Fuglede's conjecture holds for cyclic groups of order $pqrs$

The tile-spectral direction of the discrete Fuglede-conjecture is well-known for cyclic groups of square-free order, initiated by Laba and Meyerowitz, but the spectral-tile direction is far from being well-understood. The product of at most three primes as the order of the cyclic group was studied intensely in the last couple of years. In this paper we study the case when the order of the cyclic group is the product of four different primes and prove that Fuglede's conjecture holds in this case.


Introduction
The conjecture of Fuglede [7] was originated from a problem posed to him by Segal on the restriction of the usual differential operators ∂ ∂x i (acting in the distribution sense) to commuting self-adjoint operators on L 2 (Ω), where Ω is a bounded measurable subset of R n with 0 < m(Ω) < +∞. Fuglede showed that this happens if and only if Ω accepts an orthogonal basis of complex exponentials e 2πi λ,x , which is the definition of a spectral set. The set of frequencies λ denoted by Λ, is called the spectrum of Ω.
We say that S is a tile of R n , if there is a set T ⊂ R n such that almost every point of R n can be uniquely written as s + t, where s ∈ S and t ∈ T . In this case, we say that T is the tiling complement of S.
In the same paper, Fuglede proceeded to state the so-called Spectral Set Conjecture (that we will just call Fuglede's conjecture): Conjecture 1. Ω is spectral if and only if Ω is a tile.
Fuglede proved this conjecture when the spectrum or the tiling complement is a lattice. However, this conjecture was largely proven to be false, initially by Tao [22] in 2004 for dimensions n ≥ 5, and subsequently for n ≥ 3 [5,12]. In a recent paper [16] Fuglede's conjecture has verified for all convex bodies in all dimensions. The status of this conjecture remains open for R and R 2 . Recently, it was shown that a spectral subset of R of Lebesgue measure 1 has a spectrum that is periodic [8].
The notable feature of Tao's proof was the transition to the setting of finite abelian groups (although this had begun with Laba [13], connecting Fuglede's conjecture with the tiling results of Coven and Meyerowitz [2]). Let us now state Fuglede's conjecture for abelian groups in a more precise way. Definition 1.1. Let Z N denote the cyclic group of order N .
• S ⊂ Z N is called spectral, if there is some Λ ⊂ Z N with |S| = |Λ| and the exponential functions e λ (x) = ξ λx N form an orthogonal basis over S, that is e λ , e λ ′ S := s∈S e λ (s)e λ ′ (s) = |S|δ λλ ′ (1) for every λ, λ ′ ∈ Λ, where ξ N = exp(2πi/N ) a primitive N 'th root of unity. We call Λ the spectrum and (S, Λ) is called a spectral pair of Z N . • S ⊂ Z N is called a tile, if there is another subset T ⊂ Z N such that every element of Z N can uniquely be written as s + t, where s ∈ S, t ∈ T . We call T the tiling complement of S, we write S ⊕ T 1 = Z N and we call (S, T ) a tiling pair of Z N .

Conjecture 2.
For any N and S ⊂ Z N we have that S is spectral if and only if S tiles Z N .
Borrowing the notation from [3] and [17], we write S − T(G) (resp. T − S(G)), if the Spectral ⇒ T ile (resp. T ile ⇒ Spectral) direction of Fuglede's conjecture holds in G. The above mentioned connection between the conjecture on R, on Z and on finite cyclic groups is summarized below [3] (where T − S(Z N ) means that T − S(Z n ) holds for every n ∈ N): This close connection shows the importance of Fuglede's Conjecture for abelian groups. Since the result of Tao appeared, there have been many results on finite abelian groups by many authors, which we summarize below: Theorem 1.2. Let G be a finite abelian group and p, q, r different primes.
(1) If G = Z d p where p is an odd prime and d ≥ 4, then there is a spectral subset of G that does not tile [1,6].
(2) If G = Z d 2 where d ≥ 10, then there is a spectral subset of G that does not tile [6].
(3) If G = Z 3 8 , then there is a spectral subset of G that does not tile [12].
(4) If G = Z 3 24 , then there is a tile of G that is not spectral [5].
where p is prime and d ≤ 3, then any tile of G is spectral [1]. The converse holds if d ≤ 2 [9] or d = 3 and p ≤ 7. [20] or G = Z p ⊕ Z pq [11], where p = q are primes, then a subset of G is spectral if and only if it is a tile.
(7) If G = Z N and N is square-free or N = p m q n or N = p n d, where d is square-free, then any tile of G is spectral [14,18,19].
(8) If G = Z N and N = p n , p n q, p n q 2 , pqr, p 2 qr, then a subset of G is spectral if and only if it is a tile [10,14,17,18,19,21].
Using the fundamental theorem of finite abelian groups, the following holds for G whose minimum number of generators is d: where n i | n i+1 for 1 ≤ i ≤ d − 1. Thus, we obtain Corollary 1.3. Let G be a finite Abelian group, and d be the minimum number of generators of G.
• If d ≥ 4 and |G| is odd, then there is a spectral subset of G that does not tile.
• If d ≥ 10, then there is a spectral subset of G that does not tile.
In this article we investigate the discrete Fuglede's conjecture for cyclic groups of order of the product of 4 different primes. In particular we prove Theorem 1.4. Fuglede's conjecture is true for Z pqrs , where p, q, r, s are different primes.
Combining the results in Theorem 1.2 (8) and Theorem 1.4 we obtain the following corollary.
Corollary 1.5. Fuglede's conjecture is true for Z pqrs , where p, q, r, s are arbitrary primes.

