Theorem 1.4, we give the
proof of Corollary 1.6 in Section 4 and the proofs of Corollaries 1.5 and 1.7
in Section 5. In Section 6, we perform a complete classification of
generalized Alexander quandles of groups with their orders at most $15$.

######  Acknowledgments.

The first named author is partially supported by JSPS Grant-in-Aid for
Scientists Research (C) $\sharp$20K03513 and the second named author is
partially supported by JSPS Grant-in-Aid for Scientists Research (C)
$\sharp$20K03623.

##  2\. Terminologies and notation

###  2.1. Terminologies for quandles

Throughout this paper, we denote by $(Q,s)$ (or $Q$, for short) a quandle $Q$
equipped with the maps $\\{s_{x}:x\in Q\\}$ satisfying (Q1’)–(Q3’) in
Introduction.

Let $(Q,s)$ and $(Q^{\prime},s^{\prime})$ be quandles. A map $f:Q\rightarrow
Q^{\prime}$ is called a quandle homomorphism if $f$ satisfies

$f\circ s_{x}=s_{f(x)}^{\prime}\circ f\;\;\text{ for any }x\in Q.$

Moreover, $f$ is said to be a quandle isomorphism when $f$ is bijective. If
there is a quandle isomorphism between $Q$ and $Q^{\prime}$, then we say that
$Q$ and $Q^{\prime}$ are isomorphic as quandles, denoted by
$Q\cong_{\mathrm{qu}}Q^{\prime}$.

We collect the notation on quandles used in the paper:

  * •

Let $\operatorname{Aut}_{\mathrm{qu}}(Q,s)$ (or
$\operatorname{Aut}_{\mathrm{qu}}(Q)$, for short) be the set of all quandle
automorphisms from $(Q,s)$ to $(Q,s)$ itself, which is called the automorphism
group of $Q$. Note that $s_{x}\in\operatorname{Aut}_{\mathrm{qu}}(Q)$ for any
$x\in Q$.

  * •

Let $\operatorname{Inn}_{\mathrm{qu}}(Q,s)$ (or
$\operatorname{Inn}_{\mathrm{qu}}(Q)$, for short) be the subgroup of
$\operatorname{Aut}_{\mathrm{qu}}(Q)$ generated by $\\{s_{x}:x\in Q\\}$, which
is called the inner automorphism group of $Q$. It is easy to see that
$\operatorname{Inn}_{\mathrm{qu}}(Q)$ is a normal subgroup of
$\operatorname{Aut}_{\mathrm{qu}}(Q)$.

  * •

For a quandle $Q$, we call $Q$ homogeneous if for any $x,y\in Q$, there exists
a quandle automorphism $f$ on $Q$ such that $f(x)=y$. This means that
$\operatorname{Aut}_{\mathrm{qu}}(Q)$ acts transitively on $Q$.

  * •

For a quandle $Q$, we call $Q$ connected if for any $x,y\in Q$, there exists a
quandle inner automorphism $f$ on $Q$ such that $f(x)=y$. This means that
$\operatorname{Inn}_{\mathrm{qu}}(Q)$ acts transitively on $Q$.

  * •

For a quandle $(Q,s)$, let
$\operatorname{ord}(s_{x})=\min\\{n\in{\mathbb{Z}}_{>0}:\underbrace{s_{x}\circ\cdots\circ
s_{x}}_{n}=\operatorname{id}\\}$. For a homogeneous quandle $Q$, we see that
$\operatorname{ord}(s_{x})$ is constant for any $x\in Q$. In fact, for any
$x,y\in Q$, since there is $f\in\operatorname{Aut}_{\mathrm{qu}}(Q)$ such that
$y=f(x)$, we see that if $s_{x}^{m}=\operatorname{id}$, then $f\circ
s_{x}^{m}=s_{f(x)}^{m}\circ f$, so we obtain $s_{f(x)}^{m}=\operatorname{id}$,
as required. Thus, we use the notation $\operatorname{ord}(Q)$ instead of
$\operatorname{ord}(s_{x})$ for $x\in Q$ in the case $Q$ is homogeneous.

###  2.2. Terminologies for groups

Let $G,G^{\prime}$ be groups with its units $e,e^{\prime}$, respectively. We
denote by $G\cong_{\mathrm{gr}}G^{\prime}$ if $G$ and $G^{\prime}$ are
isomorphic as groups.

We collect the notation on groups used in the paper:

  * •

We denote $H<G$ (resp. $H\triangleleft G$) if $H$ is a subgroup (resp. normal
subgroup) of $G$.

  * •

Let $\operatorname{Aut}_{\mathrm{gr}}(G)$ denote the automorphism group of
$G$.

  * •

Let $\operatorname{Inn}_{\mathrm{gr}}(G)$ denote the inner automorphism group
of $G$, i.e., the set of the group automorphisms defined by $x\mapsto
axa^{-1}$ for each $a\in G$, denoted by $(\cdot)^{a}$. Note that
$\operatorname{Inn}_{\mathrm{gr}}(G)$ is a normal subgroup of
$\operatorname{Aut}_{\mathrm{gr}}(G)$. We call $\psi$ an outer automorphism if
$\psi\in\operatorname{Aut}_{\mathrm{gr}}(G)\setminus\operatorname{Inn}_{\mathrm{gr}}(G)$.

  * •

Let $Z(G)=\\{z\in G:xz=zx\text{ for any }x\in G\\}$ denote the center of $G$.
It is well known that
$\operatorname{Inn}_{\mathrm{gr}}(G)\cong_{\mathrm{gr}}G/Z(G)$.

  * •

Let $\operatorname{ord}_{G}(x)$ denote the order of $x\in G$.

  * •

Given $\psi\in\operatorname{Aut}_{\mathrm{gr}}(G)$, let

$\operatorname{Fix}(\psi,G)=\\{x\in G:\psi(x)=x\\}.$

  * •

Given a subset $K$ of $G$, let $G/K=\\{[x]:x\in G\\}$ denote the set of left
cosets with respect to $K$.

  * •

Let $C_{n}$ be a cyclic group of order $n$. We set
$C_{n}=\\{0,1,\ldots,n-1\\}$ as a set and use $+$ as an operation. We perform
the addition in $C_{n}$ modulo $n$.

  * •

Let $C_{n}^{\times}$ denote the multiplicative group of $C_{n}$, i.e.,
$C_{n}^{\times}=\\{a\in C_{n}:\gcd(a,n)=1\\}$. It is well known that
$\operatorname{Aut}_{\mathrm{gr}}(C_{n})\cong_{\mathrm{gr}}C_{n}^{\times}$.

  * •

Let $D_{n}$ be a dihedral group of order $2n$. We will give our notation for
dihedral groups in the beginning of Section 4.

  * •

Let $\mathfrak{S}_{n}$ (resp. $\mathfrak{A}_{n}$) be the symmetric group
(resp. the alternating group) on $\\{1,2,\ldots,n\\}$. It is well known that
$\mathfrak{A}_{n}$ is a normal subgroup of $\mathfrak{S}_{n}$.

For the terminologies or fundamental facts on group theory, consult, e.g.,
[12, 13].

###  2.3. Quandle triplets and generalized Alexander quandles

Let $G$ be a group and let $K$ be a subgroup of $G$. Take
$\psi\in\operatorname{Aut}_{\mathrm{gr}}(G)$. We say that a triple
$(G,K,\psi)$ is a quandle triplet if $K\subset\operatorname{Fix}(\psi,G)$.
Given a quandle triplet $(G,K,\psi)$, we can construct a quandle $Q(G,K,\psi)$
as follows: $Q(G,K,\psi)=G/K$ as a set and we define $s_{[x]}$ for each
$[x]\in G/K$ by setting

$s_{[x]}([y])=[x\psi^{-1}(x^{-1}y)]\;\;\text{for }[y]\in G/K.$

It is proved in [5] that the set $G/K$ equipped with this point symmetry
becomes a homogeneous quandle.

For the relationship between homogeneous quandles and quandle triplets, the
following is known:

######  Proposition 2.1 (see [4, Section 3]).

For any homogeneous quandle $Q$, there is a quandle triplet $(G,K,\psi)$ such
that $Q\cong_{\mathrm{qu}}Q(G,K,\psi)$.

We notice that for a given a group automorphism
$\psi\in\operatorname{Aut}_{\mathrm{gr}}(G)$, a triplet $(G,\\{e\\},\psi)$
trivially becomes a quandle triplet. Let $Q(G,\psi)=Q(G,\\{e\\},\psi)$, i.e.,
$Q(G,\psi)$ is the homogeneous quandle $(G,s)$, where $s$ is defined by

$s_{x}(y)=x\psi(x^{-1}y)\;\;\text{ for any }x,y\in G.$

This quandle $Q(G,\psi)$ is known as a generalized Alexander quandle. Note
that in the case $G$ is abelian, $Q(G,\psi)$ is called an Alexander quandle
(see [11]).

Towards a solution of Problem 1.1, we collect some invariants on generalized
Alexander quandles developed in [7]. Let
$\psi,\psi^{\prime}\in\operatorname{Aut}_{\mathrm{gr}}(G)$ and let
$Q=Q(G,\psi)$ and let $Q^{\prime}=Q(G,\psi^{\prime})$. The following
assertions are claimed in [7, Theorem 4.5]:

  1. $($a$)$

If $\psi^{\prime}=\psi^{\tau}$ for some
$\tau\in\operatorname{Aut}_{\mathrm{gr}}(G)$, then
$Q\cong_{\mathrm{qu}}Q^{\prime}$.

  2. $($b$)$

If $Q\cong_{\mathrm{qu}}Q^{\prime}$, then

     * •

$\operatorname{Inn}_{\mathrm{qu}}(Q)\cong_{\mathrm{gr}}\operatorname{Inn}_{\mathrm{qu}}(Q^{\prime})$;

     * •

$\operatorname{ord}_{\operatorname{Aut}_{\mathrm{gr}}(G)}\psi=\operatorname{ord}_{\operatorname{Aut}_{\mathrm{gr}}(G)}\psi^{\prime}$;

     * •

$|\operatorname{Fix}(\psi,G)|=|\operatorname{Fix}(\psi^{\prime},G)|$.

######  Remark 2.2.

In the above argument, we may obtain the similar results when the source group
and the target group are different, i.e., for $Q=Q(G,\psi)$ and
$Q^{\prime}=Q(G^{\prime},\psi^{\prime})$, if $Q\cong_{\mathrm{qu}}Q^{\prime}$
then

  * •

$\operatorname{Inn}_{\mathrm{qu}}(Q)\cong_{\mathrm{gr}}\operatorname{Inn}_{\mathrm{qu}}(Q^{\prime})$;

  * •

$\operatorname{ord}_{\operatorname{Aut}_{\mathrm{gr}}(G)}\psi=\operatorname{ord}_{\operatorname{Aut}_{\mathrm{gr}}(G^{\prime})}\psi^{\prime}$;

  * •

$|\operatorname{Fix}(\psi,G)|=|\operatorname{Fix}(\psi^{\prime},G^{\prime})|$.

##  3\. Proofs of the main results

In this section, we develop a new invariant $P$ (see (1)) on generalized
Alexander quandles.

Let $Q=Q(G,\psi)$ be a generalized Alexander quandle, where $G$ is a finite
group and $\psi\in\operatorname{Aut}_{\mathrm{gr}}(G)$. Given $a\in G$, let

$L_{a}:G\rightarrow G\text{ defined by }L_{a}(x)=ax\text{ for }x\in Q.$

Then we see that $L_{a}\in\operatorname{Aut}_{\mathrm{qu}}(Q)$. In fact,
$L_{a}$ is a bijection on $Q$ since $Q=G$ as a set, and

$L_{a}\circ
s_{x}(y)=ax\psi(x^{-1}y)=ax\psi(x^{-1}a^{-1}ay)=s_{ax}(ay)=s_{L_{a}(x)}\circ
L_{a}(y),$

i.e., $L_{a}$ is a quandle homomorphism. Note that $L_{a}^{-1}=L_{a^{-1}}$.

###  3.1. Normality for $P$ and $P^{2}$

Given a generalized Alexander quandle $Q=Q(G,\psi)$, let us recall the
definition of $P(Q)$ given in Introduction. We discuss some fundamental
properties of $P$.

######  Lemma 3.1.

Let $x\in P$. Then there exist $a_{1},\ldots,a_{r}\in G$ such that
$L_{x}=s_{a_{1}}\circ\cdots\circ s_{a_{r}}$.

###### Proof.

In the definition of $P$, we can add the assumption $r\equiv
0\pmod{\operatorname{ord}_{\operatorname{Aut}_{\mathrm{gr}}(G)}\psi}$ since
$s_{a_{1}}\circ\cdots\circ s_{a_{t}}(e)=s_{a_{1}}\circ\cdots\circ
s_{a_{t}}\circ s_{e}^{m}(e)$ holds for any $m\in{\mathbb{Z}}$. Hence, for
$x\in P$, there exist $a_{1},\ldots,a_{r}\in G$ such that
$x=s_{a_{1}}\circ\cdots\circ s_{a_{r}}(e)$, where we let $r\equiv
0\pmod{\operatorname{ord}_{\operatorname{Aut}_{\mathrm{gr}}(G)}\psi}$. Thus,
for any $y\in G$, we see the following:

$\displaystyle L_{x}(y)=xy=s_{a_{1}}\circ\cdots\circ
s_{a_{r}}(e)\cdot\psi^{r}(y)=s_{a_{1}}\circ\cdots\circ s_{a_{r}}(y).$

∎

In what follows, for a map $f:A\rightarrow B$ and a subset $A^{\prime}\subset
A$, let $f|_{A^{\prime}}$ denote the restriction of $f$ into $A^{\prime}$.

The following proposition claims a property on $P$ as quandles.

######  Proposition 3.2.

(i) $P$ becomes a subquandle of $Q(G,\psi)$.
(ii) $\psi|_{P}\in\operatorname{Aut}_{\mathrm{qu}}(P)$.

###### Proof.

(i) Take $x,y\in P$ arbitrarily. Then there exist $a_{1},\ldots,a_{r}\in G$
such that $y=s_{a_{1}}\circ\cdots\circ s_{a_{r}}(e)$. Thus
$s_{x}(y)=s_{x}\circ s_{a_{1}}\circ\cdots\circ s_{a_{r}}(e)$, i.e.,
$s_{x}(y)\in P$, as required.

(ii) It is enough to show that $\psi(P)\subset P$, but this can be shown
straightforwardly from $\psi=s_{e}$ as maps. ∎

Since $e=s_{e}(e)\in P$ and by Proposition 3.2, $P$ becomes a subquandle of
$Q(G,\psi)$ containing $e$. Thus we can also define the following:

(2) $\begin{split}P^{2}=P(P(Q))&:=\operatorname{Inn}_{\mathrm{qu}}(P(Q))e\\\
&=\\{x\in P:\exists p_{1},\ldots,\exists p_{t}\in P\text{ such that
}x=s_{p_{1}}\circ\cdots\circ s_{p_{t}}(e)\\}.\end{split}$

The following proposition claims a property on $P$ and $P^{2}$ as groups.

######  Proposition 3.3.

(i) $P$ is a normal subgroup of $G$.
(ii) $P^{2}$ is a normal subgroup of $P$.

###### Proof.

The statement (ii) immediately follows from (i) together with Proposition 3.2.
In fact, since $P$ is a group by (i) and $P$ also becomes a quandle by
Proposition 3.2, the statement (ii) is nothing but the statement for the
generalized Alexander quandle $Q(P,\psi|_{P})$. Thus, it suffices to show (i).

$e\in P$: We have already seen before the definition of (2).

$x,y\in P\ \Rightarrow\ xy\in P$: Take $x,y\in P$. Then there exist
$b_{1},b_{2},\ldots,b_{t}\in G$ such that $y=s_{b_{1}}\circ\cdots\circ
s_{b_{t}}(e)$. Moreover, by Lemma 3.1, there exist
$a_{1},a_{2},\ldots,a_{r}\in G$ such that $L_{x}=s_{a_{1}}\circ\cdots\circ
s_{a_{r}}$. Then

$xy=L_{x}(y)=s_{a_{1}}\circ\cdots\circ s_{a_{r}}\circ
s_{b_{1}}\circ\cdots\circ s_{b_{t}}(e).$

Hence we have $xy\in P$.

$x\in P\ \Rightarrow\ x^{-1}\in P$: Let
$m=\operatorname{ord}_{\operatorname{Aut}_{\mathrm{gr}}(G)}(\psi)$. Take $x\in
P$. Then there exist $a_{1},a_{2},\ldots,a_{r}\in G$ such that
$L_{x}=s_{a_{1}}\circ\cdots\circ s_{a_{r}}$. Since
$s_{a}^{m}=\operatorname{id}$ for any $a\in G$, we have
$L_{x}^{-1}=s_{a_{r}}^{m-1}\circ\cdots\circ s_{a_{1}}^{m-1}$. By using this,
we have

$x^{-1}=x^{-1}\cdot
e=L_{x^{-1}}(e)=L_{x}^{-1}(e)=s_{a_{r}}^{m-1}\circ\cdots\circ
s_{a_{1}}^{m-1}(e).$

Hence we have $x^{-1}\in P$.

