# Complementation of Subquandles

K.J. Amsberry J.A. Bergquist T.A. Horstkamp M.H. Lee D.N. Yetter

(August 2022)

###### Abstract

Saki and Kiani proved that the subrack lattice of a rack $R$ is necessarily
complemented if $R$ is finite but not necessarily complemented if $R$ is
infinite. In this paper, we investigate further avenues related to the
complementation of subquandles. Saki and Kiani’s example of an infinite rack
without complements is a quandle, which is neither ind-finite nor profinite.
We provide an example of an ind-finite quandle whose subobject lattice is not
complemented, and conjecture that profinite quandles have complemented
subobject lattices. Additionally, we provide a complete classification of
subquandles whose set-theoretic complement is also a subquandle, which we call
strongly complemented, and provide a partial transitivity criterion for the
complementation in chains of strongly complemented subquandles. One technical
lemma used in establishing this is of independent interest: the inner
automorphism group of a subquandle is always a subquotient of the inner
automorphism group of the ambient quandle.

##  1 Introduction

Quandles were first introduced as a complete classical knot invariant up to
orientation reversal by Joyce [1] in 1982. In recent years, quandles and racks
have found numerous applications in knot theory and algebraic topology. Aside
from these applications, quandles are also interesting as an algebraic
structure. Indeed, quandles are an abstraction of groups, as any group is a
quandle under the operation of right conjugation. Quandles can also be
constructed from abelian groups using the Takasaki operation [6].

Like other algebraic structures, quandles have subobjects. These subobjects,
or subquandles, are difficult to study because quandles lack constants such as
an identity element, so that there is no notion of a kernel for a quandle
homomorphism, and thus no analogue of normal subgroups. Nonetheless, some
progress has been made towards understanding interesting properties of
subquandles. Recently Saki and Kiani [4] showed that the lattice of
subquandles of a finite quandle is necessarily complemented, but that infinite
quandles may have non-complemented subquandles. The present work investigates
several problems related to subquandle lattices, subquandle inner automorphism
groups, and subquandles whose set-theoretic complements are also subquandles.

##  2 Background and Definitions

######  Definition 2.1.

[1] A quandle is a set $Q$ equipped with binary operations $\rhd$ and
$\rhd^{-1}$ satisfying the following axioms for all $x,y,z\in Q$:

  1. Q1.

$x\rhd x=x$.

  2. Q2.

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

  3. Q3.

$(x\rhd y)\rhd z=(x\rhd z)\rhd(y\rhd z)$.

In other words, $\rhd$ is idempotent, right-invertible, and right-
distributive. Note that the mention of $\rhd^{-1}$ is not necessary if one is
content to describe quandles as models of a first order theory, rather than an
equational theory, as Q2 is equivalent to the bijectivity of $(-)\rhd y$ for
all $y\in Q$. Models of the algebraic structure $(Q,\rhd)$ satisfying only Q2
and Q3 are called racks.

######  Example 2.2.

The Tait quandle $(\mathbf{T}_{3},\rhd)$ is a quandle with underlying set
$\\{1,2,3\\}$ and operation given by:

$\rhd$ | $1$ | $2$ | $3$
---|---|---|---
$1$ | $1$ | $3$ | $2$
$2$ | $3$ | $2$ | $1$
$3$ | $2$ | $1$ | $3$
Figure 1: The multiplication table for $\mathbf{T}_{3}$

For more about multiplication tables of quandles, see [2].

######  Example 2.3.

The transpositions in the symmetric group $S_{n}$ form a quandle under the
operation of conjugation. In fact, the Tait quandle can be realized as the
transpositions in $S_{3}$.

######  Example 2.4.

For any group $G$, defining $\rhd$ by $x\rhd y=x^{-1}yx$ gives a quandle
structure on the same underlying set. This quandle is denoted
$\mathrm{Conj}(G)$.

$\mathrm{Conj}$ is, in fact, a functor from the category of groups to the
category of quandles, and admits a left adjoint, denoted $\mathrm{Adconj}$,
given by quotienting the free group on the underlying set of the quandle by
relations which equate the generator $q\rhd r$ with the right conjugate of the
generator $q$ by the generator $r$.

The following quandle is of particular relevance to us, as it will appear in
several later examples.

######  Example 2.5.

[6] Suppose $A$ is an abelian group. Then, if we define
$x\rhd_{\mathrm{dih}}y=2y-x$ for any $x,y\in A$, $(A,\rhd_{\mathrm{dih}})$
forms a quandle. We call this quandle a Takasaki quandle, denoted $T(A)$. The
operation $\rhd_{\mathrm{dih}}$ of $T(A)$ is called the dihedral action.

These last are examples of a special class of quandles studied by Takasaki [6]
long before Joyce’s work [1]:

######  Definition 2.6.

A kei or involutory quandle is a quandle $(K,\rhd)$ which satisfies $(x\rhd
y)\rhd y=x$ for all $x,y\in K$.

######  Definition 2.7.

Given quandles $(Q_{1},\rhd_{1}),(Q_{2},\rhd_{2})$, a quandle homomorphism is
a map $h:Q_{1}\to Q_{2}$ satisfying $h(x\rhd_{1}y)=h(x)\rhd_{2}h(y)$ for all
$x,y\in Q_{1}$.

A quandle isomorphism $\rho$ is a bijective quandle homomorphism, and $\rho$
is a quandle automorphism if it maps $Q\to Q$. The set of automorphisms of $Q$
form a group $\mathrm{Aut}(Q)$ under composition.

######  Definition 2.8.

For any quandle $Q$ and fixed $y\in Q$, the map $S_{y}:x\to x\rhd y$ is called
the symmetry at $y$. By the second and third axioms, symmetries are quandle
automorphisms. The subgroup of $\mathrm{Aut}(Q)$ generated by the symmetries
is denoted $\mathrm{Inn}(Q)$ and is called the inner automorphism group of
$Q$. Its elements — composites of symmetries and their inverses — are called
inner automorphisms.

By a result of Joyce [1], $\mathrm{Inn}(Q)$ is always a normal subgroup of
$\mathrm{Aut}(Q)$ .

######  Definition 2.9.

The orbit of $s\in Q$ is
$s\cdot\mathrm{Inn}(Q):=\\{\rho(s):\rho\in\mathrm{Inn}(Q)\\}$. Note
$\mathrm{Inn}(Q)$ acts on $Q$ on the right via functional application, i.e.,
$x\cdot\rho=\rho(x)$.

######  Definition 2.10.

A quandle $Q$ is connected if the action of $\mathrm{Inn}(Q)$ on $Q$ has only
one orbit, namely $Q$ itself. That is, for all $x\in Q$,
$x\cdot\mathrm{Inn}(Q)=Q.$

Both $\mathrm{Aut}(Q)$ and $\mathrm{Inn}(Q)$ have an additional relationship
with the quandle $Q$ as augmentation groups (as defined by Joyce [1]).

######  Definition 2.11.

An augmentation of a quandle $Q$ is a map $|\;|:Q\rightarrow G$ from $Q$ to a
group $G$, together with a right action of $G$ on $Q$ by quandle
homomorphisms, satisfying

  * •

$q\rhd r=q|r|$, and

  * •

$|\;|$ is $G$-equivariant with respect to this action and the action of $G$ on
itself by right conjugation.

Joyce [1] actually defines an “augmented quandle” as a $G$-set with the
equivariance condition and the condition that $p|p|=p$, then induces a quandle
structure on the set. For our purposes, where the quandle is already given,
the definition above is equivalent to the instance of Joyce’s notion which
induce the given quandle.

The map $q\rightarrow S_{q}$ gives an augmentation of any quandle $Q$ in
$\mathrm{Inn}(Q)$ and in $\mathrm{Aut}(Q)$. Also, any quandle $Q$ admits an
augmentation in $\mathrm{Adconj}(Q)$, with the map
$|\;|:Q\rightarrow\mathrm{Adconj}(Q)$ given by inclusion of generators. As
shown in [1], this is in fact the initial augmentation of $Q$, so that the
fifth author has always favored calling $\mathrm{Adconj}(Q)$, “the universal
augmentation group of $Q$.”

######  Definition 2.12.

A subset $Q^{\prime}$ of a quandle $(Q,\rhd)$ is a subquandle of $Q$ if it is
also a quandle under the restriction of the same operation. In particular, to
show $Q^{\prime}$ is a subquandle of $Q$, it suffices to show $Q^{\prime}$ is
closed under $\rhd|_{Q^{\prime}}.$ We write $Q^{\prime}\preccurlyeq Q$ and
$Q^{\prime}\prec Q$ to denote that $Q^{\prime}$ is a subquandle and a proper
subquandle of $Q$, respectively.

It is clear that orbits under the action of the inner automorphism group are
subquandles. In fact, such orbits are of prime importance to quandle theory,
as Ehrman et. al. [3] proved that they encode the structure of the ambient
quandle. We shall briefly reiterate their definitions for a more explicit
description of this result.

