# The second quandle homology group of the knot $n$-quandle

Kokoro Tanaka  Department of Mathematics, Tokyo Gakugei University, Nukuikita
4-1-1, Koganei, Tokyo 184-8501, Japan<EMAIL_ADDRESS>and  Yuta
Taniguchi  Department of Mathematics, Graduate School of Science, Osaka
University, 1-1, Machikaneyama, Toyonaka, Osaka, 560-0043, Japan
<EMAIL_ADDRESS>
###### Abstract.

The knot quandle is a complete invariant for oriented classical knots in the
$3$-sphere up to orientation. Eisermann computed the second quandle homology
group of the knot quandle and showed that it characterizes the unknot. In this
paper, we compute the second quandle homology group of the knot $n$-quandle
completely, where the knot $n$-quandle is a certain quotient of the knot
quandle for each integer $n$ greater than one. Although the knot $n$-quandle
is weaker than the knot quandle, the second quandle homology group of the
former is found to have more information than that of the latter. As one of
the consequences, it follows that the second quandle homology group of the
knot $3$-quandle characterizes the unknot, the trefoil and the cinquefoil.

###### Key words and phrases:

quandle; twist spun knot; $n$-quandle.

###### 2020 Mathematics Subject Classification:

57K10, 57K12.

##  1\. Introduction

Quandles [Joyce1982quandle, Matveev1982distributive] are algebraic structures
whose axioms correspond to Reidemeister moves for classical knot diagrams.
They are compatible with knot theory and various knot invariants have been
defined using them. Among such invariants, the knot quandle of an oriented
classical knot, which is also called an oriented $1$-knot for short, in the
$3$-sphere is very powerful and known to be a complete invariant up to
orientation. However, it is difficult to treat the knot quandle $Q(K)$ of an
oriented $1$-knot $K$ itself, and hence it is necessary to extract information
from $Q(K)$. When extracting information, extrinsic information of $Q(K)$ such
as the set of all quandle homomorphisms from $Q(K)$ to a chosen fixed finite
quandle are often examined, but in this paper we focus on intrinsic
information of $Q(K)$ such as quandle homology groups of $Q(K)$ and the knot
$n$-quandle obtained from $Q(K)$ for each integer $n$ greater than $1$.

The homology theory for quandles was developed in [Carter2003quandle] to
define knot invariants. The first quandle homology group of any connected
quandle, such as the knot quandle $Q(K)$ of an oriented $1$-knot $K$, is known
to be isomorphic to the infinite cyclic group $\mathbb{Z}$. Eisermann
[Eisermann2003unknot] computed the second quandle homology group
$H^{Q}_{2}(Q(K))$ of $Q(K)$ and showed that it characterizes the unknot. More
precisely, he showed that $H^{Q}_{2}(Q(K))$ is trivial if $K$ is trivial, and
that $H^{Q}_{2}(Q(K))$ is group isomorphic to $\mathbb{Z}$ if $K$ is
nontrivial.

The knot $n$-quandle $Q_{n}(K)$ of an oriented $1$-knot $K$ is defined by
taking a certain quotient of $Q(K)$ for each integer $n>1$. Joyce
[Joyce1982quandle] introduced the knot quandle $Q(K)$ of $K$ and obtained
basic results such as “complete invariance” for oriented $1$-knots. In the
last chapter of [Joyce1982quandle], the knot $2$-quandle $Q_{2}(K)$ of $K$ was
also introduced. In that chapter, the knot $2$-quandles for some concrete
examples were computed, but properties of $Q_{2}(K)$ remained unknown. Winker
[Winker1984quandles] introduced the knot $n$-quandle $Q_{n}(K)$ of $K$ as a
generalization of $Q_{2}(K)$ for each integer $n>1$, and found a relation
between $Q_{n}(K)$ and the fundamental group of the $n$-fold cyclic branched
covering space $M_{K}^{n}$ branched over $K$. Hoste and Shanahan
[Hoste2017links] further deepened Winker’s observation and showed that the
finiteness of $Q_{n}(K)$ is equivalent to that of $\pi_{1}(M_{K}^{n})$. It
follows, for example, that the knot $2$-quandle of any $2$-bridge knot is
always finite. On the other hand, the knot quandle of any nontrivial $1$-knot
is known to be always infinite as in Subsection 6.2, and hence $Q_{n}(K)$ can
be considered a more tractable invariant than $Q(K)$.

In this paper, we investigate how much information the knot $n$-quandle
$Q_{n}(K)$ still has about $K$. More specifically, we compute the second
quandle homology group $H_{2}^{Q}(Q_{n}(K))$ of the knot $n$-quandle
$Q_{n}(K)$ completely in Subsection 5.3, and reveal information that
$H_{2}^{Q}(Q_{n}(K))$ has about $K$ in Subsection 5.4. Although $Q_{n}(K)$ is
weaker than $Q(K)$, the second quandle homology group of $Q_{n}(K)$ is found
to have more information about $K$ than that of $Q(K)$ for $n=3,4$ and $5$.
Here is a remarkable consequence of our computations.

######  Theorem 5.16.

The second quandle homology group of the knot $n$-quandle characterizes $(1)$
the unknot, the trefoil and the cinquefoil for $n=3$, $(2)$ the unknot and the
trefoil for $n=4$ or $5$, $(3)$ the unknot for $n>5$, and $(4)$ the unknot up
to $2$-bridge knot summands for $n=2$.

###  1.1. Strategy to compute

We outline here our strategy to compute the second quandle homology group of a
connected quandle, such as the knot $n$-quandle $Q_{n}(K)$ of an oriented
$1$-knot for each integer $n>1$. Instead of computing according to the
original definition in [Carter2003quandle], we use the quandle extention
theory developed in [Carter2003extension, Eisermann2003unknot,
Eisermann2014quandle] and the quandle covering theory developed in
[Eisermann2003unknot, Eisermann2014quandle]. For a connected quandle $X$, let
$\Lambda\curvearrowright\tilde{X}\to X$ be an extension of $X$ by a group
$\Lambda$ consisting of a group action $\Lambda\curvearrowright\tilde{X}$ on a
quandle $\tilde{X}$ and a surjective quandle homomorphism $\tilde{X}\to X$.
Now suppose that the surjection $\tilde{X}\to X$ is a universal covering. Then
it follows from [Eisermann2014quandle] that $H_{2}^{Q}(X)$ is group isomorphic
to the abelianization $\Lambda^{\mathrm{ab}}$of $\Lambda$. More precisely, we
have that $H_{2}^{Q}(X)$ is group isomorphic to $\pi_{1}(X)^{\mathrm{ab}}$ and
that $\pi_{1}(X)$ is group isomorphic to $\Lambda$, where $\pi_{1}(X)$ is the
fundamental group [Eisermann2014quandle] of the connected quandle $X$. We note
that the first statement follows from [Eisermann2014quandle, Theorem 1.15] and
the second statement follows from [Eisermann2014quandle, Proposition 5.9].

With these in mind, in order to compute $H_{2}^{Q}(Q_{n}(K))$, we will focus
on a certain fibered $2$-knot, called the $n$-twist spin of $K$ and denoted by
$\tau^{n}K$, which is an embedded $2$-sphere in the $4$-sphere defined by
Zeeman [Zeeman1965twisting]. The fiber of $\tau^{n}K$ is known to be the once
punctured closed $3$-manifold $M^{n}_{K}$, where $M^{n}_{K}$ is the $n$-fold
cyclic branched covering of $S^{3}$ branched over $K$. Let $A^{n}_{K}$ be a
cyclic group generated by a lift of the longitude of $K$ in
$\pi_{1}(M^{n}_{K})$. Then we will show that there is an extension
$A^{n}_{K}\curvearrowright Q(\tau^{n}K)\to Q_{n}(K)$ in Theorem 3.2, and that
$Q(\tau^{n}K)\to Q_{n}(K)$ is a universal covering in Theorem 4.6. Combining
these two theorems, we conclude that $H_{2}^{Q}(Q_{n}(K))$ is group isomorphic
to the cyclic group $A^{n}_{K}$. Finally, by geometric arguments, we can
determine the order of the cyclic group $A^{n}_{K}$ explicitly in Section 5.

###  1.2. Remarks

The aforementioned theorems, Theorem 3.2 and 4.6, have their own interests. We
mention a relation with [Eisermann2003unknot] and that with [Inoue2021knot]
below.

Eisermann [Eisermann2003unknot, Theorem 35] showed that, for an oriented
nontrivial $1$-knot $K$, there is an extension $\mathbb{Z}\curvearrowright
Q(\widehat{K})\to Q(K)$ coming from the long knot $\widehat{K}$ obtained by
cutting $K$ open, and also that $Q(\widehat{K})\to Q(K)$ is a universal
covering. Since the knot quandle of $\widehat{K}$ is naturally isomorphic to
that of the $0$-twist spin $\tau^{0}K$ and the “knot $0$-quandle” is the same
as the knot quandle by definition, his extension is rephrased as
$\mathbb{Z}\curvearrowright Q(\tau^{0}K)\to Q_{0}(K)$. There is no definition
for a “space $M^{0}_{K}$,” but a “group $\pi_{1}(M^{0}_{K})$” should be group
isomorphic to the commutator subgroup of the knot group of $K$; see the first
paragraph of Subsection 3.1. Hence it is natural to take a space $M^{0}_{K}$
as the infinite cyclic covering space of $S^{3}\setminus K$, and then a cyclic
group $A^{0}_{K}$ generated by a lift of the longitude of $K$ in
$\pi_{1}(M^{0}_{K})$ is group isomorphic to $\mathbb{Z}$. Thus, substituting
$0$ for $n$ in our two theorems would yield his results formally; however, it
does not give an alternative proof.