Normal subgroup: Now we will show $axa^{-1}\in P$ for any $x\in P$ and $a\in
G$. By Lemma 3.1, there exist $a_{1},\ldots,a_{r}\in G$ such that
$L_{x}=s_{a_{1}}\circ\cdots\circ s_{a_{r}}$. Since

$axa^{-1}=L_{a}\circ L_{x}(a^{-1})=L_{a}\circ s_{a_{1}}\circ\cdots\circ
s_{a_{r}}(a^{-1})=s_{a\cdot a_{1}}\circ\cdots\circ s_{a\cdot a_{r}}(a\cdot
a^{-1})=s_{a\cdot a_{1}}\circ\cdots\circ s_{a\cdot a_{r}}(e),$

we have $axa^{-1}\in P$. ∎

######  Remark 3.4.

Let $Q=Q(G,\psi)$ be a generalized Alexander quandle.
(i) By direct computation, $P(Q)=\\{e\\}$ holds if and only if
$\operatorname{Fix}(\psi,G)=G$, i.e., $\psi=\operatorname{id}$.
(ii) By definition of (1), $P(Q)=G$ holds if and only if $Q$ is connected.

######  Remark 3.5.

In [1], the notion of the _displacement group_
$\operatorname{Dis}(Q):=\langle\\{s_{x}\circ s_{y}^{-1}:x,y\in Q\\}\rangle$
for quandles $Q$ and its importance are mentioned. In the context of
generalized Alexander quandles $Q=Q(G,\psi)$, where $G$ is a finite group and
$\psi\in\operatorname{Aut}_{\mathrm{gr}}(G)$, the following is proved:

$\operatorname{Dis}(Q)\cong_{\mathrm{gr}}[G,\psi]=\langle\\{x\psi(x)^{-1}:x\in
G\\}\rangle.$

By Proposition 3.3, we can prove that $[G,\psi]\cong_{\mathrm{gr}}P(Q)$. In
fact, since $x\psi(x)^{-1}=s_{x}(e)$ for $x\in G$ and $P(Q)$ is a subgroup of
$G$, it follows that $[G,\psi]\subset P(Q)$. Moreover, since $s_{x}\circ
s_{y}(e)=s_{s_{x}(y)}(s_{x}(e))=s_{s_{x}(y)}(e)\cdot s_{\psi(x)}(e)$ for
$x,y\in G$, we obtain the other inclusion.

######  Corollary 3.6.

If $G$ is simple and $\psi\neq\operatorname{id}$, then $P=G$.

###### Proof.

Since $\psi\neq\operatorname{id}$, we have $P\neq\\{e\\}$ by Remark 3.4 (i).
Since $G$ is simple and $P$ is a normal subgroup of $G$ by Proposition 3.3
(i), we obtain that $P=G$. ∎

######  Remark 3.7.

By Remark 3.4 (ii) and Corollary 3.6, generalized Alexander quandles obtained
from finite simple groups and non-identity automorphisms are connected.

###  3.2. Description of $\operatorname{Inn}_{\mathrm{qu}}(Q(G,\psi))$

Next, we describe $\operatorname{Inn}_{\mathrm{qu}}(Q(G,\psi))$ by using $P$.

######  Proposition 3.8 (cf. [1, Lemma 2.7]).

Let $G$ be a finite group and let
$\psi\in\operatorname{Aut}_{\mathrm{gr}}(G)$. Let $P=P(Q(G,\psi))$ and
$m=\operatorname{ord}_{\operatorname{Aut}_{\mathrm{gr}}(G)}(\psi)$. Then we
have

$\operatorname{Inn}_{\mathrm{qu}}(Q(G,\psi))\cong_{\mathrm{gr}}P\rtimes_{\phi}C_{m},$

which is a semidirect product of $P$ and $C_{m}$ with
$\phi:C_{m}\rightarrow\operatorname{Aut}_{\mathrm{gr}}(G)$,
$i\mapsto\psi^{i}$, i.e.,

$(x,i)\cdot(y,j)=(x\psi^{i}(y),i+j).$

As in [7, Proposition 5.6], the semi-direct product in Proposition 3.8 may
become a direct product in some cases. The following proposition shows the
relationship between the structure of the semi-direct product and the property
of $\psi|_{P}$ when $P$ is centerless, i.e., $Z(P)=\\{e\\}$.

######  Proposition 3.9.

Work with the notation as above. Assume that $P$ is centerless. Then the
following assertions hold.
(i) If $\psi|_{P}\in\operatorname{Inn}_{\mathrm{gr}}(P)$, i.e., $\psi|_{P}$ is
an inner automorphism, then we have

$\operatorname{Inn}_{\mathrm{qu}}(Q(G,\psi))\cong_{\mathrm{gr}}P\rtimes_{\phi}C_{m}\cong_{\mathrm{gr}}P\times
C_{m}.$

(ii) If $\psi|_{P}\notin\operatorname{Inn}_{\mathrm{gr}}(P)$, i.e.,
$\psi|_{P}$ is an outer automorphism, then we have

$\operatorname{Inn}_{\mathrm{qu}}(Q(G,\psi))\cong_{\mathrm{gr}}P\rtimes_{\phi}C_{m}\not\cong_{\mathrm{gr}}P\times
C_{m}.$

###### Proof.

(i) By $\psi|_{P}\in\operatorname{Inn}_{\mathrm{gr}}(P)$, there exists $a\in
P$ such that $\psi|_{P}=(\cdot)^{a}$. Since $P$ is centerless,
$\operatorname{Inn}_{\mathrm{gr}}(P)\cong_{\mathrm{gr}}P$ holds. Then we have
$\operatorname{ord}_{\operatorname{Aut}_{\mathrm{gr}}(P)}(\psi|_{P})=\operatorname{ord}_{P}(a)$.
On the other hand,
$\operatorname{ord}_{\operatorname{Aut}_{\mathrm{gr}}(P)}(\psi|_{P})$ is a
divisor of $m=\operatorname{ord}_{\operatorname{Aut}_{\mathrm{gr}}(G)}(\psi)$.
Hence $\operatorname{ord}_{P}(a)$ is a divisor of $m$. We define the map
$\Psi:P\rtimes_{\phi}C_{m}\rightarrow P\times C_{m}$ by

$\Psi((x,i))=(xa^{i},i).$

Then it is straightforward to check that this gives a group isomorphism:

  * •

Since $a\in P$, we see that $xa^{i}\in P$ for any $x\in P$ and $i\in C_{m}$.
Also since $\operatorname{ord}_{P}(a)$ is a divisor of $m$, we see that
$xa^{i}=xa^{j}$ whenever $i\equiv j\pmod{m}$. Hence, the well-definedness
follows.

  * •

One has
$\Psi((x,i)\cdot(y,j))=\Psi((xa^{i}ya^{-i},i+j))=(xa^{i}ya^{j},i+j)=\Psi((x,i))\Psi((y,j)).$

  * •

One has
$\Psi((x,i))=\Psi((y,j))\;\Longleftrightarrow\;(xa^{i},i)=(ya^{j},j)\;\Longleftrightarrow\;x=y\text{
and }i\equiv j\pmod{m}$.

  * •

For any $(x,i)\in P\times C_{m}$, since $xa^{-i}\in P$, we have
$\Psi((xa^{-i},i))=(x,i)$, as required.

(ii) Suppose that there exists a group isomorphism $\Omega\colon
P\rtimes_{\phi}C_{m}\to P\times C_{m}$. Let $X=\\{(x,0):x\in P\\}\subset
P\rtimes_{\phi}C_{m}$. Then $X$ is a subgroup of $P\rtimes_{\phi}C_{m}$. Let
$X^{\prime}=\Omega(X)$. Then $X^{\prime}$ becomes a subgroup of $P\times
C_{m}$. For each $(x,0)\in X$, we write $(a_{x},j_{x})=\Omega(x,0)$.

Let $Y^{\prime}=\\{(e,i):i\in C_{m}\\}\subset P\times C_{m}$. Then
$Y^{\prime}$ is a subgroup of $P\times C_{m}$. Let
$Y=\Omega^{-1}(Y^{\prime})$. Then $Y$ becomes a subgroup of
$P\rtimes_{\phi}C_{m}$ and $|Y|=|Y^{\prime}|=m$ since $\Omega$ is a group
isomorphism. For each $(e,i)\in Y^{\prime}$, we write
$(b_{i},k_{i})=\Omega^{-1}((e,i))$.

Since $\Omega^{-1}$ is also a group isomorphism, we see that

$\displaystyle\Omega^{-1}((e,i)(a_{x},j_{x}))$
$\displaystyle=\Omega^{-1}((e,i))\cdot\Omega^{-1}((a_{x},j_{x}))=(b_{i},k_{i})\cdot(x,0)=(b_{i}\psi^{k_{i}}(x),k_{i})\text{
and }$ $\displaystyle\Omega^{-1}((e,i)(a_{x},j_{x}))$
$\displaystyle=\Omega^{-1}((a_{x},i+j_{x}))=\Omega^{-1}((a_{x},j_{x})(e,i))=\Omega^{-1}((a_{x},j_{x}))\cdot\Omega^{-1}((e,i))$
$\displaystyle=(x,0)\cdot(b_{i},k_{i})=(xb_{i},k_{i}).$

Hence, we obtain that
$b_{i}\psi^{k_{i}}(x)=xb_{i}\;\Longleftrightarrow\psi^{k_{i}}(x)=b_{i}^{-1}xb_{i}$
for any $x\in P$, i.e., $\psi^{k_{i}}|_{P}$ is an inner automorphism on $P$.
Note that $b_{i}$ is determined by $\psi$ and $k_{i}$ since $P$ is centerless.
We put $b_{i}=b_{k_{i}}$. This implies that

$Y=\\{(b_{i},k_{i}):i\in C_{m}\\}=\\{(b_{k},k):k\in C_{m}\text{ and
}\psi^{k}|_{P}\in\operatorname{Inn}_{\mathrm{gr}}(P)\\}.$

On the other hand, $\psi|_{P}\not\in\operatorname{Inn}_{\mathrm{gr}}(P)$. This
implies $k\neq 1$. Thus, we have $|Y|\leq m-1$. This is a contradiction to
$|Y|=m$.

Therefore, $P\rtimes_{\phi}C_{m}\not\cong_{\mathrm{gr}}P\times C_{m}$, as
desired. ∎

When $P$ is not centerless, Proposition 3.9 does not hold in general. In fact,
let $G$ be a special linear group $\mathrm{SL}(2,3)$ of degree $2$ over
$\mathbb{F}_{3}$ and let $\psi$ be the inner automorphism on $G$ with respect
to $A=\begin{pmatrix}0&-1\\\ 1&0\end{pmatrix}$. Note that
$\operatorname{ord}_{\operatorname{Aut}(G)}(\psi)=2$, nevertheless
$\operatorname{ord}_{G}(A)=4$. Then $P=P(Q(G,\psi))\cong_{\mathrm{gr}}Q_{8}$,
where $Q_{8}$ is the quaternion group. (For the detail of $Q_{8}$, see Section
6.4.) It is well known that $Z(Q_{8})\cong_{\mathrm{gr}}C_{2}$. Moreover,
$\psi|_{P}\in\operatorname{Inn}_{\mathrm{gr}}(P)$. On the other hand, we can
verify that

$\operatorname{Inn}_{\mathrm{qu}}(Q(G,\psi))\cong_{\mathrm{gr}}P\rtimes
C_{2}\not\cong_{\mathrm{gr}}P\times C_{2}.$

###  3.3. Proofs of Theorem 1.2 and Corollary 1.3

Now, we prove the following theorem, which will play a crucial role for the
proof of Theorem 1.2.

######  Theorem 3.10.

Let $G,G^{\prime}$ be finite groups with the identity elements $e,e^{\prime}$,
respectively. Let $\psi\in\operatorname{Aut}_{\mathrm{gr}}(G)$ and
$\psi^{\prime}\in\operatorname{Aut}_{\mathrm{gr}}(G^{\prime})$ and
$P=P(Q(G,\psi))$ and $P^{\prime}=P(Q(G^{\prime},\psi^{\prime}))$. Assume that
$Q(G,\psi)\cong_{\mathrm{qu}}Q(G^{\prime},\psi^{\prime})$ and let $f$ be a
quandle isomorphism $f:Q(G,\psi)\rightarrow Q(G^{\prime},\psi^{\prime})$ with
$f(e)=e^{\prime}$. Then the following assertions hold:

  * (i)

The restriction $f|_{P}$ gives a quandle isomorphism between $Q(P,\psi|_{P})$
and $Q(P^{\prime},\psi^{\prime}|_{P^{\prime}})$.

  * (ii)

The equality $f\circ\psi=\psi^{\prime}\circ f$ holds.

  * (iii)

A quandle isomorphism $f|_{P}$ is a group homomorphism. In particular,
$P\cong_{\mathrm{gr}}P^{\prime}$ and
$f|_{P}\in\operatorname{Aut}_{\mathrm{gr}}(P)$.

  * (iv)

For any $x\in G$, $f(xP)=f(x)P^{\prime}$ holds.

Note that when $Q(G,\psi)\cong_{\mathrm{qu}}Q(G^{\prime},\psi^{\prime})$, we
can always choose a quandle isomorphism $f$ between them with
$f(e)=e^{\prime}$ (see, e.g., [7]). We also remark that $Q(P,\psi|_{P})$
(resp. $Q(P^{\prime},\psi^{\prime}|_{P^{\prime}})$) becomes a subquandle of
$Q(G,\psi)$ (resp. $Q(G^{\prime},\psi^{\prime})$) by Proposition 3.2 (i).

###### Proof of Theorem 3.10.

Let $s$ (resp. $s^{\prime}$) be the point symmetry of $Q(G,\psi)$ (resp.
$Q(G^{\prime},\psi^{\prime})$). (i) Since $f|_{P}$ is a quandle homomorphism,
it is enough to show that $f|_{P}(P)=P^{\prime}$. This can be shown by

$f(s_{a_{1}}\circ\cdots\circ
s_{a_{r}}(e))=s_{f(a_{1})}^{\prime}\circ\cdots\circ
s_{f(a_{r})}^{\prime}(e^{\prime})\;\text{ and
}s^{\prime}_{a^{\prime}_{1}}\circ\cdots\circ
s_{a^{\prime}_{t}}(e^{\prime})=f(s_{f^{-1}(a^{\prime}_{1})}\circ\cdots\circ
s_{f^{-1}(a^{\prime}_{t})}(e))$

since $f$ is a quandle isomorphism with $f(e)=e^{\prime}$.

(ii) Note that $\psi=s_{e}$ holds as maps. Since $f$ is a quandle
homomorphism, we have $f\circ s_{e}=s_{f(e)}^{\prime}\circ f$, i.e.,
$f\circ\psi=\psi^{\prime}\circ f$.

(iii) Since $f|_{P}$ is a bijection between $P$ and $P^{\prime}$ by (i), it is
enough to show that $f|_{P}$ becomes a group homomorphism. For any $x\in P$,
there exist $a_{1},\ldots,a_{r}\in G$ such that $x=s_{a_{1}}\circ\cdots\circ
s_{a_{r}}(e)$. As we have already seen,

$\displaystyle f(x)=f\circ s_{a_{1}}\circ\cdots\circ
s_{a_{r}}(e)=s_{f(a_{1})}^{\prime}\circ\cdots\circ
s_{f(a_{r})}^{\prime}(e^{\prime})$

holds. Thus, for any $x,y\in P$, we have the following:

$\displaystyle f(xy)$ $\displaystyle=f\circ L_{x}(y)=f\circ
s_{a_{1}}\circ\cdots\circ s_{a_{r}}(y)\;\;\text{ (by Lemma~{}\ref{lem_Lx})}$
$\displaystyle=s_{f(a_{1})}^{\prime}\circ\cdots\circ
s_{f(a_{r})}^{\prime}(f(y))$
$\displaystyle=s_{f(a_{1})}^{\prime}\circ\cdots\circ
s_{f(a_{r})}^{\prime}(e^{\prime})\cdot f(y)\;\;\text{
($\operatorname{ord}_{\operatorname{Aut}_{\mathrm{gr}}(G)}\psi=\operatorname{ord}_{\operatorname{Aut}_{\mathrm{gr}}(G^{\prime})}\psi^{\prime}$
by Remark~{}\ref{rem:GneqG'})}$ $\displaystyle=f(x)f(y).$

(iv) Firstly, we show $f(xP)\subset f(x)P^{\prime}$. For any $x\in G$ and
$p=s_{a_{1}}\circ\cdots\circ s_{a_{r}}(e)\in P$, we have

$\displaystyle f(xp)=$ $\displaystyle f\circ L_{x}\circ
s_{a_{1}}\circ\cdots\circ s_{a_{r}}(e)=f\circ s_{xa_{1}}\circ\cdots\circ
s_{xa_{r}}(x)$ $\displaystyle=$ $\displaystyle
s^{\prime}_{f(xa_{1})}\circ\cdots\circ
s^{\prime}_{f(xa_{r})}(f(x))=L_{f(x)}\circ
s^{\prime}_{f(x)^{-1}f(xa_{1})}\circ\cdots\circ
s^{\prime}_{f(x)^{-1}f(xa_{r})}(e^{\prime})$ $\displaystyle=$ $\displaystyle
f(x)\cdot s^{\prime}_{f(x)^{-1}f(xa_{1})}\circ\cdots\circ
s^{\prime}_{f(x)^{-1}f(xa_{r})}(e^{\prime})\in f(x)P^{\prime}.$

Next, we show $f(xP)\supset f(x)P^{\prime}$. For any $x\in G$ and
$p^{\prime}=s^{\prime}_{a^{\prime}_{1}}\circ\cdots\circ
s^{\prime}_{a^{\prime}_{r}}(e^{\prime})\in P^{\prime}$, we have

$\displaystyle f(x)p^{\prime}=$ $\displaystyle L_{f(x)}\circ
s^{\prime}_{a^{\prime}_{1}}\circ\cdots\circ
s^{\prime}_{a^{\prime}_{r}}(e^{\prime})=s^{\prime}_{f(x)a^{\prime}_{1}}\circ\cdots\circ
s^{\prime}_{f(x)a^{\prime}_{r}}(f(x))$ $\displaystyle=$ $\displaystyle
s^{\prime}_{f(x)a^{\prime}_{1}}\circ\cdots\circ
s^{\prime}_{f(x)a^{\prime}_{r}}\circ f(x)=f\circ
s_{f^{-1}(f(x)a^{\prime}_{1})}\circ\cdots\circ
s_{f^{-1}(f(x)a^{\prime}_{r})}(x)$ $\displaystyle=$ $\displaystyle f\circ
L_{x}\circ s_{x^{-1}f^{-1}(f(x)a^{\prime}_{1})}\circ\cdots\circ
s_{x^{-1}f^{-1}(f(x)a^{\prime}_{r})}(e)\in f(xP).$

∎

By Theorem 3.10, we can prove Theorem 1.2.