Let $(Q,\rhd)$ be any quandle, and let $\\{Q_{1},\dots,Q_{n}\\}$ be the set of
orbits of the action of $\mathrm{Inn}(Q)$ on $Q$. Suppose $Q$ is not
connected, as otherwise the following idea is not interesting. Then, Theorem
3.1 of Ehrman, et al. [3] states that $Q$ may be expressed uniquely as
$\\#(Q_{1},\dots,Q_{n},M)$ for some $n\times n$ matrix of quandle
homomorphisms $M$, where $\rhd_{i}=\rhd|_{Q_{i}\times Q_{i}}$, and the “semi-
disjoint union” operation $\\#$ is defined in Definition 2.13 below.

For each quandle $(Q_{i},\rhd_{i})$, let the universal augmentation map
$|\;|_{Q_{i}}:Q_{i}\to\mathrm{Adconj}(Q_{i})$ be such that
$x|y|_{Q_{i}}:=x\rhd_{i}y$ for any $x,y\in Q_{i}.$ Then, for each $1\leq i\leq
n$, by the universal property of $\mathrm{Adconj}(Q_{i})$, there is a unique
group homomorphism $\varphi_{i}:\mathrm{Adconj}(Q_{i})\to\mathrm{Aut}(Q_{i})$
such that the following diagram commutes (note $S:Q_{i}\to\mathrm{Inn}(Q_{i})$
via $S(q_{i})=S_{q_{i}}$ is the symmetry mapping):

${Q_{i}}$${\mathrm{Adconj}(Q_{i})}$${\mathrm{Aut}(Q_{i})}$$\scriptstyle{S}$$\scriptstyle{|\;|_{Q_{i}}}$$\scriptstyle{\varphi_{i}}$

Figure 2: Canonical group homomorphism $\varphi_{i}$

Call $\varphi_{i}$ the canonical group homomorphism from
$\mathrm{Adconj}(Q_{i})$ to $\mathrm{Aut}(Q_{i}).$

######  Definition 2.13.

[3] For each $1\leq i,j\leq n$, let
$g_{i,j}:\mathrm{Adconj}(Q_{i})\to\mathrm{Aut}(Q_{j})$ be a group homomorphism
such that $g_{i,i}$ is the canonical group homomorphism $\varphi_{i}$. Also
let $M=(g_{i,j})$ be the $n\times n$ matrix of group homomorphisms with
entries $g_{i,j}$. Then,

$\\#(Q_{1},\dots,Q_{n},M):=\left(\coprod_{i=1}^{n}Q_{i},\rhd\right),$

where $x\rhd y:=xg_{i,j}(|y|)$ for $x\in Q_{j}$ and $y\in Q_{i}.$

For arbitrary quandles $Q_{1},\dots,Q_{n}$, if $M$ satisfies compatibility
conditions given in [3], then $M$ is called a mesh, and
$\\#(Q_{1},\dots,Q_{n},M)$ with $\rhd$ as just defined is a quandle, called
the semi-disjoint union of $Q_{1},\ldots Q_{n}$ with respect to $M$.

######  Theorem 2.14.

[3] A matrix of group homomorphisms
$M=(g)_{ij}:\mathrm{Adconj}(Q_{i})\to\mathrm{Aut}(Q_{j})$ is a mesh if and
only if the following conditions are satisfied are met for all $i,j,k$
distinct with $1\leq i,j,k\leq n$,

  1. 1.

$(xg_{ji})(|y|_{Q_{j}})\rhd_{i}z=(x\rhd_{i}z)g_{ji}(|yg_{ij}(|z|_{Q_{i}})|)_{Q_{j}}$,

  2. 2.

$(xg_{ji}(|y|_{Q_{j}}))g_{ki}(|z|_{Q_{k}})=(xg_{k,i}(|z|_{Q_{k}}))(g_{ji}(|yg_{k,j}(|z|_{Q_{k}})|_{Q_{j}})).$

More importantly, the same authors proved that there is always a suitable mesh
to decompose a quandle into the semisdisjoint union of its orbits under the
inner automorphism group,

######  Theorem 2.15.

[3] For a quandle $Q$ with $n$ orbit decompositions under $\mathrm{Inn}(Q)$,
labeled $Q_{1},\dots,Q_{n}$, there exists a mesh $M$ such that

$Q\cong\\#(Q_{1},\dots,Q_{n},M).$

##  3 Subquandle Lattices

For a quandle $Q$ and $A\subseteq Q$, we write $\langle\langle
A\rangle\rangle$ to denote the subquandle of $Q$ generated by $A$. As shown by
Saki and Kiani in [4], the set of subquandles of any quandle $Q$ under
inclusion forms a lattice, which we denote by $\mathcal{L}(Q)$. In particular,
the two lattice operations meet and join, which will be described momentarily,
are well-defined. Given two subquandles $Q_{1},Q_{2}\preccurlyeq Q$, their
meet is $Q_{1}\wedge Q_{2}=Q_{1}\cap Q_{2}$ and their join is $Q_{1}\vee
Q_{2}=\langle\langle Q_{1}\cup Q_{2}\rangle\rangle$. Note that $Q_{1}\wedge
Q_{2}$ is the largest subquandle of in $Q$ contained in $Q_{1}$ and $Q_{2}$,
while $Q_{1}\vee Q_{2}$ is the smallest subquandle containing $Q_{1}$ and
$Q_{2}$.

It is also worth noting that $Q_{1}\vee Q_{2}$ is not necessarily $Q_{1}\cup
Q_{2}$, as $Q_{1}\cup Q_{2}$ may not be a subquandle of $Q$. For example, if
$Q$ is the Tait quandle on $\\{1,2,3\\}$, $\\{1\\}\preccurlyeq Q$ and
$\\{2\\}\preccurlyeq Q$ but $\\{1\\}\vee\\{2\\}=Q$.

######  Definition 3.1.

The subquandle lattice $\mathcal{L}(Q)$ of a quandle $Q$ is complemented if
for any $Q_{1}\preccurlyeq Q$, there exists some $Q_{2}\preccurlyeq Q$ such
that $Q_{1}\vee Q_{2}=Q$ and $Q_{1}\wedge Q_{2}=\emptyset$.

In [4], Saki and Kiani proved the following results.

######  Theorem 3.2.

[4] The subrack lattice of a finite rack is complemented.

###  3.1 Strongly Complemented Subquandles

In this section, we investigate a special class of complemented subquandles:
subquandles whose set-theoretic complement is also a subquandle. As we will
show, these subracks are particularly interesting in light of [3].

######  Definition 3.3.

Let $Q$ be a quandle and $Q^{\prime}\preccurlyeq Q$. Then $Q^{\prime}$ is
strongly complemented in $Q$ if $Q\setminus Q^{\prime}\preccurlyeq Q$.

To motivate this definition, let us consider three examples.

######  Example 3.4.

  1. 1.

Consider the integers $\mathbb{Z}$ under the dihedral operation
$\rhd_{\mathrm{dih}}$ as described in Example 2.5. Here, the even integers
form a subquandle, as do the odd integers. Thus, both the even integers and
the odd integers are strongly complemented subquandles of $\mathbb{Z}$.

  2. 2.

Let $G$ be a group, and let $\\{C_{\alpha}\\}_{\alpha\in\Lambda}$ be any
collection of conjugacy classes of $G$. Because the conjugacy class of $G$
form a partition of $G$,
$\bigcup_{\alpha\in\Lambda}C_{\alpha}\preccurlyeq\mathrm{Conj}(G)$ is a
strongly complemented subquandle of $G$.

  3. 3.

For a quandle $Q$, any union of orbits under the action of the inner
automorphism group $\mathrm{Inn}(Q)$ is strongly complemented in $Q$.

We will prove that strongly complemented subquandles of a quandle $Q$ can also
be described in terms of actions of $\mathrm{Inn}(Q)$ — this description will
be made explicit in Theorem 3.7. Before proceeding, we describe the two
different actions of $\mathrm{Inn}(Q)$ which will be used throughout this
paper. We shall make extensive use of both of these actions, often together,
so it is essential to have an efficient way of distinguishing between them.

The first action is the usual right action $\cdot:Q\times\mathrm{Inn}(Q)\to Q$
given by $q\cdot\sigma=\sigma(q).$ As described previously, the action of
$\mathrm{Inn}(Q)$ creates an orbit for each $q\in Q$, which we denote by
$q\cdot\mathrm{Inn}(Q)$. In a similar way, for $Q^{\prime}\preccurlyeq Q$, we
use the notation $Q^{\prime}\cdot\mathrm{Inn}(Q)$ to refer to the union of the
orbits of elements of $Q^{\prime}$, i.e.,
$Q^{\prime}\cdot\mathrm{Inn}(Q)=\bigcup_{q\in
Q^{\prime}}q\cdot\mathrm{Inn}(Q).$ We refer to this as the orbit closure of
$Q^{\prime}$. There is more to this name than simply the fact that the orbit
closure is closed under the action of $\mathrm{Inn}(Q)$. Recall:

######  Definition 3.5.

A closure operation on a partially ordered set $(X,\leq)$ is a function
$\mathrm{cl}:X\rightarrow X$ which is inflationary, that is
$x\leq\mathrm{cl}(x)$, and idempotent, that is
$\mathrm{cl}(\mathrm{cl}(x))=\mathrm{cl}(x)$.

