# Spectral sets in $\mathbb{Z}_{p^{2}qr}$ tile

Gábor Somlai  Eötvös Loránd University, Departement of Algebra and Number
Theory
<EMAIL_ADDRESS>
###### Abstract

We prove the every spectral set in $\mathbb{Z}_{p^{2}qr}$ tiles, where $p$,
$q$ and $r$ are primes, which is a special case of Fuglede’s conjecture for
cyclic groups.

##  1 Introduction

A Lebesgue measurable set $\Omega\subset\mathbb{R}^{n}$ is called a tile if
there is set $T$ in $\mathbb{R}^{n}$, called the tiling complement of $\Omega$
such that almost every element of $\mathbb{R}^{n}$ can uniquely written as
$\omega+t$, where $\omega\in\Omega$ and $t\in T$. We say that $\Omega$ is
spectral if $L^{2}(\Omega)$ admits an orthogonal basis consisting of
exponential functions of the form $\\{e^{i(\lambda
x)}\mid\lambda\in\Lambda,x\in\Omega\\}$, where $\Lambda\subset\mathbb{R}^{n}$.
In this case $\Lambda$ is called a spectrum  for $\Omega$.

Fuglede conjectured [3] that a bounded domain $S\subset\mathbb{R}^{d}$ tiles
the $d$-dimensional Euclidean space if and only if the set of $L^{2}(S)$
functions admits an orthogonal basis of exponential functions. The conjecture
might have been motivated by Fuglede’s result [3] that it is true when the
tiling complement or the spectrum is a lattice in $\mathbb{R}^{n}$. The
conjecture was disproved in [12]. Tao considered a discrete version of the
original conjecture and constructed spectral sets in $\mathbb{Z}_{2}^{11}$ and
$\mathbb{Z}_{3}^{5}$, which are not tiles. The latter example was lifted to
$\mathbb{R}^{5}$ to refute the spectral-tile direction of Fuglede’s
conjecture. Matolcsi [6] proved that the spectral-tile direction of the
conjecture fails in $\mathbb{R}^{4}$. Kolountzakis and Matolcsi [8, 9] and
Farkas, Matolcsi and Móra [2] provided counterexamples in $\mathbb{R}^{3}$ for
both direction of the conjecture. The case of elementary abelian groups is

It is important to note that every (finite) tile of the integers is periodic.
Moreover, the tile-spectral direction of the conjecture is true for
$\mathbb{R}$ if and only it holds for $\mathbb{Z}$, which is further
equivalent for the conjecture to hold for every finite cyclic group. If the
spectral-tile direction of Fuglede’s conjecture holds for $\mathbb{R}$, then
it holds for $\mathbb{Z}$, which again implies that it holds for every
(finite) cyclic group. On the other hand, the converse of this statements is
not necessarily true.

The investigation of Fuglede’s conjecture for finite cyclic groups started
with a result of Kolountzakis and Malikiosis [10] by proving that the
conjecture holds for $\mathbb{Z}_{p^{n}q}$, where $p^{n}$ is an arbitrary
power of the prime $p$ and $q$ is a prime. As a strengthening it was proved in
[7] that Fuglede’s conjecture holds for $\mathbb{Z}_{p^{n}q^{2}}$, where $p$
and $q$ are prime. Another important result was proved by Shi [11] who showed
that Fuglede’s conjecture is true for $\mathbb{Z}_{k}$ if $k$ is the product
of $3$ (distinct) primes.

The Coven-Meyerowitz conjecture states that if $n$ is square-free, then every
tile of $\mathbb{Z}_{n}$ is complete set of residues $\pmod{k}$, where $k$ is
a divisor of $n$. This was originally settled by Laba and Meyerowitz on Tao’s
blog and a self-contained proof of this fact was provided by Shi [11].

Coven and Meyerowitz [1] proved that a subset $A$ of $\mathbb{Z}_{n}$ tiles if
two properties called (T1) and (T2), defined in the next section, are
satisfied. The converse also holds if $n$ has at most two different prime
divisors and for every $n\in\mathbb{N}$ the tiles of $\mathbb{Z}_{n}$ satisfy
(T1).

##  2 Fuglede’s conjecture for cyclic groups

Fuglede’s conjecture is still open in the $1$ and $2$ dimension. We will focus
on the one dimensional case which is heavily connected to the discrete version
of the conjecture for cyclic groups.

Let $S$ be a subset of $\mathbb{Z}_{n}$. We say that $S$ is a tile if and only
if there is a $T\subset\mathbb{Z}_{n}$ such that $S+T=\mathbb{Z}_{n}$ and
$|S||T|=n$. We say that $S$ is spectral if and only if the vector space of
complex functions on $S$ is spanned by pairwise orthogonal functions, which
are the restrictions of some irreducible representations of $\mathbb{Z}_{n}$.
The irreducible representations of $\mathbb{Z}_{n}$ are of the following form:

$\chi_{k}(x)=e^{\frac{2\pi in}{k}x},$

so these are parametrized by the elements of $\mathbb{Z}_{n}$. It is easy to
verify that $\chi_{k}$ and $\chi_{l}$ are orthogonal if and only if
$\chi_{k-l}$ is orthogonal to the trivial representation, which can also be
written as

$\sum_{s\in S}\chi_{k-l}(s)=0.$

One can assign a polynomial $m_{S}$ to $S$ by $\sum_{s\in S}x^{s}$, which s
called the mask polynomial of $S$. It is easy to see that $\sum_{s\in
S}\chi_{k}(s)=0$ if and only if $\xi_{k}$ is a root of $m_{S}$, where
$\xi_{k}$ is a primitive $k$’th root of unity. This can also be said as
$\Phi_{k}\mid m_{S}$, where $\Phi_{k}$ is the $k$’th cyclotomic polynomial.
Note that mask polynomials can also be defined for any element of the group
ring $\mathbb{Z}[\mathbb{Z}_{n}]$.