###### Proof of Theorem 1.2.

Let $\psi,\psi^{\prime}\in\operatorname{Aut}_{\mathrm{gr}}(G)$ and let
$Q=Q(G,\psi),Q^{\prime}=Q(G,\psi^{\prime})$, respectively. Assume that $Q$ and
$Q^{\prime}$ are isomorphic as quandles. By Remark 3.4, we do not consider the
case where $\psi$ or $\psi^{\prime}$ is identity. Since $G$ is simple, we see
that $G=P(Q)=P(Q^{\prime})$ by Corollary 3.6. Thus, $Q=P(Q)$ (resp.
$Q^{\prime}=P(Q^{\prime})$) as sets.

Let $f:Q\rightarrow Q^{\prime}$ be a quandle isomorphism with $f(e)=e$. Then
it follows from Theorem 3.10 (iii) that $f$ is a group automorphism of $G$.
Moreover, by Theorem 3.10 (ii), we have $f\circ\psi=\psi^{\prime}\circ f$,
i.e., $f\circ\psi\circ f^{-1}=\psi^{\prime}$. This says that $\psi$ and
$\psi^{\prime}$ are conjugate in $\operatorname{Aut}_{\mathrm{gr}}(G)$. ∎

###### Proof of Corollary 1.3.

When $n\leq 4$, it was already proved that ${\mathcal{Q}}(\mathfrak{S}_{n})$
one-to-one corresponds to the set of conjugacy classes of
$\operatorname{Aut}_{\mathrm{gr}}(\mathfrak{S}_{n})$ in [7]. Thus, we may
assume $n\geq 5$.

Let
$\psi,\psi^{\prime}\in\operatorname{Aut}_{\mathrm{gr}}(\mathfrak{S}_{n})\setminus\\{\operatorname{id}\\}$
and let
$Q=Q(\mathfrak{S}_{n},\psi),Q^{\prime}=Q(\mathfrak{S}_{n},\psi^{\prime})$,
respectively. Since $x$ and $\psi(x)$ have the same parity for any
$x\in\mathfrak{S}_{n}$, the parity of $x\psi(x^{-1})$ is even, i.e.,
$P=P(Q)\subset\mathfrak{A}_{n}$. It is known that $\mathfrak{A}_{n}$ is the
unique non-trivial normal subgroup of $\mathfrak{S}_{n}$ if $n\geq 5$. Hence,
$P=\mathfrak{A}_{n}$ holds if $\psi\neq\operatorname{id}$ (see Remark 3.4).
Similarly, we also have $P^{\prime}=P(Q^{\prime})=\mathfrak{A}_{n}$.

Assume that $Q$ and $Q^{\prime}$ are isomorphic as quandles. Since
$\mathfrak{A}_{n}$ is simple for $n\geq 5$, $\psi|_{\mathfrak{A}_{n}}$ and
$\psi^{\prime}|_{\mathfrak{A}_{n}}$ are conjugate in
$\operatorname{Aut}_{\mathrm{gr}}(\mathfrak{A}_{n})$ by Theorem 1.2. Hence, we
can take $h\in\operatorname{Aut}_{\mathrm{gr}}(\mathfrak{A}_{n})$ such that
$h\circ\psi|_{\mathfrak{A}_{n}}=\psi^{\prime}|_{\mathfrak{A}_{n}}\circ h$. It
is known that the restriction map $f\mapsto f|_{\mathfrak{A}_{n}}$ from
$\operatorname{Aut}_{\mathrm{gr}}(\mathfrak{S}_{n})$ to
$\operatorname{Aut}_{\mathrm{gr}}(\mathfrak{A}_{n})$ gives an isomorphism
between them (e.g. [14, 11.4.1]). Thus, there exists
$f\in\operatorname{Aut}_{\mathrm{gr}}(\mathfrak{S}_{n})$ such that
$f|_{\mathfrak{A}_{n}}=h$. Therefore for any $x\in\mathfrak{A}_{n}$, we have

$\displaystyle
h\circ\psi|_{\mathfrak{A}_{n}}(x)=\psi^{\prime}|_{\mathfrak{A}_{n}}\circ
h(x)\;\Longleftrightarrow f\circ\psi(x)=\psi^{\prime}\circ
f(x)\Longleftrightarrow f\circ\psi\circ f^{-1}(x)=\psi^{\prime}(x).$

Once we can prove that $f\circ\psi\circ f^{-1}(x)=\psi^{\prime}(x)$ holds for
any $x\in\mathfrak{S}_{n}$, we obtain that $\psi$ and $\psi^{\prime}$ are
conjugate. Let $\varphi=\psi^{\prime-1}\circ f\circ\psi\circ
f^{-1}\in\operatorname{Aut}_{\mathrm{gr}}(\mathfrak{S}_{n})$. It suffices to
show that $\varphi(x)=x$ holds for any
$x\in\mathfrak{S}_{n}\setminus\mathfrak{A}_{n}$.

Given $x,y\in\mathfrak{S}_{n}\setminus\mathfrak{A}_{n}$, i.e., $x$ and $y$ are
odd, we know that $x^{-1}y$ is even. Thus we have

$x^{-1}y=\varphi(x^{-1}y)\Longleftrightarrow\varphi(x)x^{-1}=\varphi(y)y^{-1}.$

This means that $\varphi(x)x^{-1}$ is constant for any
$x\in\mathfrak{S}_{n}\setminus\mathfrak{A}_{n}$. Thus, by letting
$a=\varphi(x_{0})\cdot x_{0}^{-1}\in\mathfrak{A}_{n}$ for some
$x_{0}\in\mathfrak{S}_{n}\setminus\mathfrak{A}_{n}$, we have $\varphi(x)=ax$
for any $x\in\mathfrak{S}_{n}\setminus\mathfrak{A}_{n}$.

Fix $x\in\mathfrak{S}_{n}\setminus\mathfrak{A}_{n}$ and we take
$y\in\mathfrak{A}_{n}$ arbitrarily. Note
$yx\in\mathfrak{S}_{n}\setminus\mathfrak{A}_{n}$. Then we see that
$\varphi(yx)=ayx$. On the other hand, we also have
$\varphi(yx)=\varphi(y)\varphi(x)=yax$. Hence, $a\in Z(\mathfrak{A}_{n})$. It
is well known that $Z(\mathfrak{A}_{n})=\\{e\\}$. Therefore, $a=e$. This means
that $\varphi=\operatorname{id}$, as required. ∎

###  3.4. More properties for $P$ and $P^{2}$

######  Lemma 3.11.

Let $G$ and $G^{\prime}$ be finite groups and let
$\psi\in\operatorname{Aut}_{\mathrm{gr}}(G)$ and
$\psi^{\prime}\in\operatorname{Aut}_{\mathrm{gr}}(G^{\prime})$. Assume that
there is a group isomorphism $h:P\rightarrow P^{\prime}$ satisfying
$h\circ\psi|_{P}=\psi^{\prime}|_{P^{\prime}}\circ h$. Then $h|_{P^{2}}$ gives
a group isomorphism between $P^{2}$ and $P^{\prime 2}$, i.e.,
$P^{2}\cong_{\mathrm{gr}}P^{\prime 2}$.

###### Proof.

Let $h$ be a group isomorphism satisfying
$h\circ\psi|_{P}=\psi^{\prime}|_{P^{\prime}}\circ h$. Then

$\displaystyle h(s_{a}(b))$ $\displaystyle=h(a\psi(a^{-1}b))=h(a)\cdot
h\circ\psi(a^{-1}b)=h(a)\cdot\psi^{\prime}\circ
h(a^{-1}b)=h(a)\psi^{\prime}(h(a)^{-1}h(b))$
$\displaystyle=s_{h(a)}^{\prime}(h(b))$

for $a,b\in P$. Hence, $h$ is a quandle automorphism between $P$ and
$P^{\prime}$. Therefore, by applying Theorem 3.10 (iii) for $Q(P,\psi|_{P})$
and $Q(P^{\prime},\psi|_{P^{\prime}})$, we conclude the assertion. ∎

######  Lemma 3.12.

Work with the same notation as in Lemma 3.11. Then
$|P\cap\operatorname{Fix}(\psi,G)|=|P^{\prime}\cap\operatorname{Fix}(\psi^{\prime},G^{\prime})|$
holds.

###### Proof.

We see this assertion as follows:

$\displaystyle x\in P\cap\operatorname{Fix}(\psi,G)\;$
$\displaystyle\Longleftrightarrow\;x\in P\text{ and
}x=\psi(x)\;\Longleftrightarrow\;h(x)\in P^{\prime}\text{ and
}h(x)=h\circ\psi(x)$ $\displaystyle\Longleftrightarrow\;h(x)\in
P^{\prime}\text{ and }h(x)=\psi^{\prime}\circ
h(x)\;\Longleftrightarrow\;h(x)\in
P^{\prime}\cap\operatorname{Fix}(\psi^{\prime},G^{\prime}).$

∎

For a finite group $G$ and its subset $H$ and
$\psi\in\operatorname{Aut}_{\mathrm{gr}}(G)$, we define the twisted normalizer
of $H$ with respect to $\psi$ as follows:

$\operatorname{TN}^{\psi}_{G}(H):=\\{x\in G:xH\psi(x)^{-1}=H\\}.$

We see that $\operatorname{TN}^{\psi}_{G}(H)$ is a subgroup of $G$. In fact,

  * •

$e\in\operatorname{TN}^{\psi}_{G}(H)$ since $eH\psi(e)^{-1}=H$;

  * •

for $x,y\in\operatorname{TN}^{\psi}_{G}(H)$, we have
$xyH\psi(xy)^{-1}=x(yH\psi(y)^{-1}))\psi(x)^{-1}=H$;

  * •

for $x\in G$, we see that $xH\psi(x)^{-1}=H$ holds if and only if
$x^{-1}H\psi(x)=H$ holds.

For a while, let $F=\operatorname{Fix}(\psi,G)$. Since $P$ is a normal
subgroup of $G$ (by Proposition 3.3), we have $FP=PF$.

######  Lemma 3.13.

$P$ is a normal subgroup of $\operatorname{TN}^{\psi}_{G}(P^{2})$.

###### Proof.

$P<\operatorname{TN}^{\psi}_{G}(P^{2})$: Since $P\subset PF$ is trivial, for
the proof that $P$ is a subgroup of $\operatorname{TN}^{\psi}_{G}(P^{2})$, it
is enough to show the inclusion
$PF\subset\operatorname{TN}^{\psi}_{G}(P^{2})$.

For any $p_{1},p_{2}\in P$, any $x\in F$ and any $y\in G$, we know that

$\displaystyle s_{p_{1}xp_{2}}(y)$
$\displaystyle=s_{p_{1}\tilde{p_{2}}x}(y)\;\;\text{(for some $\tilde{p_{2}}\in
P$ since $P\triangleleft G$)}$
$\displaystyle=p_{1}\tilde{p_{2}}x\psi(x^{-1}\tilde{p_{2}}^{-1}p_{1}^{-1}y)=p_{1}\tilde{p_{2}}\psi(\tilde{p_{2}}^{-1}p_{1}^{-1}y)\;\;\text{(since
}x\in F)$ $\displaystyle=s_{p_{1}\tilde{p_{2}}}(y).$

Fix $p\in P$ and $x\in F$. For each $a\in P$, let $\tilde{a}$ denote the
element of $P$ satisfying $s_{pxa}=s_{p\tilde{a}}$.

Take $p\in P$ and $x\in F$ arbitrarily. By Lemma 3.1, for an arbitrary $q\in
P^{2}$, write $L_{q}=s_{p_{1}}\circ\cdots\circ s_{p_{r}}$, where $p_{i}\in P$.
Then we see that

$\displaystyle(px)q\psi(px)^{-1}$ $\displaystyle=L_{px}\circ
L_{q}(\psi(px)^{-1})=L_{px}\circ s_{p_{1}}\circ\cdots\circ
s_{p_{r}}(\psi(px)^{-1})$ $\displaystyle=s_{pxp_{1}}\circ\cdots\circ
s_{pxp_{r}}(px\psi(px)^{-1})=s_{p\tilde{p_{1}}}\circ\cdots\circ
s_{p\tilde{p}_{r}}(p\psi(p^{-1}))$
$\displaystyle=s_{p\tilde{p_{1}}}\circ\cdots\circ s_{p\tilde{p}_{r}}\circ
s_{p}(e)\in P^{2}.$

Hence $(px)P^{2}\psi(px)^{-1}\subset P^{2}$ holds. Since
$|(px)P^{2}\psi(px)^{-1}|=|P^{2}|$, this implies that
$(px)P^{2}\psi(px)^{-1}=P^{2}$. Therefore, one has
$px\in\operatorname{TN}^{\psi}_{G}(P^{2})$.

$P\triangleleft\operatorname{TN}^{\psi}_{G}(P^{2})$: Take $p\in P$ and
$x\in\operatorname{TN}^{\psi}_{G}(P^{2})$ arbitrarily. Since $P$ is a normal
subgroup of $G$ and $x\in\operatorname{TN}^{\psi}_{G}(P^{2})\subset G$, we see
that $xpx^{-1}\in P$, as required. ∎

######  Lemma 3.14.

Let $Q=Q(G,\psi)$ and $Q^{\prime}=Q(G^{\prime},\psi^{\prime})$ be two
generalized Alexander quandles and assume that
$Q\cong_{\mathrm{qu}}Q^{\prime}$. Then
$|\operatorname{TN}^{\psi}_{G}(P^{2})|=|\operatorname{TN}^{\psi^{\prime}}_{G^{\prime}}({P^{\prime}}^{2})|$.

###### Proof.

Let $f:Q\rightarrow Q^{\prime}$ be a quandle automorphism with
$f(e)=e^{\prime}$. Then, by Theorem 3.10, we know that $f|_{P}$ is a group
homomorphism between $P$ and $P^{\prime}$.

Once we prove that
$f(x)\in\operatorname{TN}^{\psi^{\prime}}_{G^{\prime}}(P^{\prime 2})$ for any
$x\in\operatorname{TN}^{\psi}_{G}(P^{2})$, since $f$ is injective, we obtain
that
$|\operatorname{TN}^{\psi}_{G}(P^{2})|\leq|\operatorname{TN}^{\psi^{\prime}}_{G^{\prime}}(P^{\prime
2})|$. By applying the same discussion, we also obtain the other inequality,
as required.