Basic properties of Spectral and Tiling pairs
In this section we list some properties of spectral pairs that we use during the proof. Our first remark is that we will always identify the dual group of a cyclic group Z N with itself.
• First note that spectral pairs satisfy the following duality property.
Lemma 2.1. Let G be a finite abelian group. Assume that S ⊂ G is a spectral set having Λ as a spectrum. Then S is also a spectrum for Λ.
There is a similar property for spectral pairs involving the difference set of a spectrum that we introduce now. Let (S, Λ) be a spectral pair of Z N . Let us denote by 1 S the characteristic function of S, and the Fourier transform of a function f : we have for any λ, λ ′ ∈ Λ, where λ = λ ′ , that is, the difference set of a spectrum is a subset of the zero set of the Fourier transform of 1 S along with {0}.
• Relation (2) can also be expressed with the notion of the mask polynomial of a multiset.
Definition 2.2. For a cyclic group G, a field K and a function f : For a (multi)set S on G we denote the multiplicity function of S by f S (that is a function from G to N) and we denote by m S the mask polynomial of f S .
The basic connection between the mask polynomial and the Fourier transform of (the characteristic function of) a set S is1 so we may rewrite (2) as We define Z(S) , the following holds: if 2.2 Geometric interpretation of cyclic groups of square-free order and the cube rule • For m, n ∈ N with m | n let us define the natural projection of Z n onto Z m ≤ Z n , Z m denotes the only subgroup of Z n of order m. Let k ∈ N with n m | k and k ≡ 1 (mod m). Then x → k · x defines a surjective homomorphism from Z n to Z m . Note that the function defined in this way is independent on the choice of k. If U is a subset of Z n , then its projection to Z m , which we denote by U m , is a multiset defined as {k · u | u ∈ U }.
In this subsection our goal will be to introduce a geometric interpretation of cyclic groups of square-free order and an important observation that we call the cube rule. We use the 'geometric language' introduced below throughout the proof.
• Note that the cyclic group Z N with N = p 1 p 2 . . . p k (p ′ i s are different prime numbers) can be written as the direct sum ⊕ k i=1 Z p i , so its elements can also be considered as k-tuples. Let the Hamming distance of two elements x, y ∈ ⊕ k i=1 Z p i be the number of coordinates they differ. Let us denote by d H (x, y) the Hamming distance of x and y. We consider ⊕ k i=1 Z p i as the subset of the k-dimensional integer grid, so we identify it with In this interpretation every coset of a subgroup of order d = p i 1 . . . p is of Z N can be identified with a proper s-dimensional affine subspace of ⊕ k i=1 Z p i . For instance, Z p 1 -cosets are lines that are the set of vertices with the same Z p 2 -, . . . , Z p k -coordinates.
• Now we recall an important tool that was introduced in [10]. It describes a condition that a multiset B ⊆ Z N should satisfy, in case N is square-free and Φ N | m B . For an integer N we denote by ω(N ) the number of (different) prime divisors of N . If N is square-free, then we can consider where B(c) denotes the multiplicity of c in B.
• In general, for square-free N = If it is clear form the context, then we simply denote this cube by C(x, y).