We then have:

######  Proposition 3.6.

Orbit closure is a closure operation on the lattice of subquandles.

For the second action, suppose $Q^{\prime}\preccurlyeq Q$. The second action
is the right action
$\cdot:\mathcal{L}(Q)\times\mathrm{Inn}(Q)\to\mathcal{L}(Q)$ given by
$Q^{\prime}\cdot\sigma=\sigma[Q^{\prime}]$, or the image of $Q^{\prime}$ under
$\sigma$. Thus, we use the notation $[Q^{\prime}]\cdot\mathrm{Inn}(Q)$ to
refer to the set of images of $Q^{\prime}$ under elements of
$\mathrm{Inn}(Q)$, i.e.,
$[Q^{\prime}]\cdot\mathrm{Inn}(Q)=\\{Q^{\prime}\cdot\sigma:q\in
Q^{\prime}\\}.$ We use this notation to capture the action on
$\mathrm{Inn}(Q)$ on the subquandle lattice $\mathcal{L}(Q).$

Finally, for ease of notation, we will write $q\sigma$ instead of
$q\cdot\sigma$ and $Q^{\prime}\sigma$ instead of $Q^{\prime}\cdot\sigma$,
where $q\in Q$ and $Q^{\prime}\preccurlyeq Q$. With our two actions defined,
we state our classification of strongly complemented subquandles:

######  Theorem 3.7.

Let $Q$ be a quandle, and let $Q^{\prime}\preccurlyeq Q$. Denote the
subquandle lattice of $Q$ by $\mathcal{L}(Q)$. The following are equivalent:

  1. 1.

$Q\setminus Q^{\prime}\preccurlyeq Q$,

  2. 2.

$Q^{\prime}$ is a union of orbits under the action of $\mathrm{Inn}(Q)$ on
$Q$,

  3. 3.

$Q^{\prime}$ is a fixed point of the action of $\mathrm{Inn}(Q)$ on
$\mathcal{L}(Q)$,

  4. 4.

$Q=\\#(Q^{\prime},Q\setminus Q^{\prime},M)$ for a mesh $M$.

###### Proof.

First, suppose that $Q^{\prime}=Q\cdot\mathrm{Inn}(Q)$, so
$Q^{\prime}=\bigcup_{i\in\mathcal{I}}Q_{i}$ for some indexing set
$\mathcal{I}$ and each $Q_{i}$ an orbit of the action of $\mathrm{Inn}(Q)$ on
$Q$. Now, pick any $x\in Q^{\prime}$ and any $\sigma\in\mathrm{Inn}(Q)$. Then,
$x\in Q_{i}$ for some $i\in\mathcal{I}$, and $x\sigma\in Q_{i}\subseteq
Q^{\prime}$ because $Q_{i}$ is an orbit. Since $x\in Q^{\prime}$ was
arbitrary, $Q^{\prime}\sigma\subseteq Q^{\prime}$. Also, since $\sigma$ was an
arbitrary inner automorphism, $Q^{\prime}\sigma^{-1}\subseteq Q^{\prime}$,
which implies $x\sigma^{-1}\in Q^{\prime}$. Also, $x=(x\sigma^{-1})\sigma$, so
$Q^{\prime}\subseteq Q^{\prime}\sigma$. Thus, since $\sigma\in\mathrm{Inn}(Q)$
was arbitrary, $[Q^{\prime}]\cdot\mathrm{Inn}(Q)=\\{Q^{\prime}\sigma:q\in
Q^{\prime}\\}=\\{Q^{\prime}\\}$, as desired.

Now suppose that $[Q^{\prime}]\cdot\mathrm{Inn}(Q)=\\{Q^{\prime}\\}$, and pick
any $x\in Q^{\prime}$. Because $x\sigma\in Q^{\prime}$ for any
$\sigma\in\mathrm{Inn}(Q)$,
$x\cdot\mathrm{Inn}(Q)=\\{x\sigma:\sigma\in\mathrm{Inn}(Q)\\}\subseteq
Q^{\prime}$. Since $x\in Q^{\prime}$ was arbitrary,
$Q^{\prime}\cdot\mathrm{Inn}(Q)\subseteq Q^{\prime}.$ Moreover,
$Q^{\prime}\subseteq Q\cdot\mathrm{Inn}(Q),$ as each
$x=x\cdot\mathrm{id}_{\mathrm{Inn}(Q)}\in x\cdot\mathrm{Inn}(Q)$. Thus,
$Q^{\prime}=Q\cdot\mathrm{Inn}(Q)$, as desired.

Before proceeding, we establish a useful lemma:

######  Lemma 3.8.

If $Q^{\prime}$ is not a fixed point under the action of $\mathrm{Inn}(Q)$ on
$\mathcal{L}(Q)$, then there exists some $x\in Q\setminus Q^{\prime}$ such
that $Q^{\prime}\cdot S_{x}\neq Q^{\prime}$. In other words, there exists some
$x\in Q\setminus Q^{\prime}$ and some $y\in Q^{\prime}$ such that $y\rhd
x\not\in Q^{\prime}$.

###### Proof.

Since $[Q^{\prime}]\cdot\mathrm{Inn}(Q)\neq\\{Q^{\prime}\\}$, by definition,
there exists some $\sigma\in\mathrm{Inn}(Q)$ such that
$Q^{\prime}\cdot\sigma\neq Q^{\prime}$. Now, assume for the sake of
contradiction that all of the symmetries of elements in $Q$ fix $Q^{\prime}$.
Then the composition of any number of symmetries and symmetry inverses must
also fix $Q^{\prime}$. Hence, all elements of $\mathrm{Inn}(Q)$ must fix
$Q^{\prime}$, a contradiction. So, there must exist some $x$ whose symmetry
does not fix $Q^{\prime}$. Accordingly, there exists some $y\in Q^{\prime}$
such that $y\rhd x\not\in Q^{\prime}$. Clearly, $x$ cannot be in $Q^{\prime}$
as $Q^{\prime}$ is a subquandle and is closed under $\rhd$. Thus, $x\in
Q\setminus Q^{\prime}$, as desired. ∎

(2) $\Leftrightarrow$ (1): If $Q^{\prime}$ is a union of orbits, so is
$Q\setminus Q^{\prime}$. Since unions of orbits are subquandles, $Q^{\prime}$
is strongly complemented. For the converse, assume for the sake of
contradiction that $Q^{\prime}\preccurlyeq Q$ is strongly complemented but not
a union of orbits. Because $Q^{\prime}$ is not a union of orbits, $Q^{\prime}$
is not a fixed point under the action of $\mathrm{Inn}(Q)$ on
$\mathcal{L}(Q)$. Hence, by the previous lemma, there exists some $q\in
Q^{\prime}$ and some $y\in Q\setminus Q^{\prime}$ such that $q\rhd y\in
Q\setminus Q^{\prime}$. Because $Q\setminus Q^{\prime}$ is a subquandle, it is
closed under $\rhd^{-1}$, so $q=(q\rhd y)\rhd^{-1}y\in Q\setminus Q^{\prime}$.
Hence, $q\in Q^{\prime}\cap(Q\setminus Q^{\prime}),$ a contradiction.

(1) $\Leftrightarrow$ (4): If $Q^{\prime}$ is strongly complemented,
$Q^{\prime}$ is a union of orbits. Because Theorem 2.15 also holds for
semidisjoint unions of unions of orbits, we obtain
$Q=\\#(Q^{\prime},Q\setminus Q^{\prime},M)$ for some mesh $M$. The converse
follows directly from the definition of the semidisjoint union. ∎

As an immediate corollary, we have that the only strongly complemented
subquandles of a connected quandle $Q$ are $Q$ and the empty quandle, as $Q$
is only orbit of the action of $\mathrm{Inn}(Q)$ on $Q$. Another easy
corollary is that the the sublattice consisting of strongly complemented
subquandles is boolean algebra.

###  3.2 Chains of Strongly Complemented Subquandles

We now consider chains of strongly complemented quandles. Theorem 4.9 below
shows that any subquandle nested between a strongly complemented subquandle
and its ambient quandle is also strongly complemented. We also also provide a
transitivity criterion for strong complementedness in Theorem 4.13. First, we
establish a useful lemma of independent interest:

######  Lemma 3.9.

Let $Q$ be a quandle and suppose $Q^{\prime}\preccurlyeq Q$. Then, there
exists some subgroup $S_{Q^{\prime}}\leq\mathrm{Inn}(Q)$ and some normal
subgroup $K_{Q^{\prime}}\trianglelefteq S_{Q^{\prime}}$ such that
$\mathrm{Inn}(Q^{\prime})\cong S_{Q^{\prime}}/K_{Q^{\prime}}$.

In other words, the inner automorphism group of a subquandle is a subquotient
of the inner automorphism group of the ambient quandle.

###### Proof.

Define $S_{Q^{\prime}}:=\langle\\{S_{q}:q\in
Q^{\prime}\\}\rangle\leq\mathrm{Inn}(Q)$, where each
$S_{q}\in\mathrm{Inn}(Q).$ Consider the restriction map
$\tau:S_{Q^{\prime}}\to\mathrm{Inn}(Q^{\prime})$ defined by
$\tau(f)=f|_{Q^{\prime}}.$

######  Claim 3.10.

$\tau$ is well-defined.