Inoue [Inoue2021knot] introduced the notion of the Schläfli quandle $X_{n}$
that is related to the regular tessellation $\\{3,n\\}$ of geometric
$2$-spaces in the sense of the Schläfli symbol. Then, for the trefoil $3_{1}$,
he showed in [Inoue2021knot, Theorem 6.3] that there is an extension
$A_{n}\curvearrowright Q(\tau^{n}\,3_{1})\to X_{n}$ by a cyclic group $A_{n}$.
On the other hand, when we take an oriented $1$-knot $K$ as the trefoil
$3_{1}$ in Theorem 3.2, we have the extension $A^{n}_{3_{1}}\curvearrowright
Q(\tau^{n}\,3_{1})\to Q_{n}(3_{1})$. Since the quandle $X_{n}$ turns out to be
naturally isomorphic to $Q_{n}(3_{1})$ in Lemma 6.1, we see that the two
extensions are essentially the same, and in particular, we have that $A_{n}$
is group isomorphic to $A^{n}_{3_{1}}$. He computed $A_{n}$ for $n\leq 5$ and
asked what $A_{n}$ is for $n>5$. As a byproduct, we can answer his question;
in fact, we have that $A_{n}$ is group isomorphic to $\mathbb{Z}$ for $n>5$.

###  1.3. Organization

This article is organized as follows. In Section 2, we review basics of
quandle theory including the knot quandles and the knot $n$-quandles. In
Section 3, we study algebraic structures of the knot $n$-quandle and prove
that the knot quandle of the $n$-twist spin of $K$ is a central extension of
the knot $n$-quandle of $K$ (Theorem 3.2). In Section 4, we review the
covering theory for quandles and prove that the quandle covering associated
with the central extension is a universal covering (Theorem 4.6). Section 5 is
devoted to computing the second quandle homology group of the knot
$n$-quandle. Finally, in Section 6, we collect three related results: a
relation with an extension in [Inoue2021knot], the types of the knot
$n$-quandles, and the cardinalities of the finite knot $n$-quandles.

##  2\. Quandles

A quandle [Joyce1982quandle, Matveev1982distributive] is a non-empty set $X$
with a binary operation $\ast$ that satisfies the following conditions:

  * •

For any $x\in X$, we have $x\ast x=x$.

  * •

For any $y\in X$, the map $S_{y}:X\to X;x\mapsto x\ast y$ is bijective.

  * •

For any $x,y,z\in X$, we have $(x\ast y)\ast z=(x\ast z)\ast(y\ast z)$.

We denote $S_{y}^{n}(x)$ by $x\ast^{n}y$ for any $x,y\in X$ and
$n\in\mathbb{Z}$. Then we set
$\textrm{type}(X):=\textrm{min}\\{n\in\mathbb{N}\mid x\ast^{n}y=x\ \textrm{for
any }x,y\in X\\},$ where $\textrm{min}\ \emptyset=\infty$. We call
$\textrm{type}(X)$ the type of a quandle $X$.

A map $f:X\to X^{\prime}$ between quandles is said to be a quandle
homomorphism if $f(x\ast y)=f(x)\ast f(y)$ for any $x,y\in X$. A quandle
homomorphism $f:X\to X^{\prime}$ is said to be a quandle isomorphism if $f$ is
a bijection.

The associated group of $X$, denoted by $\textrm{As}(X)$, is the group
generated by the elements of $X$ subject to the relations $x\ast y=y^{-1}xy$
for $x,y\in X$. We note that ${\rm As}$ is a functor from the category of
quandles to the category of groups. The associated group ${\rm As}(X)$ acts on
$X$ from the right by $x\cdot y:=x\ast y$ for any $x,y\in X$. A quandle $X$ is
connected if the action of $\textrm{As}(X)$ on $X$ is transitive.

Let $G$ be a group and $\varphi:G\to G$ a group automorphism. We define the
binary operation $\ast$ on $G$ by $x\ast y:=\varphi(xy^{-1})y$. Then
$\textrm{GAlex}(G,\varphi)=(G,\ast)$ becomes a quandle, which is called the
generalized Alexander quandle.

Let $N(K)$ be a tubular neighborhood of $K$ and $E(K)=S^{3}\backslash{\rm
int}N(K)$ an exterior of an oriented $1$-knot $K$ in the $3$-sphere $S^{3}$.
We fix a base point $p\in E(K)$. Let $Q(K,p)$ be the set of homotopy classes
of all paths in $E(K)$ from a point in $\partial E(K)$ to the base point $p$.
The set $Q(K,p)$ is a quandle with an operation defined by
$[\alpha]\ast[\beta]:=[\alpha\cdot\beta^{-1}\cdot m_{\beta(0)}\cdot\beta]$,
where $m_{\beta(0)}$ is a meridian loop starting from $\beta(0)$ and going
along in the positive direction. We call $Q(K,p)$ the knot quandle of $K$,
which is known to be connected. Since the quandle isomorphism class of
$Q(K,p)$ does not depend on the base point $p$, we denote the knot quandle
$Q(K,p)$ simply by $Q(K)$. Note that, for an oriented $1$-knot $K$, the
associated group ${\rm As}(Q(K))$ is group isomorphic to the knot group
$G(K):=\pi_{1}(E(K))$.

Let $K$ be an oriented $1$-knot and $n$ a positive integer. We consider a
equivalence relation, denoted by $\sim_{n}$, on the knot quandle $Q(K)$
generated by $x\sim x\ast^{n}y$ for all $x,y\in Q(K)$. Then we define
$Q_{n}(K)$ by the quotient set $Q(K)/\sim_{n}$. This quotient set $Q_{n}(K)$
has a quandle operation inherited from the knot quandle $Q(K)$. We call this
quandle $Q_{n}(K)$ the knot $n$-quandle of $K$, which is also known to be
connected. In some papers [Crans2019finite, Hoste2017links,
Winker1984quandles], the knot $n$-quandle of $K$ is called the $n$-quandle of
$K$. Joyce [Joyce1982quandle] used the term “$n$-quandle” to mean a quandle
whose type is a divisor of $n$. To avoid confusion, we use the term “knot
$n$-quandle of $K$” for $Q_{n}(K)$ in this paper.

##  3\. Central extensions of the knot $n$-quandles

We discuss algebraic structures of the knot $n$-quandle $Q_{n}(K)$ of an
oriented $1$-knot $K$ in Subsection 3.1, and show that the knot quandle of the
$n$-twist spin of $K$ is a central extension of $Q_{n}(K)$ in Subsection 3.2.

###  3.1. Algebraic structures of the knot $n$-quandles

For an oriented $1$-knot $K$, we denote by $m_{K}\in G(K)$ a meridian of $K$
and by $l_{K}\in G(K)$ the preferred longitude of $K$, where $G(K)$ is the
knot group of $K$. Let $M^{n}_{K}$ be the $n$-fold cyclic branched covering of
$S^{3}$ branched over $K$ for an integer $n$ greater than $1$. Then the group
$\pi_{1}(M^{n}_{K})$ is group isomorphic to the quotient group
$\textrm{Ker}(G(K)\to\mathbb{Z}/n\mathbb{Z};m_{K}\mapsto 1)/\langle\langle
m_{K}^{n}\rangle\rangle$, where $\langle\langle m^{n}_{K}\rangle\rangle$ is
the normal closure of $m^{n}_{K}$. Since $l_{K}$ is an element of
$\textrm{Ker}(G(K)\to\mathbb{Z}/n\mathbb{Z};m_{K}\mapsto 1)$, it can be
regarded as an element of $\pi_{1}(M^{n}_{K})$. We notice that
$l_{K}\in\pi_{1}(M^{n}_{K})$ is represented by the branching set of
$M^{n}_{K}$, where this fact will be used in Subsection 5.1.

We define the group $G_{n}(K)$ by $G(K)/\langle\langle
m^{n}_{K}\rangle\rangle$. It holds that the abelianization of $G_{n}(K)$ is
$\mathbb{Z}/n\mathbb{Z}$, and it follows from [Winker1984quandles, Theorem
5.22] that we have an exact sequence

$1\to\pi_{1}(M^{n}_{K})\to
G_{n}(K)\xrightarrow{\textrm{abelianization}}\mathbb{Z}/n\mathbb{Z}\to 1.$

Since this sequence splits, the group $G_{n}(K)$ is group isomorphic to the
semidirect product $\pi_{1}(M^{n}_{K})\rtimes\mathbb{Z}/n\mathbb{Z}$. Let
$\varphi$ be the group automorphism of $\pi_{1}(M^{n}_{K})$ which is the
restriction of the group automorphism of $G_{n}(K)$ sending $g$ to
$m_{K}^{-1}gm_{K}$ for any $g$ in $G_{n}(K)$. Since $m_{K}$ commutes with
$l_{K}$ in $G(K)$, we have $\varphi(l_{K})=l_{K}$.

Let $P$ be the subgroup of $G_{n}(K)$ generated by $m_{K}$ and $l_{K}$. Let us
define the quandle operation $\ast$ on $P\backslash G_{n}(K)$ by $Px\ast
Py:=Pxy^{-1}m_{K}y$ for $Px,Py\in P\backslash G_{n}(K)$. Hoste and Shanahan
[Hoste2017links] showed that the knot $n$-quandle $Q_{n}(K)$ is quandle
isomorphic to $P\backslash G_{n}(K)$.

Let $A^{n}_{K}$ be the subgroup of $\pi_{1}(M^{n}_{K})$ generated by $l_{K}$.
Then we can define the quandle operation $\ast$ on
$A^{n}_{K}\backslash\pi_{1}(M^{n}_{K})$ by $A^{n}_{K}x\ast
A^{n}_{K}y:=A^{n}_{K}\varphi(xy^{-1})y$ for any $A^{n}_{K}x,A^{n}_{K}y\in
H\backslash\pi_{1}(M^{n}_{K})$.

######  Proposition 3.1.

The knot $n$-quandle $Q_{n}(K)$ of an oriented $1$-knot $K$ is quandle
isomorphic to $A^{n}_{K}\backslash\pi_{1}(M^{n}_{K})$ for an integer $n>1$.