Now using the character table of $\mathbb{Z}_{n}$ we have that $(S,\Lambda)$
is a spectral pair if and only if the submatrix of the character table whose
rows are indexed by the elements of $\Lambda$ and columns with those of $S$ is
a Hadamard matrix. In fact, the transpose of a Hadamard matrix is also a
Hadamard matrix so if $(S,\Lambda)$ is a spectral pair, then $(\Lambda,S)$ is
a spectral pair too.

Now we introduce the properties that are needed to formulate the Coven-
Meyerowitz conjecture for any natural number. Let $H_{S}$ be the set of prime
powers $p^{k}$ dividing $N$ such that $\Phi_{p^{k}}(x)\mid S(x)$.

  1. (T1)

$m_{S}(1)=\prod_{d\in H_{S}}\Phi_{d}(1)$,

  2. (T2)

for pairwise relative prime elements $q_{i}$ of $H_{S}$, we have $\Phi_{\prod
q_{i}}\mid m_{S}(x)$.

We remind that if (T1) and (T2) hold for some $S\subset\mathbb{Z}_{n}$, then
$S$ is a tile and if $S$ is a tile, then (T1) holds. Further we mention that
Łaba proved that a set having (T1) and (T2) properties also is a spectral set.

##  3 Preliminary lemmas

For the sake of simplicity let $n=p^{2}qr$. First, we collect the results
obtained in [7] that applies in our case. We first note that in our case for
every subgroup or factor group of $\mathbb{Z}_{n}$ we have that spectral sets
coincide with tiles. The results of Section 4 were summarized in a statement
called Reduction 1 but it is important to note that $n$ have more than $2$
different prime divisors so we only have the following:

######  Proposition 3.1.

Let us assume that $(S,\Lambda)$ is a spectral pair for an abelian group whose
subgroups and factor groups satisfies the spectral-tile direction of Fuglede’s
conjecture. Then we may assume $0\in S$, $0\in\Lambda$. Further $S$ is a tile
if one of the following holds.

  1. (a)

$S$ or $\Lambda$ does not generate $\mathbb{Z}_{n}$.

  2. (b)

$S$ can be written as the union of $\mathbb{Z}_{u}$-cosets, where $u$ is a
prime dividing $n$.

######  Lemma 3.2.

Let $0\in T\subset\mathbb{Z}_{N}$ a generating set and assume $x$ and $y$ are
different prime divisor of $N$. Then there are elements $t_{1}\neq t_{2}$ of
$T$ such that $x\nmid t_{1}-t_{2}$ and $y\nmid t_{1}-t_{2}$ for any pair of
prime divisors of $N$.

###### Proof.

$T$ is not contained in any proper coset of $\mathbb{Z}_{N}$ so it contains an
element $t_{1}$ not divisible by $x$ and $t_{2}$ not divisible by $y$. If
$y\nmid t_{1}$, then $t_{1}-0\in(T-T)$ is not divisible by either $x$ or $y$,
when we are done so we may assume $y\mid t_{1}$. Similar argument shows that
we may assume $x\mid t_{2}$. Then $x\nmid t_{1}-t_{2}$ and $y\nmid
t_{1}-t_{2}$, as required. ∎

Another important tool is the following lemma. This is the same as Proposition
3.4 in [7] formulated in a bit different language. Let $m$ be a square-free
integer, where $m$ is the product of $d$ primes. Then
$\mathbb{Z}_{m}\cong\prod\mathbb{Z}_{p_{i}}$ a direct sum of $d$ cyclic groups
of prime order so the elements of $\mathbb{Z}_{m}$ are encoded by $d$-tuples.
This allows us to introduce Hamming distance of $\mathbb{Z}_{m}$. Further we
say that $P$ is a cuboid in $\mathbb{Z}_{m}$ if it can be written as $\prod
H_{i}$, where $H_{i}\subset\mathbb{Z}_{p_{i}}$ with $|H_{i}|=2$.

######  Proposition 3.3.

Let $w$ be an element of the group ring $\mathbb{Z}[\mathbb{Z}_{m}]$ with
nonnegative coefficients, where $m=\prod_{i=1}^{d}p_{i}$, a product of $d$
different primes. Assume $\Phi_{m}\mid m_{w}$. Let $P$ be a $d$-dimensional
cuboid and $p$ a vertex of $P$. Then

$\sum_{c\in P}(-1)^{d_{H}(p,c)}w(c)=0.$ (1)

We will refer to the previous proposition or more precisely to equation (1) as
the $d$-dimensional cube-rule. Note that this lemma is a corollary of the fact
that if $\omega$ is above then $w$ can be written as the weighted sum of
$\mathbb{Z}_{p_{i}}$-cosets with rational coefficients, see Corollary 3.4 in
[7].

It is also important to know what happens if $\omega$ is a multiset on
$\mathbb{Z}_{m}$, with $\Phi_{m}\mid m_{\omega}$, where $m$ is not square
free. This case is also described in [7]. Let $m^{\prime}$ be the square free
radical of $m$. Then $\omega$ is the weighted sum of
$\mathbb{Z}_{p_{i}}$-cosets with rational coefficients again, implying that
the cube-rule holds for the restriction of $\omega$ to each
$\mathbb{Z}_{m^{\prime}}$-coset.

######  Lemma 3.4.

yLet $T\subset\mathbb{Z}_{pqr}$. Assume $\Phi_{pqr}\mid m_{T}$ and
$T\cap((t+\mathbb{Z}_{q})\cup(t+\mathbb{Z}_{r}))=\\{t\\}$ for all $t\in T$.
Then $T$ is a union of $\mathbb{Z}_{p}$-cosets.

Note that for the statements holds verbatim for any permutation of the primes
$p,q,r$.