###### Proof.

For $z\in Q$ and $\epsilon\in\\{-1,1\\}$, we write $S_{z}^{\epsilon}$ to mean
the symmetry $(x\mapsto x\rhd^{\epsilon}z)\in\mathrm{Inn}(Q).$ We need two
fairly obvious properties of bijections which preserve a given subset: First,
let $n\in\mathbb{N}$, and let $Q^{\prime}\subseteq Q$ be sets. For each $1\leq
i\leq n$, let $f_{i}:Q\to Q$ satisfy $f_{i}[Q^{\prime}]\subseteq Q^{\prime}.$
Then,

$(f_{n}\circ f_{n-1}\circ\cdots\circ
f_{1})|_{Q^{\prime}}=(f_{n}|_{Q^{\prime}})\circ(f_{n-1}|_{Q^{\prime}})\cdots\circ(f_{1}|_{Q^{\prime}}).$

Note that each restriction $f_{i}|_{Q^{\prime}}$ is well-defined by the
assumption $f_{i}[Q^{\prime}]\subseteq Q^{\prime}.$ Second, let
$Q^{\prime}\subseteq Q$ be sets, and suppose $f:Q\to Q$ is a bijection such
that $f[Q^{\prime}]=Q^{\prime}$. Then
$(f|_{Q^{\prime}})^{-1}=f^{-1}|_{Q^{\prime}}.$

The first fact allows us to freely move restrictions to $Q^{\prime}$ in and
out of compositions of symmetries of $Q$. The second fact allows us to freely
move inverses of symmetries of $Q$ in and out of restrictions to $Q^{\prime}$.
Now, to show $\tau$ maps every element of $S_{Q^{\prime}}$ into
$\mathrm{Inn}(Q^{\prime})$, suppose $f\in S_{Q^{\prime}}$ has a decomposition
$f=S_{x_{n}}^{\epsilon_{n}}\circ S_{x_{n-1}}^{\epsilon_{n-1}}\cdots\circ
S_{x_{1}}^{\epsilon_{1}}$ for some $n\in\mathbb{N}$ and each $x_{i}\in
Q^{\prime}$, $\epsilon_{i}\in\\{-1,1\\}$. Then,

$\displaystyle\tau(f)$ $\displaystyle=(S_{x_{n}}^{\epsilon_{n}}\circ
S_{x_{n-1}}^{\epsilon_{n-1}}\circ\cdots\circ
S_{x_{1}}^{\epsilon_{1}})|_{Q^{\prime}}$
$\displaystyle=(S_{x_{n}}^{\epsilon_{n}}|_{Q^{\prime}})\circ\cdots\circ(S_{x_{1}}^{\epsilon_{1}}|_{Q^{\prime}})$
$\displaystyle=(S_{x_{n}}|_{Q^{\prime}})^{\epsilon_{n}}\circ\cdots\circ(S_{x_{1}}|_{Q^{\prime}})^{\epsilon_{1}}\in\mathrm{Inn}(Q^{\prime}).$

Hence, $\tau[S_{Q^{\prime}}]\subseteq\mathrm{Inn}(Q^{\prime})$. ∎

######  Claim 3.11.

$\tau$ is surjective.

###### Proof.

Fix $\alpha\in\mathrm{Inn}(Q^{\prime})$, and suppose
$\alpha=T_{x_{n}}^{\epsilon_{n}}\circ\cdots\circ T_{x_{1}}^{\epsilon_{1}}$,
where each $T_{x_{i}}^{\epsilon_{i}}\in\mathrm{Inn}(Q^{\prime})$ is defined
via $T_{x_{i}}^{\epsilon_{i}}(x):=x\rhd^{\epsilon_{i}}x_{i}$. Note that
$S_{x_{n}}^{\epsilon_{n}}\circ\cdots\circ S_{x_{1}}^{\epsilon_{1}}$ is an
element of $\mathrm{Inn}(Q).$ Moreover,

$\displaystyle\tau(S_{x_{n}}^{\epsilon_{n}}\circ\cdots\circ
S_{x_{1}}^{\epsilon_{1}})$
$\displaystyle=(S_{x_{n}}^{\epsilon_{n}}\circ\cdots\circ
S_{x_{1}}^{\epsilon_{1}})|_{Q^{\prime}}$
$\displaystyle=S_{x_{n}}^{\epsilon_{n}}|_{Q^{\prime}}\circ\cdots\circ
S_{x_{1}}^{\epsilon_{1}}|_{Q^{\prime}}$
$\displaystyle=(S_{x_{n}}|_{Q^{\prime}})^{\epsilon_{n}}\circ\cdots\circ(S_{x_{1}}|_{Q^{\prime}})^{\epsilon_{1}}$
$\displaystyle=T_{x_{n}}^{\epsilon_{n}}\circ\cdots\circ
T_{x_{1}}^{\epsilon_{1}}=\alpha.$

∎

######  Claim 3.12.

$\tau$ is a group homomorphism.

###### Proof.

Let $f,g\in S_{Q^{\prime}}$ be arbitrary, and suppose
$f=S_{x_{n}}^{e_{n}}\circ\cdots\circ S_{x_{1}}^{e_{1}}$ and
$g=S_{y_{m}}^{f_{m}}\circ\cdots\circ S_{y_{1}}^{f_{1}},$ such that
$m,n\in\mathbb{N}$, and each $x_{i},y_{j}\in Q^{\prime}$,
$e_{i},f_{j}\in\\{-1,1\\}$. Since each symmetry fixes $Q^{\prime}$, $f$ and
$g$ must also fix $Q^{\prime}$, as they are compositions of symmetries and
inverse symmetries. Hence, by Fact 1, $\tau(f\circ g)=(f\circ
g)|_{Q^{\prime}}=f|_{Q^{\prime}}\circ g|_{Q^{\prime}}=\tau(f)\circ\tau(g)$. ∎

Therefore, $\tau$ is a surjective group homomorphism. By the First Isomorphism
Theorem, $S_{Q^{\prime}}/\ker(\tau)\cong\mathrm{Inn}(Q^{\prime}),$ as desired.
∎

######  Theorem 3.13.

Suppose $Q^{\prime\prime}\preccurlyeq Q^{\prime}\preccurlyeq Q$ and that
$Q^{\prime\prime}$ is strongly complemented within $Q$. Then,
$Q^{\prime\prime}$ is strongly complemented in $Q^{\prime}$.

###### Proof.

By Theorem 3.7, it suffices to show that $Q^{\prime\prime}$ is a fixed point
under the action of $\mathrm{Inn}(Q^{\prime})$ on $\mathcal{L}(Q^{\prime})$.
To that end, fix arbitrary $\sigma\in\mathrm{Inn}(Q^{\prime})$. Consider
$S_{Q^{\prime}}:=\langle S(Q^{\prime})\rangle\leq\mathrm{Inn}(Q)$, where
$S(Q^{\prime})$ denotes the image of $Q^{\prime}$ under the function which
maps elements to their symmetries. Since the restriction map
$\tau:S_{Q^{\prime}}\to\mathrm{Inn}(Q^{\prime})$ from Lemma 3.9 is surjective,
there exists some $\sigma^{\prime}\in S_{Q^{\prime}}$ extending $\sigma.$
Then, because $Q^{\prime\prime}$ is strongly complemented in $Q$, by Theorem
3.7, it is a fixed point of the action of $\mathrm{Inn}(Q)$ on
$\mathcal{L}(Q).$ In particular,
$Q^{\prime\prime}\sigma^{\prime}=Q^{\prime\prime}$. But, since
$\sigma^{\prime}$ extends $\sigma:Q^{\prime}\to Q^{\prime}$ and since
$Q^{\prime\prime}\subseteq Q^{\prime}$, we must also have
$Q^{\prime\prime}\sigma=Q^{\prime\prime}$. Thus, since
$\sigma\in\mathrm{Inn}(Q)$ was arbitrary, $Q^{\prime\prime}$ is indeed a fixed
point under the action of $\mathrm{Inn}(Q^{\prime})$ on
$\mathcal{L}(Q^{\prime})$, as desired. ∎

To establish our transitivity criterion, we will need several technical
lemmas.

######  Lemma 3.14.

Suppose $Q^{\prime\prime}\preccurlyeq Q^{\prime}\preccurlyeq Q$ such that
$Q^{\prime\prime}$ is strongly complemented within $Q^{\prime}$ and
$Q^{\prime}$ is strongly complemented within $Q$. Then,
$Q^{\prime\prime}\cdot\mathrm{Inn}(Q)\subseteq Q^{\prime}$.

