Equi-distributed property and spectral set conjecture on $\mathbb{Z}_{p^2}\times \mathbb{Z}_{p}$

TL;DR

This paper proves that in the finite abelian group _{p^2} imes \u007_{p}, a set is spectral if and only if it tiles, by establishing an equidistributed property and confirming Fuglede's conjecture for this group.

Contribution

The paper introduces an equidistributed property in rac{p^2}{p} groups and uses it to prove Fuglede's spectral set conjecture in rac{p^2}{p} groups.

Findings

Fuglede's spectral set conjecture holds on rac{p^2}{p} groups.

Established an equidistributed property in rac{p^2}{p} groups.

Spectral sets in rac{p^2}{p} groups are exactly the tiles.

Abstract

In this paper, we show an equi-disctributed property in $2$-dimensional finite abelian groups $\mathbb{Z}_{p^2}\times \mathbb{Z}_{p}$ where $p$ is a prime number. By using this equi-disctributed property, we prove that Fuglede's spectral set conjecture holds on groups $\mathbb{Z}_{p^2}\times \mathbb{Z}_{p}$, namely, a set in $\mathbb{Z}_{p^2}\times \mathbb{Z}_{p}$ is a spectral set if and only if it is a tile.

Equations

$$\hat{G}=\{{\chi }_g:g\in G\}$$

$${\chi }_g\cdot {\chi }_{g^{\prime }}={\chi }_{g+g^{\prime }},\forall g,g^{\prime }\in G.$$

$$\{{\chi }_b:b\in B\}$$

$$♯A=♯B\text{and}\sum _{a\in A}{\chi }_b\cdot \overline{{\chi }_{b^{\prime }}}(a)=0,\forall b,b^{\prime }\in B,b\ne b^{\prime }.$$

$${\hat{1}}_A(g)=\sum _{a\in A}\overline{{\chi }_g}(a),$$

$$\underset{t\in T}{⨆}(A+t)=G.$$

$$\sum _{b\in B}{1}_A(x-b)=1,\forall x\in G,$$

$$1_A\ast 1_B=1.$$

$${\hat{1}}_A\cdot {\hat{1}}_B=(♯G){\delta }_0,$$

$$⟨\mathbf{d},{\mathbf{d}}^{\prime }⟩=d_1{d}_1^{\prime }+p^{n-m}{d}_2{d}_2^{\prime }\in {\mathbb{Z}}_{p^n}.$$

$$H_{n,m}(\text{d},t):=\{\text{x}\in {\mathbb{Z}}_{p^n}\times {\mathbb{Z}}_{p^m}:⟨\text{x},\text{d}⟩=t\},$$

$$r\mathbf{d}=(\widetilde{d_1},\widetilde{d_2})\in {\mathbb{Z}}_{p^n}\times {\mathbb{Z}}_{p^m},$$

$$\sum _{(a_1,a_2)\in A}{e}^{2\pi i(\frac{a_1{d}_1}{p^n}+\frac{a_2{d}_2}{p^m})}=0,$$

$$\sum _{(a_1,a_2)\in A}{e}^{\frac{2\pi i(a_1{d}_1+p^{n-m}{a}_2{d}_2)}{p^n}}=0.$$

$$\sum _{(a_1,a_2)\in A}{e}^{\frac{2\pi ir(a_1{d}_1+p^{n-m}{a}_2{d}_2)}{p^n}}=0,\text{for any}r\in {\mathbb{Z}}_{p^n}^{\ast },$$

$$\sum _{(a_1,a_2)\in A}{e}^{2\pi i(\frac{a_1(r{d}_1)}{p^n}+\frac{a_2(r{d}_2)}{p^m})}=0,\text{for any}r\in {\mathbb{Z}}_{p^n}^{\ast }.$$

$${\hat{1}}_A(\text{d})=\frac{1}{p^{n+m}}\sum _{\text{x}\in {\mathbb{Z}}_{p^n}\times {\mathbb{Z}}_{p^m}}{e}^{\frac{2\pi i⟨\text{x},\text{d}⟩}{p^n}}{1}_A(\text{x}).$$

$$0=\sum _{\text{x}\in {\mathbb{Z}}_{p^n}\times {\mathbb{Z}}_{p^m}}{\xi }_n^{⟨\text{x},\text{d}⟩}{1}_A(\text{x})=\sum _{t\in {\mathbb{Z}}_{p^n}}{\xi }_n^t\sum _{⟨\text{x},\text{d}⟩=t}{1}_A(\text{x}).$$

$$n(t)=♯(A\cap H_{n,m}(\mathbf{d},t))=\sum _{⟨\text{x},\text{d}⟩=t}{1}_A(\text{x})\in \mathbb{Z}.$$

$$\sum _{t\in {\mathbb{Z}}_{p^n}}n(t){\xi }_n^t=0.$$

$$P(X)=\sum _{t\in {\mathbb{Z}}_{p^n}}n(t)X^t\in \mathbb{Z}[X].$$

$${\Phi }_{p^n}(X)={\Phi }_p(X^{p^{n-1}})=1+X^{p^{n-1}}+X^{2p^{n-1}}+\cdots +X^{(p-1)p^{n-1}}.$$

$$P(X)=Q(X){\Phi }_{p^n}(X).$$

$$(0,1),(1,0),(p,0),(p,c)\text{and}(1,c)\text{for all}c\in {\mathbb{Z}}_p\setminus \{0\}.$$

$$x_i+c{y}_i\not\equiv x_j+c{y}_j\mathrm{mod}p,\text{for all}i\ne j,$$

$$\{x_0+c{y}_0,x_1+c{y}_1,\cdots ,x_{p-1}+c{y}_{p-1}\}\equiv \{0,1,\dots ,p-1\}\mathrm{mod}p.$$

$$u=(k-j)(\alpha -\beta {)}^{-1}\in {\mathbb{Z}}_p.$$

$$x_j+u{y}_j\equiv j+(k-j)(\alpha -\beta {)}^{-1}\cdot \alpha \equiv (k\alpha -j\beta )(\alpha -\beta {)}^{-1}\mathrm{mod}p,$$

$$x_k+u{y}_k\equiv k+(k-j)(\alpha -\beta {)}^{-1}\cdot \beta \equiv (k\alpha -j\beta )(\alpha -\beta {)}^{-1}\mathrm{mod}p.$$

$$x_j+u{y}_j\equiv x_k+u{y}_k\mathrm{mod}p.$$

Full text

1. Introduction

Let $G$ be a locally compact abelian group. Let $\widehat{G}$ be the dual group of $G$ which consists of all continuous group characters. Let $\Omega$ be a Borel measurable subset in $G$ with positive finite Haar measure. The set $\Omega$ is called a spectral set if there exists a set $\Lambda\subset\widehat{G}$ which is an Hilbert basis of $L^{2}(\Omega)$, and that $\Omega$ is called a tile of $G$ by translation if there exists a set $T\subset G$ of translates such that almost surely $\Omega\oplus T=G$. In 1974, Fuglede [4] formulated the following so-called spectral set conjecture concerning the equivalence relation between tiles and spectral sets:

A Borel subset $\Omega\subset G$ of positive and finite Haar measure is a spectral set if and only if it is a tile.