Let $x\in\operatorname{TN}^{\psi}_{G}(P^{2})$. For a given $q^{\prime}\in
P^{\prime 2}$, by Lemma 3.1, let
$L_{q^{\prime}}=s_{p_{1}^{\prime}}^{\prime}\circ\cdots\circ
s_{p_{r}^{\prime}}^{\prime}$, where $p_{i}^{\prime}\in P^{\prime}$ and
$r\equiv
0\pmod{\operatorname{ord}_{\operatorname{Aut}_{\mathrm{gr}}(G^{\prime})}\psi^{\prime}}$.
Then we see the following:

$\displaystyle f(x)q^{\prime}\psi^{\prime}(f(x))^{-1}$
$\displaystyle=L_{f(x)}\circ
L_{q^{\prime}}(\psi^{\prime}(f(x))^{-1})=L_{f(x)}\circ
s_{p_{1}^{\prime}}^{\prime}\circ\cdots\circ
s_{p_{r}^{\prime}}^{\prime}(\psi^{\prime}(f(x))^{-1})$
$\displaystyle=s_{f(x)p_{1}^{\prime}}^{\prime}\circ\cdots\circ
s_{f(x)p_{r}^{\prime}}^{\prime}(f(x)\psi^{\prime}(f(x))^{-1})=s_{f(x)p_{1}^{\prime}}^{\prime}\circ\cdots\circ
s_{f(x)p_{r}^{\prime}}^{\prime}\circ s^{\prime}_{f(x)}(e^{\prime})$
$\displaystyle=f\left(s_{xp_{1}}\circ\cdots\circ s_{xp_{r}}\circ
s_{x}(e)\right)$ (where we let $f(xp_{i})=f(x)p_{i}^{\prime}$ for each $i$
with $p_{i}\in P$, see Theorem 3.10 (iv)) $\displaystyle=f\left(L_{x}\circ
s_{p_{1}}\circ\cdots\circ s_{p_{r}}\circ s_{e}(x^{-1})\right)=f\left(x\cdot
s_{p_{1}}\circ\cdots\circ s_{p_{r}}(e)\cdot s_{e}(x^{-1})\right)$
$\displaystyle=f(x\cdot\underbrace{s_{p_{1}}\circ\cdots\circ
s_{p_{r}}(e)}_{\in P^{2}}\cdot\psi(x)^{-1})\in f(P^{2})=P^{\prime
2}\;\;\text{(since $x\in\operatorname{TN}^{\psi}_{G}(P^{2})$)}.$

This implies that
$f(x)\in\operatorname{TN}^{\psi^{\prime}}_{G^{\prime}}(P^{\prime 2})$. ∎

We introduce the following two conditions (P1) and (P2).

  1. (P1)

$P^{2}$ is a normal subgroup of $G$.

  2. (P2)

$P^{2}=\\{s_{p}(e):p\in P\\}$ holds.

######  Remark 3.15.

(i) For (P1), note that $H\triangleleft G$ and $K\triangleleft H$ do not imply
$K\triangleleft G$ in general. For (P2), note that the inclusion $\supset$ is
always true, but $\subset$ does not necessarily hold.

(ii) The conditions (P1) and (P2) are independent. In fact,
$Q(Q_{8},\psi_{4})$ in Table 5 in Section 6.4 satisfies (P1) and does not
satisfy (P2). On the other hand, the following $Q^{\prime}$ does not satisfy
(P1) and satisfies (P2); $Q^{\prime}=Q(G^{\prime},\psi^{\prime})$ with
$G^{\prime}=\mathfrak{S}_{3}\times\mathfrak{S}_{3}$ and $\psi^{\prime}\colon
G^{\prime}\to G^{\prime}$ defined by $(x,y)\mapsto(y,x)$.

The following proposition claims that (P1) and (P2) are quandle invariants.

######  Proposition 3.16.

Let $Q=Q(G,\psi)$ and $Q^{\prime}=Q(G^{\prime},\psi^{\prime})$ be two
generalized Alexander quandles and assume that
$Q\cong_{\mathrm{qu}}Q^{\prime}$.

  * (i)

$Q$ satisfies (P1) if and only if $Q^{\prime}$ satisfies (P1).

  * (ii)

$Q$ satisfies (P2) if and only if $Q^{\prime}$ satisfies (P2).

###### Proof.

Take a quandle isomorphism $f\colon Q\to Q^{\prime}$ with $f(e)=e^{\prime}$.

(i) Assume that $Q$ satisfies (P1). For $a^{\prime}\in G^{\prime}$ and
$q^{\prime}\in P^{\prime 2}$, we prove $a^{\prime}q^{\prime}a^{\prime-1}\in
P^{\prime 2}$. Then there exists $a\in G$ such that $f(a)=a^{\prime}$. By
Lemma 3.1, let $L_{q}^{\prime}=s^{\prime}_{p^{\prime}_{1}}\circ\cdots\circ
s^{\prime}_{p^{\prime}_{r}}$, where $p^{\prime}_{i}\in P^{\prime}$ and
$r\equiv
0\pmod{\operatorname{ord}_{\operatorname{Aut}_{\mathrm{gr}}(G^{\prime})}\psi^{\prime}}$.
Then we have

$\displaystyle a^{\prime}q^{\prime}a^{\prime-1}$
$\displaystyle=f(a)q^{\prime}f(a)^{-1}=L_{f(a)}\circ
L_{q^{\prime}}(f(a)^{-1})$ $\displaystyle=L_{f(a)}\circ
s^{\prime}_{p^{\prime}_{1}}\circ\cdots\circ
s^{\prime}_{p^{\prime}_{r}}(f(a)^{-1})$
$\displaystyle=s^{\prime}_{f(a)p^{\prime}_{1}}\circ\cdots\circ
s^{\prime}_{f(a)p^{\prime}_{r}}(e^{\prime})$
$\displaystyle=s^{\prime}_{f(ap_{1})}\circ\cdots\circ
s^{\prime}_{f(ap_{r})}(e^{\prime})\quad(\text{by Theorem~{}\ref{breakthrough}
(iv)})$ $\displaystyle=f(s_{ap_{1}}\circ\cdots\circ s_{ap_{r}}(e))$
$\displaystyle=f(a\cdot s_{p_{1}}\circ\cdots\circ s_{p_{r}}(e)\cdot a^{-1}).$

Since $Q$ satisfies (P1), we have $a\cdot s_{p_{1}}\circ\cdots\circ
s_{p_{r}}(e)\cdot a^{-1}\in P^{2}$. By Lemma 3.11, we have
$a^{\prime}q^{\prime}a^{\prime-1}\in P^{\prime 2}$, i.e., $Q^{\prime}$
satisfies (P1).

(ii) Assume that $Q$ satisfies (P2). For $q^{\prime}\in P^{\prime 2}$, we
prove that there exists $p^{\prime}\in P^{\prime}$ such that
$q^{\prime}=s^{\prime}_{p^{\prime}}(e^{\prime})$. By Lemma 3.11, there exists
$q\in P^{2}$ such that $q^{\prime}=f(q)$. Since $Q$ satisfies (P2), there
exists $p\in P$ such that $q=s_{p}(e)$. Then we have
$q^{\prime}=f(s_{p}(e))=s^{\prime}_{f(p)}(e^{\prime})$. By Theorem 3.10 (i),
we have $f(p)\in P^{\prime}$. Hence $Q^{\prime}$ satisfies (P2). ∎

######  Lemma 3.17.

Assume that $Q$ satisfies (P1) and (P2). Then
$\operatorname{TN}^{\psi}_{G}(P^{2})=PF$.

###### Proof.

Since $PF\subset\operatorname{TN}^{\psi}_{G}(P^{2})$ is always true by the
proof of Lemma 3.13, we may prove the other inclusion.

Take $x\in\operatorname{TN}^{\psi}_{G}(P^{2})$. Then
$xP^{2}\psi(x)^{-1}=P^{2}$. Since $P^{2}$ is a normal subgroup of $G$ by (P1),
we have $P^{2}\psi^{-1}(x)=\psi^{-1}(x)P^{2}$. Hence, $x\psi(x)^{-1}\in
P^{2}$, i.e., there is $q\in P^{2}$ with $x\psi(x)^{-1}=q$. By (P2), there is
$p\in P$ such that $q=s_{p}(e)$, i.e., $x\psi(x^{-1})=s_{p}(e)=p\psi(p^{-1})$.
Hence, $p^{-1}x=\psi(p^{-1}x)$, i.e., $p^{-1}x\in F$. Therefore, $x\in PF$, as
desired. ∎

###  3.5. Proof of Theorem 1.4

Now, we are in the position to give a proof of Theorem 1.4. The whole part of
this subsection is devoted to giving its proof.

First, we prove the “Only if” part. Assume that
$Q\cong_{\mathrm{qu}}Q^{\prime}$. Then $|G|=|G^{\prime}|$ trivially follows
and
$|\operatorname{Fix}(\psi,G)|=|\operatorname{Fix}(\psi^{\prime},G^{\prime})|$
also follows by Remark 2.2. Let $f:Q\rightarrow Q^{\prime}$ be a quandle
automorphism with $f(e)=e^{\prime}$ and let $h=f|_{P}$. Then $h$ becomes a
group homomorphism satisfying
$h\circ\psi|_{P}=\psi^{\prime}|_{P^{\prime}}\circ h$ (see Theorem 3.10) and
(C-2) easily follows from $f(s_{a}(e))=f|_{P}(s_{a}(e))=h(s_{a}(e))$ and
$f(s_{a}(e))=s_{f(a)}^{\prime}(e^{\prime})$, where $f(a)\in G^{\prime}$.

In the remaining part, we give the proof of the “If” part. Let us assume the
conditions (A), (B) and (C). We define the set $S$ (resp. $S^{\prime}$) of
coset representatives of $P^{2}$ (resp. $P^{\prime 2}$) as follows:

$S=\\{s_{a}(e)P^{2}:a\in G\\}\text{ and
}S^{\prime}=\\{s_{a^{\prime}}^{\prime}(e^{\prime})P^{\prime 2}:a^{\prime}\in
G^{\prime}\\}.$

The first step: First, we compute $|S|$. For this, let

$\pi:G/P\rightarrow S,\;\;\pi(aP)=s_{a}(e)P^{2}.$

The well-definedness is verified as follows: for any $p\in P$ and $q\in
P^{2}$, we see that

$\displaystyle s_{ap}(e)q$
$\displaystyle=ap\psi(ap)^{-1}q=ap\psi(p^{-1})\psi(a)^{-1}q=a\underbrace{s_{p}(e)}_{\in
P^{2}}\psi(a)^{-1}q$ $\displaystyle=a\psi(a)^{-1}\tilde{q}q,\;\text{where
$\tilde{q}\in P^{2}$ (by \ref{G-normal})}$ $\displaystyle\in s_{a}(e)P^{2}.$

Since $\pi$ is surjective, for the computation of $|S|$, we may compute
$|\pi^{-1}(s_{a}(e)P^{2})|$. We see the following:

$\displaystyle
bP\in\pi^{-1}(s_{a}(e)P^{2})\;\Longleftrightarrow\;\pi(bP)=s_{a}(e)P^{2}\;\Longleftrightarrow\;s_{b}(e)P^{2}=s_{a}(e)P^{2}$
$\displaystyle\Longleftrightarrow\;\;$ $\displaystyle
b\psi(b)^{-1}P^{2}=a\psi(a)^{-1}P^{2}\;\Longleftrightarrow\;bP^{2}\psi(b)^{-1}=aP^{2}\psi(a)^{-1}\text{
(by \ref{G-normal})}$ $\displaystyle\Longleftrightarrow\;\;$ $\displaystyle
a^{-1}bP^{2}\psi(a^{-1}b)^{-1}=P^{2}\;\Longleftrightarrow\;a^{-1}b\in\operatorname{TN}^{\psi}_{G}(P^{2})=PF\text{
(by Lemma~{}\ref{lem:PF})}$ $\displaystyle\Longleftrightarrow\;\;$
$\displaystyle a^{-1}bP\subset PF\;\Longleftrightarrow\;bP\subset aPF.$

This implies that $|\pi^{-1}(s_{a}(e)P^{2})|=|PF/P|=|F/P\cap F|$. Hence,

$|S|=\frac{|G/P|}{|\pi^{-1}(s_{a}(e)P^{2})|}=\frac{|G/P|}{|F/P\cap
F|}=\frac{|G|\cdot|P\cap F|}{|P|\cdot|F|}.$

The second step: We construct a bijection between $S$ and $S^{\prime}$ using
$h$. Firstly, we define the map $\tilde{h}\colon P/P^{2}\to
P^{\prime}/P^{\prime 2}$ by setting $\tilde{h}(pP^{2}):=h(p)P^{\prime 2}$.
Since

$\displaystyle p_{1}P^{2}=p_{2}P^{2}$
$\displaystyle\;\Longleftrightarrow\;p_{2}^{-1}p_{1}\in
P^{2}\;\Longleftrightarrow\;h(p_{2}^{-1}p_{1})\in P^{\prime 2}\quad(\text{by
Lemma~{}\ref{lem:P2_1}})$
$\displaystyle\;\Longleftrightarrow\;h(p_{1})P^{\prime 2}=h(p_{2})P^{\prime
2},$

we see that $\tilde{h}$ is well-defined and injective. The surjectivity of
$\tilde{h}$ follows by the the surjectivity of $h|_{P^{2}}\colon P^{2}\to
P^{\prime 2}$. Hence, $\tilde{h}$ is a bijection between $P/P^{2}$ and
$P^{\prime}/P^{\prime 2}$. Next, we can show that the restriction
$\tilde{h}|_{S}$ of $\tilde{h}$ into $S$ gives a bijection between $S$ and
$S^{\prime}$.

  * •

We have $\tilde{h}(S)\subset S^{\prime}$ since
$\tilde{h}(s_{a}(e)P^{2})=h(s_{a}(e))P^{\prime
2}=s_{a^{\prime}}^{\prime}(e^{\prime})P^{\prime 2}$ by (C-2).

  * •

Since $\tilde{h}$ is injective, $\tilde{h}|_{S}$ is also injective.

  * •

Moreover, we have $|S|=|S^{\prime}|$. In fact, by our assumptions, we have
$|G|=|G^{\prime}|$, $|F|=|F^{\prime}|$ and $|P|=|P^{\prime}|$ and $|P\cap
F|=|P^{\prime}\cap F^{\prime}|$ by Lemma 3.12. Hence, by the first step, we
see that

$\displaystyle|S|=\frac{|G|\cdot|P\cap
F|}{|P|\cdot|F|}=\frac{|G^{\prime}|\cdot|P^{\prime}\cap
F^{\prime}|}{|P^{\prime}|\cdot|F^{\prime}|}=|S^{\prime}|.$

Therefore, $\tilde{h}$ gives a bijection between $S$ and $S^{\prime}$.

The third step: Let $A\subset G$ be a set of coset representatives of $G/P$.
We construct a map $k\colon A\to G^{\prime}$ such that $k(A)$ is a set of
coset representatives of $G^{\prime}/P^{\prime}$ and
$h(s_{a}(e))=s^{\prime}_{k(a)}(e^{\prime})$ holds for any $a\in A$.

Let $\pi^{\prime}:G^{\prime}/P^{\prime}\rightarrow S^{\prime}$ similarly to
$\pi$. Firstly, we construct a bijective map $k\colon G/P\to
G^{\prime}/P^{\prime}$ such that
$\tilde{h}|_{S}(\pi(aP))=\pi^{\prime}(k(aP^{\prime}))$ holds for any $aP\in
G/P$.

$\textstyle{G/P\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\simeq}$$\scriptstyle{k}$$\scriptstyle{\pi}$$\textstyle{G^{\prime}/P^{\prime}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\pi^{\prime}}$$\textstyle{S\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\simeq}$$\scriptstyle{\tilde{h}|_{S}}$$\textstyle{S^{\prime}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\circlearrowright}$

Since $\displaystyle|\pi^{-1}(\pi(aP))|=\frac{|F|}{|P\cap
F|}=\frac{|F^{\prime}|}{|P^{\prime}\cap
F^{\prime}|}=|\pi^{\prime-1}(\tilde{h}(\pi(aP)))|$, we can construct a
bijection $k_{a}$ between $\pi^{-1}(\pi(aP))$ and
$\pi^{\prime-1}(\tilde{h}(\pi(aP)))$ for each fixed $aP\in G/P$. Thus, we can
construct a bijection $k$ between $G/P$ and $G^{\prime}/P^{\prime}$ by
attaching $k_{a}$’s for $aP\in G/P$.

Next, we apply $k$ to $A$ using the same symbol. Then for any $a\in A$, we
have $h(s_{a}(e))P^{\prime 2}=s^{\prime}_{k(a)}(e^{\prime})P^{\prime 2}$,
i.e., $h(s_{a}(e))=s_{k(a)}^{\prime}(e^{\prime})q^{\prime}$ for some
$q^{\prime}\in P^{\prime 2}$. Here, we see that

$\displaystyle s^{\prime}_{k(a)}(e^{\prime})q^{\prime}$
$\displaystyle=k(a)\psi^{\prime}(k(a))^{-1}q^{\prime}=k(a)\tilde{q}^{\prime}\psi^{\prime}(k(a))^{-1}\text{
(for some }\tilde{q}^{\prime}\text{ since $Q^{\prime}$ satisfies
\ref{G-normal})}$
$\displaystyle=k(a)s_{p^{\prime}}^{\prime}(e^{\prime})\psi^{\prime}(k(a))^{-1}\text{
(since $Q^{\prime}$ satisfies \ref{iikanji})}$
$\displaystyle=k(a)p^{\prime}\psi^{\prime}(p^{\prime})^{-1}\psi^{\prime}(k(a))^{-1}=k(a)p^{\prime}\psi^{\prime}((k(a)p^{\prime})^{-1})$
$\displaystyle=s_{k(a)p^{\prime}}^{\prime}(e^{\prime}).$

Note that $k(a)$ and $k(a)p^{\prime}$ belong to the same coset of
$G^{\prime}/P^{\prime}$. By replacing $k(a)$ into $k(a)p^{\prime}$, we can
construct the desired map $k:A\rightarrow G^{\prime}$.