###### Proof.

By the definition of $Q^{\prime\prime}\cdot\mathrm{Inn}(Q),$ it suffices to
show that $q\cdot\mathrm{Inn}(Q)\subseteq Q^{\prime}$ for any $q\in
Q^{\prime\prime}.$ To that end, fix arbitrary $q\in Q^{\prime\prime}$ and
$\sigma\in\mathrm{Inn}(Q)$. Since $Q^{\prime}$ is strongly complemented within
$Q$, by Theorem 3.7, $[Q^{\prime}]\cdot\mathrm{Inn}(Q)=\\{Q^{\prime}\\},$ so
$Q^{\prime}f=Q^{\prime}$ for any $f\in\mathrm{Inn}(Q)$. In particular, since
$q\in Q^{\prime}$ and $\sigma\in\mathrm{Inn}(Q)$, $q\sigma\in Q^{\prime}$. But
since $\sigma\in\mathrm{Inn}(Q)$ was arbitrary, it follows that
$q\cdot\mathrm{Inn}(Q)\subseteq Q^{\prime}$, as desired. ∎

It turns out that $Q\setminus(Q^{\prime\prime}\cdot\mathrm{Inn}(Q))$ will
complement $Q^{\prime\prime}$ in $\mathcal{L}(Q)$, which the above lemma will
help show.

######  Lemma 3.15.

Suppose $Q^{\prime\prime}\preccurlyeq Q^{\prime}\preccurlyeq Q$ such that
$Q^{\prime\prime}$ is strongly complemented within $Q^{\prime}$ and
$Q^{\prime}$ is strongly complemented within $Q$. Then, for any
$\gamma\in\mathrm{Inn}(Q)$, $Q^{\prime\prime}\gamma$ is strongly complemented
within $Q^{\prime}$.

###### Proof.

Since $Q^{\prime\prime}$ is strongly complemented in $Q^{\prime}$ and
$Q^{\prime}$ is strongly complemented in $Q$, by Theorem 3.7, we have
$Q^{\prime}\setminus Q^{\prime\prime}\preccurlyeq Q^{\prime}$ and
$[Q^{\prime}]\cdot\mathrm{Inn}(Q)=\\{Q^{\prime}\\}$. Fix arbitrary
$\gamma\in\mathrm{Inn}(Q)$. Since the images of subquandles of $Q$ under inner
automorphisms of $Q$ are still subquandles of $Q$, $(Q^{\prime}\setminus
Q^{\prime\prime})\gamma\preccurlyeq Q$. Also, because $Q^{\prime}\setminus
Q^{\prime\prime}\subset Q^{\prime}$ and $\gamma:Q^{\prime}\to Q^{\prime},$
$(Q^{\prime}\setminus Q^{\prime\prime})\gamma\subset Q^{\prime}$. But, since
we showed $(Q^{\prime}\setminus Q^{\prime\prime})\gamma$ is a quandle,
$(Q^{\prime}\setminus Q^{\prime\prime})\gamma\preccurlyeq Q^{\prime}$. We also
know $Q^{\prime}\gamma=Q^{\prime}$ because
$[Q^{\prime}]\cdot\mathrm{Inn}(Q)=\\{Q^{\prime}\\}$. Also, since $\gamma$ is a
bijection, $(Q^{\prime}\setminus
Q^{\prime\prime})\gamma=Q^{\prime}\gamma\setminus Q^{\prime\prime}\gamma$.
Combining the previous two statements, we have $(Q^{\prime}\setminus
Q^{\prime\prime})\gamma=Q^{\prime}\setminus Q^{\prime\prime}\gamma\preccurlyeq
Q^{\prime}$. Thus, since its set-theoretic complement is a subquandle of
$Q^{\prime}$, $Q^{\prime\prime}\gamma$ is strongly complemented within
$Q^{\prime}$. ∎

######  Lemma 3.16.

Suppose $Q^{\prime\prime}\preccurlyeq Q^{\prime}\preccurlyeq Q$ such that
$Q^{\prime\prime}$ is strongly complemented within $Q^{\prime}$ and
$Q^{\prime}$ is strongly complemented within $Q$. Let
$\sigma\in\mathrm{Inn}(Q)$. Then there exists some
$\sigma^{\prime}\in\langle\\{S_{x}:x\in Q\setminus
Q^{\prime}\\}\rangle\leq\mathrm{Inn}(Q)$ such that
$Q^{\prime\prime}\sigma=Q^{\prime\prime}\sigma^{\prime}$.

###### Proof.

Since $\sigma\in\mathrm{Inn}(Q)$, we may fix a representation
$\sigma=S_{x_{1}}S_{x_{2}}\cdots S_{x_{n}}$ of $\sigma$ as a string of
symmetries, where each $x_{i}\in Q$ and $\epsilon_{i}\in\\{-1,1\\}$. Our
strategy will be to show that by removing all instances of symmetries at
elements in $Q^{\prime}$ which appear in $S_{x_{1}}S_{x_{2}}\cdots S_{x_{n}}$,
we obtain another element $\sigma^{\prime}\in\mathrm{Inn}(Q)$ which acts the
same as $\sigma$ upon $Q^{\prime\prime}$ in $\mathcal{L}(Q)$.

A systematic way to construct $\sigma^{\prime}$ from the string
$S_{x_{1}}S_{x_{2}}\cdots S_{x_{n}}$ is by checking if $x_{1}$ is in
$Q^{\prime}$ and removing $S_{x_{1}}$ iff it is, then similarly checking
$x_{2}$ and possibly removing $S_{x_{2}}$, and so on. Assume for the sake of
contradiction that in doing this process, there is an index $k$ such that
$x_{k}\in Q^{\prime}$ and that when $S_{x_{k}}$ is removed from the string
$S_{x_{1}}S_{x_{2}}\cdots S_{x_{n}}$, the resulting inner automorphism acts
differently than $\sigma$ does upon $Q^{\prime\prime}$. Let $i$ be the first
such index $k$, so for all $1\leq j<i$ such that $x_{j}\in Q^{\prime}$,
removing $S_{x_{j}}$ doesn’t change the resulting action on
$Q^{\prime\prime}$, but removing $S_{x_{i}}$ does.

Then, let $\gamma=S_{x_{1}}\cdots S_{x_{i-1}}$ and
$\gamma^{\prime}=S_{x_{i+1}}\cdots S_{x_{n}}$, so we have $\sigma=\gamma
S_{x_{i}}\gamma^{\prime}$. By the associativity of group action, we have
$Q^{\prime\prime}\sigma=((Q^{\prime\prime}\gamma)S_{x_{i}})\gamma^{\prime}$.
Moreover, by the previous lemma, $Q^{\prime\prime}\gamma$ is strongly
complemented within $Q^{\prime}$. Hence, $Q^{\prime\prime}\gamma$ is a fixed
point of the action of $\mathrm{Inn}(Q^{\prime})$ on
$\mathcal{L}(Q^{\prime})$, as follows from Theorem 3.7. Additionally, since
$x_{i}\in Q^{\prime}$, we know that $S_{x_{i}}|_{Q^{\prime}}$ is an element of
$\mathrm{Inn}(Q^{\prime})$, and moreover, we have
$xS_{x_{i}}|_{Q^{\prime}}=xS_{x_{i}}$ for any $x\in Q^{\prime}$. From this and
the previous statement, it follows that
$(Q^{\prime\prime}\gamma)S_{x_{i}}=(Q^{\prime\prime}\gamma)S_{x_{i}}|_{Q^{\prime}}=Q^{\prime\prime}\gamma$.
Now, substituting, we have the identity

$Q^{\prime\prime}\sigma=((Q^{\prime\prime}\gamma)S_{x_{i}})\gamma^{\prime}=(Q^{\prime\prime}\gamma)\gamma^{\prime}=Q^{\prime\prime}(\gamma\gamma^{\prime}).$

Hence, removing $S_{x_{i}}$ doesn’t change the action of the resulting string
(namely $\gamma\gamma^{\prime}$) on $Q^{\prime\prime}$, contradicting our
assumption that it did. So, no such index $i$ can exist, and hence removing
all symmetries at elements in $Q^{\prime}$ from $S_{x_{1}}S_{x_{2}}\cdots
S_{x_{n}}$ also doesn’t resulting string’s, namely $\sigma^{\prime}$’s, action
on $Q^{\prime\prime}$. ∎

######  Theorem 3.17.

Let $Q^{\prime\prime}\preccurlyeq Q^{\prime}\preccurlyeq Q$ such that
$Q^{\prime\prime}$ is strongly complemented within $Q^{\prime}$, while
$Q^{\prime}$ is strongly complemented within $Q$. Then $Q^{\prime\prime}$ is
complemented within $Q$.