Fuglede’s conjecture has attracted substantial attention over the last decades, especially in the setting $G=\mathbb{R}^{d}$. Many positive results towards the conjecture were established before Tao [25] who disproved the conjecture by showing that the direction “Spectral $\Rightarrow$ Tiling” is not tenable for $\mathbb{R}^{d}$ when $d\geq 5$. Now it is known that the conjecture does not hold in both directions for $\mathbb{R}^{d},d\geq 3$ [8, 17, 19, 23]. But the conjecture is still open in lower dimensions $\mathbb{R}$ and $\mathbb{R}^{2}$.

In its general setting, few is known about this conjecture, even for finite groups. Hence, the question becomes for which group $G$, the spectral set conjecture holds. For $N\geq 1$, the ring of integers modulo $N$ is denoted by $\mathbb{Z}_{N}=\mathbb{Z}/N\mathbb{Z}$. We only know that Fuglede’s spectral set conjecture holds on $\mathbb{Z}_{p^{n}}$ [15, 5], $\mathbb{Z}_{p}\times\mathbb{Z}_{p}$ [11], $\mathbb{Z}_{p^{n}q}$ with $n\geq 1$ [24], $p$-adic field $\mathbb{Q}_{p}$ [5, 6], $\mathbb{Z}_{pqr}$ with $p,q,r$ different primes [22] and very recently $\mathbb{Z}_{p^{n}q^{2}}$ with $n\geq 1$ [13]. Nevertheless, we either know few about for which group, the spectral set conjecture fails. We only know that the direction “spectral sets $\Rightarrow$ tiles” fails on $\mathbb{Z}_{3}^{6}$ [25], $\mathbb{Z}_{8}^{3}$ [19], $\mathbb{Z}_{p}^{4}\quad($when $p$ prime and $p\equiv 1\mod 4)$ and $\mathbb{Z}_{p}^{5}\quad($when $p$ prime and $p\equiv 3\mod 4)$ [1]; the direction “tiles $\Rightarrow$ spectral sets ” fails on $(\mathbb{Z}/6\mathbb{Z})^{5}$ [24], $(\mathbb{Z}/6\mathbb{Z})^{3}\times\mathbb{Z}_{6p}$, where $p$ is prime with gcd$(6,p)=1$ [7] and $(\mathbb{Z}/24\mathbb{Z})^{3}$ [8]. Actually, these non-tile spectral sets (resp. non-spectral tiles) in finite abelian groups lead to non-tile spectral sets (resp. non-spectral tiles) in euclidean spaces. We remark here that we still don’t know whether there exists non-spectral tiles in $\mathbb{Z}_{p^{n}}^{d}$ for $d\geq 1$ and $p$ prime. The following conjecture is therefore proposed:

Conjecture 1.1**.**

Let $p$ be a prime. For all $n,d\geq 1$, the tiles in the groups $\mathbb{Z}_{p^{n}}^{d}$ are always spectral sets.

As we have mentioned eariler, the known non-tile spectral sets on $\mathbb{R}^{d}$ ($d\geq 3$) are constructed by using the non-tile spectral sets on $\mathbb{Z}_{p^{n}}^{d}$ for some $n$ (see [8, 17, 19, 23]). Since Fuglede’s spectral set conjecture is widely open on $\mathbb{R}^{2}$, it is interesting to know whether Fuglede’s spectral set conjecture holds on $2$-dimensional finite abelian groups $\mathbb{Z}_{p^{n}}^{2}$ for all $n$. In fact, if Fuglede’s conjecture fails on $\mathbb{R}^{2}$, then the counterexample is probably constructed by a counterexample on $\mathbb{Z}_{p^{n}}^{2}$ for some $n$. We don’t know anything about the conjecture on these groups except that Iosevich, Mayeli and Pakianathan [11] proved that the spectral set conjecture is true on $\mathbb{Z}_{p}\times\mathbb{Z}_{p}$.

In this paper, our main interest is Fuglede’s spectral set conjecture on $\mathbb{Z}_{p^{n}}\times\mathbb{Z}_{p^{m}}$ for $n,m\geq 1$. We first show an equi-distribued property in $\mathbb{Z}_{p^{n}}\times\mathbb{Z}_{p^{m}}$. Then we focus on the groups $\mathbb{Z}_{p^{2}}\times\mathbb{Z}_{p}$. Our main result is the following.

Theorem 1.1**.**

A set in $\mathbb{Z}_{p^{2}}\times\mathbb{Z}_{p}$ is a spectral set if and only if it is a tile of $\mathbb{Z}_{p^{2}}\times\mathbb{Z}_{p}$ by translation.

Haessig et al. [9] proved that for $A\subset\mathbb{Z}_{p}^{d}$ and $l\in\mathbb{Z}_{p}^{d}$, the equation $\widehat{1}_{A}(l)=0$ if and only if $A$ is equi-distributed on the $p$ hyperplanes $H_{t}=\{x:x\cdot l=t\}$ for $t\in\mathbb{Z}_{p}^{d}$. Actually the $p$ hyperplanes $H_{t}$ (for $t\in\mathbb{Z}_{p}^{d}$) form a partition of $\mathbb{Z}_{p}^{d}$. Iosevich, Mayeli and Pakianathan [11] used this property to prove Fuglede’s spectral set conjecture valid on $\mathbb{Z}_{p}\times\mathbb{Z}_{p}$. But such property is not true in general or over other fields. Thus we need other kind of equi-distributed property in general case. In this paper, we prove such equi-distributed property in $\mathbb{Z}_{p^{n}}\times\mathbb{Z}_{p^{m}}$ in Lemma 3.1. It is the main tool to prove Theorem 1.1. Actually, Lemma 3.1 shows us that it is not equi-distributed on each set of a certain partition of $\mathbb{Z}_{p^{n}}\times\mathbb{Z}_{p^{m}}$, but that it is equi-distributed in some chosen planes. This is the main difference between the equi-distributed property in $\mathbb{Z}_{p^{n}}\times\mathbb{Z}_{p^{m}}$ and the one in $\mathbb{Z}_{p}\times\mathbb{Z}_{p}$.

The article is organized as follows. In Section 2, we discuss some basic properties of spectral sets and tiles in finite abelian groups. In Section 3, we prove an equi-distributed property for a set in $\mathbb{Z}_{p^{n}}\times\mathbb{Z}_{p^{m}}$ whenever $\mathcal{Z}_{A}$ is not empty. In Section 4, we define the compatible value and then discuss some properties of it. In Section 6, we prove the “Tiles $\Rightarrow$ Spectral sets” part of Theorem 1.1. In Section 7, we prove the “Spectral sets $\Rightarrow$ Tiles” part of Theorem 1.1.