The fourth step: Now, we can construct a quandle automorphism between $Q$ and
$Q^{\prime}$ by using the bijections $h:P\rightarrow P^{\prime}$ (guaranteed
by the condition (C)) and $k:A\rightarrow k(A)$ constructed in the third step
as follows. For any $x\in G$, we uniquely write $x$ for $x=a_{x}p_{x}$, where
$a_{x}\in A$ and $p_{x}\in P$. By $a_{x}p_{x}a_{x}^{-1}\in P$ since $P$ is a
normal subgroup of $G$, we can define $f(x)$ by setting
$h(a_{x}p_{x}a_{x}^{-1})k(a_{x})$. The bijectivity of $f$ is trivial since $h$
and $k$ are bijective. Since

$a_{x}\psi(p_{x}^{-1}a_{x}^{-1})=a_{x}\psi(a_{x}^{-1})\psi(a_{x}p_{x}^{-1}a_{x}^{-1})=s_{a_{x}}(e)\cdot\psi(a_{x}p_{x}^{-1}a_{x}^{-1})\in
P$

and the properties of $h$ and $k$, we have

(3) $\displaystyle h(a_{x}\psi(p_{x}^{-1}a_{x}^{-1}))$
$\displaystyle=h(s_{a_{x}}(e)\psi(a_{x}p_{x}^{-1}a_{x}^{-1}))=h(s_{a_{x}}(e))h(\psi(a_{x}p_{x}^{-1}a_{x}^{-1}))$
$\displaystyle=s^{\prime}_{k(a_{x})}(e^{\prime})\psi^{\prime}(h(a_{x}p_{x}^{-1}a_{x}^{-1}))=k(a_{x})\psi^{\prime}(k(a_{x})^{-1})\psi^{\prime}(h(a_{x}p_{x}^{-1}a_{x}^{-1}))$
$\displaystyle=k(a_{x})\psi^{\prime}(f(x)^{-1}).$

Using (3), we can check that $f$ is a quandle homomorphism as follows: For
elements $x=a_{x}p_{x}$ and $y=a_{y}p_{y}$ in $G$,

$\displaystyle f\circ s_{x}(y)$
$\displaystyle=f(s_{a_{x}p_{x}}(a_{y}p_{y}))=f(a_{y}\cdot
s_{a_{y}^{-1}a_{x}p_{x}}(p_{y}))$ $\displaystyle=h(a_{y}\cdot
s_{a_{y}^{-1}a_{x}p_{x}}(p_{y})\cdot a_{y}^{-1})k(a_{y})\quad(\text{since
$s_{a_{y}^{-1}a_{x}p_{x}}(p_{y})\in P$})$ $\displaystyle=h(a_{y}\cdot
a_{y}^{-1}a_{x}p_{x}\psi(p_{x}^{-1}a_{x}^{-1}a_{y}p_{y})\cdot
a_{y}^{-1})k(a_{y})$ $\displaystyle=h(a_{x}p_{x}a_{x}^{-1}\cdot
a_{x}\psi(p_{x}^{-1}a_{x}^{-1})\cdot\psi(a_{y}p_{y})a_{y}^{-1})k(a_{y})$
$\displaystyle=h(a_{x}p_{x}a_{x}^{-1})\cdot
h(a_{x}\psi(p_{x}^{-1}a_{x}^{-1}))\cdot
h(a_{y}\psi(p_{y}^{-1}a_{y}^{-1}))^{-1}k(a_{y})$
$\displaystyle=h(a_{x}p_{x}a_{x}^{-1})\cdot
k(a_{x})\psi^{\prime}(f(x)^{-1})\cdot\psi^{\prime}(f(y))k(a_{y})^{-1}\cdot
k(a_{y})\quad(\text{by \eqref{eq:kantan}})$
$\displaystyle=f(x)\psi^{\prime}(f(x)^{-1}f(y))$
$\displaystyle=s^{\prime}_{f(x)}(f(y)),$

as desired.

##  4\. Application of Theorem 1.4 to dihedral groups

In this section, we discuss Problem 1.1 for dihedral groups.

For this, we first collect some fundamental materials on dihedral groups and
their automorphisms. Let $D_{n}$ be the dihedral group of order $2n$. We use
the notation $\sigma$ and $\tau$ as follows:

$D_{n}=\\{\sigma^{i}:i=0,1,\ldots,n-1\\}\cup\\{\tau\sigma^{i}:i=0,1,\ldots,n-1\\}$

equipped with the relations $\tau^{2}=\sigma^{n}=e$ and
$\sigma\tau=\tau\sigma^{-1}$.

We recall the structure of $\operatorname{Aut}_{\mathrm{gr}}(D_{n})$. Let

$\mathrm{Aff}(C_{n})=\\{\varphi_{a,b}:a\in C_{n}^{\times},b\in C_{n}\\},$

where for $\tau^{\epsilon}\sigma^{i}\in D_{n}$ with $\epsilon\in\\{0,1\\}$ and
$0\leq i\leq n-1$, we define a group homomorphism
$\varphi_{a,b}:D_{n}\rightarrow D_{n}$ by setting

$\varphi_{a,b}(\tau^{\epsilon}\sigma^{i})=\tau^{\epsilon}\sigma^{ai+\epsilon
b}.$

Then it is well known that

$\operatorname{Aut}_{\mathrm{gr}}(D_{n})\cong_{\mathrm{gr}}\mathrm{Aff}(C_{n}).$

Note that $\varphi_{c,d}^{-1}=\varphi_{c^{-1},-c^{-1}d}$ and

(4)
$\displaystyle\varphi_{c,d}\circ\varphi_{a,b}\circ\varphi_{c,d}^{-1}=\varphi_{a,bc+(1-a)d}$

hold.

In what follows, for $a\in C_{n}^{\times}$ and $b\in C_{n}$, we use the
notation $Q(D_{n},a,b)$ instead of $Q(D_{n},\varphi_{a,b})$.

###  4.1. Conjugacy classes for $\operatorname{Aut}_{\mathrm{gr}}(D_{n})$

We discuss the conjugacy classes of $\operatorname{Aut}_{\mathrm{gr}}(D_{n})$.

######  Lemma 4.1.

Let $c,d$ are integers and $m,n$ are positive integers.

  * (i)

A linear congruence $cz\equiv d\pmod{n}$ with the variable $z$ has a solution
if and only if $d$ is a multiple of $\gcd(n,c)$. Moreover if this linear
congruence has a solution $z_{0}$, then the solutions $z$ are of the form
$z\equiv z_{0}+\frac{n}{\gcd(n,c)}i$ $(i=0,1,\ldots,\gcd(n,c)-1)$.

  * (ii)

There exists $p$ with $\gcd(n,p)=1$ such that $pc\equiv\gcd(m,c)\pmod{m}$.

  * (iii)

We have

$\\{cb:b\in C_{n}\\}=\\{gy:y\in C_{n}\\},$

where $g=\gcd(n,c)$.

###### Proof.

(i) This is well known (see, e.g., [10, Theorem 2.2]).

(ii) Let $g=\gcd(m,c)$. Then there are $x,y\in{\mathbb{Z}}$ such that
$mx+cy=g$. Hence, $cy\equiv g\pmod{m}$. Let $m^{\prime}=m/g$ and
$c^{\prime}=c/g$. Then $m^{\prime}x+c^{\prime}y=1$. Note that
$\gcd(m^{\prime},y)=1$. Thus, by Dirichlet’s theorem on arithmetic
progressions, there are infinitely many primes $p$ which is of the form
$p=m^{\prime}k+y$ ($k\in{\mathbb{Z}}$). For each $p$, we have

$pc=(m^{\prime}k+y)c^{\prime}g=mka^{\prime}+cy\equiv g\pmod{m}.$

Since there are infinitely many such primes $p$, we can choose a prime $p$
with $\gcd(n,p)=1$, as required.

(iii) One inclusion is trivial since $g$ divides $c$. Thus it is enough to
show that $g$ is contained in $\\{cb:b\in C_{n}\\}$. By (i), there is $b\in
C_{n}$ such that $cb\equiv g\pmod{n}$. Hence, one has $g\in\\{cb:b\in
C_{n}\\}$, as required. ∎

######  Proposition 4.2.

Let
$\varphi_{a,b},\varphi_{a^{\prime},b^{\prime}}\in\operatorname{Aut}_{\mathrm{gr}}(D_{n})$
and put $d=\gcd(n,1-a,b)$ and $d^{\prime}=\gcd(n,1-a^{\prime},b^{\prime})$.
Then $\varphi_{a,b}$ and $\varphi_{a^{\prime},b^{\prime}}$ are conjugate if
and only if the following two conditions are satisfied:

  * •

$a\equiv a^{\prime}\pmod{n}$;

  * •

$d=d^{\prime}$,

i.e., the set of representatives of conjugacy classes of
$\operatorname{Aut}_{\mathrm{gr}}(D_{n})$ is given by

(5) $\begin{split}\\{\varphi_{a,d}:a\in C_{n}^{\times},\;d\text{ is a divisor
of }g\\},\end{split}$

where $g=\gcd(n,1-a)$.

###### Proof.

Let $g=\gcd(n,1-a)$ and $g^{\prime}=\gcd(n,1-a^{\prime})$. Note that
$d=\gcd(g,b)$ since
$\gcd(\alpha,\beta,\gamma)=\gcd(\gcd(\alpha,\beta),\gamma)$ holds for each
$\alpha,\beta,\gamma\in{\mathbb{Z}}$.

Assume that $\varphi_{a,b}$ and $\varphi_{a^{\prime},b^{\prime}}$ are
conjugate. By (4), there exist $r\in C_{n}^{\times}$ and $s\in C_{n}$ such
that $\varphi_{a^{\prime},b^{\prime}}=\varphi_{a,br+(1-a)s}$. Then $a\equiv
a^{\prime}\pmod{n}$ trivially holds. Moreover we have

$d^{\prime}=\gcd(g^{\prime},b^{\prime})=\gcd(g,br+(1-a)s)=\gcd(g,br)=\gcd(g,b)=d$

since $(1-a)s$ is a multiple of $g$ and $r\in C_{n}^{\times}$.

Assume that $a\equiv a^{\prime}\pmod{n}$ and $d=d^{\prime}$ hold. In order to
show that $\varphi_{a,b}$ and $\varphi_{a^{\prime},b^{\prime}}$ are conjugate,
we find $r\in C_{n}^{\times}$ and $s\in C_{n}$ such that $b^{\prime}\equiv
br+(1-a)s\pmod{n}$. By Lemma 4.1 (ii), there exist $p,q\in C_{n}^{\times}$
such that $pb\equiv d,\ qb^{\prime}\equiv d\pmod{g}$. Hence there exist
$k,k^{\prime}\in{\mathbb{Z}}$ such that $d\equiv pb+kg\pmod{n}$ and
$b^{\prime}\equiv q^{-1}d+k^{\prime}g\pmod{n}$. On the other hand, by Lemma
4.1 (i), there exists $z_{0}\in C_{n}$ such that $(1-a)z_{0}\equiv g\pmod{n}$.
Therefore we have

$b^{\prime}\equiv q^{-1}d+k^{\prime}g\equiv q^{-1}(pb+kg)+k^{\prime}g\equiv
q^{-1}pb+(q^{-1}k+k^{\prime})z_{0}(1-a)\pmod{n}.$

Then $q^{-1}p\in C_{n}^{\times}$ and $(q^{-1}k+k^{\prime})z_{0}\in C_{n}$ are
the desired $r$ and $s$, respectively. ∎

Note that $\varphi_{a,0}$ is conjugate to $\varphi_{a,g}$ in
$\operatorname{Aut}_{\mathrm{gr}}(D_{n})$, where $a\in C_{n}^{\times}$ and
$g=\gcd(n,1-a)$. In what follows, we will use $\varphi_{a,g}$ as a
representative of conjugacy classes of
$\operatorname{Aut}_{\mathrm{gr}}(D_{n})$ instead of $\varphi_{a,0}$, but we
will use $\varphi_{a,0}$ for the computation.

######  Example 4.3.

A set of representatives of conjugacy classes of each of
$\operatorname{Aut}_{\mathrm{gr}}(D_{4})$,
$\operatorname{Aut}_{\mathrm{gr}}(D_{5})$,
$\operatorname{Aut}_{\mathrm{gr}}(D_{6})$ and
$\operatorname{Aut}_{\mathrm{gr}}(D_{8})$ is given as follows:

$\displaystyle\operatorname{Aut}_{\mathrm{gr}}(D_{4}):\;\;$
$\displaystyle\\{\varphi_{1,0},\varphi_{1,1},\varphi_{1,2},\varphi_{3,1},\varphi_{3,2}\\}$
$\displaystyle\operatorname{Aut}_{\mathrm{gr}}(D_{5}):\;\;$
$\displaystyle\\{\varphi_{1,0},\varphi_{1,1},\varphi_{2,1},\varphi_{3,1},\varphi_{4,1}\\}$
$\displaystyle\operatorname{Aut}_{\mathrm{gr}}(D_{6}):\;\;$
$\displaystyle\\{\varphi_{1,i}:i=0,1,2,3\\}\cup\\{\varphi_{5,i}:i=1,2\\}$
$\displaystyle\operatorname{Aut}_{\mathrm{gr}}(D_{8}):\;\;$
$\displaystyle\\{\varphi_{1,i}:i=0,1,2,4\\}\cup\\{\varphi_{3,i},\varphi_{7,i}:i=1,2\\}\cup\\{\varphi_{5,i}:i=1,2,4\\}$

###  4.2. Invariants for $Q(D_{n},a,b)$

######  Proposition 4.4.

Let $a\in C_{n}^{\times}$, $b\in C_{n}$, $g=\gcd(n,1-a)$ and $d=\gcd(g,b)$.
Then

$\displaystyle\operatorname{Fix}(\varphi_{a,b},D_{n})=\begin{cases}\langle\sigma^{n/g},\;\tau\sigma^{z_{0}}\rangle\
\text{for some $z_{0}$},\quad&\text{if }d=g,\\\
\langle\sigma^{n/g}\rangle,\quad&\text{if }d\neq g.\end{cases}$

In particular,
$|\operatorname{Fix}(\varphi_{a,b},D_{n})|=\begin{cases}2g,&\text{if }d=g,\\\
g,&\text{if }d\neq g.\end{cases}$

###### Proof.

By $\varphi_{a,b}(\sigma^{i})=\sigma^{ai}$, we see that
$\varphi_{a,b}(\sigma^{i})=\sigma^{i}$ if and only if $(1-a)i\equiv 0$ (mod
$n$). By Lemma 4.1 (i), $i$ is a multiple of $n/g$.

Moreover, by $\varphi_{a,b}(\tau\sigma^{i})=\tau\sigma^{ai+b}$, we also see
that $\varphi_{a,b}(\tau\sigma^{i})=\tau\sigma^{i}$ if and only if
$(1-a)i\equiv b\pmod{n}$. By Lemma 4.1 (i), this has a solution if and only if
$b$ is a multiple of $g$, i.e., $d=g$. Moreover, when $d=g$, the solutions are
$i\equiv z_{0}+\frac{n}{g}j\pmod{n}$ ($j=0,1,\ldots g-1$). ∎

For $x,y\in Q(D_{n},a,b)$, let $x=\tau^{\epsilon}\sigma^{i}$ and
$y=\tau^{\delta}\sigma^{j}$. Then we see the following:

$s_{x}(y)=\tau^{\delta}\sigma^{(-1)^{\epsilon+\delta}(1-a)i+aj+\mu(\epsilon+\delta)b},$

where

(6) $\mu(i)=\begin{cases}1&\text{if $i$ is odd},\\\ 0&\text{if $i$ is even}\\\
\end{cases}$

for $i\in C_{2}$. In particular, we have

(7) $s_{x}(e)=\sigma^{(-1)^{\epsilon}(1-a)i+\epsilon b}.$

######  Proposition 4.5.

Let $a\in C_{n}^{\times}$, $b\in C_{n}$, $g=\gcd(n,1-a)$, $d=\gcd(g,b)$ and
$g_{2}=\gcd(n/d,1-a)$. Then

$P(Q(D_{n},a,b))=\langle\sigma^{d}\rangle\;\;\text{and}\;\;P^{2}(Q(D_{n},a,b))=\langle\sigma^{dg_{2}}\rangle.$

###### Proof.

It directly follows from (7) and Lemma 4.1 (iii) that
$P(Q(D_{n},a,b))=\langle\sigma^{d}\rangle$.

For $p\in P(Q(D_{n},a,b))$, let $p=\sigma^{dj}$ for some $j$. Then
$s_{p}(e)=\sigma^{d(1-a)j}$. Since $j$ runs over $j=0,1,\ldots,n/d-1$, we
obtain that $P^{2}(Q(D_{n},a,b))=\langle\sigma^{dg_{2}}\rangle$. ∎

###  4.3. Proof of Corollary 1.6

Now, we give a proof of Corollary 1.6.

The first step: First, we check that $Q(D_{n},a,b)$ satisfies (P1) and (P2).