###### Proof.

Since $Q^{\prime\prime}\cdot\mathrm{Inn}(Q)$ is a union of orbits by
construction, by Theorem 3.7, it follows that
$Q\setminus(Q^{\prime\prime}\cdot\mathrm{Inn}(Q))\preccurlyeq Q$. And since
$Q^{\prime\prime}\subseteq Q^{\prime\prime}\cdot\mathrm{Inn}(Q)$,
$Q^{\prime\prime}\cap
Q\setminus(Q^{\prime\prime}\cdot\mathrm{Inn}(Q))=\emptyset$. Thus, to show
$Q\setminus(Q^{\prime\prime}\cdot\mathrm{Inn}(Q))$ is a complement for
$Q^{\prime\prime}$ in $\mathcal{L}(Q)$, it remains to show that
$Q^{\prime\prime}\vee Q\setminus(Q^{\prime\prime}\cdot\mathrm{Inn}(Q))=Q$. To
do this, it suffices to show that
$Q^{\prime\prime}\cdot\mathrm{Inn}(Q)\subseteq Q^{\prime\prime}\vee
Q\setminus(Q^{\prime\prime}\cdot\mathrm{Inn}(Q))$ — if this holds, we will
have

$\underbrace{(Q\setminus(Q^{\prime\prime}\cdot\mathrm{Inn}(Q)))\cup(Q^{\prime\prime}\cdot\mathrm{Inn}(Q))}_{Q}\subseteq
Q^{\prime\prime}\vee Q\setminus(Q^{\prime\prime}\cdot\mathrm{Inn}(Q)),$

which implies $Q=Q^{\prime\prime}\vee
Q\setminus(Q^{\prime\prime}\cdot\mathrm{Inn}(Q)).$

To that end, choose any $x\in Q^{\prime\prime}\cdot\mathrm{Inn}(Q)$. Then,
there is some $\sigma\in\mathrm{Inn}(Q)$ such that $x\in
Q^{\prime\prime}\sigma$. By Lemma 3.16, there exists some
$\sigma^{\prime}\in\langle\\{S_{x}:x\in Q\setminus Q^{\prime}\\}\rangle$ such
that $Q^{\prime\prime}\sigma^{\prime}=Q^{\prime\prime}\sigma$. Fix
$q_{1},\dots,q_{n}\in Q\setminus Q^{\prime}$ and
$\epsilon_{1},\dots,\epsilon_{n}\in\\{-1,1\\}$ such that
$\sigma^{\prime}=S_{q_{1}}^{\epsilon_{1}}\dots S_{q_{n}}^{\epsilon_{n}}$. By
the definition of the action of $\mathrm{Inn}(Q)$,
$Q^{\prime\prime}\sigma^{\prime}=\\{q\rhd^{\epsilon_{1}}q_{1}\dots\rhd^{\epsilon_{n}}q_{n}:q\in
Q^{\prime\prime}\\}$. Thus,
$x=q\rhd^{\epsilon_{1}}q_{1}\dots\rhd^{\epsilon_{n}}q_{n}$ for some $q\in
Q^{\prime\prime}$. In particular, $x\in\langle\langle
Q^{\prime\prime}\cup(Q\setminus Q^{\prime})\rangle\rangle$. Additionally, we
know $Q^{\prime\prime}\cdot\mathrm{Inn}(Q)\subseteq Q^{\prime}$ by Lemma 3.14,
so $Q\setminus Q^{\prime}\subseteq
Q\setminus(Q^{\prime\prime}\cdot\mathrm{Inn}(Q))$. Therefore,

$\displaystyle x\in\langle\langle Q^{\prime\prime}\cup(Q\setminus
Q^{\prime})\rangle\rangle$ $\displaystyle\subseteq\langle\langle
Q^{\prime\prime}\cup(Q\setminus(Q^{\prime\prime}\cdot\mathrm{Inn}(Q)))\rangle\rangle=Q^{\prime\prime}\vee
Q\setminus(Q^{\prime\prime}\cdot\mathrm{Inn}(Q))$

by the definition of join. But, since $x\in
Q^{\prime\prime}\cdot\mathrm{Inn}(Q)$ was arbitrary, it follows that
$Q^{\prime\prime}\cdot\mathrm{Inn}(Q)\subseteq Q^{\prime\prime}\vee
Q\setminus(Q^{\prime\prime}\cdot\mathrm{Inn}(Q))$, as desired. ∎

This theorem not only shows that $Q^{\prime\prime}$ is complemented when it
lies two links down in a finite chain of strongly complemented subquandles of
$Q$ — it also provides an explicit complement for $Q^{\prime\prime}$, namely
$Q\setminus(Q\cdot\mathrm{Inn}(Q))$. Moreover, since this complement of
$Q^{\prime\prime}$ is a union of orbits, it is strongly complemented within
$Q$, namely by $Q\cdot\mathrm{Inn}(Q)$.

An natural question to consider is whether complementedness is carried down
further links in a finite chain of strongly complemented subquandles. More
explicitly, suppose we have a chain of strongly complemented subquandles,
$Q_{3}\preccurlyeq Q_{2}\preccurlyeq Q_{1}\preccurlyeq Q$. A natural attempt
would be to consider $Q\setminus(Q_{3}\cdot\mathrm{Inn}(Q))$ as a possible
complement of $Q_{3}$ in $\mathcal{L}(Q)$. However, the join requirement for
this to be a complement is not nearly as straightforward to verify. It is our
hope that future research will prove that complementedness is transitive
across chains of strongly complemented of arbitrary finite length, or else
find a counterexample.

###  3.3 Infinite Quandles

The question of whether the subquandle lattice of a finite quandle is
complemented has been settled by previous research. More explicitly, since
quandles are racks, the following theorem shows that the subquandle lattices
of finite quandles are necessarily complemented.

######  Theorem 3.18.

[4] Let $R$ be a finite rack. Then the following hold:

  1. 1.

The intersection of all the maximal subracks of $R$ is empty.

  2. 2.

Suppose that $R_{1},R_{2}\preccurlyeq R$. Then $R_{1}$ is complemented in
$\mathcal{R}(R_{1}\vee R_{2})$ by some subset of $R_{2}$.

######  Corollary 3.19.

[4] If $R$ is a finite rack, then $\mathcal{R}(R)$ is complemented.

The natural follow-up to this result is whether infinite racks, and hence
infinite quandles, have necessarily complemented subobject lattices. Some
infinite racks, such as trivial racks, possess complemented subobject
lattices. However, the following theorem demonstrates that this is not always
the case:

######  Theorem 3.20.

[4] The rack $(\mathbb{Q},\rhd_{\mathrm{dih}})$ does not have a complemented
subrack lattice. In particular, the singleton $\\{0\\}\preccurlyeq\mathbb{Q}$
lacks a complement.

To understand the proof of Theorem 3.20, we first need the following
definition.

######  Definition 3.21.

We say $M\prec Q$ is maximal if there is no $N\prec Q$ such that $M\prec N$.
Equivalently, $Q$ is the only proper extension of $M$ in $\mathcal{L}(Q).$

Observe that since singletons are always subquandles, any maximal subquandle
$M$ of a quandle $Q$ must be complemented by $\\{x\\}$ for any $x\in
Q\setminus M$. However, this is not true for racks, as racks may not
necessarily have singleton subracks.

The proof of Theorem 3.20 relies the containment of all subracks of a finite
rack $R$ in some maximal subrack of $R$ [4]. In particular, the “top layer” of
the subrack lattice of $R$ is composed of a set of pairwise-disjoint maximal
subracks under which every other proper subrack of $R$ lies. However, this
strategy is no longer useful for infinite quandles, as the following result
shows that maximal subquandles may not even exist.

######  Proposition 3.22.

$(\mathbb{Q},\rhd_{\mathrm{dih}})$ has no maximal subquandles.

###### Proof.

Fix arbitrary $M\prec\mathbb{Q}$. We use the fact that maximal subquandles
must be complemented by every singleton in their set-theoretic complement and
show that this cannot hold for $M$. We will also use an analagous fact for
groups:

######  Lemma 3.23.

Let $G$ be a group. Then, $M\lneq G$ is maximal iff $\langle
M\cup\\{x\\}\rangle=G$ for each $x\in G\setminus M$.

###### Proof.

The forward direction is clear. Now suppose $\langle M\cup\\{x\\}\rangle=G$
for each $x\in G\setminus M$. Fix arbitrary $N\leq G$ properly extending $M$.
Then, there is some nonempty $S\subseteq G\setminus M$ such that $N=\langle
M\cup S\rangle.$ Fixing some $s\in S$, we see

$G=\langle M\cup\\{s\\}\rangle\subseteq\langle M\cup S\rangle=N,$

so $N=G$. Hence, the only subgroup of $G$ properly extending $M$ is $G$
itself, so $M$ is maximal. ∎

By Theorem 3.20, $\\{0\\}$ has no complement in $\mathcal{L}(\mathbb{Q})$, so
$\\{0\\}$ is not a complement of $M$. Thus, if $0\notin M$, $M$ cannot be
maximal. Now suppose $0\in M.$ By Lemma 5.1 in [4], $M$ is a subgroup of
$(\mathbb{Q},+).$ Since $\mathbb{Q}$ has no maximal subgroups, by Lemma 3.23,
there is some $q\in\mathbb{Q}\setminus M$ such that $\langle
M\cup\\{q\\}\rangle$ is a proper subgroup of $\mathbb{Q}.$ Hence,
$\langle\langle M\cup\\{q\\}\rangle\rangle\subseteq\langle
M\cup\\{q\\}\rangle\subset Q$, so $\\{q\\}$ is not a complement of $M$ in
$\mathcal{L}(Q)$. Therefore, $M$ cannot be maximal. ∎

###  3.4 Ind-finite Quandles

Although $\mathbb{Q}$ is an infinite quandle with a non-complemented
sublattice, $\mathbb{Q}$ does not have any clear relationship to finite
quandles, which do have complemented sublattices. In this section, we capture
one type of relationship an infinite quandle can have with finite quandles by
examining ind-finite quandles. We also provide an example which shows that the
subquandle lattices of ind-finite quandles are not necessarily complemented.