Notation

• •

For $N\in\mathbb{N}$, we denote by $\mathbb{Z}_{N}$ the cyclic group $\mathbb{Z}/N\mathbb{Z}$ or sometimes the finite set $\{0,1,\dots,N-1\}$.

• •

For a finite abelian group $G$, denote by $G^{*}$ the set of all invertible elements in $G$.

2. Preliminary

In this section, we discuss some basic properties spectral sets and tiles in finite groups.

Let $G$ be a finite abelian group. Let $\widehat{G}$ be the dual group of $G$. It is well-known that $\widehat{G}$ is also a finite abelian group and furthermore isomorphic to $G$. Consequently, we can write

[TABLE]

with the commutative multiplicity

[TABLE]

2.1. Spectral sets in $G$

We recall that a subset $A\subset G$ is said to be spectral if there is a set $B\subset G$ such that

[TABLE]

forms an orthogonal basis in $L^{2}(A)$. In such case, the set $B$ is called a spectrum of $A$ and the pair $(A,B)$ called a spectral pair. Since the dimension of $L^{2}(A)$ is $\sharp A$, the pair $(A,B)$ being a spectral pair is equivalent to that $\sharp A=\sharp B$ and meanwhile the set $\{\chi_{b}:b\in B\}$ is orthogonal in $L^{2}(A)$, that is to say,

[TABLE]

This means that the complex matrix $M=\left(\chi_{b}(a)\right)_{b\in B,a\in A}$ is a complex Hadamard matrix, i.e. $M\overline{M}^{T}=(\sharp A)I$ where $M^{T}$ is the transpose of $M$ and $I$ is the identity matrix. Since $\sharp A=\sharp B$, it follows that $\overline{M}^{T}M=(\sharp B)I$. This deduces that $B$ is a spectral set with spectrum $A$.

Denote by $\mathcal{Z}_{A}=\{g\in G:\widehat{1}_{A}(g)=0\}$ the set of zeros of the Fourier transform of the indicator function $1_{A}$. Since

[TABLE]

we restate the above equivalent definition of spectral sets as follows.

Proposition 2.1**.**

Let $A,B\subset G$. The following are equivalent.

• (1)

$(A,B)$* is a spectral pair.*

• (2)

$(B,A)$* is a spectral pair.*

• (3)

$\sharp A=\sharp B$; $(B-B)\setminus\{0\}\subset\mathcal{Z}_{A}$.

2.2. Tiles in $G$

Recall that a subset $A\subset G$ is said to be a tile if there is a set $T\subset G$ such that

[TABLE]

In such case, the set $T$ is called a tiling complement of $A$ and the pair $(A,T)$ is called a tiling pair. Using Fourier transform, we have the following equivalent definitions of tiles in $G$.

Lemma 2.2**.**

Let $A,B$ be subsets in $G$. Then the following statements are equivalent.

• (1)

$(A,B)$* is a tiling pair.*

• (2)

$(B,A)$* is a tiling pair.*

• (3)

$A\oplus B=G$.

• (4)

$\sharp A\cdot\sharp B=\sharp G$* and $\mathcal{Z}_{A}\cup\mathcal{Z}_{B}=G\setminus\{0\}$.*

Proof.

It is trivial to see that $(1)\Leftrightarrow(2)\Leftrightarrow(3)$. It is sufficient to prove $(1)\Leftrightarrow(4)$. In fact, the pair $(A,B)$ being a tiling pair is equivalent to

[TABLE]

that is to say,

[TABLE]

Taking Fourier transformation of both sides of $(\ref{eq:AB tile})$, we get

[TABLE]

which implies $\sharp A\cdot\sharp B=\sharp G$ and $\mathcal{Z}_{A}\cup\mathcal{Z}_{B}=G\setminus\{0\}$. ∎

3. Equi-distribution on planes

In this section, we prove an equi-distributed property for a set in $\mathbb{Z}_{p^{n}}\times\mathbb{Z}_{p^{m}}$ ($n,m\in\mathbb{N}$) whenever $\mathcal{Z}_{A}$ is not empty.

Without loss of generality, we assume in this section that $n>m$. Let $\xi_{n}=e^{2\pi i\frac{1}{p^{n}}}$ be a primitive $p^{n}$-th root of unity. For ${\rm\mathbf{d}}=(d_{1},d_{2}),{\rm\mathbf{d}}^{\prime}=(d_{1}^{\prime},d_{2}^{\prime})\in\mathbb{Z}_{p^{n}}\times\mathbb{Z}_{p^{m}}$, we define the inner product in $\mathbb{Z}_{p^{n}}\times\mathbb{Z}_{p^{m}}$ by the formula

[TABLE]

We define

[TABLE]

for $\textbf{d}\in\mathbb{Z}_{p^{n}}\times\mathbb{Z}_{p^{m}}$ and $t\in\mathbb{Z}_{p^{n}}$. We call such set a plane in $\mathbb{Z}_{p^{n}}\times\mathbb{Z}_{p^{m}}$.

For $r\in\mathbb{Z}_{p^{n}}^{*}$ and ${\rm\mathbf{d}}=(d_{1},d_{2})\in\mathbb{Z}_{p^{n}}\times\mathbb{Z}_{p^{m}}$, we define the scalar product by

[TABLE]

where $\tilde{d_{1}}\equiv rd_{1}\mod\mathbb{Z}_{p^{n}}$ and $\tilde{d_{2}}\equiv rd_{2}\mod\mathbb{Z}_{p^{m}}$.

Now we state the main result in this section.

Lemma 3.1**.**

Let $A\subset\mathbb{Z}_{p^{n}}\times\mathbb{Z}_{p^{m}}$ and ${\rm\mathbf{d}}\in\mathbb{Z}_{p^{n}}\times\mathbb{Z}_{p^{m}}$. The following are equivalent.

• (1)

$\hat{1}_{A}({\rm\mathbf{d}})=0$.

• (2)

$\hat{1}_{A}(r{\rm\mathbf{d}})=0$* for any $r\in\mathbb{Z}_{p^{n}}^{*}$.*

• (3)

$\sharp(A\cap H({\rm\mathbf{d}},t))=\sharp(A\cap H({\rm\mathbf{d}},t^{\prime}))$* if $p^{n-1}\mid t-t^{\prime}$.*

Proof.

$(2)\Rightarrow(1):$ It is trivial.

$(1)\Rightarrow(2):$ Denote by ${\rm\mathbf{d}}=(d_{1},d_{2})$. The fact that $\hat{1}_{A}({\rm\mathbf{d}})=0$ is equivalent to

[TABLE]

that is to say,

[TABLE]

It follows from Galois theory that

[TABLE]

which means that

[TABLE]

We obtain that $\hat{1}_{A}(r{\rm\mathbf{d}})=0$ for any $r\in\mathbb{Z}_{p^{n}}^{*}$.