Preliminary statements
Let us start with some statements from [10].
Lemma 2.4. Let G be a finite abelian group. It is enough to prove the Spectral ⇒ Tile direction of Fuglede's conjecture for spectral pairs (S, Λ) with 0 ∈ S and 0 ∈ Λ.
Lemma 2.5. Let G be a finite abelian group and S a spectral set in G, that does not generate G. Assume that for every proper subgroup H of G we have S − T(H). Then S tiles G. Lemma 2.6. Let N be a natural number and suppose that S − T(Z N /H) holds for every 1 = H ≤ Z N . Assume that (S, Λ) is a spectral pair and Λ is not primitive. Then S tiles Z N . Lemma 2.7. Let N be a natural number, S a spectral set in Z N and p a prime divisor of N . Assume that S − T(Z N p ). If S is the union of Z p -cosets, then S tiles Z N .
Then by assumption we have that S i −i is a tile on Z N p . As N is square-free, by Proposition tiling complement for S in Z N , finishing the proof of the lemma.
By [19], we have that S − T(Z pqr ) holds, where p, q, r are different primes. Combining with Lemma 2.8 we get the following.
Corollary 2.9. Let (S, Λ) be a spectral pair in Z N , where N = pqrs with p, q, r, s different prime numbers. Suppose that Λ is the union of Z p -cosets. Then S is a tile.
Lemma 2.10. Let N = p 1 p 2 · · · p k be a square-free integer, and let S ⊆ Z N be a spectral set whose cardinality is divisible by at least k − 1 primes among p 1 , . . . , p k . Then S tiles Z N .
Proof. If N | |S|, then S = Z N and there is nothing to prove. So, without loss of generality we can suppose that p 1 . . . p k−1 | |S| and p k ∤ |S|. Let Λ ⊆ Z N be a spectrum of S. Since Thus, every element of S is unique mod N p k , yielding |S| ≤ p 1 · · · p k−1 , and combined with p 1 · · · p k−1 | |S| we finally obtain |S| = p 1 · · · p k−1 .
Therefore, S consists of a complete set of coset representatives modulo the subgroup N p k Z N , and so this subgroup is exactly a tiling complement of S. Lemma 2.10 immediately implies the following.
Corollary 2.11. Let N = pqrs with different primes p, q, r, s and (S, Λ) a spectral pair of Z N . Then, in order to prove S − T(Z pqrs ), we can assume that r ∤ |S| and s ∤ |S| (otherwise we are done by Lemma 2.10). This automatically implies that Φ r , Φ s ∤ m S , m Λ .
We also remind that if the spectral set is small, then it tiles. Let us gather the information above in the following summary. Summary 1. Let N = p 1 . . . p k be a square-free integer and (S, Λ) a spectral pair of Z N . Any of the following assertions implies that S tiles Z N .
• S or Λ does not generate Z N .
• S or Λ is the union of Z p i -cosets for some p i .
• all p ′ i s except at most one divide the cardinality of S.
We can also suppose that 0 ∈ S and 0 ∈ Λ to prove the Spectral ⇒ Tile direction of Fuglede's conjecture.

Technical lemmata for the proof
• Let N = p 1 . . . p k be a square-free natural number. For any primitive N 'th root of unity ξ N and for an integer t one can uniquely write the complex number ξ t N as ξ Let ℓ be an integer with (ℓ, p 1 ) = 1 and ℓp 1 ≡ 1 (mod N p 1 ). Then ξ ℓp 1 N is a primitive N p 1 'th root of unity and Since ξ ℓp 1 N is a primitive N p 1 'th root of unity, by equation (7) we get that Φ N Using our notion B j we can rewrite these equations as Note that Lemma 2.13 can be generalised as it is stated in Proposition B.1. As we do not use it in the proof of the main result we present it in Appendix B.
Lemma 2.14. Let N = pqrs for different prime numbers p, q, r, s and S ⊆ Z N a set and Proof. By Lemma 2.13, the assumption Φ N Φ N/s | m S implies that S satisfies the 3D cube rule on every Z N/s -coset. By the way of contradiction, assume that S is not the union of Z p -cosets. Hence, there exists an x ∈ S with (x + Z p ) \ S = ∅. Let x p ∈ (x + Z p ) \ S. Then using our assumption S ∩ (x + Z q ) = S ∩ (x + Z r ) = {x} we obtain that in every 3D cube containing x and x p in x + Z N/s there is no element z ∈ S with d H (x, z) = 1. Thus, for any y ∈ x + Z N/s with d H (x, y) = 3 and x p ∈ C(x, y) we have that y ∈ S, by applying the 3D cube rule on C(x, y).
Note that Lemma 2.14 holds verbatim for any permutation of the primes p, q, r, s. Lemma 2.15. Let N = pqrs for different prime numbers p, q, r, s, (S, Λ) a spectral pair in Z N and r, s ∤ |S| = |Λ|. Proof.
1. By the way of contradiction, assume Φ N/p | m S . By Lemma 2.13, Φ N Φ N/p | m S implies that the 3D cube rule holds for every Z N/p -coset. Since r ∤ |S| and s ∤ |S| we have S ∩ ((x + Z r ) ∪ (x + Z s )) = {x} for all x ∈ S. Applying Lemma 2.14 gives that S is a union of Z q -cosets and hence a tile by Summary 1. Similar argument applies if Φ N/q | m S .
. Similar argument as in the previous case implies that Λ is a union of Z q -cosets (resp. Z p -cosets). Then, by Summary 1, S is a tile.