For any $\tau^{\epsilon}\sigma^{i}\in D_{n}$ and any $\sigma^{dg_{2}j}\in
P^{2}$, we see the following:

$(\tau^{\epsilon}\sigma^{i})\cdot\sigma^{dg_{2}j}\cdot(\tau^{\epsilon}\sigma^{i})^{-1}=\tau^{\epsilon}\sigma^{i+dg_{2}j-i}\tau^{\epsilon}=\sigma^{(-1)^{\epsilon}dg_{2}j}\in
P^{2},$

which implies that $Q(D_{n},a,b)$ satisfies (P1).

Fix $\sigma^{dg_{2}j}\in P^{2}$. Since $\gcd(n,(1-a)d)=dg_{2}$, by Lemma 4.1
(i), there exists $z_{0}\in C_{n}$ such that $(1-a)dz_{0}\equiv
dg_{2}j\pmod{n}$. Hence, for any $\sigma^{dg_{2}j}\in P^{2}$, by letting
$p=\sigma^{dz_{0}}\in P$, we see that

$s_{p}(e)=\sigma^{(1-a)dz_{0}}=\sigma^{dg_{2}j}.$

Thus, $Q(D_{n},a,b)$ satisfies (P2).

Therefore, we can apply Theorem 1.4 for $G=G^{\prime}=D_{n}$.

The second step: Next, we prove the “Only if” part of the statement. Let
$Q=Q(D_{n},a,b)$ and $Q^{\prime}=Q(D_{n},a^{\prime},b^{\prime})$ and assume
that $Q\cong_{\mathrm{qu}}Q^{\prime}$.

  * •

Then
$|\operatorname{Fix}(\varphi_{a,b},D_{n})|=|\operatorname{Fix}(\varphi_{a^{\prime},b^{\prime}},D_{n})|$
holds.

  * •

By Proposition 4.5, we see that $P=P(Q)\cong_{\mathrm{gr}}C_{n/d}$ and
$P^{\prime}=P(Q^{\prime})\cong_{\mathrm{gr}}C_{n/d^{\prime}}$. Thus, we obtain
that $d=d^{\prime}$.

  * •

Let $h:C_{n/d}\rightarrow C_{n/d}$ be a group isomorphism satisfying the
condition (C) in Theorem 1.4. Note that $\varphi_{a,b}|_{P}$,
$\varphi_{a^{\prime},b^{\prime}}|_{P^{\prime}}$ and $h$ are multiplication
maps by some $a,a^{\prime},c\in C_{n/d}^{\times}$, respectively. By (C-1), we
have $ca=a^{\prime}c$ in $C_{n/d}$. Since $c\in C_{n/d}^{\times}$, we have
$a=a^{\prime}$ in $C_{n/d}$.

The third step: Finally, we prove the “If” part of the statement. Let
$a,a^{\prime}\in C_{n}^{\times}$ and $b,b^{\prime}\in C_{n}$ which satisfy all
the conditions in Corollary 1.6. Then the conditions (A) and (B) in Theorem
1.4 are clearly satisfied. Let us consider the identity map
$h=\operatorname{id}_{C_{n/d}}$. Then $h:P\rightarrow P^{\prime}$ is a group
isomorphism. In what follows, we verify that $h$ satisfies (C-1) and (C-2).

  * (C-1)

This is equivalent to the condition $a\equiv a^{\prime}\pmod{n/d}$ as
discussed in the second step.

  * (C-2)

We divide the discussions into two cases:

    * –

“$d=g$ and $d^{\prime}=g^{\prime}$” or “$d\neq g$ and $d^{\prime}\neq
g^{\prime}$”;

    * –

$d\neq g$ and $d^{\prime}=g^{\prime}$ (the case $d=g$ and $d^{\prime}\neq
g^{\prime}$ is similar).

For the first case, it follows from the assumption
$|\operatorname{Fix}(\varphi_{a,b},D_{n})|=|\operatorname{Fix}(\varphi_{a^{\prime},b^{\prime}},D_{n})|$
together with Proposition 4.4 that $g=g^{\prime}$ holds. By Lemma 4.1 (i),
there exists $z_{0}\in C_{n}$ such that $(1-a^{\prime})z_{0}\equiv
1-a\pmod{n}$. Then, for any $x=\tau^{\epsilon}\sigma^{i}$, by letting
$x^{\prime}=\tau^{\epsilon}\sigma^{z_{0}i}$, we see that

$\displaystyle h(s_{x}(e))$
$\displaystyle=s_{x}(e)=\sigma^{(-1)^{\epsilon}(1-a)i+\epsilon
b}=\sigma^{(-1)^{\epsilon}(1-a^{\prime})z_{0}i+\epsilon
b}=s_{x^{\prime}}^{\prime}(e).$

See (7).

For the second case, we have $g=2g^{\prime}$ (see Proposition 4.4). By Lemma
4.1 (i), there exists $z_{0}\in C_{n}$ such that $(1-a^{\prime})z_{0}\equiv
1-a\pmod{n}$. Similar to the above, we obtain that
$h(s_{x}(e))=s_{x^{\prime}}^{\prime}(e)$.

##  5\. Relationships among generalized Alexander quandles arising from
$C_{2n}$ and $D_{n}$

In this section, we first discuss Problem 1.1 for finite abelian groups. We
remark that a solution of Problem 1.1 for finite abelian groups was already
given in [11, Theorem 2.1], but this paper gives another interpretation of
[11, Theorem 2.1] from viewpoints of group theory. After those discussions, we
investigate a relationship between $Q(D_{n},a,b)$ and $Q(C_{2n},a^{\prime})$,
where $a,a^{\prime}\in C_{n}^{\times}$ and $b\in C_{n}$.

###  5.1. Proof of Corollary 1.5

###### Proof of Corollary 1.5.

We apply Theorem 1.4 to finite abelian groups. What we may prove is the
following:

  * (i)

$P=\operatorname{Im}\rho$;

  * (ii)

$Q(G,\psi)$ always satisfies (P1) and (P2) when $G$ is abelian;

  * (iii)

the condition (C-2) is automatically satisfied when $G$ is abelian;

  * (iv)

the condition (B) follows from (A) and (C).

(i) Since $G$ is abelian, we see that

$P=\\{x+\psi(-x):x\in G\\}=\\{x-\psi(x):x\in
G\\}=\operatorname{Im}(\mathrm{id}_{G}-\psi)=\operatorname{Im}\rho.$

(ii) (P1) is trivial since $P^{2}$ is a subgroup of an abelian group $G$, and
thus, $P^{2}$ is also abelian. Moreover, since
$P^{2}=\operatorname{Im}\rho^{2}$ holds, we see that $Q(G,\psi)$ satisfies
(P2).

(iii) Since $h(s_{a}(e))\in\operatorname{Im}\rho^{\prime}=P^{\prime}$ for each
$a\in G$, we see that there is $a^{\prime}\in G^{\prime}$ with
$h(s_{a}(e))=\rho^{\prime}(a^{\prime})=s_{a^{\prime}}(e^{\prime})$.

(iv) We know that $\operatorname{Fix}(\psi,G)=\\{x\in
G:\psi(x)=x\\}=\operatorname{Ker}\rho$. Hence,
$\operatorname{Im}\rho\cong_{\mathrm{gr}}G/\operatorname{Ker}\rho$. Since
$|G|=|G^{\prime}|$ by (A) and $|P|=|P^{\prime}|$ by (C), we conclude that

$|\operatorname{Fix}(\psi,G)|=\frac{|G|}{|\operatorname{Im}\rho|}=\frac{|G|}{|P|}=\frac{|G^{\prime}|}{|P^{\prime}|}=|\operatorname{Fix}(\psi^{\prime},G^{\prime})|,$

as required. ∎

######  Remark 5.1.

We see that [11, Theorem 2.1] can be rephrased as Corollary 1.5, which is
explained as follows.

Let us recall the statement of [11, Theorem 2.1]: two finite Alexander
quandles $M$ and $N$ of the same cardinality are isomorphic as quandles if and
only if there is an isomorphism of $\Lambda$-modules
$h:(1-t)M\rightarrow(1-t)N$. (Please refer [11, Section 1] for undefined
notions.) We see that $(1-t)M$ (resp. $(1-t)N$) is isomorphic to
$\operatorname{Im}\rho$ (resp. $\operatorname{Im}\rho^{\prime}$) as abelian
groups, and we can interpret the existence of an isomorphism
$h:(1-t)M\rightarrow(1-t)N$ of $\Lambda$-modules as the existence of a group
isomorphism $h:\operatorname{Im}\rho\rightarrow\operatorname{Im}\rho^{\prime}$
satisfying
$h\circ\psi|_{\operatorname{Im}\rho}=\psi^{\prime}|_{\operatorname{Im}\rho^{\prime}}\circ
h$.

We also give a proof of [11, Corollary 2.2] by using Corollary 1.5.

######  Corollary 5.2 ([11, Corollary 2.2]).

Let $a,a^{\prime}\in C_{n}^{\times}$ and $g=\gcd(n,1-a)$ and
$g^{\prime}=\gcd(n,1-a^{\prime})$. Let $Q=Q(C_{n},a)$ and let
$Q^{\prime}=Q(C_{n},a^{\prime})$. Then $Q\cong_{\mathrm{qu}}Q^{\prime}$ if and
only if $g=g^{\prime}$ and $a\equiv a^{\prime}\pmod{n/g}$.

###### Proof.

Note that
$\operatorname{Aut}_{\mathrm{gr}}(C_{n})\cong_{\mathrm{gr}}C_{n}^{\times}$ as
groups and $\operatorname{Im}\rho$ in the statement of Corollary 1.5 is equal
to $\\{gy:y\in C_{n}\\}=\langle g\rangle\cong_{\mathrm{gr}}C_{n/g}$ by Lemma
4.1 (iii).

  * •

The condition on the existence of a group isomorphism
$h:\operatorname{Im}\rho\rightarrow\operatorname{Im}\rho^{\prime}$ is
equivalent to $n/g=n/g^{\prime}$, i.e., $g=g^{\prime}$. Then $h$ corresponds
to some element of $C_{n/g}^{\times}$.

  * •

The condition
$h\circ\psi|_{\operatorname{Im}\rho}=\psi^{\prime}|_{\operatorname{Im}\rho^{\prime}}\circ
h$ is equivalent to that $ha\equiv a^{\prime}h\pmod{n/g}$, i.e., $a\equiv
a^{\prime}\pmod{n/g}$.

∎

###  5.2. Proof of Corollary 1.7

###### Proof of Corollary 1.7.

Let $Q=Q(C_{2n},a)$ and let $Q^{\prime}=Q(D_{n},\tilde{a},g)$. We have already
discussed that $Q$ and $Q^{\prime}$ satisfy (P1) and (P2) in Proof of
Corollary 1.5 and Section 4, respectively. Thus, we can apply Theorem 1.4 and
it is enough to show that $Q$ and $Q^{\prime}$ satisfy the conditions (A),(B)
and (C).

On (A), it is trivial since $|C_{2n}|=|D_{n}|=2n$. On (B), we have
$|\operatorname{Fix}(a,C_{2n})|=|\\{x\in
C_{2n}:(1-a)x=0\\}|=\gcd(2n,1-a)=2\gcd(n,k)=2g$. On the other hand, we see
that $d^{\prime}=\gcd(n,1-\tilde{a},g)=\gcd(n,2k,g)=g$ and
$g^{\prime}=\gcd(n,1-\tilde{a})=\gcd(n,-2k)$ is equal to $g$ or $2g$. If
$g^{\prime}=g$, then $d^{\prime}$ coincides with $g^{\prime}$. Hence
$|\operatorname{Fix}(\varphi_{\tilde{a},g},D_{n})|=2g^{\prime}=2g$. If
$g^{\prime}=2g$, then $d^{\prime}$ is a proper divisor of $g^{\prime}$. Hence
$|\operatorname{Fix}(\varphi_{\tilde{a},g},D_{n})|=g^{\prime}=2g$. Thus, in
all cases, we obtain $|\operatorname{Fix}(\varphi_{\tilde{a},g},D_{n})|=2g$.
This implies
$|\operatorname{Fix}(a,C_{2n})|=|\operatorname{Fix}(\varphi_{\tilde{a,g}},D_{n})|$.

Our remaining task is to check that (C) is satisfied. Before it, we have to
compute $P=P(Q)$ and $P^{\prime}=P(Q^{\prime})$.

  * •

We see that $P=P(Q)=\langle 2g\rangle<C_{2n}$.

  * •

By Proposition 4.5, we see that $P^{\prime}=\langle\sigma^{g}\rangle$.

Note that $P\cong_{\mathrm{gr}}P^{\prime}\cong_{\mathrm{gr}}C_{n/g}$.

We define $h:\langle 2g\rangle\rightarrow\langle\sigma^{g}\rangle$ as follows:

$h(2gj):=\sigma^{gj}\text{ for }j=0,1,\ldots,n/g-1.$

It is trivial that $h$ is a group isomorphism. On (C-1), we see that

$\displaystyle h\circ a(2gj)=h(2agj)=\sigma^{agj}\;\text{ and
}\;\varphi_{\tilde{a},g}\circ
h(2gj)=\varphi_{\tilde{a},g}(\sigma^{gj})=\sigma^{\tilde{a}gj}=\sigma^{agj}.$

On (C-2), for any $i\in C_{2n}$, we see that

$h(s_{i}(e))=h((1-a)i)=h\left(2g\frac{1-a}{2g}i\right)=\sigma^{\frac{1-a}{2}i}.$

On the other hand, for $x^{\prime}=\tau^{\epsilon}\sigma^{j}\in D_{n}$, we see
from (7) that

$s_{x^{\prime}}^{\prime}(e)=\sigma^{(-1)^{\epsilon}(1-\tilde{a})j+\epsilon
g}=\sigma^{(-1)^{\epsilon}(1-a)j+\epsilon g}.$

  * •

When $g^{\prime}=g$, the linear congruence
$(1-a)j\equiv\frac{(1-a)i}{2}\pmod{n}$ has a solution $j=z_{0}$ by Lemma 4.1
(i). By letting $\epsilon=0$ and $j=z_{0}$, we see that
$s_{x^{\prime}}^{\prime}(e)=\sigma^{\frac{(1-a)i}{2}}=h(s_{i}(e))$.

  * •

When $g^{\prime}=2g$, we see that $n$ must be even. Then we can define the
function $\mu$ like (6). Put $\ell=\frac{ki}{g}\in C_{n}$. Then
$\ell+\mu(\ell)$ must be even. In this case, the linear congruence
$(-1)^{\mu(\ell)}(1-a)j\equiv-(\ell+\mu(\ell))g\pmod{n}$ has a solution
$j=z_{1}$ by Lemma 4.1 (i). By letting $\epsilon=\mu(\ell)$ and $j=z_{1}$, we
see that

$s_{x^{\prime}}^{\prime}(e)=\sigma^{(-1)^{\mu(\ell)}(1-a)z_{1}+\mu(\ell)g}=\sigma^{-(\ell+\mu(\ell))g+\mu(\ell)g}=\sigma^{-\ell
g}=\sigma^{-ki}=h(s_{i}(e)).$

∎

##  6\. Generalized Alexander quandles arising from finite groups of small
orders

This section is devoted to solving Problem 1.1 in the case where the order of
$G$ is small by using Theorem 1.4 together with Corollaries 1.5, 1.6 and 1.7.
More concretely, we determine ${\mathcal{Q}}_{\operatorname{GAQ}}(n)$ for
small integers $n$. For this purpose, we investigate the generalized Alexander
quandles arising from the groups with its orders $p$ or $2p$ or $p^{2}$, where
$p$ is a prime.

###  6.1. The case $|G|=p$

Let $G$ be a group with $|G|=p$, where $p$ is a prime. Then it is very well
known that $G\cong_{\mathrm{gr}}C_{p}$. Moreover, since $C_{p}$ is simple, it
follows from Theorem 1.2 that ${\mathcal{Q}}_{\operatorname{GAQ}}(p)$
bijectively corresponds to $\operatorname{Aut}_{\mathrm{gr}}(C_{p})$, which is
isomorphic to $C_{p}^{\times}$ as groups. In particular,
$|{\mathcal{Q}}_{\operatorname{GAQ}}(p)|=p-1$.

###  6.2. The case $|G|=2p$

Let $G$ be a group with $|G|=2p$, where $p$ is an odd prime. (Note that the
case $p=2$ will be discussed in Subsection 6.3.) Then it is well known that
$G\cong_{\mathrm{gr}}C_{2p}$ or $D_{p}$ (see, e.g., [13, Theorem 4.19]).
Thanks to Corollary 1.7, it suffices to discuss only $Q(D_{p},\psi)$, where
$\psi\in\operatorname{Aut}_{\mathrm{gr}}(D_{p})$. Hence,
${\mathcal{Q}}_{\operatorname{GAQ}}(2p)={\mathcal{Q}}(D_{p})$. On the other
hand, we know by Proposition 4.2 that the conjugacy classes for
$\operatorname{Aut}_{\mathrm{gr}}(D_{p})$ bijectively correspond to
$\\{\varphi_{a,d}:a\in C_{p}^{\times},\text{$d$ is a divisor of $g$}\\}$,
where $g=\gcd(p,1-a)$. Note that $C_{p}^{\times}=\\{1,2,\ldots,p-1\\}$.

  * •

When $a=1$, we have $g=p$, so the possible $d$’s are $1$ and $p=g$.