###### Proof.

For notational simplicity, we denote $m_{K},l_{K}$ and $A^{n}_{K}$ by $m,l$
and $A$, respectively. It is sufficient to show that $P\backslash G_{n}(K)$
and $A\backslash\pi_{1}(M^{n}_{K})$ are quandle isomorphic. Since $G_{n}(K)$
is group isomorphic to $\pi_{1}(M^{n}_{K})\rtimes\mathbb{Z}/n\mathbb{Z}$, for
any $x\in G_{n}(K)$, there exist unique $i_{x}\in\\{0,\ldots,n-1\\}$ and
$g_{x}\in\pi_{1}(M^{n}_{K})$ such that $x=m^{i_{x}}g_{x}$. Define a map
$\Phi\colon P\backslash G_{n}(K)\to A\backslash\pi_{1}(M^{n}_{K})$ by
$\Phi(Px):=Ag_{x}$ for each $Px\in P\backslash G_{n}(K)$. If $Px=Py$, we have
$x=m^{i}l^{j}y$ for some $i,j$. Hence, it holds that

$x=m^{i}l^{j}y=m^{i}l^{j}m^{i_{y}}g_{y}=m^{i}m^{i_{y}}l^{j}g_{y}=m^{i+i_{y}}l^{j}g_{y}.$

Thus, we have $\Phi(Px)=Al^{j}g_{y}=Ag_{y}=\Phi(Py)$, which implies that the
map $\Phi$ is well-defined.

Next, define a map $\Psi\colon A\backslash\pi_{1}(M^{n}_{K})\to P\backslash
G_{n}(K)$ by $\Psi(Ag):=Pg$ for each $Ag\in A\backslash\pi_{1}(M^{n}_{K})$. It
is easily seen that the map $\Psi$ is well-defined. By direct computations, it
holds that $\Psi\circ\Phi={\rm Id}$ and $\Phi\circ\Psi={\rm Id}$.

For any $Ag_{1},Ag_{2}\in A\backslash\pi_{1}(M^{n}_{K})$, we have

$\displaystyle\Psi(Ag_{1}\ast Ag_{2})$ $\displaystyle=$
$\displaystyle\Psi(A\varphi(g_{1}g_{2}^{-1})g_{2})$ $\displaystyle=$
$\displaystyle P\varphi(g_{1}g_{2}^{-1})g_{2}$ $\displaystyle=$ $\displaystyle
Pm^{-1}g_{1}g_{2}^{-1}mg_{2}$ $\displaystyle=$ $\displaystyle
Pg_{1}g_{2}^{-1}mg_{2}$ $\displaystyle=$ $\displaystyle Pg_{1}\ast
Pg_{2}=\Psi(Ag_{1})\ast\Psi(Ag_{2}),$

which implies that $\Psi$ is a quandle isomorphism. ∎

###  3.2. Central extensions of the knot $n$-quandles

A quandle $\tilde{X}$ is an extension [Eisermann2003unknot] of a quandle $X$
by a group $\Lambda$ if there is a surjective quandle homomorphism
$p:\tilde{X}\to X$ and a left group action $\Lambda\curvearrowright\tilde{X}$
such that the following conditions are satisfied:

  * (E1)

For any $\tilde{x},\tilde{y}\in\tilde{X}$ and $\lambda\in\Lambda$, we have
$\lambda\cdot(\tilde{x}\ast\tilde{y})=(\lambda\cdot\tilde{x})\ast\tilde{y}$
and $\tilde{x}\ast(\lambda\cdot\tilde{y})=\tilde{x}\ast\tilde{y}$.

  * (E2)

For any $x\in X$, the group $\Lambda$ acts on the preimage $p^{-1}(x)$ freely
and transitively.

If $\Lambda$ is abelian, we call $\tilde{X}$ a central extension of $X$ by the
group $\Lambda$.

A 2-knot $F$ is a 2-sphere embedded in $S^{4}$, and is fibered if its
complement $S^{4}\setminus F$ is a fiber bundle over $S^{1}$. For an oriented
2-knot $F$, we can define the knot quandle of $F$, denoted by $Q(F)$, in a
similar manner to the knot quandle of an oriented $1$-knot. Inoue
[Inoue2019fibered] showed that the knot quandle of an oriented fibered 2-knot
$F$ is quandle isomorphic to $\textrm{GAlex}(\pi_{1}(M),\psi)$, where $M$ is a
fiber of the complement $S^{4}\setminus F$ and $\psi$ is the group
automorphism of $\pi_{1}(M)$ induced by the monodromy of $S^{4}\setminus F$.

For an oriented 1-knot $K$ and an integer $n$, Zeeman [Zeeman1965twisting]
introduced the oriented $2$-knot $\tau^{n}K$, which is called the $n$-twist
spun $K$ or the $n$-twist spin of $K$. Furthermore, for each positive integer
$n$, he showed that $\tau^{n}K$ is a fibered 2-knot whose fiber is the once
punctured $M^{n}_{K}$ and whose monodromy is the canonical generator of the
covering transformation group of $M^{n}_{K}$. We note that the group
automorphism of $\pi_{1}(M^{n}_{K})$ induced by the monodromy coincides with
$\varphi$ defined in Subsection 3.1. The aim of this section is to show the
following for an oriented $1$-knot $K$ and an integer $n>1$.

######  Theorem 3.2.

The knot quandle $Q(\tau^{n}K)$ of the oriented $2$-knot $\tau^{n}K$ is a
central extension of the knot $n$-quandle $Q_{n}(K)$ of an oriented $1$-knot
$K$ by the cyclic group $A^{n}_{K}$ for an integer $n>1$.

###### Proof.

Let $p:\pi_{1}(M^{n}_{K})\to A^{n}_{K}\backslash\pi_{1}(M^{n}_{K})$ be the
quotient map. By [Inoue2019fibered, Theorem 3.1], the quandle $Q(\tau^{n}K)$
is quandle isomorphic to $\textrm{GAlex}(\pi_{1}(M^{n}_{K}),\varphi)$. Hence,
by Proposition 3.1, the map $p$ is a surjective quandle homomorphism from
$Q(\tau^{n}K)$ to $Q_{n}(K)$. We define the action of $A^{n}_{K}$ on
$\pi_{1}(M^{n}_{K})$ by $l_{K}^{i}\cdot x=l_{K}^{i}x$ for any $i\in\mathbb{Z}$
and $x\in\pi_{1}(M^{n}_{K})$. Obviously, the condition $(\textrm{E2})$ is
satisfied. Since it holds that $\varphi(l_{K})=l_{K}$, we have

$(l_{K}\cdot x)\ast
y=\varphi(l_{K}xy^{-1})y=l_{K}\varphi(xy^{-1})y=l_{K}\cdot(x\ast y)$

and

$x\ast(l_{K}\cdot
y)=\varphi(x(l_{K}y)^{-1})(l_{K}y)=\varphi(xy^{-1}l_{K}^{-1})l_{K}y=\varphi(xy^{-1})y=x\ast
y$

for each $x,y\in\pi_{1}(M^{n}_{K})=Q(\tau^{n}K)$. This implies that the
condition $(\textrm{E1})$ is also satisfied. ∎

######  Remark 3.3.

If $l_{K}$ is the identity element in $\pi_{1}(M^{n}_{K})$, then the knot
quandle of $\tau^{n}K$ is quandle isomorphic to the knot $n$-quandle of $K$.

##  4\. Universal coverings of the knot $n$-quandles

We review covering theory for quandles and the fundamental groups of quandles
in Subsection 4.1. We investigate a relation between the types of quandles and
quandle coverings in Subsection 4.2. Then, in Subsection 4.3, we show the
surjective quandle homomorphism $Q(\tau^{n}K)\to Q_{n}(K)$ associated with the
central extension $A^{n}_{K}\curvearrowright Q(\tau^{n}K)\to Q_{n}(K)$ is a
universal covering (Theorem 4.6).

###  4.1. Covering theory for quandles

A quandle homomorphism $p:\tilde{X}\to X$ between quandles is a covering if
$p$ is surjective and $p(\tilde{x})=p(\tilde{y})$ implies
$\alpha\ast\tilde{x}=\alpha\ast\tilde{y}$ for any
$\alpha,\tilde{x},\tilde{y}\in\tilde{X}$. This property allows to define an
action of $X$ on $\tilde{X}$ by $\alpha*x:=\alpha*\tilde{x}$ with
$\tilde{x}\in p^{-1}(x)$ for $x\in X$ and $\alpha\in\tilde{X}$, where this
action will be used in the proof of Theorem 4.5. Here is an important example
of a covering. For an extension $\tilde{X}$ is of a quandle $X$ by a group
$\Lambda$, there is a surjective quandle homomorphism $p:\tilde{X}\to X$
associated with the extension. Then we can verify that $p:\tilde{X}\to X$ is a
covering.

A pointed quandle $(X,x)$ is a quandle $X$ with a chosen base element $x$ of
$X$. A map $f:(X,x)\to(X^{\prime},x^{\prime})$ between pointed quandles is a
quandle homomorphism (resp. covering) if $f$ is a quandle homomorphism (resp.
covering) from $X$ to $X^{\prime}$ such that $f(x)=x^{\prime}$. A quandle
covering $f:(Q,q)\to(X,x)$ is said to be a universal covering of $(X,x)$ if
for any quandle covering $p:(\tilde{X},\tilde{x})\to(X,x)$, there exists a
unique quandle homomorphism $\tilde{f}:(Q,q)\to(\tilde{X},\tilde{x})$ such
that $p\circ\tilde{f}=f$.

The fundamental group of a quandle $X$ based at an element $x$ of $X$, which
was introduced in [Eisermann2014quandle], is defined by
$\\{g\in\textrm{As}(X)^{\prime}\mid x\cdot g=x\\}$, where
$\textrm{As}(X)^{\prime}$ is the commutator subgroup of $\textrm{As}(X)$. The
fundamental group of $X$ based at $x$ is denoted by $\pi_{1}(X,x)$. We note
that $\pi_{1}(X,x)$ does not depend on the choice of the base point $x$
whenever $X$ is connected, and that $\pi_{1}(X,x)$ is sometimes denoted by
$\pi_{1}(X)$ for a connected quandle $X$. A quandle $X$ is said to be simply
connected if $X$ is connected and $\pi_{1}(X,x)$ is trivial for some $x$ in
$X$. We recall here [Eisermann2014quandle, Proposition 5.15], which
characterizes the simply connectedness of quandles in terms of universal
coverings. Since the treatment of base points of pointed quandles is somewhat
ambiguous in [Eisermann2014quandle, Proposition 5.15], we modify his condition
appropriately as follows.

######  Proposition 4.1.

The following are equivalent for a quandle $Q$.

  * $(1)$

The quandle $Q$ is simply connected.