###### Proof.

Assume $\Phi_{pqr}\mid m_{T}$, then $T$ satisfies the $3$ dimensional cube-
rule. Suppose $t\in T$. By our assumption $T\cap(t+\mathbb{Z}_{q})=\\{t\\}$
and $T\cap(t+\mathbb{Z}_{r})=\\{t\\}$.

By the way of contradiction, assume $T$ is not a union of
$\mathbb{Z}_{p}$-cosets so there is a $t\in T$ and
$t_{p}\in(t+\mathbb{Z}_{p})\setminus T$. Then for every cuboid containing $t$
and $t_{p}$ as its vertices, the neighbours (consider the cube as the natural
graph on $8$ vertices) of $t$ are not in $T$. Now using the cube-rule we
obtain that the vertex of the cube, which is of Hamming distance $3$ from $t$
is contained in $T$. Then for every $x\in\mathbb{Z}_{pqr}$ with $p\mid
x-t_{p}$ and $d_{H}(x,t)=3$ we have $x\in T$. Then there are elements of $T$
whose difference is divisible by $pr$ if $q>2$ and the same holds with $pq$ if
$r>2$. This contradicts the assumption that
$T\cap((x+\mathbb{Z}_{q})\cup(x+\mathbb{Z}_{r}))=\\{x\\}$. ∎

Now we prove a Lemma that have will have applications during the proof of our
main result.

######  Lemma 3.5.

Let $T\subset Z_{N}$, where $N$ is a positive integer and let $m$ and $r$ are
divisors of $N$ with $(m,r)=1$. Assume further that for every $d\mid m$ we
have that for every $\phi_{d}\mid m_{T}$ or $\phi_{dr}\mid m_{T}$. Then
$T_{m}=c\mathbb{Z}_{m}+rD$, where $c\geq 0$ an integer and $D$ is a multiset
on $\mathbb{Z}_{m}$

###### Proof.

Since $(d,r)=1$ we obtain $\Phi_{d}r(x)=\Phi_{d}(x^{r})$. This identity is can
be written as $\Phi_{d}r(x)=\Phi_{d}(x)^{r}$ in the polynomial ring
$\mathbb{Z}_{r}[x]$. Now $\Phi_{d}r(x)\mid m_{T}(x)$ implies $\Phi_{d}(x)\mid
m_{T}(x)$ in $\mathbb{Z}_{r}[x]$. These cyclotomic polynomials are pairwise
relatively primes $\mathbb{Z}_{r}[x]$ as well so we obtain $\prod_{d\mid
m}\Phi_{d}\mid m_{T}$ in $\mathbb{Z}_{r}[x]$. This implies that $m_{T}\equiv
0\pmod{x^{m}-1}$ in $\mathbb{Z}_{r}[x]$. Now let $T_{m}$ the multiset obtained
from $T$ by the natural projection to $\mathbb{Z}_{m}$ and let
$c=\min_{x\in\mathbb{Z}_{m}}t(x)$. Then the claim follows from $m_{T}\equiv
0\pmod{x^{m}-1}$ in $\mathbb{Z}_{r}[x]$. ∎

Let $n$ be a fixed natural number. For a divisor of $n$ we write $a\mid\mid b$
if $a$ divides $b$ and there is no divisor $a^{\prime}>a$ of $n$ with
$a^{\prime}\mid b$.

##  4 Proof of the main result

Let $(S,\Lambda)$ be a spectral pair. We will distinguish certain different
cases by the cardinality of $S$, which equals to those of $\Lambda$.

###  4.1 $|S|$ have 3 prime divisors

In this case we assume that $|S|$ is divisible by three of the primes $p,q,r$
counted with multiplicity. Thus the cases handled here are
$p^{2}q\mid\mid|S|$, $p^{2}r\mid\mid|S|$ and $pqr\mid\mid|S|$.

Assume $p^{2}q\mid\mid|S|$. Project $S$ to $\mathbb{Z}_{p^{2}q}$. If two
elements of $S$ project to the same element of $\mathbb{Z}_{p^{2}q}$, then we
have a pair of elements of $s_{1},s_{2}\in S$ with $p^{2}q\mid\mid
s_{1}-s_{2}$. Thus we have $\Phi_{r}\mid m_{\Lambda}$, which implies
$r\mid|\Lambda|=|S|$, a contradiction. Thus $|S|=p^{2}q$ and $S$ is a complete
set of residues $\pmod{p^{2}q}$ so is a tile. The same argument works if
$p^{2}r\mid\mid|S|$.

Let us assume now that $pqr\mid\mid|S|$. If a $\mathbb{Z}_{p^{2}}$-coset
contains at least $p+1$ elements of $S$, then $S$ contains a pair of elements
contained in the same $\mathbb{Z}_{p}$-coset and a pair of elements contained
in the same $\mathbb{Z}_{p^{2}}$-coset but in different
$\mathbb{Z}_{p}$-cosets. These imply $\Phi_{p}\mid m_{\Lambda}$ and
$\Phi_{p^{2}}\mid m_{\Lambda}$ so $p^{2}\mid|\Lambda|=|S|$, a contradiction.

It remains to investigate the case when each $\mathbb{Z}_{p^{2}}$-coset
contains exactly $p$ elements of $S$, which gives $|S|=pqr$. Moreover by
excluding $\Phi_{p}\Phi_{p^{2}}\mid m_{\Lambda}$ we obtain that the
intersection of $S$ with each $\mathbb{Z}_{p^{2}}$-coset is either a
$\mathbb{Z}_{p}$-coset or it is a complete coset representative of
$\mathbb{Z}_{p}$ for every $\mathbb{Z}_{p^{2}}$-coset. In both of these cases
$S$ is a tile.