$(1)\Leftrightarrow(3):$ It is easy to see that

[TABLE]

It follows that

[TABLE]

Let

[TABLE]

Then we have

[TABLE]

It follows that $\xi_{n}$ is a root of the polynomial

[TABLE]

Recall that the minimal polynomial over $\mathbb{Z}$ of $\xi_{n}$ is

[TABLE]

Therefore there exists $Q\in\mathbb{Z}[X]$ such that

[TABLE]

Hence, we conclude that $n(t)=n(t^{\prime})$ if $p^{n-1}\mid t-t^{\prime}$ ∎

Remark 3.1**.**

Now we are concerned with the case $\mathbb{Z}_{p^{2}}\times\mathbb{Z}_{p}$. By Lemma 3.1 (2), we define the relation ${\rm\mathbf{d}}\sim{\rm\mathbf{d}}^{\prime}$ if there exists $r\in\mathbb{Z}_{p^{2}}^{*}$ such that ${\rm\mathbf{d}}=r{\rm\mathbf{d}}^{\prime}$. Thus the equivalent classes in $\mathbb{Z}_{p^{2}}\times\mathbb{Z}_{p}$ by $\sim$ are

[TABLE]

Thus when we study the set of zeros $\mathcal{Z}_{\mathbb{Z}_{p^{2}}\times\mathbb{Z}_{p}}$, we only need to consider the elements which have the above forms.

4. Compatible values

In this section, we introduce the notion of the compatible value for $p$ pairs. This notion will be helpful to deduce some properties of spectral sets and tiles in $\mathbb{Z}_{p^{2}}\times\mathbb{Z}_{p}$, which will be shown in Section 5.

Given $p$ pairs $(x_{i},y_{i})_{0\leq i\leq p-1}\in(\mathbb{Z}_{p}\times\mathbb{Z}_{p})^{p}$, we consider the value $c\in\mathbb{Z}_{p}$ such that

[TABLE]

that is to say,

[TABLE]

If it is the case, we call the value $c$ compatible with the $p$ pairs $(x_{i},y_{i})_{0\leq i\leq p-1}$.

It is not hard to see that for every $a\in\mathbb{Z}_{p}$, each element in $\mathbb{Z}_{p}$ is compatible with $(i,a)_{0\leq i\leq p-1}$. However, given some special $(x_{i},y_{i})_{0\leq i\leq p-1}$, we will see that some element in $\mathbb{Z}_{p}$ can not be compatible with $(x_{i},y_{i})_{0\leq i\leq p-1}$.

Lemma 4.1**.**

Let $(x_{i},y_{i})_{0\leq i\leq p-1}\in(\mathbb{Z}_{p}\times\mathbb{Z}_{p})^{p}$. Suppose that $x_{i}\not=x_{j}$ for all $0\leq i\not=j\leq p-1$. Suppose that not all $y_{j}$ are the same for $0\leq j\leq p-1$. Then there exists $c\in\mathbb{Z}_{p}^{*}$ such that $c$ is not compatible with $(x_{i},y_{i})_{0\leq i\leq p-1}$.

Proof.

Without loss of generality, we assume that $x_{i}=i$ for all $0\leq i\leq p-1$. Let $j,k\in\mathbb{Z}_{p}$ such that $y_{j}=\alpha$ and $y_{k}=\beta$ with $\alpha\not=\beta$. Let

[TABLE]

It follows that

[TABLE]

and

[TABLE]

It implies that

[TABLE]

Hence, the value $u$ is not compatible with $(x_{i},y_{i})_{0\leq i\leq p-1}$. ∎

The following lemma tells that the optimum in Lemma 4.1 can be achieved in some case.

Lemma 4.2**.**

Let $r\in\mathbb{Z}_{p}^{*}$. Then the value $c\in\mathbb{Z}_{p}$ is compatible with $(ri,i)_{0\leq i\leq p-1}$ if and only if $c\not=p-r$.

Proof.

We first observe that

[TABLE]

for all $0\leq i\leq p-1$. Thus the value $(p-r)$ is not compatible with $(ri,i)_{0\leq i\leq p-1}$.

On the other hand, for $c\in\mathbb{Z}_{p}\setminus\{p-r\}$, we have $(r+c)\in\mathbb{Z}_{p}^{*}$ and

[TABLE]

It follows that

[TABLE]

Thus, for any $c\in\mathbb{Z}_{p}\setminus\{p-r\}$, it is compatible with $(ri,i)_{0\leq i\leq p-1}$. ∎

5. Properties of spectral sets and tiles

In this section, we show some properties of spectral sets and tiles in $\mathbb{Z}_{p^{2}}\times\mathbb{Z}_{p}$. These properties not only have their own interest but also are useful to the proof of Theorem 1.1. The main results in this section are summarized as follows. Let $A\subset\mathbb{Z}_{p^{2}}\times\mathbb{Z}_{p}$ with $1<\sharp A<p^{3}$ (we don’t consider the trivial situation where $\sharp A=1$ or $p^{3}$). Actually, we have already proved some condition on the cardinality of tiles and spectral sets in previous sections: if $A$ is a tile in $\mathbb{Z}_{p^{2}}\times\mathbb{Z}_{p}$, then by Lemma 2.2, $\sharp A$ divides $\sharp(\mathbb{Z}_{p^{2}}\times\mathbb{Z}_{p})=p^{3}$, that is, $\sharp A$ has to be $1,p,p^{2}$ or $p^{3}$; if $A$ is a spectral set in $\mathbb{Z}_{p^{2}}\times\mathbb{Z}_{p}$, then by Lemma 3.1, $\sharp A$ is divided by $p$. In this section, we first show that if $p^{2}<\sharp A<p^{3}$, then the set $A$ has no chance to be a spectral set (Lemma 5.2). We then prove that if $\sharp A=p$ and $\mathcal{Z}_{A}\not=\emptyset$, then $A$ is a tile and meanwhile a spectral set (Proposition 5.3). In the rest of this section, we deal with some special cases: Lemma 5.4 and Lemma 5.5 tells us that if some special elements are contained in $\mathcal{Z}_{A}$, then $\sharp A=p^{2}$ and meanwhile $A$ is a tile and a spectral set; Lemma 5.6 shows that if a tile (resp. a spectral set) is in a proper subgroup of $\mathbb{Z}_{p^{2}}\times\mathbb{Z}_{p}$, then it is a spectral set (resp. a tile). Thus for tiles, we only need to focus on the case where $\sharp A=p^{2}$; for spectral sets, we only need to consider the case where $p<\sharp A\leq p^{2}$. We leave these cases to the next two sections.

For $A\subset\mathbb{Z}_{p^{2}}\times\mathbb{Z}_{p}$, we are concerned with the difference set $(A-A)$ in the following lemma. We show that if $\sharp A$ is large (i.e. $\sharp A>p^{2}$), then each non-trivial direction is contained in $(A-A)$.