  * $(2)$

There exists an element $q$ in $Q$ such that any covering from $(Q,q)$ to any
pointed quandle $(X,x)$ is a universal covering of $(X,x)$.

###  4.2. Coverings and the types of quandles

We investigate a relation between the type of a quandle and that of another
quandle, when there is a quandle covering from one to the other.

######  Lemma 4.2.

For quandles $X$ and $\tilde{X}$, suppose that there is a surjective quandle
homomorphism $p:\tilde{X}\to X$. If the type of $\tilde{X}$ is finite, then
the type of $X$ is a divisor of that of $\tilde{X}$.

###### Proof.

For each two elements $a,b$ in $X$, we have $p^{-1}(\tilde{a})\neq\emptyset$
and $p^{-1}(\tilde{b})\neq\emptyset$ by the assumption, and choose two
elements $\tilde{a},\tilde{b}$ in $\tilde{X}$ such that $\tilde{a}\in
p^{-1}(a)$ and $\tilde{b}\in p^{-1}(b)$. Then we have

$a\ast^{\tilde{m}}b=p(\tilde{a})\ast^{\tilde{m}}p(\tilde{b})=p(\tilde{a}\ast^{\tilde{m}}\tilde{b})=p(\tilde{a})=a,$

where $\tilde{m}$ denotes the type of $\tilde{X}$. The equation
$a\ast^{\tilde{m}}b=a$ implies the assertion. ∎

######  Proposition 4.3.

For quandles $X$ and $\tilde{X}$, suppose that there is a quandle covering
$p:\tilde{X}\to X$. If the type of $X$ is finite and $X$ is connected, then
the type of $\tilde{X}$ is equal to that of $X$.

###### Proof.

For an element $x\in X$, let $\bar{p}:(\bar{X},\bar{x})\to(X,x)$ be the
universal covering of the connected pointed quandle $(X,x)$, where $\bar{X}$
is also connected by [Eisermann2014quandle, Lemma 5.2 and Theorem 5.3]. For
notational simplicity, we denote $\mathrm{type}(X)$,
$\mathrm{type}(\tilde{X})$ and $\mathrm{type}(\bar{X})$ by $m$, $\tilde{m}$
and $\bar{m}$, respectively.

We show that $\bar{m}\mid m$. For any elements $\bar{a},\bar{b}\in\bar{X}$, we
have
$\bar{p}(\bar{a}\ast^{m}\bar{b})=\bar{p}(\bar{a})\ast^{m}\bar{p}(\bar{b})=\bar{p}(\bar{a})$.
This implies that the map $S^{m}_{\bar{b}}:\bar{X}\to\bar{X}$ is a quandle
homomorphism satisfying $\bar{p}=\bar{p}\circ S^{m}_{\bar{b}}$. By
[Eisermann2014quandle, Corollary 4.11], we see that $S^{m}_{\bar{b}}={\rm
Id}_{\bar{X}}$, since $S^{m}_{\bar{b}}$ stabilizes the element
$\bar{b}\in\bar{X}$ and $\bar{X}$ is connected. Thus we have $\bar{m}\mid m$
and in particular $\bar{m}<\infty$.

We set $I:=p^{-1}(x)$. We define a quandle operation $I\times\bar{X}$ by
$(i,\bar{a})\ast(j,\bar{b})=(i,\bar{a}\ast\bar{b})$ for $i,j\in I$ and
$\bar{a},\bar{b}\in\bar{X}$. Then by [Eisermann2014quandle, Proposition 5.6
and Lemma 6.4], the fundamental group $\pi_{1}(X,x)$ acts on $\bar{X}$ from
the left, and on $I$ from the right. We consider an equivalence relation,
denoted by $\sim$, on $I\times\bar{X}$ defined by $(i\cdot
g,\bar{a})\sim(i,g\cdot\bar{a})$ for $i\in I,\bar{a}\in\bar{X}$ and
$g\in\pi_{1}(X,x)$. Then the quotient set $I\times\bar{X}/\sim$ has a quandle
operation inherited from $I\times\bar{X}$. It follows from
[Eisermann2014quandle, Theorem 6.6] that the quandle $\tilde{X}$ is quandle
isomorphic to $I\times\bar{X}/\sim$. This implies that there is a surjective
quandle homomorphism $f:I\times\bar{X}\to\tilde{X}$.

We can easily see that ${\rm type}(I\times\bar{X})={\rm
type}(\bar{X})=\bar{m}$ by the definition of the quandle $I\times\bar{X}$. It
follows from $\bar{m}<\infty$ that Lemma 4.2 can be applied for the surjective
quandle homomorphism $f\colon I\times\bar{X}\to\tilde{X}$. Then it holds that
$\tilde{m}\mid\bar{m}$, and further by $\bar{m}\mid m$ we have $\tilde{m}\mid
m$ and in particular $\tilde{m}<\infty$. It also follows from
$\tilde{m}<\infty$ that Lemma 4.2 can be applied for the covering
$p:\tilde{X}\to X$. Then it holds that $m\mid\tilde{m}$, and further by
$\tilde{m}\mid m$ we have $\tilde{m}=m$. ∎

######  Corollary 4.4.

For quandles $X$ and $\tilde{X}$, suppose that there is a quandle covering
$p:\tilde{X}\to X$. If the type of $\tilde{X}$ is finite and $\tilde{X}$ is
connected, then the type of $X$ is equal to that of $\tilde{X}$.

###### Proof.

It follows from Lemma 4.2 that the type of $X$ is finite. Since $p$ is
surjective and the source $\tilde{X}$ of $p$ is connected, the target $X$ of
$p$ is also connected. Then we can apply Proposition 4.3 for the covering $p$.
∎

###  4.3. Universal coverings of the knot $n$-quandles

For an oriented $1$-knot $K$ and an integer $n>1$, we show the following
proposition needed to prove that the quandle covering associated with the
central extension $A^{n}_{K}\curvearrowright Q(\tau^{n}K)\to Q_{n}(K)$ is a
universal covering.

######  Theorem 4.5.

The knot quandle $Q(\tau^{n}K)$ of the $2$-knot $\tau^{n}K$ is simply
connected for an oriented $1$-knot $K$ and an integer $n>1$.

###### Proof.

Let $D$ be a diagram of the long knot obtained by cutting $K$ open. Let us
recall the presentation of $Q(\tau^{n}K)$ obtained from $D$; see
[Satoh2002surface]. Traveling along $D$ from one end point to the other end
point according to the orientation of $D$, we label the arcs of $D$
consecutively from $a_{0}$ to $a_{m}$, where $m$ is the number of the
crossings of $D$. For each $i$, let $\chi_{i}$ be the crossing of $D$ dividing
$a_{i-1}$ and $a_{i}$, and $\varepsilon_{i}$ the sign of the crossing
$\chi_{i}$. We denote by $\kappa_{i}$ the subscript of the label of the over
arc at the crossing $\chi_{i}$, that is, the label of that over arc is
$a_{\kappa_{i}}$ for each $i$. We define the relator $r_{i}$ by
$a_{i-1}\ast^{\varepsilon_{i}}a_{\kappa_{i}}=a_{i}$. Then the knot quandle
$Q(\tau^{n}K)$ has a following presentation:

$\langle a_{0},a_{1},\ldots,a_{m}\mid
r_{1},\ldots,r_{m},a_{i}\ast^{n}a_{0}=a_{i}\ (i=1,\ldots,m)\rangle.$

By Proposition 4.1, in order to prove that $Q(\tau^{n}K)$ is simply connected,
it is sufficient to show that any covering from $(Q(\tau^{n}K),a_{0})$ to any
pointed quandle $(X,x)$ is a universal covering of $(X,x)$. Let
$f:(Q(\tau^{n}K),a_{0})\to(X,x)$ and $p:(\tilde{X},\tilde{x})\to(X,x)$ be two
coverings. It follows from the relator $r_{i}$ of the presentation of
$Q(\tau^{n}K)$ that we have
$f(a_{i-1})*^{\varepsilon_{i}}f(a_{\kappa_{i}})=f(a_{i})$ for all $i$. We
define a map $\widetilde{f}:\\{a_{0},a_{1},\ldots,a_{m}\\}\to\tilde{X}$ as
follows. First, we set $\widetilde{f}(a_{0}):=\tilde{x}$. Since $p$ is a
covering, we can set
$\widetilde{f}(a_{i}):=\widetilde{f}(a_{i-1})\ast^{\varepsilon_{i}}f(a_{\kappa_{i}})$
at each crossing $\chi_{i}$ inductively; recall the first paragraph of
Subsection 4.1. Since we have
$p(\widetilde{f}(a_{i-1})\ast^{\varepsilon_{i}}f(a_{\kappa_{i}}))=p((\widetilde{f}(a_{i-1}))\ast^{\varepsilon_{i}}f(a_{\kappa_{i}}),$
it holds that $p(\widetilde{f}(a_{i}))=f(a_{i})$ for all $i$ by induction.
Moreover, it follows from $p(\widetilde{f}(a_{i}))=f(a_{i})$ that we have
$\widetilde{f}(a_{i})=\widetilde{f}(a_{i-1})\ast^{\varepsilon_{i}}\widetilde{f}(a_{\kappa_{i}})$
for all $i$.

Since the type of $Q(\tau^{n}K)$ is equal to $n$ by [TanakaTaniguchi, Theorem
3.3], we see that ${\rm type}(X)\mid n$ by Lemma 4.2 for the covering $f$.
Since $f$ is surjective and the source $Q(\tau^{n}K)$ of $f$ is connected, the
target $X$ of $f$ is also connected. Then, by Proposition 4.3 for the covering
$p$, it holds that ${\rm type}(\tilde{X})={\rm type}(X)$ and hence we have
${\rm type}(\tilde{X})\mid n$. Thus, it holds that
$\widetilde{f}(a_{i})\ast^{n}\widetilde{f}(a_{0})=\widetilde{f}(a_{i})$ for
all $i$. This implies that the map
$\widetilde{f}:\\{a_{0},a_{1},\ldots,a_{m}\\}\to\tilde{X}$ can be uniquely
extended to the quandle homomorphism
$\widetilde{f}:(Q(\tau^{n}K),a_{0})\to(\tilde{X},\tilde{x})$ such that
$p\circ\widetilde{f}=f$. ∎

Using Proposition 4.1 and Theorem 4.5 for the covering associated with the
central extention in Theorem 3.2, we have the following:

######  Theorem 4.6.

The covering $p\colon Q(\tau^{n}K)\to Q_{n}(K)$ associated with the central
extension $A^{n}_{K}\curvearrowright Q(\tau^{n}K)\to Q_{n}(K)$ is a universal
covering of $(Q_{n}(K),x)$ for any $x$ in $Q_{n}(K)$ for an oriented $1$-knot
$K$ and an integer $n>1$.