######  Definition 3.24.

A set $(A,\leq)$ is directed if for all $a,b\in A$, there exists some $c\in A$
such that $a\leq c$ and $b\leq c$. In other words, every pair of elements in
$A$ has an upper bound.

######  Definition 3.25.

A nonempty quandle $Q$ is ind-finite if there is a family of finite quandles
$\\{Q_{i}\\}_{i\in\mathcal{I}}$ indexed by a directed set $(\mathcal{I},\leq)$
such that for all $i,j\in\mathcal{I}$,

  1. 1.

$Q_{i}\preccurlyeq Q$,

  2. 2.

$|Q_{i}|<\infty$,

  3. 3.

$i\leq j$ implies $Q_{i}\preccurlyeq Q_{j}$, and

  4. 4.

$Q=\bigcup_{i\in\mathcal{I}}Q_{i}.$

There is an equivalent formulation of ind-finiteness which will soon be
useful. This formulation requires the Axiom of Dependent Choice.

######  Lemma 3.26.

Suppose $Q$ is a nonempty quandle. Then, $Q$ is ind-finite if and only if
every finitely generated subquandle of $Q$ is finite.

###### Proof.

Suppose that $Q$ is ind-finite with $Q=\bigcup_{i\in\mathcal{I}}Q_{i}$, and
fix an arbitrary finitely generated subquandle $Q^{\prime}$ of $Q$. Let
$\\{g_{1},\dots,g_{n}\\}$ be a generating set for $Q^{\prime}$. Then, for each
$1\leq j\leq n$, since $g_{j}\in Q$, there exists some $i(j)\in\mathcal{I}$
such that $g_{j}\in Q_{i(j)}$. Clearly,

$Q^{\prime}=\langle\langle\\{g_{1},g_{2},\dots,g_{n}\\}\rangle\rangle\subseteq\langle\langle
Q_{i(1)},\dots,Q_{i(n)}\rangle\rangle=\langle\langle\cup_{1\leq j\leq
n}Q_{i(j)}\rangle\rangle.$

But, since $\preccurlyeq$ is a total order on $\\{Q_{i}\\}_{i\in\mathcal{I}}$,
$\bigcup_{1\leq j\leq n}Q_{i(j)}=Q_{m}$, where $m=\max_{1\leq j\leq n}i(j).$
Thus, $\langle\langle Q_{i(1)},\dots,Q_{i(n)}\rangle\rangle=\langle\langle
Q_{m}\rangle\rangle=Q_{m}$. Since $Q_{m}$ is finite by assumption and
$Q^{\prime}\subseteq Q_{m}$, $Q^{\prime}$ must also be finite.

Now suppose that every finitely generated subquandle of $Q$ is finite. Let
$x\in Q$ be arbitrary, and define $Q_{0}=\langle\langle x\rangle\rangle$ and
$\mathcal{I}=|Q|$ (which is a cardinal that isn’t necessarily finite). For
each $i\in\mathcal{I}$, while $\bigcup_{k\leq i}Q_{k}\neq Q,$ let
$Q_{i+1}=\langle\langle Q_{i},y\rangle\rangle$ for some $y\in
Q\setminus\bigcup_{k\leq i}Q_{k}.$ Since each $Q_{i}$ is finitely generated
subquandle of $Q$ by construction, each $Q_{i}$ must also be finite. Moreover,
$\bigcup_{i\in|Q|}Q_{i}=Q$, since every element of $Q$ can be reached by this
construction, and $Q_{i}\prec Q_{i+1}$ for each $i\in\mathcal{I}$. By
definition, $Q$ is ind-finite. ∎

As a corollary, we find that the infinite quandle $\mathbb{Q}$ cannot be ind-
finite.

######  Corollary 3.27.

$(\mathbb{Q},\rhd_{\mathrm{dih}})$ is not ind-finite.

###### Proof.

Fix any $x,y\in\mathbb{Q}$ such that $x<y$. By part (5) of Theorem 3.3 in [5],

$\langle\langle x,y\rangle\rangle=\\{r_{0}x+r_{1}y\\},$

where $r_{0},r_{1}\in\mathbb{Z}$, exactly one of $r_{0},r_{1}$ is odd, and
$r_{0}+r_{1}=1$. In particular,
$\\{y+j(y-x):j\in\mathbb{N}\\}\subseteq\langle\langle x,y\rangle\rangle.$
Hence, $\langle\langle x,y\rangle\rangle$ is finitely-generated and infinite
subquandle of $\mathbb{Q}$, so by the previous lemma, $\mathbb{Q}$ cannot be
ind-finite. ∎

But, it turns out that we can transform $\mathbb{Q}$ quite easily to create an
ind-finite quandle. We will also see that this transformed quandle maintains
the properties of having a non-complemented subquandle lattice and no maximal
subquandles, thus proving that the subquandle lattices of ind-finite quandles
are not necessarily complemented.

######  Proposition 3.28.

$(\mathbb{Q}/\mathbb{Z},\rhd_{\mathrm{dih}})$ is ind-finite.

###### Proof.

It is easy to see that when an abelian group is made into a quandle using the
Takasaki operation, any subgroup is a subquandle. Therefore,
$\langle\frac{1}{n!}+\mathbb{Z}\rangle$ is a (finite) subquandle of
$\mathbb{Q}/\mathbb{Z}$ for each $n\in\mathbb{N}.$ Observe then that
$\mathbb{Q}/\mathbb{Z}$ is the ascending union of the subgroups generated by
the cosets of the reciprocals of factorials, i.e.,
$\mathbb{Q}/\mathbb{Z}=\bigcup_{n\in\mathbb{N}}\langle\frac{1}{n!}+\mathbb{Z}\rangle$,
and thus is easily seen to be ind-finite. ∎

######  Proposition 3.29.

$(\mathbb{Q}/\mathbb{Z},\rhd_{\mathrm{dih}})$ does not have a complemented
subquandle lattice and has no maximal subquandles.

###### Proof.

Note that
$\mathbb{Q}/\mathbb{Z}=\\{[\frac{m}{n}]:\frac{m}{n}\in\mathbb{Q}\\},$ where
$[\frac{m}{n}]=\frac{m}{n}+\mathbb{Z}.$ We are assuming
$\mathbb{Q}/\mathbb{Z}$ is equipped with the (well-defined) operation
$[x]\rhd_{\mathrm{dih}}[y]=[2x-y]$, where $x,y\in\mathbb{Q}$. Thus, in the
proofs of Lemma 5.1 and Theorem 5.2 in [4], we may replace the elements
$x\in\mathbb{Q}$ with their corresponding cosets $[x]\in\mathbb{Q}/\mathbb{Z}$
to obtain the following analogous results for $\mathbb{Q}/\mathbb{Z}$:

  1. 1.

The nonempty subquandles of $\mathbb{Q}/\mathbb{Z}$ are exactly the cosets of
the subgroups of $\mathbb{Q}/\mathbb{Z}.$

  2. 2.

$\mathcal{L}((\mathbb{Q}/\mathbb{Z},\rhd_{\mathrm{dih}}))$ is not
complemented. In particular, $\\{\mathbb{Z}\\}$ has no complement.

To show $\mathbb{Q}/\mathbb{Z}$ has no maximal subquandles, fix an arbitrary
$M\prec\mathbb{Q}/\mathbb{Z}$ — we show that $M$ cannot be maximal. As in
Proposition 3.22, if $\mathbb{Z}\notin M$, $M$ cannot be maximal.
Additionally, if $\mathbb{Z}\in M$, note that $(\mathbb{Q}/\mathbb{Z},+)$ has
no maximal subgroups, so the same argument from Proposition 3.22 shows that
$M$ cannot be maximal. ∎

###  3.5 Profinite Quandles

Profinite quandles are the dual to ind-finite quandles — just as ind-finite
quandles are the direct limit of a directed system of finite quandles,
profinite quandles are the inverse limit of an inverse system of finite
quandles. In this section, we formalize the notion of a profinite quandle and
provide several examples of infinite profinite quandles. We leave the
complementedness of profinite quandle lattices as a question for future
research.

######  Definition 3.30.

Let $(\mathcal{I},\leq)$ be a directed set, and let
$\\{Q_{i}\\}_{i\in\mathcal{I}}$ be a family of quandles indexed by
$\mathcal{I}$. Suppose we have a family of quandle homomorphisms
$\\{f_{ij}:Q_{j}\to Q_{i}:i,j\in\mathcal{I},\ i\leq j\\}$ such that for all
$i,j,k\in\mathcal{I}$,

  1. 1.

$f_{ii}=\mathrm{id}_{Q_{i}}$, and

  2. 2.

If $i\leq j\leq k$, $f_{ik}=f_{ij}\circ f_{jk}$.