Lemma 5.1**.**

Let $A\subset\mathbb{Z}_{p^{2}}\times\mathbb{Z}_{p}$. Suppose $\sharp A>p^{2}$. Then for all ${\rm\mathbf{d}}\in\mathbb{Z}_{p^{2}}\times\mathbb{Z}_{p}\setminus\{0\}$, we have $(A-A)\cap{\rm\mathbf{d}}(\mathbb{Z}_{p^{2}}\times\mathbb{Z}_{p})^{*}\not=\emptyset$.

Proof.

Let ${\rm\mathbf{d}}=(c_{1}+c_{2}p,c_{3})\in(\mathbb{Z}_{p^{2}}\times\mathbb{Z}_{p})\setminus\{0\}$ where $c_{i}\in\mathbb{Z}_{p}$ and at least one of $c_{i}$ is non-zero for $i=1,2,3$. We first consider the case where $c_{1}\not=0$. Let

[TABLE]

It is not hard to see that every element in $\mathbb{Z}_{p^{2}}\times\mathbb{Z}_{p}$ is a $\mathbb{Z}_{p}$-combination of ${\rm\mathbf{d}},\textbf{v}$ and w, that is, for any $\textbf{x}\in\mathbb{Z}_{p^{2}}\times\mathbb{Z}_{p}$, there exists $a_{1},a_{2}$ and $a_{3}$ in $\mathbb{Z}_{p}$ such that

[TABLE]

Since $\sharp A>p^{2}$, by pigeonhole principle, there exist x and $\textbf{x}^{\prime}$ in $\mathbb{Z}_{p^{2}}\times\mathbb{Z}_{p}$ such that

[TABLE]

with $a_{1},a_{1}^{\prime},a_{2},a_{3}\in\mathbb{Z}_{p}$ and $a_{1}\not=a_{1}^{\prime}$. Thus we obtain that

[TABLE]

The proof of the case where $c_{2}\not=0$ or $c_{3}\not=0$ is similar. Thus we complete the proof. ∎

The following lemma shows that for a set $A\subset\mathbb{Z}_{p^{2}}\times\mathbb{Z}_{p}$, if $p^{2}<\sharp A<p^{3}$, then $A$ is not a spectral set in $\mathbb{Z}_{p^{2}}\times\mathbb{Z}_{p}$.

Lemma 5.2**.**

Let $A\subset\mathbb{Z}_{p^{2}}\times\mathbb{Z}_{p}$ be a spectral set. Suppose $\sharp A>p^{2}$. Then $A=\mathbb{Z}_{p^{2}}\times\mathbb{Z}_{p}$.

Proof.

Let $B$ be a spectrum of $A$. It follows that $\sharp B=\sharp A>p^{2}$ and

[TABLE]

By Lemma 5.1 and Lemma 3.1 $(2)$, we have

[TABLE]

which means that

[TABLE]

By elementary Fourier analysis, we have

[TABLE]

where $\xi_{n}$ is a primitive $n$-th root of unity. It follows that

[TABLE]

Since ${1}_{A}(x)$ is an indicator of a set, implying that ${1}_{A}(x)$ takes value in $\{0,1\}$, we conclude that $\sharp A=p^{3}$. Thus, we obtain that the set $A$ has to be $\mathbb{Z}_{p^{2}}\times\mathbb{Z}_{p}$. ∎

Generally, Lemma 5.1 and Lemma 5.2 are also valid for all $2$-dimensional finite abelian groups $\mathbb{Z}_{p^{n}}\times\mathbb{Z}_{p^{m}}$ ($n,m\geq 0$):

Let $A\subset\mathbb{Z}_{p^{n}}\times\mathbb{Z}_{p^{m}}$. Suppose $\sharp A>p^{n+m-1}$. Then for all ${\rm\mathbf{d}}\in(\mathbb{Z}_{p^{n}}\times\mathbb{Z}_{p^{m}})\setminus\{\textbf{0}\}$, we have $(A-A)\cap{\rm\mathbf{d}}(\mathbb{Z}_{p^{n}}\times\mathbb{Z}_{p^{m}})^{*}\not=\emptyset$. Moreover, if $A$ is a spectral set, then $A=\mathbb{Z}_{p^{n}}\times\mathbb{Z}_{p^{m}}$.

The proof is similar with the proof of Lemma 5.1 and Lemma 5.2. We leave it to the readers to work out the details.

In what follows, we write simply $H(\textbf{d},t)=H_{2,1}(\textbf{d},t)$ for every $t\in\mathbb{Z}_{p^{2}}$ and ${\rm\mathbf{d}}\in\mathbb{Z}_{p^{2}}\times\mathbb{Z}_{p}$. Moreover, for any subset $B\subset\mathbb{Z}_{p^{2}}\times\mathbb{Z}_{p}$, we write

[TABLE]

A simple fact which we will use several times later is that the statement $H((p,c),t)\cap B=\emptyset$ for all $c\in\mathbb{Z}_{p}$ and all $t\in\mathbb{Z}_{p^{2}}$ with $p\not|t$ implies that $h_{B}((p,c),t)=0$ for all $c\in\mathbb{Z}_{p}$ and all $t\in\mathbb{Z}_{p^{2}}$ with $p\not|t$.

Now we are concerned with the case where $\sharp A=p$.

Proposition 5.3**.**

Let $A\subset\mathbb{Z}_{p^{2}}\times\mathbb{Z}_{p}$ with $\sharp A=p$. Suppose that $\mathcal{Z}_{A}\not=\emptyset$. Then $A$ is a tile and also a spectral set.

Proof.

We first prove that $A$ is a spectral set by constructing its spectrum. Suppose that ${\rm\mathbf{d}}\in\mathcal{Z}_{A}$. Define

[TABLE]

Since ${\rm\mathbf{d}}\not=\textbf{0}$, we have $\sharp B=p$. It is easy to see that

[TABLE]

It follows that

[TABLE]

By Lemma 2.1, we obtain that $B$ is a spectrum of $A$.

It remains to prove that $A$ is a tile. Denote by

[TABLE]

Due to Remark 3.1, we decompose the proof in the following four cases.

Case 1: $(0,1)\in\mathcal{Z}_{A}$. We define

[TABLE]

We will prove that $B_{1}$ is a tiling complement of $A$. It is easy to see that $\sharp B_{1}=p^{2}$. A simple computation shows that for all $c\in\mathbb{Z}_{p}$ and for all $t\in\mathbb{Z}_{p^{2}}$, we have

[TABLE]

By Lemma 3.1, we have

[TABLE]

On the other hand, by similar computation with $(\ref{eq:B_1(0)})$, we obtain that for all $c\in\mathbb{Z}_{p}$ and $z\in\mathbb{Z}_{p^{2}}$,

[TABLE]

By Lemma 3.1, we have

[TABLE]

Combining (5.3), (5.4) and Remark 3.1, we get

[TABLE]

By Lemma 2.2, we conclude that $B_{1}$ is a tiling complement of $A$.