###  4.4. Induced group homomorphisms

We observe the induced group homomorphism $p_{\\#}:{\rm
As}(Q(\tau^{n}K))\to{\rm As}(Q_{n}(K))$ in Proposition 4.7, which will not be
used in this paper but has its own interest. Although it follows from
[Eisermann2014quandle, Example 1.25] that there exists a universal covering
$f:(\tilde{X},\tilde{x})\to(X,x)$ such that $\mathrm{As}(\tilde{X})$ is group
isomorphic to $\mathrm{As}(X)$ but the induced group homomorphism
$f_{\\#}:{\rm As}(\tilde{X})\to{\rm As}(X)$ is not a group isomorphism, we
have the following for the universal covering $p\colon Q(\tau^{n}K)\to
Q_{n}(K)$.

######  Proposition 4.7.

The map $p_{\\#}:{\rm As}(Q(\tau^{n}K))\to{\rm As}(Q_{n}(K))$ induced from the
universal covering $p\colon Q(\tau^{n}K)\to Q_{n}(K)$ is a group isomorphism.

In order to prove Proposition 4.7, we need the following:

######  Proposition 4.8.

Let $p:(\tilde{X},\tilde{x})\to(X,x)$ be a covering between pointed quandles.
If $X$ is connected and $p$ is a universal covering, then the following are
equivalent$:$

  * •

The quandle $\tilde{X}$ is simply connected.

  * •

The induced map $p_{\\#}:{\rm As}(\tilde{X})\to{\rm As}(X)$ is a group
isomorphism.

###### Proof.

We remark that if a quandle homomorphism $f:X\to Y$ is surjective, the induced
group homomorphism $f_{\\#}:{\rm As}(X)\to{\rm As}(Y)$ is also surjective. It
is shown in [Eisermann2014quandle, Proposition 5.10] that each covering
$p:(\tilde{X},\tilde{x})\to(X,x)$ induces the group homomorphism
$p_{\ast}:\pi_{1}(\tilde{X},\tilde{x})\to\pi_{1}(X,x)$ and that ${\rm
Ker}(p_{\\#}:{\rm As}(\tilde{X})\to{\rm As}(X))$ is equal to ${\rm
Ker}(p_{\ast})$.

By [Eisermann2014quandle, Theorem 5.22 and Proposition 5.23], we see that the
induced group homomorphism
$p_{\ast}:\pi_{1}(\tilde{X},\tilde{x})\to\pi_{1}(X,x)$ is the $0$-map, since
$X$ is connected and $p$ is universal. This implies that ${\rm Ker}(p_{\ast})$
is equal to $\pi_{1}(\tilde{X},\tilde{x})$. Hence, the quandle $\tilde{X}$ is
simply connected if and only if $p_{\\#}$ is a group isomorphism. ∎

###### Proof of Proposition 4.7.

By Theorem 4.6, the covering $p:Q(\tau^{n}K)\to Q_{n}(K)$ is a universal
covering. Since the knot quandle $Q(\tau^{n}K)$ is simply connected, it holds
that the induced group homomorphism $p_{\\#}:{\rm As}(Q(\tau^{n}K))\to{\rm
As}(Q_{n}(K))$ is a group isomorphism by Proposition 4.8. ∎

##  5\. The second quandle homology group of an $n$-quandle

We compute the second quandle homology group $H^{Q}_{2}(Q_{n}(K))$ of the knot
$n$-quandle $Q_{n}(K)$ for each oriented $1$-knot $K$ and each integer $n>1$.
Instead of computing according to the original definition in
[Carter2003quandle], we compute it by using the central extension
$A^{n}_{K}\curvearrowright Q(\tau^{n}K)\to Q_{n}(K)$, where $A^{n}_{K}$ is the
cyclic subgroup of $\pi_{1}(M^{n}_{K})$ generated by the preferred longitude
$l_{K}$ of $K$ as in Section 3. We recall here [Eisermann2014quandle,
Proposition 5.9], which connects an extension of $X$ with the fundamental
group of $X$ for a connected quandle $X$:

######  Proposition 5.1.

For a connected quandle $X$, if a quandle $\tilde{X}$ is an extension of $X$
by a group $\Lambda$, then the associated covering $p:\tilde{X}\to X$ induces
a natural group homomorphism $h:\pi_{1}(X,x)\to\Lambda$ for any $x\in X$. In
particular, $h$ is a group isomorphism if and only if $p$ is a universal
covering.

By this proposition, we have the following:

######  Proposition 5.2.

The fundamental group $\pi_{1}(Q_{n}(K),x)$ is group isomorphic to $A^{n}_{K}$
for an oriented $1$-knot $K$ and an integer $n>1$.

###### Proof.

Theorem 3.2 states that $Q(\tau^{n}K)$ is an extension of the connected
quandle $Q_{n}(K)$ by $A^{n}_{K}$. Since the associated covering
$p:Q(\tau^{n}K)\to Q_{n}(K)$ is universal by Theorem 4.6, it follows from
Proposition 5.1 that $\pi_{1}(Q_{n}(K),x)$ is group isomorphic to $A^{n}_{K}$
for any $x\in Q_{n}(K)$. ∎

It is known in [Eisermann2014quandle, Theorem 1.15] that the second quandle
homology group $H^{Q}_{2}(X)$ of $X$ is group isomorphic to the abelianization
of the fundamental group $\pi_{1}(X,x)$ for a connected pointed quandle
$(X,x)$. Hence we have:

######  Corollary 5.3.

The second quandle homology group $H^{Q}_{2}(Q_{n}(K))$ is group isomorphic to
$A^{n}_{K}$ for an oriented $1$-knot $K$ and an integer $n>1$.

By Corollary 5.3, in order to compute $H^{Q}_{2}(Q_{n}(K))$, it is sufficient
to determine the order of $l_{K}$, which is the generator of the cyclic
subgroup $A^{n}_{K}$ of $\pi_{1}(M^{n}_{K})$. We discuss the order of $l_{K}$
when $K$ is prime in Subsection 5.1 and when $K$ is composite in Subsection
5.2, respectively.

###  5.1. The case of prime knots

We compute the order of $l_{K}$ in $\pi_{1}(M^{n}_{K})$ when $K$ is prime.
First, we discuss the case where $\pi_{1}(M^{n}_{K})$ is finite. Next, we
discuss the case where $\pi_{1}(M^{n}_{K})$ is infinite.

####  5.1.1. The case where $\pi_{1}(M^{n}_{K})$ is finite

We compute the order of $l_{K}$ in $\pi_{1}(M^{n}_{K})$ when
$\pi_{1}(M^{n}_{K})$ is finite. By the elliptization theorem, the group
$\pi_{1}(M^{n}_{K})$ is finite if and only if the universal covering space of
$M^{n}_{K}$ is $S^{3}$; see [Aschenbrenner2015, Subsection 1.7]. According to
the Dunbar’s classification [Dunbar1988geometric] of spherical orbifolds, we
have the following:

######  Lemma 5.4.

For an oriented prime $1$-knot $K$ and an integer $n>1$, if the universal
covering space of $M^{n}_{K}$ is $S^{3}$, then one of the following holds:

  * $(1)$

A knot $K$ is a $2$-bridge knot $S(\alpha,\beta)$ and $n=2$.

  * $(2)$

A knot $K$ is a Montesinos knot
$M(b;\frac{1}{2},\frac{\beta_{2}}{3},\frac{\beta_{3}}{3})$ and $n=2$.

  * $(3)$

A knot $K$ is a Montesinos knot
$M(b;\frac{1}{2},\frac{\beta_{2}}{3},\frac{\beta_{3}}{5})$ and $n=2$.

  * $(4)$

A knot $K$ is the trefoil $3_{1}$ and $n=3,4,5$.

  * $(5)$

A knot $K$ is the cinquefoil $5_{1}$ and $n=3$.

Here we follow convention of the textbook [Burde2014knots] on knot theory for
notations of a $2$-bridge knot $S(\alpha,\beta)$ and a Montesinos knot
$M(b;\frac{\beta_{1}}{\alpha_{1}},\frac{\beta_{2}}{\alpha_{2}},\frac{\beta_{3}}{\alpha_{3}})$,
where the integers $\alpha,\alpha_{i},\beta$ and $\beta_{i}$ satisfy
$\alpha\geq 2,\alpha_{i}\geq 2$ and ${\rm gcd}(\alpha,\beta)={\rm
gcd}(\alpha_{i},\beta_{i})=1$ for $i=1,2,3$.

######  Proposition 5.5.

For an oriented prime $1$-knot $K$ and an integer $n>1$, if
$\pi_{1}(M^{n}_{K})$ is finite, then we have the following table for the order
of $l_{K}$.

$\begin{array}[]{c || c c c | c c c | c}\hline\cr\hline\cr\text{\rm Lemma~{}\ref{lemm:finite}}&(1)&(2)&(3)&&(4)&&(5)\\\ \hline\cr K&S(\alpha,\beta)&M(b;\frac{1}{2},\frac{\beta_{2}}{3},\frac{\beta_{3}}{3})&M(b;\frac{1}{2},\frac{\beta_{2}}{3},\frac{\beta_{3}}{5})&3_{1}&3_{1}&3_{1}&5_{1}\\\ n&2&2&2&3&4&5&3\\\ \hline\cr\text{\rm order of}\ l_{K}&1&2&4&2&4&10&6\\\ \hline\cr\hline\cr\end{array}$

###### Proof.

First, we discuss the cases (4) and (5) in Lemma 5.4.

Case (4). According to [Inoue2021knot], if $n$ is 3, 4 or 5, then it holds
that the order of $l_{3_{1}}$ is equal to 2, 4 or 10, respectively in
$\pi_{1}(M^{n}_{3_{1}})$.