Note that these arguments also work if $|S|>p^{2}\min\\{q,r\\}$ or $|S|>pqr$
so from now on we assume $|S|<\min\\{p^{2}q,p^{2}r,pqr\\}$.

###  4.2 $|S|$ has two prime divisors

Now we handle the cases when $p^{2}\mid\mid|S|$, $pq\mid\mid|S|$,
$pr\mid\mid|S|$ and $qr\mid\mid|S|$.

Case 1. Let us assume first that $p^{2}\mid\mid|S|$.
If every $\mathbb{Z}_{qr}$-coset contains exactly one element of $S$, then $S$
is a tile. Thus we may assume there are $s_{1}\neq s_{2}\in S$ with $p^{2}\mid
s_{1}-s_{2}$. If $p^{2}q\mid\mid s_{1}-s_{2}$ or $p^{2}r\mid\mid s_{1}-s_{2}$,
then $r\mid|S|$ or $q\mid|S|$, respectively. Both of these cases contradict
$p^{2}\mid\mid|S|$. Thus we have $pq\mid\mid s_{1}-s_{2}$ so $\Phi_{qr}\mid
m_{\Lambda}$.

Assume first that $\Phi_{n}\mid m_{S}$. Then we may apply the cube-rule on
every $\mathbb{Z}_{pqr}$-coset. By Lemma 3.4 we obtain that $S$ is the union
of $\mathbb{Z}_{p}$-cosets, which case is handled in Proposition 3.1 (b). Thus
we may assume $\Phi_{n}\nmid m_{S}$.

For $k\mid n$ let us denote the image of $\Lambda$ and $T$ via the natural
projection to $\mathbb{Z}_{k}$ by $\Lambda_{k}$ and $T_{k}$, respectively. We
may consider $\Lambda_{k}$ as a set or a multiset i.e. an element of
$\mathbb{Z}[\mathbb{Z}_{k}]$ with nonegative coefficients. It follows from
$\Phi_{qr}\mid m_{\Lambda}$ that $\Lambda_{qr}$ is the weighted sum of
$\mathbb{Z}_{q}$ and $\mathbb{Z}_{r}$-cosets with nonnegative weights. Both
type of cosets appear since otherwise we would have $q\mid|S|$ or $r\mid|S|$,
a contradiction.

Assume $r>q$ (the case when $q>r$ can be handled similarly). Clearly $r\geq
3$. Then there are $\lambda,\lambda^{\prime}\in\Lambda$ with
$q\mid\lambda-\lambda^{\prime}$ but $q\nmid\lambda-\lambda^{\prime}$ and there
is $\lambda^{\prime\prime}\in\Lambda$, whose $q$ and $r$ coordinates differ
from those of $\lambda$ and $\lambda^{\prime}$. By this we mean
$q\nmid\lambda-\lambda"$, $q\nmid\lambda^{\prime}-\lambda"$,
$r\nmid\lambda-\lambda"$ and $r\nmid\lambda^{\prime}-\lambda"$. Then since
$\Phi_{n}\nmid m_{S}$ we have $p\mid\lambda-\lambda^{\prime\prime}$ and
$p\mid\lambda^{\prime}-\lambda^{\prime\prime}$ so we have
$pq\mid\lambda-\lambda^{\prime}$ and $r\nmid\lambda-\lambda^{\prime}$. Thus we
either have $\Phi_{r}\mid m_{S}$, which is excluded since $r\nmid|S|$ or we
have $\Phi_{pr}\mid m_{S}$.

We may define a graph $\Gamma$ whose vertices are the elements of $\Lambda$
and two vertices are adjacent if and only if both the $q$ and $r$-coordinates
of the vertices are different. If $\Gamma$ is connected, then we have
$\Lambda$ is contained in a $\mathbb{Z}_{pqr}$-coset, which is excluded by
Proposition 3.1. Then since the multiset $\Lambda_{qr}$ is the sum of
$\mathbb{Z}_{q}$ and $\mathbb{Z}_{r}$-cosets and both types appear, we can
only have that $\Lambda_{qr}$ as a set is the union of a
$\mathbb{Z}_{q}$-coset $Q$ and $\mathbb{Z}_{r}$-coset $R$. In this case,
$\Gamma$ has a large connected component consisting of those elements of
$\Lambda$ which do not project to the intersection of $Q$ and $R$ denoted by
$x\in\mathbb{Z}_{qr}$. Note that the number of elements of $\Lambda$
projecting to $x$ is less than $\frac{|\Lambda|}{2}$.

We claim that $\Phi_{n}\nmid m_{\Lambda}$. By the way of contradiction let us
assume $\Phi_{n}\mid m_{\Lambda}$. Then using the same argument when we
excluded $\Phi_{n}\mid m_{S}$ we have that $\Lambda$ is the union of
$\mathbb{Z}_{p}$-cosets. But then $(\Lambda,S)$ is a spectral pair and by
Proposition 3.1 we have that $\Lambda$ is a tile whence $|\Lambda|=p^{2}$.
Theorem B1 in [1] shows that (T1) holds for $\Lambda$. Further it is clear
from (T1) and our condition $p^{2}\mid\mid|S|$ that $|\Lambda|=|S|=p^{2}$.
Thus it follows from (T1) that $\Phi_{p}\Phi_{p^{2}}\mid m_{\Lambda}$. Since
$\Phi_{p}\Phi_{p^{2}}=1+x+\ldots+x^{p^{2}-1}$ and $|\Lambda|=p^{2}$ we have
$\Lambda$ is a complete set of coset representatives of $\mathbb{Z}_{qr}$ in
$\mathbb{Z}_{n}$. This contradicts the fact that all but the elements of
$\Lambda$ projecting to $x$ give the same remainder $\pmod{p}$.