Case 2: $\exists c\in\mathbb{Z}_{p}$ such that $(1,c)\in\mathcal{Z}_{A}$. By Lemma 3.1 and the fact that $\sharp A=p$, we obtain that there exists $\bar{x}\in\mathbb{Z}_{p}$ such that

[TABLE]

Using (5.1), we rewrite (5.5) as follows (taking a permutation of the index set $\{i\}_{0\leq i\leq p-1}$ if necessary):

[TABLE]

It follows that

[TABLE]

Thus $A$ has to be the form

[TABLE]

Let

[TABLE]

Following the results in [11], the set $\widetilde{A}$ is a tile. Denoting by $\widetilde{B}_{2}$ a tiling complement of $\widetilde{A}$, we define

[TABLE]

It follows that $B_{2}$ is a tiling complement of $A$. In fact, for any $(x+py,z)\in\mathbb{Z}_{p^{2}}\times\mathbb{Z}_{p}$, there exists unique decomposition

[TABLE]

where $(\bar{x}+py^{\prime},z^{\prime})\in A$, $((x-\bar{x})+py^{\prime\prime},z^{\prime\prime})\in B_{2}$ and $(y,z)=(y^{\prime},z^{\prime})+(y^{\prime\prime},z^{\prime\prime})\in\mathbb{Z}_{p}\times\mathbb{Z}_{p}$ in the unique way according to the tiling pair $(\widetilde{A},\widetilde{B})$.

Case 3: $\exists c\in\mathbb{Z}_{p}^{*}$ such that $(p,c)\in\mathcal{Z}_{A}$. We define

[TABLE]

It is easy to see that $\sharp B_{3}=p^{2}$. The statement that the set $B_{3}$ is a tiling complement of $A$ is a direct consequence by Lemma 2.2 and the following claim.

Claim: $\mathcal{Z}_{B_{3}}=(\mathbb{Z}_{p^{2}}\times\mathbb{Z}_{p})\setminus\{(0,0),(p,c)\}$.

Proof.

According to Remark 3.1, we will prove the following three kinds of zeros are belonging to $\mathcal{Z}_{B_{3}}$.

1. $(0,1)$. The result is followed directly from Lemma 3.1 and the fact that for $z\in\mathbb{Z}_{p^{2}}$,

[TABLE]

Thus we get $(0,1)\in\mathcal{Z}_{B_{3}}$.

2. $(1,d)$ for all $d\in\mathbb{Z}_{p}$. A simple calculation shows that for all $d\in\mathbb{Z}_{p}$ and for all $t\in\mathbb{Z}_{p^{2}}$, we have

[TABLE]

By Lemma 3.1, we obtain that $(1,d)\in\mathcal{Z}_{B_{3}}$ for all $d\in\mathbb{Z}_{p}$.

3. $(p,d)$ for all $d\in\mathbb{Z}_{p}\setminus\{c\}$. By Lemma 4.2, we obtain that for all $d\in\mathbb{Z}_{p}\setminus\{c\}$ and for $t\in\mathbb{Z}_{p^{2}}$,

[TABLE]

By Lemma 3.1, we have $(p,d)\in\mathcal{Z}_{B_{3}}$ for all $d\in\mathbb{Z}_{p}\setminus\{c\}$. ∎

Case 4: $(p,0)\in\mathcal{Z}_{A}$. Define

[TABLE]

We will prove that $B_{4}$ is a tiling complement of $A$. It is easy to see that $\sharp B_{4}=p^{2}$. A simple computation shows that for all $c\in\mathbb{Z}_{p}$, we have

[TABLE]

By Lemma 3.1, we have

[TABLE]

On the other hand, we obtain that for all $c\in\mathbb{Z}_{p}^{*}$,

[TABLE]

Moreover, we have

[TABLE]

By Lemma 3.1, we have

[TABLE]

Combining (5.6), (5.7) and Remark 3.1, we get

[TABLE]

By Lemma 2.2, we conclude that $B_{4}$ is a tiling complement of $A$.

To sum up, we conclude that the set $A$ is a tile in $\mathbb{Z}_{p^{2}}\times\mathbb{Z}_{p}$.

∎

According to Remark 3.1, we only need to study some particular zeros. The following lemma shows that if $\mathcal{Z}_{B}$ contains two special kinds of zeros, then $\sharp B=p^{2}$ and $B$ has a certain form.

Lemma 5.4**.**

Let $B$ be a subset of $\mathbb{Z}_{p^{2}}\times\mathbb{Z}_{p}$. Suppose that $\sharp B\leq p^{2}$ and $(p,0),(1,c)\in\mathcal{Z}_{B}$ for some $c\in\mathbb{Z}_{p}$. Then $\sharp B=p^{2}$ and $B$ has the form

[TABLE]

Proof.

By the fact that $(p,0)\in\mathcal{Z}_{B}$ and Lemma 3.1, we have

[TABLE]

We observe that

[TABLE]

In fact, supposing $(x+py,z)\in H((1,c),i+pj)$, we obtain that

[TABLE]

which implies

[TABLE]

that is to say, $(x+py,z)\in H((p,0),pi)$. By the fact that $(1,c)\in\mathcal{Z}_{B}$ and Lemma 3.1, we have

[TABLE]

for all $0\leq i\leq p-1$. Combining (5.9) and (5.11), we obtain that $\frac{1}{p}\sharp B$ is divided by $p$ and

[TABLE]

Since $\sharp B\leq p^{2}$, we have $\sharp B=p^{2}$ and $h_{B}((1,c),i+pj)=1$ for all $0\leq i,j\leq p-1$. By (5.10), the point $(x+py,z)$ lies in the plane $H((1,c),i+pj)$ if and only if

[TABLE]

Denoting by $(i+py_{ij},z_{ij})$ the only point lying in $H((1,c),i+pj)\cap B$, we conclude that the set $B$ has to have the form (5.8). ∎

We show that the set having the form in Lemma 5.4 is exactly a tile and also a spectral set.

Lemma 5.5**.**

Let $B$ be a subset of $\mathbb{Z}_{p^{2}}\times\mathbb{Z}_{p}$. Suppose that $B$ has the form (5.8). Then $B$ is a tile and also a spectral set.

Proof.

Let

[TABLE]

It is easy to see that $\sharp A=p^{2}$. We will prove that $A$ is a spectrum of $B$. In fact, a simple computation shows that

[TABLE]

Since $(1,c),(p,0)\in\mathcal{Z}_{B}$, by Lemma 2.1, we conclude that $A$ is a spectrum of $B$.