  * •

When $a\neq 1$, we have $g=1$, so the possible $d$ is only $1=g$.

Therefore, $\operatorname{Aut}_{\mathrm{gr}}(D_{p})$ bijectively corresponds
to $\\{(a,0):a\in C_{p}^{\times}\\}\cup\\{(1,1)\\}\subset C_{p}^{\times}\times
C_{p}$. Those give all different generalized Alexander quandles by (the third
condition of) Corollary 1.6. In particular,
$|{\mathcal{Q}}_{\operatorname{GAQ}}(2p)|=p$.

###  6.3. The case $|G|=p^{2}$

Let $G$ be a group with $|G|=p^{2}$, where $p$ is a prime. Then it is well
known that $G\cong_{\mathrm{gr}}C_{p^{2}}$ or $G\cong_{\mathrm{gr}}C_{p}\times
C_{p}$ (see, e.g., [13, Corollary 4.5 and Exercise 2.68]).

First, let us consider ${\mathcal{Q}}(C_{p^{2}})$ and take
$Q\in{\mathcal{Q}}(C_{p^{2}})$. Since $P=P(Q)$ is a subgroup of $C_{p^{2}}$,
we see that $P$ is isomorphic to $\\{0\\}$ or $C_{p}$ or $C_{p^{2}}$ as
groups. Here, we note that $s_{e}$ is nothing but a group automorphism of
$C_{p^{2}}$, which corresponds to an element of $C_{p^{2}}^{\times}$. Let
$a\in C_{p^{2}}^{\times}$ correspond to $s_{e}$. Here, we see from the
description $s_{x}(e)=(1-a)x$ and Lemma 4.1 (iii) that $P=\\{gy:y\in
C_{p^{2}}\\}\cong_{\mathrm{gr}}C_{p^{2}/g}$, where $g=\gcd(1-a,p^{2})$.

  * •

Note that $P=\\{0\\}$ if and only if $Q$ is a trivial quandle (see Remark
3.4).

  * •

Let $P\cong_{\mathrm{gr}}C_{p}$. Then $g=p$. Thus, $a=kp+1$ for some $1\leq
k\leq p-1$. For $a=kp+1$ and $a^{\prime}=k^{\prime}p+1$ with $1\leq
k,k^{\prime}\leq p-1$, let $g=\gcd(1-a,n=p^{2})$ and
$g^{\prime}=\gcd(1-a^{\prime},n)$, respectively. Then we see that
$g=g^{\prime}=p$ and $a\equiv a^{\prime}\pmod{n/g(=p)}$. Thus,
$Q(C_{p^{2}},a)\cong_{\mathrm{qu}}Q(C_{p^{2}},a^{\prime})$ for any $a=kp+1$
and $a^{\prime}=k^{\prime}p+1$ by Corollary 5.2.

  * •

Let $P\cong_{\mathrm{gr}}C_{p^{2}}$. Then $g=1$. Thus, each $a\in C_{p^{2}}$
with $g=1$ defines different quandles.

Therefore, in this case, ${\mathcal{Q}}(C_{p^{2}})$ bijectively corresponds to

$\\{1\\}\cup\\{p+1\\}\cup\\{kp+r:0\leq k\leq p-1,2\leq r\leq p-1\\}\subset
C_{p^{2}}^{\times}.$

Next, let us consider ${\mathcal{Q}}(C_{p}\times C_{p})$. Let $G=C_{p}\times
C_{p}$ and regard $G$ as an additive group of a vector space $C_{p}^{2}$ over
a prime field $C_{p}$. Then it is known that
$\operatorname{Aut}_{\mathrm{gr}}(G)\cong_{\mathrm{gr}}\operatorname{GL}(2,p)$
and a set of representatives of conjugacy classes for
$\operatorname{Aut}_{\mathrm{gr}}(G)$ is determined by rational canonical
forms as follows:

(8) $\displaystyle\\{\lambda I:\lambda\in
C_{p}^{\times}\\}\cup\left\\{\begin{pmatrix}0&a\\\ 1&b\end{pmatrix}:a\in
C_{p}^{\times},\;b\in C_{p}\right\\},$

where $I=\begin{pmatrix}1&0\\\ 0&1\end{pmatrix}$ (cf. [8, Chapter 6], [2, 9]).
Note that $I\in\operatorname{Aut}_{\mathrm{gr}}(G)$ corresponds to a trivial
quandle.

  * •

For $\lambda\in C_{p}^{\times}$ with $\lambda\neq 1$, since $1-\lambda\in
C_{p}^{\times}$, we see that $\operatorname{Im}(1-\lambda)I=C_{p}\times
C_{p}$. Hence, by Corollary 1.5, we obtain that each $\lambda\in
C_{p}^{\times}$ defines different quandles.

  * •

For $M=\begin{pmatrix}0&a\\\ 1&b\end{pmatrix}$ with $a\in C_{p}^{\times}$ and
$b\in C_{p}$, let us consider $N:=I-M$. Then

$\displaystyle N=\begin{pmatrix}1&-a\\\
-1&1-b\end{pmatrix}\rightarrow\begin{pmatrix}1&-a\\\
0&1-a-b\end{pmatrix},\text{ i.e.,
}\operatorname{rank}N=\begin{cases}1&\text{if }b=1-a,\\\ 2&\text{if }b\neq
1-a.\end{cases}$

If $b\neq 1-a$, then $N$ is regular, so
$\operatorname{Im}\rho=\operatorname{Im}N=C_{p}\times C_{p}$. Hence, each $M$
defines different quandles. If $b=1-a$, then $\operatorname{Im}\rho$ is the
image of $\begin{pmatrix}1\\\ -1\end{pmatrix}$, i.e.,
$P\cong_{\mathrm{gr}}C_{p}$. Moreover, since $M\begin{pmatrix}1\\\
-1\end{pmatrix}=-a\begin{pmatrix}1\\\ -1\end{pmatrix}$, we see that
$Q(P,M)\cong_{\mathrm{qu}}Q(C_{p},-a)$. Hence, each $a\in C_{p}^{\times}$
defines different quandles.

Therefore, in this case, ${\mathcal{Q}}(C_{p}\times C_{p})$ bijectively
corresponds to (8) itself.

Furthermore, from the above discussions, we see the following:

  * •

$Q(C_{p^{2}},1)\cong_{\mathrm{qu}}Q(C_{p}\times C_{p},I)$ holds;

  * •

$Q(C_{p^{2}},p+1)\cong_{\mathrm{qu}}Q(C_{p}\times C_{p},M)$ holds, where
$M=\begin{pmatrix}0&-1\\\ 1&2\end{pmatrix}$;

  * •

$Q(C_{p^{2}},a)\cong_{\mathrm{qu}}Q(C_{p}\times C_{p},M)$ does not hold for
any $a\in C_{p^{2}}^{\times}$ and $M$ in (8) for other cases. In fact, we have
$P(Q(C_{p^{2}},a))=C_{p^{2}}$, while $C_{p^{2}}$ cannot be a normal subgroup
of $C_{p}\times C_{p}$ (cf. Proposition 3.3 (i)).

In particular,

$|{\mathcal{Q}}_{\operatorname{GAQ}}(p^{2})|=|{\mathcal{Q}}(C_{p^{2}})|+|{\mathcal{Q}}(C_{p}\times
C_{p})|-(1+1)=2p^{2}-2p-1.$

###  6.4. $n=8$

Let $G$ be a group with $|G|=8$. Then $G$ is isomorphic to $C_{8}$ or
$C_{4}\times C_{2}$ or $C_{2}\times C_{2}\times C_{2}$ or $D_{4}$ or the
quaternion group $Q_{8}$ (see, e.g., [12, p. 250]). By Corollary 1.7, we do
not need to discuss ${\mathcal{Q}}(C_{8})$.

In what follows, we discuss ${\mathcal{Q}}(G)$ case-by-case.

The case $G=C_{4}\times C_{2}$: It is known that
$\operatorname{Aut}_{\mathrm{gr}}(C_{4}\times C_{2})\cong_{\mathrm{gr}}D_{4}$
whose is automorphism is given as follows: for $\sigma\in D_{4}$, we let
$\psi_{\sigma}:C_{4}\times C_{2}\rightarrow C_{4}\times C_{2}$ by setting
$\psi_{\sigma}((i,j))=(i+2j,i+j)$, and for $\tau\in D_{4}$, we let
$\psi_{\tau}:C_{4}\times C_{2}\rightarrow C_{4}\times C_{2}$ by setting
$\psi_{\tau}((i,j))=(-i,i+j)$. Then the set of representatives of the
conjugacy classes for $\operatorname{Aut}_{\mathrm{gr}}(C_{4}\times
C_{2})\cong_{\mathrm{gr}}D_{4}$ is as follows:

$\operatorname{id},\;\;\psi_{\sigma},\;\;\psi_{\sigma}^{2},\;\;\psi_{\tau}\;\text{
and }\;\psi_{\tau}\circ\psi_{\sigma}.$

For each
$\psi\in\\{\operatorname{id},\psi_{\sigma},\psi_{\sigma}^{2},\psi_{\tau},\psi_{\tau}\circ\psi_{\sigma}\\}$,
we compute the invariants of $Q(C_{4}\times C_{2},\psi)$ as in Table 2:

| $\psi$ | $\operatorname{ord}\psi$ | $|\operatorname{Fix}(\psi,G)|$ | $P$ | $\psi|_{P}$
---|---|---|---|---|---
$Q^{8}_{1}$ | $\operatorname{id}$ | 1 | 8 | $\\{e\\}$ | $\operatorname{id}$
$Q^{8}_{2}$ | $\psi_{\sigma}$ | 4 | 2 | $C_{2}\times C_{2}$ | $\begin{pmatrix}0&1\\\ 1&0\end{pmatrix}$
$Q^{8}_{3}$ | $\psi_{\sigma}^{2}$ | 2 | 4 | $C_{2}$ | $\operatorname{id}$
$Q^{8}_{4}$ | $\psi_{\tau}$ | 2 | 4 | $C_{2}$ | $\operatorname{id}$
$Q^{8}_{5}$ | $\psi_{\sigma}\circ\psi_{\tau}$ | 2 | 4 | $C_{2}$ | $\operatorname{id}$
Table 2. Invariants for $Q(C_{4}\times C_{2},\psi)$

By Corollary 1.5, we can check that
$Q^{8}_{3}\cong_{\mathrm{qu}}Q^{8}_{4}\cong_{\mathrm{qu}}Q^{8}_{5}$.

The case $G=C_{2}\times C_{2}\times C_{2}$: It is known that
$\operatorname{Aut}_{\mathrm{gr}}(C_{2}\times C_{2}\times
C_{2})\cong_{\mathrm{gr}}\operatorname{GL}(3,2)$ and a set of representatives
of its conjugacy classes (i.e., rational canonical forms) is as follows:

$\begin{pmatrix}1&0&0\\\ 0&1&0\\\ 0&0&1\end{pmatrix},\;\begin{pmatrix}1&0&0\\\
0&0&1\\\ 0&1&0\end{pmatrix},\;\begin{pmatrix}0&0&1\\\ 1&0&0\\\
0&1&0\end{pmatrix},\;\begin{pmatrix}0&0&1\\\ 1&0&1\\\
0&1&1\end{pmatrix},\;\begin{pmatrix}0&0&1\\\ 1&0&0\\\
0&1&1\end{pmatrix},\;\begin{pmatrix}0&0&1\\\ 1&0&1\\\ 0&1&0\end{pmatrix}.$

Let $M_{1},\ldots,M_{6}$ be the above matrices from left to right,
respectively. Then we can compute the invariants of $Q(C_{2}\times C_{2}\times
C_{2},M_{i})$ for $i=1,\ldots,6$ as in Table 3:

| $\psi$ | $\operatorname{ord}\psi$ | $|\operatorname{Fix}(\psi,G)|$ | $P$ | $\psi|_{P}$
---|---|---|---|---|---
$Q^{8}_{6}$ | $M_{1}=\operatorname{id}$ | 1 | 8 | $\\{e\\}$ | $\operatorname{id}$
$Q^{8}_{7}$ | $M_{2}$ | 2 | 4 | $C_{2}$ | $\mathrm{id}$
$Q^{8}_{8}$ | $M_{3}$ | 3 | 2 | $C_{2}\times C_{2}$ | $\begin{pmatrix}0&1\\\ 1&1\end{pmatrix}$
$Q^{8}_{9}$ | $M_{4}$ | 4 | 2 | $C_{2}\times C_{2}$ | $\begin{pmatrix}0&1\\\ 1&0\end{pmatrix}$
$Q^{8}_{10}$ | $M_{5}$ | 7 | 1 | $C_{2}\times C_{2}\times C_{2}$ | $M_{5}$
$Q^{8}_{11}$ | $M_{6}$ | 7 | 1 | $C_{2}\times C_{2}\times C_{2}$ | $M_{6}$
Table 3. Invariants for $Q(C_{2}\times C_{2}\times C_{2},M_{i})$

By Corollary 1.5, we conclude that those quandles are all different.

The case $G=D_{4}$: A set of representatives of conjugacy classes is
$\varphi_{1,0},\varphi_{1,1},\varphi_{1,2},\varphi_{3,1}$ and $\varphi_{3,2}$
as in Example 4.3. We see from Corollary 1.6 that
$Q(D_{4},\varphi_{1,2})\cong_{\mathrm{qu}}Q(D_{4},\varphi_{3,2})$ and other
quandles are all different. We can compute the invariants for
$Q(D_{4},\varphi_{a,b})$ as in Table 4:

| $\psi$ | $\operatorname{ord}\psi$ | $|\operatorname{Fix}(\psi,G)|$ | $P$ | $\psi|_{P}$
---|---|---|---|---|---
$Q^{8}_{12}$ | $\varphi_{1,0}=\operatorname{id}$ | 1 | 8 | $\\{e\\}$ | $\operatorname{id}$
$Q^{8}_{13}$ | $\varphi_{3,1}$ | 2 | 2 | $C_{4}$ | $3\in C_{4}^{\times}$
$Q^{8}_{14}$ | $\varphi_{1,2},\varphi_{3,2}$ | 2 | 4 | $C_{2}$ | $\mathrm{id}$
$Q^{8}_{15}$ | $\varphi_{1,1}$ | 4 | 4 | $C_{4}$ | $\mathrm{id}$
Table 4. Invariants for $Q(D_{4},\varphi_{a,b})$

The case $G=Q_{8}$: Let $Q_{8}=\\{\pm 1,\pm i\pm j,\pm
k\\}\subset{\mathbb{H}}$. Then it is known that
$\operatorname{Aut}_{\mathrm{gr}}(Q_{8})\cong_{\mathrm{gr}}\mathfrak{S}_{4}$
and its set of representatives of conjugacy classes is as follows:

$\displaystyle\psi_{1}\colon$ $\displaystyle\pm 1\mapsto\pm 1,\;\pm
i\mapsto\pm i,\ \pm j\mapsto\pm j,\ \pm k\mapsto\pm
k\quad(\psi_{1}=\mathrm{id}),$ $\displaystyle\psi_{2}\colon$ $\displaystyle\pm
1\mapsto\pm 1,\;\pm i\mapsto\pm i,\ \pm j\mapsto\mp j,\ \pm k\mapsto\mp
k\quad(\psi_{2}=(\cdot)^{i}),$ $\displaystyle\psi_{3}\colon$ $\displaystyle\pm
1\mapsto\pm 1,\;\pm i\mapsto\pm j,\ \pm j\mapsto\pm i,\ \pm k\mapsto\mp k,$
$\displaystyle\psi_{4}\colon$ $\displaystyle\pm 1\mapsto\pm 1,\;\pm
i\mapsto\pm j,\ \pm j\mapsto\pm k,\ \pm k\mapsto\pm i,$
$\displaystyle\psi_{5}\colon$ $\displaystyle\pm 1\mapsto\pm 1,\;\pm
i\mapsto\pm j,\ \pm j\mapsto\mp i,\ \pm k\mapsto\pm
k\quad(\psi_{5}=\psi_{3}\circ\psi_{2}).$

We can compute the invariants for $Q(Q_{8},\psi_{i})$ for $i=2,\ldots,5$ as in
Table 5. We also check whether (P1) and (P2) are satisfied or not.

| $\psi$ | $\operatorname{ord}\psi$ | $|\operatorname{Fix}(\psi,G)|$ | $P$ | $\psi|_{P}$ | (P1) | (P2)
---|---|---|---|---|---|---|---
$Q^{8}_{16}$ | $\psi_{1}$ | 1 | 8 | 1 | $\mathrm{id}$ | T | T
$Q^{8}_{17}$ | $\psi_{2}$ | 2 | 4 | $C_{2}$ | $\mathrm{id}$ | T | T
$Q^{8}_{18}$ | $\psi_{3}$ | 2 | 2 | $C_{4}$ | $3\in C_{4}^{\times}$ | T | T
$Q^{8}_{19}$ | $\psi_{4}$ | 3 | 2 | $Q_{8}$ | $\psi_{4}$ | T | F
$Q^{8}_{20}$ | $\psi_{5}$ | 4 | 4 | $C_{4}$ | $\mathrm{id}$ | T | T
Table 5. Invariants for $Q(Q_{8},\psi_{i})$