Thus we have that there is $\mu$ in $\Lambda$ projecting to
$x\in\mathbb{Z}_{qr}$ with $p\nmid\mu-\lambda$ and
$p\nmid\mu-\lambda^{\prime\prime}$, where $\lambda$ projects to
$R\setminus\\{x\\}$ and $\lambda^{\prime}$ to $Q\setminus\\{x\\}$. Thus we
obtain $\Phi_{p^{2}q}\mid m_{S}$ and $\Phi_{p^{2}r}\mid m_{S}$ since
$\Phi_{n}\nmid m_{S}$.

We claim that $p>r$. Otherwise there are more than $p$ elements of $\Lambda$,
projecting to mutually different points of $R\setminus\\{x\\}$. Their
difference is divisible by $p$ since they are in the same connected component
of $\Gamma$ but then there would be a pair whose difference is divisible by
$p^{2}$ as well, implying $\Phi_{r}\mid m_{S}$ and $r\mid|S|$, a
contradiction. Similar argument shows that we may assume $p>q$.

A simple calculation shows that the number of elements $m$ of $\Lambda$
projecting to $x$ exceeds $p$. This follows from $p^{2}\leq kq+lr$ and $m=k+l$
since $p>q,r$, where $k$ and $l$ are defined by $\Lambda_{qr}=kQ+lR$. Thus
there are elements $\lambda_{3},\lambda_{4},\lambda_{5}$ of $\Lambda$ with
$qr\mid\mid\lambda_{3}-\lambda_{4}$ and $pqr\mid\mid\lambda_{3}-\lambda_{5}$
thus implying $\Phi_{p}\Phi_{p^{2}}\mid m_{S}$.

We remind that $\phi_{pr}\mid m_{S}$. The fact that $\Phi_{p}\Phi_{p^{2}}\mid
m_{S}$ implies that every $\mathbb{Z}_{qr}$-coset contains the same amount of
elements of $S$ since $\Phi_{p}\Phi_{p^{2}}=\sum_{i=0}^{p^{2}-1}x^{i}$. Denote
this number by $a$. If $a=1$, then $S$ is a tile so we assume $a\geq 2$. We
have already assumed $|S|<p^{2}q$ so each $\mathbb{Z}_{qr}$ contains at most
$q-1$ elements of $S$. Now project $S$ to $\mathbb{Z}_{p^{2}r}$. The image of
$S$ is a set denoted by $S_{p^{2}r}$. This is a set since otherwise
$\Phi_{q}\mid m_{\Lambda}$, contradicting the fact that $q\nmid|\Lambda|=|S|$.
Then each $\mathbb{Z}_{r}$-coset in $\mathbb{Z}_{p^{2}r}$ contains $2\leq a<q$
elements of $S_{p^{2}r}$. By Lemma 3.2 we have $\Phi_{p^{2}r}\mid m_{S}$ or
$\Phi_{p^{2}qr}\mid m_{S}$, but the latter is excluded. Using
$\Phi_{p^{2}r}\mid m_{S}$ we have that the intersection of $S_{p^{2}r}$ with
each $\mathbb{Z}_{pr}$-coset is the union of $\mathbb{Z}_{p}$-cosets or
$\mathbb{Z}_{r}$-cosets. Since $a<q{\color[rgb]{1,0,0}<r}$ it is the union of
$\mathbb{Z}_{p}$-cosets only.

Then we build up a graph $\Gamma^{\prime}$ whose vertices are the elements of
$\Lambda$ and two vertices are adjacent if and only if both their difference
is not divisible by either $p$ or $r$. It follows from the previous
observations using $p\geq 3$ that $\Gamma^{\prime}$ is connected. Then since
$\Phi_{n}\nmid m_{\Lambda}$ we have that the $q$ coordinates of these elements
of $\Lambda$ is the same so they are all contained in a proper coset of
$\mathbb{Z}_{n}$. Thus by Proposition 3.1 we have that $S$ is a tile.

Case 2. Assume $pq\mid\mid|S|$.

We may exclude the case, when $S$ is complete set of residues $\pmod{pq}$,
which is the same $S$ contains exactly one element from each
$\mathbb{Z}_{pr}$-coset. Then it is clear, that there are elements of $S$
projecting to the same element of $\mathbb{Z}_{pq}$. Then we either have
$\Phi_{pr}\mid m_{\Lambda}$ or $\Phi_{p}\mid m_{\Lambda}$ or $\Phi_{r}\mid
m_{\Lambda}$. The latter case is impossible since $r\nmid|S|$.

Now we project $S$ to $\mathbb{Z}_{p^{2}q}$. The projection $S_{p^{2}q}$ is a
set, otherwise $\Phi_{r}\mid m_{\Lambda}$ and thus $r\mid|S|$, a
contradiction. Then if there is a $\mathbb{Z}_{p^{2}}$-coset of
$\mathbb{Z}_{p^{2}q}$ containing more than $p$ elements of $S_{p^{2}q}$, then
there are two of them, whose difference is not divisible by $p$, implying
$\Phi_{p^{2}}\mid m_{\Lambda}$ or $\Phi_{p^{2}r}\mid m_{\Lambda}$. If every
$\mathbb{Z}_{p^{2}}$-coset of $\mathbb{Z}_{p^{2}q}$ contains exactly $p$
elements of $S_{p^{2}q}$ and for each $\mathbb{Z}_{p^{2}}$-coset all of these
elements are contained in the same $\mathbb{Z}_{p}$-coset, then we have that
$S$ is a tile. Thus we may assume $\Phi_{p^{2}}\mid m_{\Lambda}$ or
$\Phi_{p^{2}r}\mid m_{\Lambda}$.