${Q_{k}}$${Q_{j}}$${Q_{i}}$$\scriptstyle{f_{ik}}$$\scriptstyle{f_{jk}}$$\scriptstyle{f_{ij}}$

Then, the pair $(\\{Q_{i}\\}_{i\in\mathcal{I}},\\{f_{ij}\\}_{i\leq
j\in\mathcal{I}})$ is called an inverse system of quandles and morphisms over
$\mathcal{I}.$

######  Definition 3.31.

Given an inverse system $(\\{Q_{i}\\}_{i\in\mathcal{I}},\\{f_{ij}\\}_{i\leq
j\in\mathcal{I}})$, we define the inverse limit of the system as follows:

$Q=\lim_{\overleftarrow{i\in\mathcal{I}}}Q_{i}=\left\\{\vec{q}\in\prod_{i\in\mathcal{I}}Q_{i}:q_{i}=f_{ij}\left(q_{j}\right)\text{
for all }i\leq j\text{ in }\mathcal{I}\right\\}.$

We say a quandle $Q$ is profinite it is the inverse limit of the inverse
system composed of a family of finite quandles and their morphisms. Indeed,
this construction always yields a quandle:

######  Proposition 3.32.

Profinite quandles are quandles under coordinatewise operations.

###### Proof.

Suppose that $Q=\displaystyle\lim_{\overleftarrow{i\in\mathcal{I}}}Q_{i},$
where $(\\{Q_{i}\\}_{i\in\mathcal{I}},\\{f_{ij}\\}_{i\leq j\in\mathcal{I}})$
is an inverse system of finite quandles. Also suppose the associated quandle
operation for each $Q_{i}$ is $\rhd_{i}.$ Because each $Q_{i}$ is a quandle
under $\rhd_{i}$, axioms Q1, Q2, and Q3 hold for $Q$ since they hold in each
of its coordinates. But, since $Q$ may be infinite, we must also verify that
$Q$ is closed under its operation $\rhd$. Without loss of generality, let
$\mathcal{I}=\mathbb{N}$. Also let $(q_{1},q_{2},\dots),(r_{1},r_{2},\dots)\in
Q$ be arbitrary. Then,

$\displaystyle(q_{1},q_{2},\dots)\rhd(r_{1},r_{2},\dots)$
$\displaystyle=(q_{1}\rhd_{1}r_{1},q_{2}\rhd_{2}r_{2},\dots),$

so, by the definition of $Q$, it suffices to show that for arbitrary
$i,j\in\mathcal{I}$, $f_{ij}(q_{j}\rhd_{j}r_{j})=q_{i}\rhd_{i}r_{i}$. But,
because $f_{ij}$ is a quandle homomorphism,
$f_{ij}(q_{j}\rhd_{j}r_{j})=f_{ij}(q_{j})\rhd_{i}f_{ij}(r_{j})=q_{i}\rhd_{i}r_{i}$.
Hence, $Q$ is closed under $\rhd$, as desired. ∎

Next, we provide a canonical way to construct a profinite quandle from a
profinite abelian group: consider the group as a Takasaki kei, as defined in
Example 2.5.

######  Proposition 3.33.

Suppose $(\\{Q_{i}\\}_{i\in\mathcal{I}},\\{f_{ij}\\}_{i\leq j\in\mathcal{I}})$
is an inverse system of finite abelian groups and their morphisms. Then, its
inverse limit $Q$ is a profinite Takasaki kei under coordinatewise operations.

###### Proof.

Since each $Q_{i}$ is an abelian group, it is a Takasaki kei. Additionally,
each group homomorphism $f_{ij}$ is a quandle homomorphism by following lemma:

######  Lemma 3.34.

Suppose $(Q_{1},\rhd_{\mathrm{dih}})$ and $(Q_{2},\rhd_{\mathrm{dih}})$ are
Takasaki quandles, and suppose $(Q_{1},+)$ and $(Q_{2},+)$ are the
corresponding groups. Then, any group homomorphism $f:Q_{1}\to Q_{2}$ is also
a quandle homomorphism.

This lemma proves that regarding an abelian group as a Takasaki quandle is a
functor into $\mathbf{Kei}$, the full subcategory of $\mathbf{Quand}$ whose
objects are keis. We shall give it a name.

######  Definition 3.35.

The Takasaki functor, denoted
$\mathrm{Tak}:\mathbf{Ab}\rightarrow\mathbf{Kei}$, assigns to each abelian
group the Takesaki kei.

Thus, $(\\{Q_{i}\\}_{i\in\mathcal{I}},\\{f_{ij}\\}_{i\leq j\in\mathcal{I}})$
is also an inverse system of finite Takasaki quandles. To show that the
inverse limit of this inverse system is a profinite quandle, it suffices to
show that the functor $\mathrm{Tak}$ has a left adjoint, as this will show it
preserves all limits. We shall construct this left adjoint explicitly.

######  Definition 3.36.

Let $K$ be a kei. Then we define $\mathrm{AdTak}(K)$ to be the quotient group

$\mathrm{AdTak}(K)=\langle K\rangle_{Ab}/\langle x\rhd y-2y+x:x,y\in
K\rangle_{Ab}$

It is easy to see that $\mathrm{AdTak}$ is a left adjoint to $\mathrm{Tak}$.

######  Proposition 3.37.

Let $\eta$ be the inclusion $x\mapsto\overline{x}$. Also let $A$ be an abelian
group, and let $\varphi$ be a morphism of keis $\varphi:K\to\mathrm{Tak}(A)$.
Then $\varphi$ factors uniquely through $\mathrm{Tak}(\varphi^{\prime})$ for a
unique morphism of abelian groups $\varphi^{\prime}$, as in the commutative
diagram below.

${K}$${{\mathrm{Tak}(\mathrm{AdTak}(K))}}$${{\mathrm{Tak}(A)}}$$\scriptstyle{\varphi}$$\scriptstyle{\eta}$$\scriptstyle{\mathrm{Tak}(\varphi^{\prime})}$

###### Proof.

The map $\eta$ can be thought of as the composition (in sets) of the inclusion
$K\hookrightarrow\mathrm{AdTak}(K)$ along with $p:\langle
K\rangle_{Ab}\to\langle K\rangle_{Ab}/\langle x\rhd y-2y+x:x,y\in
K\rangle_{Ab}$ (regarded as a function between underlying sets). By the
universal property of free abelian groups, the function $\varphi$ induces a
unique group homomorphism $\varphi^{\prime}:\langle K\rangle_{Ab}\to A$. By
definition of a Takasaki kei, for any elements $x,y\in K$, we have
$\varphi(x\rhd y)=\varphi(x)\rhd\varphi(y)=2\varphi(y)-\varphi(x)$. Hence
$\varphi(x\rhd y)-2\varphi(y)+\varphi(x)=\varphi(x\rhd y-2y+x)=0$. Therefore,
$\langle x\rhd y-2y+x:x,y\in K\rangle_{Ab}\subseteq\ker(\varphi^{\prime})$,
which implies $\langle x\rhd y-2y+x:x,y\in
K\rangle_{Ab}\trianglelefteq\ker(\varphi^{\prime})$. By the universal property
of quotients in abelian groups, $\varphi^{\prime}$ induces yet another
(unique) homomorphism $\mathrm{Tak}(\varphi^{\prime}):\mathrm{AdTak}(K)\to A$
of abelian groups. ∎

Therefore, by Proposition 3.33, the inverse limit of
$(\\{Q_{i}\\}_{i\in\mathcal{I}},\\{f_{ij}\\}_{i\leq j\in\mathcal{I}})$ is a
profinite quandle under coordinatewise operations. ∎

Because $\mathrm{Tak}$ admits a left adjoint, it preserves all limits, in
particular the limits giving a profinite abelian group as the limit of the
diagram of all its finite quotient groups. This observation gives rise to a
large class of profinite keis:

######  Example 3.38.

  1. 1.

Suppose $p\in\mathbb{N}$ is a prime. Then, the $p$-adic integers
$\mathbb{Z}_{p}$ form a profinite Takasaki quandle.

  2. 2.

The additive group of the ring of profinite integers
$\hat{\mathbb{Z}}=\prod_{p}\mathbb{Z}_{p}$ forms a profinite Takasaki quandle.

The same argument applied to the functor $\mathrm{Conj}$ gives:

######  Example 3.39.

If $G$ is a profinite group, then $\mathrm{Conj}(G)$ is a profinite quandle.

Clearly some pro-finite quandles have complemented subquandle lattices. For
example, besides finite quandles [4], trivial quandles with an infinite
profinite set (that is an infinite totally disconnected compact Hausdorff
space) as an underlying set are profinite quandles, being the inverse limit of
a diagram of finite trivial quandles. Moreover, these quandles, as do all
trivial quandles, have the (necessarily complemented) Boolean algebra of all
subsets equipped with trivial quandle structures as a subquandle lattice.

However, in general, if $Q$ is infinite, it is hard to determine if
$\mathcal{L}(Q)$ is complemented, as even the lattice of subquandles of
$\mathcal{L}(Q)$ is difficult to examine because the underlying sets of either
class of examples above are uncountable.

Nonetheless, we conjecture that the subquandle lattice of a profinite quandle
is complemented.

##  4 Acknowledgments

The authors wish to thank the National Science Foudation for financial support
for the conduct of this research under NSF Grant DMS-1659123. The first four
authors wish to thank the fifth author, along with Marianne Korten and Kim
Klinger-Logan, for their mentorship and support during and after the SUMaR
2023, and Kansas State University for its hospitality during the REU.

##  5 Bibliography

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