Furthermore, by Theorem 1.4, we can verify the following isomorphisms:

$\displaystyle
Q^{8}_{1}\cong_{\mathrm{qu}}Q^{8}_{6}\cong_{\mathrm{qu}}Q^{8}_{12}\cong_{\mathrm{qu}}Q^{8}_{16};\quad
Q^{8}_{13}\cong_{\mathrm{qu}}Q^{8}_{18};$ $\displaystyle
Q^{8}_{3}\cong_{\mathrm{qu}}Q^{8}_{4}\cong_{\mathrm{qu}}Q^{8}_{5}\cong_{\mathrm{qu}}Q^{8}_{7}\cong_{\mathrm{qu}}Q^{8}_{14}\cong_{\mathrm{qu}}Q^{8}_{17};\quad
Q^{8}_{2}\cong_{\mathrm{qu}}Q^{8}_{9};\quad
Q^{8}_{15}\cong_{\mathrm{qu}}Q^{8}_{20}.$

By summarizing the above discussions, we obtain the following list (see Table
6):

| $\operatorname{ord}\psi$ | $|\operatorname{Fix}(\psi,G)|$ | $P$ | $\psi|_{P}$
---|---|---|---|---
$Q^{8}_{1}$ | 1 | 8 | 1 | $\mathrm{id}$
$Q^{8}_{13}$ | 2 | 2 | $C_{4}$ | $\times 3$
$Q^{8}_{3}$ | 2 | 4 | $C_{2}$ | $\mathrm{id}$
$Q^{8}_{8}$ | 3 | 2 | $C_{2}\times C_{2}$ | $\begin{pmatrix}0&1\\\ 1&1\end{pmatrix}$
$Q^{8}_{19}$ | 3 | 2 | $Q_{8}$ | $\psi_{4}$
$Q^{8}_{2}$ | 4 | 2 | $C_{2}\times C_{2}$ | $\begin{pmatrix}0&1\\\ 1&0\end{pmatrix}$
$Q^{8}_{15}$ | 4 | 4 | $C_{4}$ | $\mathrm{id}$
$Q^{8}_{10}$ | 7 | 1 | $C_{2}\times C_{2}\times C_{2}$ | $M_{5}$
$Q^{8}_{11}$ | 7 | 1 | $C_{2}\times C_{2}\times C_{2}$ | $M_{6}$
Table 6. List of ${\mathcal{Q}}_{\operatorname{GAQ}}(8)$

###  6.5. $n=12$

It is known that $G$ is isomorphic to $C_{12}$, $C_{6}\times C_{2}$, $D_{6}$,
$\mathrm{Dic}_{3}$ or $\mathfrak{A}_{4}$ (see, e.g., [12, P. 252]), where
$\mathrm{Dic}_{3}$ is a dicyclic group whose presentation is as follows:

$\mathrm{Dic}_{3}=\langle a,b:a^{6}=1,\ b^{2}=a^{3},\ b^{-1}ab=a^{-1}\rangle.$

By Corollary 1.7, we do not need to discuss ${\mathcal{Q}}(C_{12})$.

Our classification can be performed in the similar way to the case $n=8$.

The case $G=C_{6}\times C_{2}$: It is known that
$\operatorname{Aut}_{\mathrm{gr}}(C_{6}\times C_{2})\cong_{\mathrm{gr}}D_{6}$
whose automorphism is given as follows: for $\sigma\in D_{6}$, we let
$\alpha_{\sigma}:C_{6}\times C_{2}\rightarrow C_{6}\times C_{2}$ by setting
$\alpha_{\sigma}((i,j))=(2i+3j,i+j)$, and for $\tau\in D_{6}$, we let
$\alpha_{\tau}:C_{6}\times C_{2}\rightarrow C_{6}\times C_{2}$ by setting
$\alpha_{\tau}((i,j))=(-i,i+j)$. Then the set of the conjugacy classes for
$\operatorname{Aut}_{\mathrm{gr}}(C_{6}\times C_{2})\cong_{\mathrm{gr}}D_{6}$
is as follows:

$\operatorname{id},\;\;\alpha_{\sigma},\;\;\alpha_{\sigma}^{2},\;\;\alpha_{\sigma}^{3},\;\;\alpha_{\tau}\;\text{
and }\;\alpha_{\tau}\circ\alpha_{\sigma}.$

For each
$\psi\in\\{\operatorname{id},\alpha_{\sigma},\alpha_{\sigma}^{2},\alpha_{\sigma}^{3},\alpha_{\tau},\alpha_{\tau}\circ\alpha_{\sigma}\\}$,
we compute the invariants of $Q(C_{6}\times C_{2},\psi)$ as in Table 7:

| $\psi$ | $\operatorname{ord}\psi$ | $|\operatorname{Fix}(\psi,G)|$ | $P$ | $\psi|_{P}$
---|---|---|---|---|---
$Q^{12}_{1}$ | $\operatorname{id}$ | 1 | 12 | 1 | $\mathrm{id}$
$Q^{12}_{2}$ | $\alpha_{\tau}$ | 2 | 2 | $C_{6}$ | $\times 5$
$Q^{12}_{3}$ | $\alpha_{\sigma}^{3}$ | 2 | 4 | $C_{3}$ | $\times 2$
$Q^{12}_{4}$ | $\alpha_{\tau}\circ\alpha_{\sigma}$ | 2 | 6 | $C_{2}$ | $\mathrm{id}$
$Q^{12}_{5}$ | $\alpha_{\sigma}^{2}$ | 3 | 3 | $C_{2}\times C_{2}$ | $\begin{pmatrix}0&1\\\ 1&1\end{pmatrix}$
$Q^{12}_{6}$ | $\alpha_{\sigma}$ | 6 | 1 | $C_{6}\times C_{2}$ | $\alpha_{\sigma}$
Table 7. List of $\mathcal{Q}(C_{6}\times C_{2})$

The case $G=D_{6}$: Along Section 4, we compute the invariants of
${\mathcal{Q}}(D_{6})$ as in Table 8:

| $\psi$ | $\operatorname{ord}\psi$ | $|\operatorname{Fix}(\psi,G)|$ | $P$ | $\psi|_{P}$
---|---|---|---|---|---
$Q^{12}_{7}$ | $\varphi_{1,0}$ | 1 | 12 | 1 | $\mathrm{id}$
$Q^{12}_{8}$ | $\varphi_{5,1}$ | 2 | 2 | $C_{6}$ | $\times 5$
$Q^{12}_{9}$ | $\varphi_{5,2}$ | 2 | 4 | $C_{3}$ | $\times 2$
$Q^{12}_{10}$ | $\varphi_{1,3}$ | 2 | 6 | $C_{2}$ | $\mathrm{id}$
$Q^{12}_{11}$ | $\varphi_{1,2}$ | 3 | 6 | $C_{3}$ | $\mathrm{id}$
$Q^{12}_{12}$ | $\varphi_{1,1}$ | 6 | 6 | $C_{6}$ | $\mathrm{id}$
Table 8. List of $\mathcal{Q}(D_{6})$

The case $G=\mathrm{Dic}_{3}$: Note that each element of $\mathrm{Dic}_{3}$
can be of the form $a^{i}b^{\varepsilon}$ for $0\leq i\leq 5$ and
$\varepsilon\in\\{0,1\\}$. It is known that
$\operatorname{Aut}_{\mathrm{gr}}(\mathrm{Dic}_{3})\cong_{\mathrm{gr}}D_{6}$
whose isomorphism is given as follows: for $\sigma\in D_{6}$, we let
$\beta_{\sigma}:\mathrm{Dic}_{3}\rightarrow\mathrm{Dic}_{3}$ by setting
$\beta_{\sigma}(a^{i}b^{\varepsilon})=a^{i+\varepsilon}b^{\varepsilon}$, and
for $\tau\in D_{6}$, we let
$\beta_{\tau}:\mathrm{Dic}_{3}\rightarrow\mathrm{Dic}_{3}$ by setting
$\beta_{\tau}(a^{i}b^{\varepsilon})=a^{-i}b^{\varepsilon}$. Then the set of
representatives of the conjugacy classes for
$\operatorname{Aut}_{\mathrm{gr}}(\mathrm{Dic}_{3})\cong_{\mathrm{gr}}D_{6}$
is as follows:

$\operatorname{id},\;\;\beta_{\sigma},\;\;\beta_{\sigma}^{2},\;\;\beta_{\sigma}^{3},\;\;\beta_{\tau}\;\text{
and }\;\beta_{\tau}\circ\beta_{\sigma}.$

For each
$\psi\in\\{\operatorname{id},\beta_{\sigma},\beta_{\sigma}^{2},\beta_{\sigma}^{3},\beta_{\tau},\beta_{\tau}\circ\beta_{\sigma}\\}$,
we compute the invariants of $Q(\mathrm{Dic}_{3},\psi)$ as in Table 9:

| $\psi$ | $\operatorname{ord}\psi$ | $|\operatorname{Fix}(\psi,G)|$ | $P$ | $\psi|_{P}$ | (P1) | (P2)
---|---|---|---|---|---|---|---
$Q^{12}_{13}$ | $\operatorname{id}$ | 1 | 12 | 1 | $\mathrm{id}$ | T | T
$Q^{12}_{14}$ | $\beta_{\tau}\circ\beta_{\sigma}$ | 2 | 2 | $C_{6}$ | $\times 5$ | T | T
$Q^{12}_{15}$ | $\beta_{\tau}$ | 2 | 4 | $C_{3}$ | $\times 2$ | T | T
$Q^{12}_{16}$ | $\beta_{\sigma}^{3}$ | 2 | 6 | $C_{2}$ | $\operatorname{id}$ | T | T
$Q^{12}_{17}$ | $\beta_{\sigma}^{2}$ | 3 | 6 | $C_{3}$ | $\operatorname{id}$ | T | T
$Q^{12}_{18}$ | $\beta_{\sigma}$ | 6 | 6 | $C_{6}$ | $\operatorname{id}$ | T | T
Table 9. List of $\mathcal{Q}(\mathrm{Dic}_{3})$

The case $G=\mathfrak{A}_{4}$: It is known that
$\operatorname{Aut}_{\mathrm{gr}}(\mathfrak{A}_{4})\cong\mathfrak{S}_{4}$. We
compute the invariants of ${\mathcal{Q}}(\mathfrak{A}_{4})$ as follows:

| $\psi$ | $\operatorname{ord}\psi$ | $|\operatorname{Fix}(\psi,G)|$ | $P$ | $\psi|_{P}$ | (P1) | (P2)
---|---|---|---|---|---|---|---
$Q^{12}_{19}$ | $\mathrm{id}$ | 1 | 12 | 1 | $\mathrm{id}$ | T | T
$Q^{12}_{20}$ | $(\cdot)^{(12)}$ | 2 | 2 | $\mathfrak{A}_{4}$ | $(\cdot)^{(12)}$ | T | F
$Q^{12}_{21}$ | $(\cdot)^{(12)(34)}$ | 2 | 4 | $C_{2}\times C_{2}$ | $\operatorname{id}$ | T | T
$Q^{12}_{22}$ | $(\cdot)^{(123)}$ | 3 | 3 | $C_{2}\times C_{2}$ | $\begin{pmatrix}0&1\\\ 1&1\end{pmatrix}$ | T | T
$Q^{12}_{23}$ | $(\cdot)^{(1234)}$ | 4 | 2 | $\mathfrak{A}_{4}$ | $(\cdot)^{(1234)}$ | T | F
Table 10. List of $\mathcal{Q}(\mathfrak{A}_{4})$

Furthermore, by Theorem 1.4 and using GAP [3], we can verify the following
isomorphisms:

$\displaystyle
Q^{12}_{1}\cong_{\mathrm{qu}}Q^{12}_{7}\cong_{\mathrm{qu}}Q^{12}_{13}\cong_{\mathrm{qu}}Q^{12}_{19};\quad
Q^{12}_{2}\cong_{\mathrm{qu}}Q^{12}_{8}\cong_{\mathrm{qu}}Q^{12}_{14};$
$\displaystyle
Q^{12}_{3}\cong_{\mathrm{qu}}Q^{12}_{9}\cong_{\mathrm{qu}}Q^{12}_{15};\quad
Q^{12}_{4}\cong_{\mathrm{qu}}Q^{12}_{10}\cong_{\mathrm{qu}}Q^{12}_{16};\quad
Q^{12}_{5}\cong_{\mathrm{qu}}Q^{12}_{22}.$

By summarizing the above discussions, we obtain the following list (see Table
11):

| $\operatorname{ord}\psi$ | $|\operatorname{Fix}(\psi,G)|$ | $P$ | $\psi|_{P}$
---|---|---|---|---
$Q_{1}^{12}$ | 1 | 12 | 1 | $\mathrm{id}$
$Q_{2}^{12}$ | 2 | 2 | $C_{6}$ | $\times 5$
$Q^{12}_{20}$ | 2 | 2 | $\mathfrak{A}_{4}$ | $(\cdot)^{(12)}$
$Q_{3}^{12}$ | 2 | 4 | $C_{3}$ | $\times 2$
$Q^{12}_{21}$ | 2 | 4 | $C_{2}\times C_{2}$ | $\mathrm{id}$
$Q_{4}^{12}$ | 2 | 6 | $C_{2}$ | $\mathrm{id}$
$Q^{12}_{5}$ | 3 | 3 | $C_{2}\times C_{2}$ | $\begin{pmatrix}0&1\\\ 1&1\end{pmatrix}$
$Q^{12}_{11}$ | 3 | 6 | $C_{3}$ | $\mathrm{id}$
$Q^{12}_{23}$ | 4 | 2 | $\mathfrak{A}_{4}$ | $(\cdot)^{(1234)}$
$Q^{12}_{6}$ | 6 | 1 | $C_{6}\times C_{2}$ | $\alpha_{\sigma}$
$Q^{12}_{12}$ | 6 | 6 | $C_{6}$ | $\mathrm{id}$
Table 11. List of ${\mathcal{Q}}_{\operatorname{GAQ}}(12)$

###  6.6. $n=15$

Let $G$ be a group with $|G|=15$. Then $G$ is isomorphic to $C_{15}$ (see,
e.g., [12, P. 252]), i.e.,
${\mathcal{Q}}_{\operatorname{GAQ}}(15)={\mathcal{Q}}(C_{15})$. By Corollary
5.2, it is straightforward to check that
$\operatorname{Aut}_{\mathrm{gr}}(C_{15})\cong_{\mathrm{gr}}C_{15}^{\times}=\\{1,2,4,7,8,11,13,14\\}$
bijectively corresponds to ${\mathcal{Q}}(C_{15})$. In particular,
$|{\mathcal{Q}}_{\operatorname{GAQ}}(15)|=8$.

###  6.7. $n=16$

In the case $n=16$, we encounter the problem. Namely, there are two groups $G$
and $G^{\prime}$ together with their automorphisms
$\psi\in\operatorname{Aut}_{\mathrm{gr}}(G)$ and
$\psi^{\prime}\in\operatorname{Aut}_{\mathrm{gr}}(G^{\prime})$ such that we
cannot distinguish whether
$Q(G,\psi)\cong_{\mathrm{qu}}Q(G^{\prime},\psi^{\prime})$ or not by using
Theorem 1.4. In fact, let $G=C_{2}\times Q_{8}$ and let
$G^{\prime}=(C_{4}\times C_{2})\rtimes C_{2}$. We take
$\phi\in\operatorname{Aut}_{\mathrm{gr}}(Q_{8})\cong_{\mathrm{gr}}\mathfrak{S}_{4}$
with $\operatorname{ord}\phi=3$ and let
$\psi\in\operatorname{Aut}_{\mathrm{gr}}(G)$ be defined by
$\psi((i,x))=(i,\phi(x))$, and we also take
$\psi^{\prime}\in\operatorname{Aut}_{\mathrm{gr}}(G^{\prime})$ with
$\operatorname{ord}\psi^{\prime}=3$. Actually, we see that both $Q(G,\psi)$
and $Q(G^{\prime},\psi^{\prime})$ do not satisfy (P2), so we cannot apply
Theorem 1.4, while all of the invariants we have developed coincide.

Hence, we cannot verify whether those are isomorphic as quandles or not by the
current methods.

######  Remark 6.1.

From Tables 2–11, it seems that the isomorphism among finite generalized
Alexander quandles with the same cardinality can be determined only by the
pair of $P$ and $\psi|_{P}$. However, there exist quandles that are not
isomorphic even if the pair of $P$ and $\psi|_{P}$ are equal. When $n=16$,
$Q(D_{8},1,2)$ and $Q(D_{8},5,2)$ have $P=P^{\prime}\cong_{\mathrm{gr}}C_{4}$
(i.e., $d=d^{\prime}=2$) and
$\varphi_{1,2}|_{P}=\varphi_{5,2}|_{P^{\prime}}=\mathrm{id}$ (i.e., $1\equiv
5\pmod{\frac{8}{2}}$). On the other hand, since
$|\operatorname{Fix}(\varphi_{1,2},D_{4})|=8$ and
$|\operatorname{Fix}(\varphi_{5,2},D_{4})|=4$, the two quandles $Q(D_{8},1,2)$
and $Q(D_{8},5,2)$ are not isomorphic.

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