Applying Lemma 3.5 using the conditions that $\Phi_{p}\mid m_{\Lambda}$ or
$\Phi_{pr}\mid m_{\Lambda}$ and $\Phi_{p^{2}}\mid m_{\Lambda}$ or
$\Phi_{p^{2}r}\mid m_{\Lambda}$ we obtain that the projection of $\Lambda$ to
$\mathbb{Z}_{p^{2}}$ is of the following form:

$\Lambda_{p^{2}}=c\mathbb{Z}_{p^{2}}+rD,$ (2)

where $c$ is a nonnegative constant and $D$ is a multiset on
$\mathbb{Z}_{p^{2}}$. If $c=0$, then $r\mid|\Lambda|$ and if $D=0$, then
$p^{2}\mid|\Lambda|$. Both cases contradict our assumption that
$pq\mid\mid|S|=|\Lambda|$. Thus there are at least $r+1$ elements of $\Lambda$
projecting to the same element of $\mathbb{Z}_{p^{2}}$. If $q<r$, then there
are two of them, whose difference is divisible by $q$ as well, implying
$\Phi_{r}\mid m_{S}$, a contradiction. Thus we may assume $r>q$ and we also
obtain $\Phi_{q}\mid m_{S}$.

Assume $\Phi_{p^{2}r}\mid m_{\Lambda}$. Then $\Lambda_{p^{2}r}$ is a multiset,
which is the sum of $\mathbb{Z}_{p}$-cosets and $\mathbb{Z}_{r}$-cosets. Since
$c>0$ and $D\neq 0$, there is a $\mathbb{Z}_{pr}$-coset of
$\mathbb{Z}_{p^{2}q}$, whose intersection with the multiset $\Lambda_{p^{2}r}$
is the sum of $k$ $\mathbb{Z}_{p}$-cosets and $l$ $\mathbb{Z}_{r}$-coset with
$k+l\geq 2$.

Now we argue that $\Lambda_{p^{2}r}$ contains a $\mathbb{Z}_{r}$-coset. Assume
this is not the case, thus we can write $\Lambda_{p^{2}r}$ as the sum of
$\mathbb{Z}_{p}$-cosets. Then the number of elements of $\Lambda_{p^{2}r}$
contained in each $\mathbb{Z}_{pq}$-coset is divisible by $p$. If
$\Phi_{p}\mid m_{\Lambda}$, then these numbers are the same so we would have
$p^{2}\mid m_{\Lambda}$, a contradiction. If $\Phi_{pr}\mid m_{\Lambda}$, then
$\Lambda_{pr}$ is the sum of $\mathbb{Z}_{p}$ and $\mathbb{Z}_{r}$-cosets. If
$\Lambda_{pr}$ contains a $\mathbb{Z}_{r}$-coset, then
$\Lambda_{p^{2}r}$-contains a $\mathbb{Z}_{pr}$-coset since it is the union of
$\mathbb{Z}_{p}$-cosets so $\Lambda_{p^{2}r}$ contains a
$\mathbb{Z}_{r}$-coset as required. If $\Lambda_{pr}$ is the union of
$\mathbb{Z}_{p}$-cosets only, then we again have $p^{2}\mid|\Lambda|$, a
contradiction.

Since $c>0$ in equation (2) and since we have a $\mathbb{Z}_{p^{2}r}$-coset
containing such a $\mathbb{Z}_{r}$-coset, which is also contained in
$\Lambda_{p^{2}r}$, we have two elements
$\lambda_{1},\lambda_{2}\in\Lambda_{p^{2}r}$ with
$pq\mid\mid\lambda_{1}-\lambda_{2}$. It is not hard to see from the
description of $\Lambda$ in this case that for every $d\mid p^{2}r$ we have
$\lambda,\lambda^{\prime}\in\Lambda_{p^{2}r}$ with
$d\mid\mid\lambda-\lambda^{\prime}$. Then we have
$\Phi_{p}\Phi_{r}\Phi_{pr}\Phi_{p^{2}}\Phi_{p^{2}r}\mid m_{S}$ in
$\mathbb{Z}_{q}[x]$ so by projecting $S$ to $\mathbb{Z}_{p^{2}r}$ we obtain a
multiset of the form $c^{\prime}\mathbb{Z}_{p^{2}r}+qD^{\prime}$ ($c,D\geq
0$). If $c^{\prime}=0$, then $S$ is spectral set, which is the union of
$\mathbb{Z}_{q}$-cosets, hence a tile. If $c^{\prime}>0$, then $|S|>p^{2}r$,
which we have already handled.

Thus we may assume $\Phi_{p^{2}r}\nmid m_{\Lambda}$ so we have
$\Phi_{p^{2}}\mid m_{\Lambda}$. Then since $p^{2}\nmid|S|$ we must have
$\Phi_{p}\nmid m_{\Lambda}$ so we have $\Phi_{pr}\mid m_{\Lambda}$. We remind
that we have already seen that $\Phi_{q}\mid m_{S}$ so $S$ is equidistributed
in the $\mathbb{Z}_{p^{2}r}$-cosets. Now we apply this to obtain information
about the structure of $S$.

We investigate the intersection of $S$ with each $\mathbb{Z}_{p^{2}r}$-cosets.
Assume $s_{1},s_{2}\in S$ are contained in a $\mathbb{Z}_{p^{2}r}$-coset but
are not contained in a $\mathbb{Z}_{pr}$-coset. If their $r$-coordinate is
different, then we would have $\Phi_{p^{2}r}\mid m_{\Lambda}$, which we have
excluded above. Similarly, if $s_{3}\neq s_{4}\in S$ are contained in a
$\mathbb{Z}_{pr}$-coset, then they need to have different $r$-coordinates,
otherwise we would have $\Phi_{p}\mid m_{\Lambda}$. Their difference is not
divisible by $p^{2}$ since we would have $\Phi_{r}\mid m_{\Lambda}$, which is
impossible since $r\nmid|\Lambda|=|S|$. Each $\mathbb{Z}_{p^{2}r}$-coset
contains the same amount of elements of $S$ by $\Phi_{q}\mid m_{S}$, which is
then at least $p$ and the previous argument shows that each
$\mathbb{Z}_{p^{2}r}$-coset contains exactly $p$ elements of $S$ and either
they lie in different $\mathbb{Z}_{pr}$-cosets or they are all contained in
one $\mathbb{Z}_{pr}$-coset with $p^{2}$ not dividing their differences. It is
easy to see that if for each $\mathbb{Z}_{p^{2}r}$-coset only one of the two
types appear, then $S$ is a tile.