It remains to show that $B$ is also a tile. Let $\Omega$ be a subset of $\mathbb{Z}_{p}\times\mathbb{Z}_{p}$ which has the form

[TABLE]

According to [11], we see that $\Omega$ is a tile. Let $T$ be a tiling complement of $\Omega$. Let

[TABLE]

We will prove that $C$ is actually a tiling complement of $B$. In fact, for any $(x+py,z)\in\mathbb{Z}_{p^{2}}\times\mathbb{Z}_{p}$, there exists the unique decomposition

[TABLE]

where $(x+py^{\prime},z^{\prime})\in B$, $(py^{\prime\prime},z^{\prime\prime})\in C$ and $(y,z)=(y^{\prime},z^{\prime})+(y^{\prime\prime},z^{\prime\prime})\in\mathbb{Z}_{p}\times\mathbb{Z}_{p}$ in the unique way according to the tiling pair $(\Omega,T)$.

∎

Lemma 5.4 and Lemma 5.5 gives us a sufficient condition for tiles (or spectral sets) and is useful to handle the case where $p<\sharp B\leq p^{2}$, which will be shown in Section 6 and Section 7.

We finish this section by the following lemma concerning tiles or spectral sets in a proper subgroup of $\mathbb{Z}_{p^{2}}\times\mathbb{Z}_{p}$.

Lemma 5.6**.**

Let $G$ be a proper subgroup of $\mathbb{Z}_{p^{2}}\times\mathbb{Z}_{p}$. Let $A\subset G$. Then $A$ is a tile if and only if it is a spectral set.

Proof.

It is not hard to check that up to an isomorphism, the proper subgroups of $\mathbb{Z}_{p^{2}}\times\mathbb{Z}_{p}$ are

[TABLE]

We know that Fuglede’s spectral set conjecture holds on these groups (see [15], [5] and [11]). Thus we complete the proof. ∎

6. Tiles $\Rightarrow$ Spectral sets

Let $(A,B)$ be a tiling pair in $\mathbb{Z}_{p^{2}}\times\mathbb{Z}_{p}$. In this section, we will prove that both $A$ and $B$ are spectral sets.

If $\sharp A=1$ or $\sharp B=1$, then it is easy to see that $A$ and $B$ are spectral sets. Now suppose that $\sharp A>1$ and $\sharp B>1$. Without loss of generality, we may assume that $\sharp A=p$ and $\sharp B=p^{2}$. Due to Proposition 5.3, we already know that $A$ is a spectral set. It remains to prove that $B$ is a spectral set.

We observe that $(1,0)$ and $(p,0)$ can not be in $\mathcal{Z}_{A}$ at the same time. In fact, if $(1,0),(p,0)\in\mathcal{Z}_{A}$, then $p^{2}$ divides $\sharp A$. This is contradict to the hypothesis $\sharp A=p$. By Lemma 2.2, either $(1,0)$ or $(p,0)$ is in $\mathcal{Z}_{B}$. We thus decompose the proof into the following three cases.

Case 1: $(1,0),(p,0)\in\mathcal{Z}_{B}$. By Lemma 5.4 and Lemma 5.5, we obtain that $B$ is a spectral set.

Case 2: $(1,0)\in\mathcal{Z}_{B}$ and $(p,0)\notin\mathcal{Z}_{B}$. It follows that $(p,0)\in\mathcal{Z}_{A}$. By Lemma 5.4 and the fact that $\sharp A=p$, we have that $(1,a)\notin\mathcal{Z}_{A}$ for all $0\leq a\leq p-1$. It follows that $(1,a)\in\mathcal{Z}_{B}$ for all $0\leq a\leq p-1$.

On the other hand, by Lemma 3.1 and the fact that $\sharp A=p$, we obtain that

[TABLE]

Denote by

[TABLE]

Due to (6.1), we may assume that

[TABLE]

It follows that

[TABLE]

This implies that $x_{i}\equiv i\mod p$, for all $0\leq i\leq p-1.$ Thus $A$ has to be the form

[TABLE]

We consider the following two cases.

• (1)

All $z_{i}$ are the same for $0\leq i\leq p-1$.

It follows that not all $h_{A}((0,1),p_{i})$ are the same for $0\leq i\leq p-1$. By Lemma 3.1, we have $(0,1)\notin\mathcal{Z}_{A}$. Then $(0,1)\in\mathcal{Z}_{B}$. We define

[TABLE]

It is easy to see that $\sharp B=p^{2}$ and $(\overline{B}-\overline{B})\setminus\{0\}$ is contained in

[TABLE]

which is a subset of $\mathcal{Z}_{B}.$ Thus we obtain that $\overline{B}$ is a spectrum of $B$.

• (2)

Not all $z_{i}$ are the same for $0\leq i\leq p-1$.

By Lemma 4.1, there exists $1\leq c\leq p-1$ such that the value $c$ is not compatible with $(i,z_{i})_{0\leq i\leq p-1}$. It follows that not all $h_{A}((p,c),pi)$ are the same for $0\leq i\leq p-1$. Thus we have $(p,c)\notin\mathcal{Z}_{A}$. Then we have $(p,c)\in\mathcal{Z}_{B}$. Define

[TABLE]

It is easy to see that $\sharp B=p^{2}$. A simple computation shows that the set $(\widetilde{B}-\widetilde{B})\setminus\{0\}$ is contained in

[TABLE]

which is a subset of $\mathcal{Z}_{B}.$ Thus we conclude that $\widetilde{B}$ is a spectrum of $B$.

Case 3: $(p,0)\in\mathcal{Z}_{B}$ and $(1,0)\notin\mathcal{Z}_{B}$. It follows that $(1,0)\in\mathcal{Z}_{A}$. According to Case 2 in Proposition 5.3, we obtain that $A$ has the form

[TABLE]

for some $x\in\mathbb{Z}_{p}$. If not all $z_{i}$ are the same for $0\leq i\leq p-1$, then by Lemma 4.1, there exists $c\in\mathbb{Z}_{p}^{*}$ such that the value $c$ is not compatible with $(i,z_{i})_{0\leq i\leq p-1}$. It follows that not all $h_{A}((1,c),x+pi)$ are the same for $0\leq i\leq p-1$. Thus we have $(1,c)\notin\mathcal{Z}_{A}$. By Lemma 2.2, we have $(1,c)\in\mathcal{Z}_{B}$. By Lemma 5.4 and Lemma 5.5, we obtain that $B$ is a spectral set. Now suppose that all $z_{i}$ are the same for $0\leq i\leq p-1$. It is not hard to check that

[TABLE]

for all $0\leq a\leq p-1$ and all $0\leq i\leq p-1$. By Lemma 3.1, we have $(0,1),(p,a)\notin\mathcal{Z}_{A}$ for all $0\leq a\leq p-1$. Then $(0,1),(p,a)\in\mathcal{Z}_{B}$ for all $0\leq a\leq p-1$. We define

[TABLE]

It is easy to see that $\sharp B=p^{2}$ and

[TABLE]

Thus we obtain that $\overline{B}$ is a spectrum of $B$.