Case (5). It is shown in [Crans2019finite] that the order of $Q_{3}(5_{1})$ is
equal to $20$. By Proposition 3.1 and Corollary 5.3, the group
$H^{Q}_{2}(Q_{3}(5_{1}))$ is the cyclic group whose order is equal to
$|Q(\tau^{3}(5_{1}))|/|Q_{3}(5_{1})|$. Since
$|Q(\tau^{3}(5_{1}))|=|\pi_{1}(M^{3}_{5_{1}})|=120$, it holds that
$H^{Q}_{2}(Q_{3}(5_{1}))$ is group isomorphic to $\mathbb{Z}/6\mathbb{Z}$,
which implies that the order of $l_{5_{1}}$ is equal to $6$ in
$\pi_{1}(M^{3}_{5_{1}})$.

Next we discuss the cases (1), (2) and (3) in Lemma 5.4. In these cases, we
have $n=2$ and the universal covering space of $M^{2}_{K}$ is $S^{3}$. Let
$p:S^{3}\to M^{2}_{K}$ be the universal covering of $M^{2}_{K}$. Then, the
inverse image $L:=p^{-1}(\tilde{K})$ of the branching set $\tilde{K}$ of
$M^{n}_{K}$ is an oriented $1$-link in $S^{3}$, where the orientation of $L$
is induced by the orientation of $\tilde{K}$. For a component $L^{\prime}$ of
$L$, we see that the restriction map $p|_{L^{\prime}}:L^{\prime}\to\tilde{K}$
is a covering and that the degree of $p|_{L^{\prime}}$ coincides with the
order of $l_{K}=[\tilde{K}]\in\pi_{1}(M^{2}_{K})$. Since $p$ is a covering, it
holds that the degree of $p|_{L^{\prime}}$ is equal to
$|\pi_{1}(M^{2}_{K})|/c_{L}$, where $c_{L}$ denote the number of the
components of $L$. Thus, it is sufficient to compute the number $c_{L}$ for
the oriented $1$-link $L=p^{-1}(\tilde{K})$ in $S^{3}$, where $L$ has been
studied in detail by Sakuma in [Sakuma1990geometries, Section 5].

Case (1). Suppose that $K$ is a 2-bridge knot $S(\alpha.\beta)$. Then we have
$c_{L}=\alpha$ by [Sakuma1990geometries, Theorem 5.3]. Since
$|\pi_{1}(M^{2}_{K})|=\alpha$, we see that the order of $l_{K}$ is equal to
$1$ in $\pi_{1}(M^{2}_{K})$.

Case (2). Suppose that $K$ is a Montesinos knot
$M(b;\frac{1}{2},\frac{\beta_{2}}{3},\frac{\beta_{3}}{3})$. We set
$\mu=|6(-b+\frac{1}{2}+\frac{\beta_{2}}{3}+\frac{\beta_{3}}{3})|$. Then we
have $c_{L}=12\mu$ and $|\pi_{1}(M^{2}_{K})|=24\mu$ by [Sakuma1990geometries,
Theorem 5.3 and Theorem 2.10], respectively. Hence we see that the order of
$l_{K}$ is equal to $2$ in $\pi_{1}(M^{2}_{K})$.

Case (3). Suppose that $K$ is a Montesinos knot
$M(b;\frac{1}{2},\frac{\beta_{2}}{3},\frac{\beta_{3}}{5})$. We set
$\mu=|30(-b+\frac{1}{2}+\frac{\beta_{2}}{3}+\frac{\beta_{3}}{5})|$. Then we
have $c_{L}=30\mu$ and $|\pi_{1}(M^{2}_{K})|=120\mu$ by [Sakuma1990geometries,
Theorem 5.3 and Theorem 2.10], respectively. Hence we see that the order of
$l_{K}$ is equal to $4$ in $\pi_{1}(M^{2}_{K})$. ∎

####  5.1.2. The case where $\pi_{1}(M^{n}_{K})$ is infinite

We compute the order of $l_{K}$ in $\pi_{1}(M^{n}_{K})$ when
$\pi_{1}(M^{n}_{K})$ is infinite.

######  Lemma 5.6.

For an oriented prime $1$-knot $K$ and an integer $n$ greater than $1$, if the
fundamental group $\pi_{1}(M^{n}_{K})$ is infinite, then the universal
covering space of $M^{n}_{K}$ is homeomorphic to $\mathbb{R}^{3}$.

###### Proof.

We note that by the equivalent sphere theorem [Meeks1980topology], the
branched covering space $M^{n}_{K}$ is irreducible. See also
[Sakuma1982regular, Theorem 2].

Suppose that $M^{n}_{K}$ contains an essential torus. Then we see that
$M^{n}_{K}$ is a Haken manifold. Waldhausen [Waldhausen1968irreducible] showed
that the universal covering space of a Haken manifold is $\mathbb{R}^{3}$.
Thus, the universal covering space of $M^{n}_{K}$ is $\mathbb{R}^{3}$.

Suppose that $M^{n}_{K}$ does not contain an essential torus. According to the
geometrization theorem, we see that $M^{n}_{K}$ admits one of the following
eight geometries:
$S^{3},S^{2}\times\mathbb{E},\mathbb{E}^{3},\mathbb{H}^{3},\mathbb{H}^{2}\times\mathbb{E},{\rm
Nil},{\rm Sol}$ and $\widetilde{{\rm SL}_{2}(\mathbb{R})}$. We note that each
underlying space of six geometries except the first two of the eight is
$\mathbb{R}^{3}$. Thus, the universal covering space of $M^{n}_{K}$ is either
$S^{3},S^{2}\times\mathbb{R}$ or $\mathbb{R}^{3}$. Since $M^{n}_{K}$ is
irreducible, we see that $\pi_{2}(M^{n}_{K})=0$ by
[Papakyriakopoulos1957Dehn]. This implies that the universal covering space of
$M^{n}_{K}$ is not homeomorphic to $S^{2}\times\mathbb{R}$. As we mentioned
just before Lemma 5.4, if the universal covering space of $M^{n}_{K}$ is
homeomorphic to $S^{3}$, the fundamental group $\pi_{1}(M^{n}_{K})$ is finite.
Hence, we see that the universal covering space of $M^{n}_{K}$ is
$\mathbb{R}^{3}$. ∎

######  Proposition 5.7.

For an oriented prime $1$-knot $K$ and an integer $n$ greater than $1$, if the
group $\pi_{1}(M^{n}_{K})$ is infinite, then $l_{K}$ is not a torsion element
in $\pi_{1}(M^{n}_{K})$, that is, the order of $l_{K}$ is infinite in
$\pi_{1}(M^{n}_{K})$.

###### Proof.

Since $M^{n}_{K}$ is irreducible and $\pi_{1}(M^{n}_{K})$ is infinite, we have
that $\pi_{1}(M^{n}_{K})$ is torsion free; see [Aschenbrenner2015, Subsection
3.2 (C.3)]. Thus, it is sufficient to show that $l_{K}$ is nontrivial in
$\pi_{1}(M^{n}_{K})$. Assume that $l_{K}$ is the identity element in
$\pi_{1}(M^{n}_{K})$. By Lemma 5.6, let $p:\mathbb{R}^{3}\to M^{n}_{K}$ be the
universal covering of $M^{n}_{K}$. By the assumption, the inverse image
$p^{-1}(\tilde{K})$ of the branching set $\tilde{K}$ of $M^{n}_{K}$ is a
$1$-link in $\mathbb{R}^{3}$, that is, each connected component of
$p^{-1}(\tilde{K})$ is homeomorphic to a circle, not a line. For a fixed
element $x$ in $p^{-1}(\tilde{K})$, let
$\widetilde{\varphi}:\mathbb{R}^{3}\to\mathbb{R}^{3}$ be a lift of the
canonical covering transformation $\varphi$ of $M^{n}_{K}$ such that
$\widetilde{\varphi}(x)=x$, and $G$ a subgroup of
$\langle\widetilde{\varphi}\rangle$ such that the order $|G|$ of $G$ is prime.
We denote $\textrm{Fix}(G)$ the fixed point set of $\mathbb{R}^{3}$ by $G$.
Since the connected component of $p^{-1}(\tilde{K})$ containing $x$ is fixed
by $G$, the set $\textrm{Fix}(G)$ must contain a circle. It follows from Smith
theory of group actions that $H_{1}(\textrm{Fix}(G);\mathbb{F}_{|G|})$ is
trivial, where $\mathbb{F}_{|G|}$ is a finite field of prime order $|G|$; see
[Bredon1972introduction, Theorem 5.2 in Chapter III] for example. This
contradicts the fact that $\textrm{Fix}(G)$ contains a circle. Thus, we see
that $l_{K}$ is nontrivial in $\pi_{1}(M^{n}_{K})$. ∎

###  5.2. The case of composite knots

We discuss the order of $l_{K}$ in $\pi_{1}(M^{n}_{K})$ when $K$ is composite.
Before that, we summarize when $l_{K}$ is trivial in $\pi_{1}(M^{n}_{K})$ for
an oriented prime $1$-knot $K$. It follows from Proposition 5.5 and 5.7 that
we have the following:

######  Corollary 5.8.

The element $l_{K}$ for an oriented prime $1$-knot $K$ is trivial in
$\pi_{1}(M^{2}_{K})$ if and only if $K$ is $2$-bridge knot. The element
$l_{K}$ for $K$ is always nontrivial in $\pi_{1}(M^{n}_{K})$ for each interger
$n>2$.

For oriented $1$-knots $K_{1}$ and $K_{2}$, let $K$ be the connected sum of
$K_{1}$ and $K_{2}$. Then we see that the fundamental group
$\pi_{1}(M^{n}_{K})$ is group isomorphic to the free product
$\pi_{1}(M^{n}_{K_{1}})\ast\pi_{1}(M^{n}_{K_{2}})$ for each integer $n>1$.
Moreover, by the definition of $l_{K}$, it holds that $l_{K}=l_{K_{1}}\cdot
l_{K_{2}}$ in $\pi_{1}(M^{n}_{K})$. Hence, we have the following:

######  Lemma 5.9.

Let $K$ be the connected sum of oriented $1$-knots $K_{1}$ and $K_{2}$. If
$l_{K_{2}}$ is trivial in $\pi_{1}(M^{n}_{K_{2}})$, then the order of $l_{K}$
in $\pi_{1}(M^{2}_{K})$ is equal to that of $l_{K_{1}}$ in
$\pi_{1}(M^{2}_{K_{1}})$. If $l_{K_{1}}$ and $l_{K_{2}}$ are nontrivial in
$\pi_{1}(M^{n}_{K_{1}})$ and $\pi_{1}(M^{n}_{K_{2}})$ respectively, then the
order of $l_{K}$ is infinite in $\pi_{1}(M^{n}_{K})$ for each integer $n>2$.