Now we argue that $\Phi_{p^{2}q}\mid m_{S}$ or $\Phi_{n}\mid m_{S}$. By
Proposition 3.1 we may assume $0\in\Lambda$ and that $\Lambda$ is not
contained in a proper subgroup of $\mathbb{Z}_{p^{2}qr}$. Then our claim
follows from Lemma 3.2.

Since $r\nmid|S|=|\Lambda|$ we have $S_{p^{2}q}$ is a set. Assume
$\Phi_{p^{2}q}\mid m_{S}$, then $S_{p^{2}q}$ is the union of $\mathbb{Z}_{p}$
and $\mathbb{Z}_{q}$-cosets. Since there is at least one
$\mathbb{Z}_{p^{2}r}$-coset such that each of its $\mathbb{Z}_{pr}$-cosets
contains exactly one element of $S$ we have that $S_{p^{2}q}$ is the union of
$\mathbb{Z}_{q}$-cosets all contained in different $\mathbb{Z}_{pq}$-cosets.
This contradicts the existence of $\mathbb{Z}_{p^{2}r}$-cosets of
$\mathbb{Z}_{n}$, which contains a $\mathbb{Z}_{pr}$-coset containing exactly
$p$ elements of $S$ (any pair of different these elements have different
$r$-coordinate).

Assume now that $\Phi_{n}\mid m_{S}$. It is clear from our previous discussion
that there are $x,y$ with $p\mid x-y$ and $q\nmid x-y$ such that
$\mathbb{Z}_{pr}+x$ and $\mathbb{Z}_{pr}+y$ contains $1$ and $p$ elements of
$S$, respectively. We remind that if it contains $p$, then in that
$\mathbb{Z}_{pq}$-coset, the difference of the elements of $S$ lying in this
$\mathbb{Z}_{pr}$-coset is not divisible by either $p^{2}$ or with $r$. Then
one can build up a $3$-dimensional cube in $\mathbb{Z}_{pqr}$, which contains
exactly one element of $S$, which contradicts the fact that $S$ satisfies the
$3$-dimensional cube-rule in each $\mathbb{Z}_{pqr}$-cosets.

A similar argument works if $pr\mid\mid|S|$ since the role of $q$ and $r$ is
symmetric.

Case 3. Let us assume that $qr\mid\mid|S|$.
Then either $S$ is a complete set of residues $\pmod{qr}$, whence $S$ is a
tile or there are two different elements of $S$, whose difference is divisible
by $qr$. This would imply $\Phi_{p}\mid m_{\Lambda}$ or $\Phi_{p^{2}}\mid
m_{\Lambda}$. In both of these cases we have $p\mid|\Lambda|=|S|$, a
contradiction.

###  4.3 $|S|$ has at most one prime divisors among $p,q,r$

Let us assume $1\mid\mid|S|$ or $p\mid\mid|S|$ or $q\mid\mid|S|$ or
$r\mid\mid|S|$.

If $\Phi_{n}\mid m_{S}$, then the intersection of $S$ with each
$\mathbb{Z}_{pqr}$-coset satisfies the $3$-dimensional cube-rule. Then we
cannot have $1\mid|S|$ and if $|S|=p$ or $|S|=q$ or $|S|=r$, then $S$ is the
union of $\mathbb{Z}_{p}$-cosets, $\mathbb{Z}_{q}$-cosets or
$\mathbb{Z}_{r}$-cosets, respectively by Lemma 3.4. These cases are excluded
by Proposition 3.1.

A similar argument works for $\Lambda$ so if $\Phi_{n}\mid m_{\Lambda}$, then
$\Lambda$ is the union of $\mathbb{Z}_{p}$-coset, $\mathbb{Z}_{q}$-cosets or
$\mathbb{Z}_{r}$-cosets. Then since $(\Lambda,S)$ is also a spectral pair we
have by Proposition 3.1 that $\Lambda$ is a tile. Then $|\Lambda|=|S|\mid n$
so we have that $|S|=|\Lambda|=p$ or $|S|=|\Lambda|=q$ or $|S|=|\Lambda|=r$.
The fact that $\Lambda$ is $\mathbb{Z}_{\Lambda}$-coset implies that
$\Phi_{|\Lambda|}\mid m_{S}$. Then it is easy to see that (T1) and (T2) are
satisfied for $S$ so it is a tile. Thus we may assume $\Phi_{n}\nmid m_{S}$
and $\Phi_{n}\nmid m_{\Lambda}$.

By Lemma 3.2 we have $\Phi_{p^{2}q}\mid m_{S}$ since $\Phi_{n}\nmid m_{S}$.

Without loss of generality we may assume $r\nmid|S|$ since the role of $q$ and
$r$ are symmetric. Then $S_{p^{2}q}$ is a set so it is the disjoint union of
$\mathbb{Z}_{p}$-cosets and $\mathbb{Z}_{q}$-cosets. By Proposition 3.1 we
have that $\langle S\rangle=\mathbb{Z}_{n}$ so $\langle
S_{p^{2}q}\rangle=\mathbb{Z}_{p^{2}q}$. Then it follows that at least two
$\mathbb{Z}_{pq}$-cosets of $\mathbb{Z}_{p^{2}q}$ contain elements of
$S_{p^{2}q}$. Note that the same argument works for $\Lambda_{p^{2}q}$ as
well. We remind that $S_{p^{2}q}$ is the union of $\mathbb{Z}_{p}$-cosets and
$\mathbb{Z}_{q}$-cosets.