7. Spectral sets $\Rightarrow$ Tiles

Let $A$ be a spectral set in $\mathbb{Z}_{p^{2}}\times\mathbb{Z}_{p}$. In this section, we will prove that $A$ is also a tile.

If $\sharp A=1$ or $p^{3}$, then $A$ is a tile trivially. We suppose that $1<\sharp A<p^{3}$. By Lemma 5.2, we have $\sharp A\leq p^{2}$. Let $B$ be a spectrum of $A$. By Lemma 2.1, we have

[TABLE]

Since $\sharp B=\sharp A>1$, we have $\mathcal{Z}_{A}\not=\emptyset$. By Lemma 3.1, we get $\sharp A\geq p$. If $\sharp A=p$, then by Proposition 5.3, we know that $A$ is a tile. Thus in what follows, we assume that

[TABLE]

By pigeon principle, we have

[TABLE]

The same is for $\mathcal{Z}_{B}$.

Without loss of generality, we assume that $(0,0)\in A$ and $(0,0)\in B$. If $(A-A)\cap(\mathbb{Z}_{p^{2}}^{*}\times\mathbb{Z}_{p})=\emptyset$ or $(B-B)\cap(\mathbb{Z}_{p^{2}}^{*}\times\mathbb{Z}_{p})=\emptyset$, then we obtain

[TABLE]

It follows that either $A$ or $B$ is in a non-trivial proper subgroup of $\mathbb{Z}_{p^{2}}\times\mathbb{Z}_{p}$. By Lemma 5.6, we deduce that $A$ and $B$ are tiles.

Now we suppose that $(A-A)\cap(\mathbb{Z}_{p^{2}}^{*}\times\mathbb{Z}_{p})\not=\emptyset$ and $(B-B)\cap(\mathbb{Z}_{p^{2}}^{*}\times\mathbb{Z}_{p})\not=\emptyset$. This implies that there exists $c_{1},c_{2}\in\mathbb{Z}_{p}$ such that

[TABLE]

Lemma 7.1**.**

The sets $A$ and $B$ are as above. Then $(p,0)\in\mathcal{Z}_{A}$ or $(p,0)\in\mathcal{Z}_{B}$.

Proof.

We prove it by contradiction. We assume that $(p,0)\notin\mathcal{Z}_{A}$ and $(p,0)\notin\mathcal{Z}_{B}$. By (7.1), we have $(1,0)\in\mathcal{Z}_{A}$ and $(1,0)\in\mathcal{Z}_{B}$. If there are two different elements $(x+py,z),(x+py^{\prime},z)\in A$, then

[TABLE]

Since $(y-y^{\prime})\in\mathbb{Z}_{p^{2}}^{*}$, by Lemma 2.1, we have $(p,0)\in\mathcal{Z}_{B}$ which is not possible. Thus there exists $E_{A}\subset\mathbb{Z}_{p}\times\mathbb{Z}_{p}$ and $f:E_{A}\to\mathbb{Z}_{p}$ such that

[TABLE]

Observe that

[TABLE]

for all $0\leq i\leq p-1$. Since $(1,0)\in\mathcal{Z}_{A}$, we have that $\{z\in\mathbb{Z}_{p}:(x,z)\in E_{A},f(x,z)=i\}$ are constant for all $0\leq i\leq p-1$. By Lemma 3.1, we obtain that

[TABLE]

and

[TABLE]

This means that

[TABLE]

The same is for $B$, that is to say,

[TABLE]

for some $\widetilde{E}_{B}\subset\mathbb{Z}_{p}$ and $g:\widetilde{E}_{B}\times\mathbb{Z}_{p}\to\mathbb{Z}_{p}$. Since $\sharp A>p$, we have $\sharp\widetilde{E}_{A}>1$. It follows that

[TABLE]

By Lemma 2.1, we have

[TABLE]

This implies that $c$ is compatible with $(g(x,z),z)_{0\leq z\leq p-1}$ for any $x\in\widetilde{E}_{B}$ and for any $0\leq c\leq p-1$. For any $x\in\widetilde{E}_{B}$, since the map $g(x,\cdot):\mathbb{Z}_{p}\to\mathbb{Z}_{p}$ is subjective, it is bijective. This is contradict to Lemma 4.1. Therefore, we conclude that $(p,0)\in\mathcal{Z}_{A}$ or $(p,0)\in\mathcal{Z}_{B}$. ∎

Due to Lemma 7.1, without loss of generality, we suppose $(p,0)\in\mathcal{Z}_{B}$. Since $(1,c_{2})\in\mathcal{Z}_{B}$, by Lemma 5.4 and Lemma 5.5, the set $B$ is a tile and $\sharp B=p^{2}$. If we also have $(p,0)\in\mathcal{Z}_{A}$, then repeating the same reason, we obtain that $A$ is a tile, which completes the proof. Now we suppose that $(p,0)\notin\mathcal{Z}_{A}$. By the proof of Lemma 7.1, we have

[TABLE]

for some $g:\mathbb{Z}_{p}\times\mathbb{Z}_{p}\to\mathbb{Z}_{p}$ with $g(x,\cdot)$ bijective for all $x\in\mathbb{Z}_{p}$. It follows that

[TABLE]

By Lemma 2.1, we have

[TABLE]

Since $\sharp A=\sharp B>p$, by pigeon principle, we obtain that

[TABLE]

or there exists $d\in\mathbb{Z}_{p}^{*}$ such that

[TABLE]

If $(p,d)\in\mathcal{Z}_{A}$ for some $d\in\mathbb{Z}_{p}^{*}$, then we define

[TABLE]

It is easy to see that $\sharp C=p$. Observe that $h_{C}((0,1),pi)=1$ for all $0\leq i\leq p-1$. By Lemma 3.1, we have $(0,1)\in\mathcal{Z}_{C}$. On the other hand, by Lemma 4.2, the value $d^{\prime}$ is compatible with $(j,(-d)^{-1}j)_{0\leq j\leq p-1}$ for all $0\leq d^{\prime}\leq p-1$ with $d^{\prime}\not=d$. Then we have $h_{C}((p,d^{\prime}),pi)=1$ for all $0\leq i\leq p-1$ and for all $0\leq d^{\prime}\leq p-1$ with $d^{\prime}\not=d$. By Lemma 3.1, we have $(p,d^{\prime})\in\mathcal{Z}_{C}$ for all $0\leq d^{\prime}\leq p-1$ with $d^{\prime}\not=d$. By Lemma 2.2, we conclude that $C$ is a tiling complement of $A$.

If $(0,1)\in\mathcal{Z}_{A}$, then we define

[TABLE]

It is easy to see that $\sharp C=p$ and $h_{C}((p,d),pi)=1$ for all $0\leq i\leq p-1$ and all $0\leq d\leq p-1$. By Lemma 3.1, we have

[TABLE]

By Lemma 2.2, we conclude that $D$ is a tiling complement of $A$. This completes the proof.