Combining Lemma 5.9 with Corollary 5.8, we have the following:

######  Proposition 5.10.

Let $K$ be an oriented composite $1$-knot. If $K$ is equivalent to an oriented
$1$-knot $K^{\prime}$ up to $2$-bridge knot summands, then the order of
$l_{K}$ in $\pi_{1}(M^{2}_{K})$ is equal to that of $l_{K^{\prime}}$ in
$\pi_{1}(M^{2}_{K^{\prime}})$. If an integer $n$ is greater than $2$, then the
order of $l_{K}$ is always infinite in $\pi_{1}(M^{n}_{K})$.

###  5.3. Second quandle homology groups

Summarizing the results (Proposition 5.5, 5.7 and 5.10), we can compute
$H^{Q}_{2}(Q_{n}(K))$ completely as follows.

######  Theorem 5.11.

The following hold for the knot $2$-quandles.

  * •

If an oriented $1$-knot $K$ is equivalent to the unknot up to $2$-bridge knot
summands, then $H^{Q}_{2}(Q_{2}(K))$ is trivial.

  * •

If $K$ is equivalent to a Montesinos knot
$M(b;\frac{1}{2},\frac{\beta_{2}}{3},\frac{\beta_{3}}{3})$ up to $2$-bridge
knot summands, then $H^{Q}_{2}(Q_{2}(K))$ is group isomorphic to
$\mathbb{Z}/2\mathbb{Z}$.

  * •

If $K$ is equivalent to a Montesinos knot
$M(b;\frac{1}{2},\frac{\beta_{2}}{3},\frac{\beta_{3}}{5})$ up to $2$-bridge
knot summands, then $H^{Q}_{2}(Q_{2}(K))$ is group isomorphic to
$\mathbb{Z}/4\mathbb{Z}$.

  * •

Otherwise, it holds that $H^{Q}_{2}(Q_{2}(K))$ is group isomorphic to
$\mathbb{Z}$.

######  Theorem 5.12.

The following hold for the knot $3$-quandles.

  * •

If an oriented $1$-knot $K$ is the unknot, then $H^{Q}_{2}(Q_{3}(K))$ is
trivial.

  * •

If $K$ is the trefoil, then $H^{Q}_{2}(Q_{3}(K))$ is group isomorphic to
$\mathbb{Z}/2\mathbb{Z}$.

  * •

If $K$ is the cinquefoil, then $H^{Q}_{2}(Q_{3}(K))$ is group isomorphic to
$\mathbb{Z}/6\mathbb{Z}$.

  * •

Otherwise, it holds that $H^{Q}_{2}(Q_{3}(K))$ is group isomorphic to
$\mathbb{Z}$.

######  Theorem 5.13.

The following hold for the knot $4$-quandles.

  * •

If an oriented $1$-knot $K$ is the unknot, then $H^{Q}_{2}(Q_{4}(K))$ is
trivial.

  * •

If $K$ is the trefoil, then $H^{Q}_{2}(Q_{4}(K))$ is group isomorphic to
$\mathbb{Z}/4\mathbb{Z}$.

  * •

Otherwise, it holds that $H^{Q}_{2}(Q_{4}(K))$ is group isomorphic to
$\mathbb{Z}$.

######  Theorem 5.14.

The following hold for the knot $5$-quandles.

  * •

If an oriented $1$-knot $K$ is the unknot, then $H^{Q}_{2}(Q_{5}(K))$ is
trivial.

  * •

If $K$ is the trefoil, then $H^{Q}_{2}(Q_{5}(K))$ is group isomorphic to
$\mathbb{Z}/10\mathbb{Z}$.

  * •

Otherwise, it holds that $H^{Q}_{2}(Q_{5}(K))$ is group isomorphic to
$\mathbb{Z}$.

######  Theorem 5.15.

The following hold for the knot $n$-quandles for each $n>5$.

  * •

If an oriented $1$-knot $K$ is the unknot, then $H^{Q}_{2}(Q_{n}(K))$ is
trivial.

  * •

Otherwise, it holds that $H^{Q}_{2}(Q_{n}(K))$ is group isomorphic to
$\mathbb{Z}$.

###  5.4. Characterizations of some knots

Eisermann [Eisermann2003unknot] showed that the second quandle homology group
of the knot quandle characterizes the unknot. As a consequence of our results
(Theorem 5.11, 5.12, 5.13, 5.14 and 5.15), an analogous result holds for the
knot $n$-quandle when $n>2$. Moreover, we can characterize some nontrivial
$1$-knots for $n=3,4$ and $5$.

######  Theorem 5.16.

The second quandle homology group of the knot $n$-quandle characterizes $(1)$
the unknot, the trefoil and the cinquefoil for $n=3$, $(2)$ the unknot and the
trefoil for $n=4$ or $5$, $(3)$ the unknot for $n>5$, and $(4)$ the unknot up
to $2$-bridge knot summands for $n=2$.

##  6\. Related results

In the final section, we collect the three related results: a relation of our
extension in Theorem 3.2 with the extension in [Inoue2021knot], the types of
the knot $n$-quandles, and the cardinalities of the finite knot $n$-quandles.

###  6.1. Relation with Inoue’s result

Inoue [Inoue2021knot] introduced the notion of the Schläfli quandles and
showed that the knot quandle $Q(\tau^{n}3_{1})$ of the $n$-twist spun trefoil
is a central extension of the Schläfli quandle $\\{3,n\\}$ by a cyclic group
$A_{n}$ for each $n>1$. He also computed the cyclic group $A_{n}$ for $n\leq
5$ and asked what $A_{n}$ is for $n>5$. We confirm that the central extension
in Theorem 3.2 can be considered as a generalization of his extension, and
also give an answer to his question.

We review the definition of the Schläfli quandle. Let us consider the regular
tessellation $\\{3,n\\}$ (of the spherical $2$-space for $2\leq n\leq 5$, the
Euclidean $2$-space for $n=6$ and the hyperbolic $2$-space for $n\geq 7$) in
the sence of Schläfli symbols. We denote the set of all vertices of
$\\{3,n\\}$ by $X_{n}$. The set $X_{n}$ is a quandle with an operation by
$v\ast w:=r_{w}(v)$, where $r_{w}$ is the rotation about $w$ by the angle
$2\pi/n$. We call the quandle $X_{n}$ the Schläfli quandle related to
$\\{3,n\\}$. Here we find a relation of $X_{n}$ with $Q_{n}(3_{1})$.

######  Lemma 6.1.

The Schläfli quandle $X_{n}$ is quandle isomorphic to the knot $n$-quandle
$Q_{n}(3_{1})$ of the trefoil $3_{1}$.

###### Proof.

By [Inoue2021knot, Lemma 6.4], the Schläfli quandle $X_{n}$ has a presentation

$\langle v,w\mid(v\ast w)\ast v=w,(w\ast v)\ast w=v,w\ast^{n}v=w\rangle.$

Since $r_{w}$ is the rotation by $2\pi/n$, it holds that $v\ast^{n}w=v$; see
also Remark 6.2 below. Thus, the Schläfli quandle $X_{n}$ has a presentation

$\langle v,w\mid(v\ast w)\ast v=w,(w\ast v)\ast
w=v,w\ast^{n}v=w,v\ast^{n}w=v\rangle.$

Since $\langle v,w\mid(v\ast w)\ast v=w,(w\ast v)\ast w=v\rangle$ is a
presentation of the knot quandle $Q(3_{1})$, the second presentation of
$X_{n}$ is equal to a presentation of the knot $n$-quandle $Q_{n}(3_{1})$. ∎

######  Remark 6.2.

Although we show $v*^{n}w=v$ in the proof of Lemma 6.1 by a geometric property
of the rotation $r_{w}$, we can also prove it as a consequence obtained from
the three relators $(v\ast w)\ast v=w$, $(w\ast v)\ast w=v$ and $w\ast^{n}v=w$
of the first presentation of $X_{n}$. In fact, we have

$\begin{split}v*^{n}w&=v*^{n}((v*w)*v)=(v*v)*^{n}((v*w)*v)=\Big{(}v*^{n}(v*w)\Big{)}*v\\\
&=\Big{(}((v*^{-1}w)*w)*^{n}(v*w)\Big{)}*v=\Big{(}((v*^{-1}w)*^{n}v)*w\Big{)}*v\\\
&=\Big{(}\big{(}(v*^{n}v)*^{-1}(w*^{n}v)\big{)}*w\Big{)}*v=\Big{(}(v*^{-1}w)*w\Big{)}*v\\\
&=v*v=v,\end{split}$

where the relator $(v\ast w)\ast v=w$ implies the first equality, and the
relator $w\ast^{n}v=w$ implies the third equality from the end. We note that
the relator $(w\ast v)\ast w=v$ is not used in the computation.

It follows from Lemma 6.1 that the extension $A^{n}_{3_{1}}\curvearrowright
Q(\tau^{n}3_{1})\to Q_{n}(3_{1})$ in Theorem 3.2 for the trefoil $3_{1}$ is
essentially the same as his extension $A_{n}\curvearrowright
Q(\tau^{n}\,3_{1})\to X_{n}$. Hence, by Proposition 5.7, we have that the
cyclic group $A_{n}$ is group isomorphic to $\mathbb{Z}$ for $n>5$, answering
Inoue’s question. Here we give an alternative direct proof.