Assume there is a $\mathbb{Z}_{pq}$-cosets in $\mathbb{Z}_{p^{2}q}$, whose
intersection with $S_{p^{2}q}$ contains a $\mathbb{Z}_{q}$-coset and another
$\mathbb{Z}_{pq}$-coset containing a $\mathbb{Z}_{p}$-coset also contained in
$S_{p^{2}q}$. Then for every $1\neq d\mid p^{2}q$ there are
$s_{d,1},s_{d,2}\in S$ such that
$d\mid\mid\pi_{p^{2}q}(s_{d,1})-\pi_{p^{2}q}(s_{d,2})$, where $\pi_{p^{2}q}$
is the natural projection of $\mathbb{Z}_{n}$ to $\mathbb{Z}_{p^{2}q}$. Hence
$\Phi_{d}\mid m_{\Lambda}$ or $\Phi_{dr}\mid m_{\Lambda}$ for every $1\neq
d\mid p^{2}q$. Thus by Lemma 3.5 we have that
$\Lambda=c\mathbb{Z}_{p^{2}q}+rD$. If $c>0$, then $|S|\geq p^{2}q$ and if
$c=0$, then $r\mid|S|$. Both cases have already excluded. Again, the same
argument works for $\Lambda$.

Assume now that the intersection of $S_{p^{2}q}$ with each
$\mathbb{Z}_{pq}$-coset is the union of (possible 0) $\mathbb{Z}_{q}$-cosets
and this intersection is nonempty for at least two cosets of
$\mathbb{Z}_{pq}$. Then it follows from $\Phi_{n}\nmid m_{\Lambda}$ that if
$x,y\in S$, whose natural projections to $\mathbb{Z}_{p^{2}q}$ are not
contained in a proper coset of $\mathbb{Z}_{p^{2}q}$, then their difference
divisible by $r$. If $q>2$, then it follows that the difference of any two
elements of $S$ is divisible by $r$ so it is contained in a proper coset of
$\mathbb{Z}_{n}$. Proposition 3.1 gives that $S$ is a tile in this case.

If $q=2$ but there are more than two $\mathbb{Z}_{pq}$-cosets containing
elements of $S_{p^{2}q}$, then we build up a graph $\Gamma$ having vertex set
$S$ and two vertices are adjacent if and only if their difference is not
divisible by either $p$ or $q$. Again, the difference of two adjacent vertices
is divisible by $r$. It is not hard to verify that $\Gamma$ is connected and
then $S$ is contained in a $\mathbb{Z}_{p^{2}q}$-coset of $\mathbb{Z}_{n}$ so
it is a tile.

If there is a $\mathbb{Z}_{p}q$-coset in $\mathbb{Z}_{p^{2}q}$, whose
intersection with $S_{p^{2}q}$ contains at least two $\mathbb{Z}_{q}$-cosets
and another one which contains at least one $\mathbb{Z}_{q}$-coset, then again
we have that for every $1\neq d\mid p^{2}q$ there are elements $s_{1},s_{2}$
of $S_{p^{2}q}$ such that $d\mid\mid s_{1}-s_{2}$, which case has already been
handled above.

The $|S|=4$ case follows from a theorem of Kolountzakis and Matolcsi [4],
which says that spectral sets of cardinality at most $5$ in finite abelian
groups are tiles.

Thus it remains that $S_{p^{2}q}$ is the union of $\mathbb{Z}_{p}$-cosets
only. We also have that these cosets are not contained in a
$\mathbb{Z}_{pq}$-coset or a $\mathbb{Z}_{p^{2}}$-coset of
$\mathbb{Z}_{p^{2}q}$. For every $s\in S_{p^{2}q}$ the unique element of $S$
projecting to $s$ is denoted by $\bar{s}$ Assume that for every $x\in
S_{p^{2}q}$ there is an $y\in S_{p^{2}q}$ such that $p\nmid x-y$ and $q\nmid
x-y$. Then for every $x^{\prime}\in x+Z_{p}\subset S$, we have $p\nmid
x^{\prime}-y$ and $q\nmid x^{\prime}-y$. Since $\Phi_{n}\nmid m_{\Lambda}$ we
have that $r\mid\bar{x}-\bar{y}$ and $\bar{x^{\prime}}-\bar{y}$ so
$r\mid\bar{x}-\bar{x^{\prime}}$. The same holds for every element of
$x+\mathbb{Z}_{p}$ so $\overline{x+u}\colon~{}u\in\mathbb{Z}_{p}$ is a
$\mathbb{Z}_{p}$-coset. Then $S$ is the union of $\mathbb{Z}_{p}$-cosets,
which is handled by Proposition 3.1.

If there is a $x\in S_{p^{2}q}$ such that $p\mid x-y$ or $q\mid x-y$ for every
$y\in S_{p^{2}q}$, then $S_{p^{2}q}$ is contained in
$(x+\mathbb{Z}_{pq})\cup(x+\mathbb{Z}_{p^{2}})$. Since $S_{p^{2}q}$ is not
contained in any of this two sets appearing in the union we again have that
for every $d\mid p^{2}q$ there are $x,y\in S_{p^{2}q}$ with $d\mid\mid x-y$,
which case has already been settled.

## Acknowledgement

I am grateful for Gergely Kiss, Romanos Diogenes Malikiosis and Máté Vizer.
Many of the ideas appearing in the manuscript come up during fruitful
discussions aiming to prove other results concerning Fuglede’s conjecture or
in our common paper [7]. For example I learned a version of Lemma 3.5 from
Romanos Diogenes Malikiosis and an earlier version of Lemma 3.4 was worked out
by Gergely Kiss and Máté Vizer.

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