###### A direct proof of $A^{n}_{3_{1}}\cong\mathbb{Z}$ for $n>5$.

Since $M^{n}_{3_{1}}$ is an irreducible 3-manifold with infinite fundamental
group for $n>5$, the group $\pi_{1}(M^{n}_{3_{1}})$ is torsion-free; see
[Aschenbrenner2015, Subsection 3.2 (C.3)]. Thus, it is sufficient to show that
$l_{3_{1}}$ is not equal to the identity element $e$ in
$\pi_{1}(M^{n}_{3_{1}})$. It is known in [rolfsen1976knots] that
$\pi_{1}(M^{n}_{3_{1}})$ has the following presentation

$\langle x_{1},\ldots,x_{n}\mid x_{i-1}=x_{i}x_{i-2}\ (i=1,\ldots,n)\rangle,$

where the indices of the relations are taken modulo $n$. It is also known in
[Inoue2021knot] that $l_{3_{1}}=x_{i}x_{i-1}^{-1}x_{i}^{-1}x_{i-1}$ in
$\pi_{1}(M^{n}_{3_{1}})$ for each $i$.

Assume that $l_{3_{1}}=e$ in $\pi_{1}(M^{n}_{3_{1}})$. For any $i$, we have

$x_{i}x_{i-1}^{-1}x_{i}^{-1}x_{i-1}=e\Leftrightarrow
x_{i}x_{i-1}^{-1}=x_{i-1}^{-1}x_{i}\Leftrightarrow x_{i}x_{i-1}=x_{i-1}x_{i}$

and

$x_{i}=x_{i+1}x_{i-1}\Leftrightarrow x_{i}=x_{i+2}x_{i}x_{i-1}\Leftrightarrow
x_{i}=x_{i+2}x_{i-1}x_{i}\Leftrightarrow e=x_{i+2}x_{i-1}.$

Furthermore, it holds that

$x_{i}=x_{i+1}x_{i-1}\Leftrightarrow
x_{i-1}x_{i}x_{i-1}^{-1}=x_{i-1}x_{i+1}\Leftrightarrow x_{i}=x_{i-1}x_{i+1}.$

It follows that $x_{i}$ commutes with $x_{j}$ for any $i$ and $j$, and hence
$\pi_{1}(M^{n}_{3_{1}})$ is abelian. Then the fundamental group
$\pi_{1}(M^{n}_{3_{1}})$ is group isomorphic to the first homology group
$H_{1}(M^{n}_{3_{1}})$. If $n$ is equal to $6k,6k\pm 1,6k\pm 2$ or $6k-3$,
then it is known in [rolfsen1976knots, Section 10.D] that
$H_{1}(M^{n}_{3_{1}})$ is group isomorphic to
$\mathbb{Z}^{2},0,\mathbb{Z}/3\mathbb{Z}$ or $(\mathbb{Z}/2\mathbb{Z})^{2}$,
respectively. On the other hand, for $n>5$, it is known in [Aschenbrenner2015]
that $\pi_{1}(M^{n}_{3_{1}})$ is infinite and is not group isomorphic to
$\mathbb{Z}^{2}$. This is a contradiction. ∎

###  6.2. Types of the knot $n$-quandles

We discuss the types of the knot $n$-quandles. By definition, we can see that
the type of $Q_{n}(K)$ is a divisor of $n$ for an oriented nontrivial $1$-knot
$K$ and an integer $n>1$. Although the type of $Q_{n}(K)$ has not been
determined exactly until now, we can prove the following:

######  Theorem 6.3.

The type of $Q_{n}(K)$ is equal to $n$ for any oriented nontrivial $1$-knot
$K$ and any integer $n>1$.

###### Proof.

The authors proved in [TanakaTaniguchi, Theorem 3.3] that the type of the knot
quandle $Q(\tau^{n}K)$ of $n$-twist spin $\tau^{n}K$ is equal to $n$. We
recall that $Q(\tau^{n}K)$ is connected and there is a covering from
$Q(\tau^{n}K)$ to $Q_{n}(K)$. By Corollary 4.4, it holds that ${\rm
type}(Q_{n}(K))={\rm type}(Q(\tau^{n}K))=n$. ∎

We give an alternative proof for [Winker1984quandles, Theorem 5.2.5] by
Theorem 6.3.

######  Theorem 6.4.

If the knot $n$-quandle $Q_{n}(K)$ of an oriented $1$-knot $K$ is trivial for
some integer $n>1$, then $K$ is trivial.

###### Proof.

If $K$ is nontrivial, then by Theorem 6.3 we have that the type of $Q_{n}(K)$
is equal to $n$ for any integer $n>1$, and hence that $Q_{n}(K)$ is nontrivial
for any integer $n>1$, which implies the contraposition. ∎

Using Theorem 6.4, we give a proof for the following fact, where it may be
folklore and a reference that contains its proof may not be found.

######  Theorem 6.5.

The knot quandle $Q(K)$ is infinite for any oriented nontrivial $1$-knot $K$.
Moreover, the type of $Q(K)$ is infinite.

###### Proof.

If a quandle is finite, then type of the quandle is also finite, since the
order of a bijection from the finite set to itself is finite. Thus it is
sufficient to show the latter statement. Suppose that the type of $Q(K)$ is
finite. Then, for a positive integer $n$ relatively prime to the type of
$Q(K)$, it follows from the Euclidean algorithm that the knot $n$-quandle
$Q_{n}(K)$ is trivial, and hence we have that $K$ is trivial by Theorem 6.4.
This is a contradiction. ∎

###  6.3. The finite knot $n$-quandles

We compute the cardinalities of all finite knot $n$-quandles completely. Hoste
and Shanahan [Hoste2017links] showed that the knot $n$-quandle $Q_{n}(K)$ is
finite if and only if $\pi_{1}(M^{n}_{K})$ is finite for an oriented $1$-knot
$K$ and an integer $n>1$. Since the finite knot $n$-quandles of torus knots
and $2$-bridge knots have already been studied in [Crans2019finite], it is
sufficient to compute the cardinalities of the finite knot $2$-quandles of
Montesinos knots except for $2$-bridge knots, that is, for Montesinos knots
$M(b;\frac{1}{2},\frac{\beta_{2}}{3},\frac{\beta_{3}}{3})$ and Montesinos
knots $M(b;\frac{1}{2},\frac{\beta_{2}}{3},\frac{\beta_{3}}{5})$; see Lemma
5.4.

######  Theorem 6.6.

The following hold for the finite knot $2$-quandles of Montesinous knots
except for $2$-bridge knots.

  * $(1)$

If $K$ is a Montesinos knot
$M(b;\frac{1}{2},\frac{\beta_{2}}{3},\frac{\beta_{3}}{3})$, then we have that
$|Q_{2}(K)|$ is equal to $12\mu_{1}$, where we set
$\mu_{1}=|6(-b+\frac{1}{2}+\frac{\beta_{2}}{3}+\frac{\beta_{3}}{3})|$.

  * $(2)$

If $K$ is a Montesinos knot
$M(b;\frac{1}{2},\frac{\beta_{2}}{3},\frac{\beta_{3}}{5})$, then we have that
$|Q_{2}(K)|$ is equal to $30\mu_{2}$, where we set
$\mu_{2}=|30(-b+\frac{1}{2}+\frac{\beta_{2}}{3}+\frac{\beta_{3}}{5})|$.

###### Proof.

Since $Q_{n}(K)$ is quandle isomorphic to the quandle
$A^{n}_{K}\backslash\pi_{1}(M^{n}_{K})$, the order of $Q_{n}(K)$ is equal to
$|\pi_{1}(M^{n}_{K})|/|A^{n}_{K}|$ if $Q_{n}(K)$ is finite.

We prove the case (1). Let $K$ be a Montesinos knot
$M(b;\frac{1}{2},\frac{\beta_{2}}{3},\frac{\beta_{3}}{3})$. Then it follows
from [Sakuma1990geometries, Theorem 2.10] that we have
$|\pi_{1}(M^{2}_{K})|=24\mu_{1}$. Since $|A^{2}_{K}|=2$ by Proposition 5.5, we
have $|Q_{2}(K)|=12\mu_{1}$.

We prove the case (2). Let $K$ be a Montesinos knot
$M(b;\frac{1}{2},\frac{\beta_{2}}{3},\frac{\beta_{3}}{5})$. Then it follows
from [Sakuma1990geometries, Theorem 2.10] that we have
$|\pi_{1}(M^{2}_{K})|=120\mu_{2}$. Since $|A^{2}_{K}|=4$ by Proposition 5.5,
we have $|Q_{2}(K)|=30\mu_{2}$. ∎

The cardinalities of the finite knot $n$-quandles for torus knots and 2-bridge
knots have already been computed in [Crans2019finite]; see also
[Hoste2017links] for the trefoil. Combining their works and our result, we
obtain the complete list of the cardinalities of all finite knot $n$-quandles
for oriented $1$-knots as follows.

$\begin{array}[]{c | c c c c c c c}\hline\cr\hline\cr K&S(\alpha,\beta)&M(b;\frac{1}{2},\frac{\beta_{2}}{3},\frac{\beta_{3}}{3})&M(b;\frac{1}{2},\frac{\beta_{2}}{3},\frac{\beta_{3}}{5})&3_{1}&3_{1}&3_{1}&5_{1}\\\ n&2&2&2&3&4&5&3\\\ \hline\cr|Q_{n}(K)|&\alpha&12\mu_{1}&30\mu_{2}&4&6&12&20\\\ \hline\cr\textrm{source}&\textrm{\cite[cite]{[\@@bibref{}{Crans2019finite}{}{}]}}&\textrm{Theorem~{}\ref{theo:order_Montesinos_n-quandle}}&\textrm{Theorem~{}\ref{theo:order_Montesinos_n-quandle}}&\textrm{\cite[cite]{[\@@bibref{}{Hoste2017links}{}{}]}}&\textrm{\cite[cite]{[\@@bibref{}{Hoste2017links}{}{}]}}&\textrm{\cite[cite]{[\@@bibref{}{Hoste2017links}{}{}]}}&\textrm{\cite[cite]{[\@@bibref{}{Crans2019finite}{}{}]}}\\\ \hline\cr\hline\cr\end{array}$

## Acknowledgement

The authors would like to thank Seiichi Kamada and Makoto Sakuma for their
helpful comments. The first author was supported by JSPS KAKENHI Grant Number
17K05242 and 21K03220. The second author was supported by JSPS KAKENHI Grant
Number 21J21482.

## References