Reduction mod p
• The lemma following below is due to Lam and Leung [15], written in polynomial notation.
, then the multiset S pq is the sum of Z p -and Z q -cosets. In polynomial notation, . Then the cardinality of S satisfies |S| = pk+qℓ, for some k, ℓ ≥ 0.
• In what follows, it will be very useful to reduce the coefficients of mask polynomials mod p, for some primes p | N (i.e., we will consider them as polynomials in F p [x]). For square-free integer m not divisible by p the following equation holds.
. Then at least one of the following conditions holds: If in addition, m S (ξ p 1 )m S (ξ p 4 )m S (ξ p 1 p 4 ) = 0, then either (ii) holds or (iii) holds along with p 1 | |S|.
Proof. It follows from our condition that Thus If the polynomial P is identically zero, then obviously (ii) holds. If (ii) fails, there is some j with |S j mod p 2 p 3 | ≥ c + p 1 ≥ p 1 . If there is some j such that |S j mod p 2 p 3 | > p 1 , then there are two distinct elements of S j mod p 2 p 3 having the same remainder modulo p 1 , hence (S − S) ∩ N p 4 Z ⋆ N = ∅ and (iii) holds by (4). If both (ii) and (iii) fail, then we must have |S j mod p 2 p 3 | ≤ p 1 for every j, i.e., c = 0 and |S j mod p 2 p 3 | = 0 or p 1 for every j. Moreover, is a proper set, which implies that | ≤ 1, for every j and k. For those j that |S j mod p 2 p 3 | = p 1 holds, we must for all k, which in polynomial form is written as which easily yields (i).
The latter is greater than p 2 p 3 when c ≥ 1, since P is not identically zero, otherwise (ii) holds. Hence c = 0, and obviously p 1 | |S|, as desired. 3 The main result and the structure of the proof Theorem 3.1. Fuglede's conjecture is true for Z pqrs , where p, q, r, s are different primes.
Proof. Note that the Tile ⇒ Spectral direction follows from an argument of Meyerowitz and Laba written on Tao's blog 2 and also by Ruxi Shi [19]. Now we consider the Spectral ⇒ Tile direction. Let (S, Λ) be a spectral pair of Z pqrs . Let m S and m Λ denote their mask polynomials. In the proof we distinguish the following cases.

Proof of Case
We also assume that r, s ∤ |S| = |Λ| hence, by Corollary 2.11, Φ r , Φ s ∤ m S , m Λ . Lemma 3.2. Let (S, Λ) be a spectral pair of Z N . Assume that for every x ∈ S we have x + Z p ⊆ S or x + Z q ⊆ S. Then S is a tile.
Proof. By Summary 1, if S is the union of Z p -cosets (or Z q -cosets), then S is a tile. Since S is primitive, S contains x + Z p and y + Z q with (x + Z p ) ∩ (y + Z q ) = ∅.
First, assume that (x + Z p ) ∪ (y + Z q ) is contained in a Z N/s -coset. Then it is easy to find x 1 ∈ (x + Z p ), x 2 ∈ (y + Z q ) with x 1 − x 2 ∈ Z r , implying r | |Λ|, a contradiction. Note that a similar argument works if (x + Z p ) ∪ (y + Z q ) is contained in a Z N/r -coset. Now assume that (x + Z p ) ∪ (y + Z q ) is not contained in any proper coset of Z N . Then we may write (x + Z p ) = {(a, b 1 , c 1 , d 1 Then by suitable choice of a and b we obtain that Φ N | m Λ and Φ N p | m Λ . Using Lemma 2.15.2 we obtain that S is a tile. Lemma 3.3. Let (S, Λ) be a spectral pair of Z N . If Φ N | m S and there exists an x ∈ S such that x + Z p ⊆ S and x + Z q ⊆ S, then S is tile.
Proof. Let us assume that there are x, y, z ∈ Z N such that x ∈ S, y, z ∈ S and y ∈ x + Z p , z ∈ x + Z q . As Φ N | m S , the 4D cube rule holds for S. Then, by taking any 4D cube C having vertices x, y and z, we obtain that S contains at least one element of C of Hamming distance 3 from x. Indeed, points of Hamming distance 1 are excluded by y, z ∈ S and r ∤ |S|, s ∤ |S|. Without loss of generality we may assume r > s.
Fix a 4D cube C containing x, y, z. Let u = u(x, y, z) be the point of C such that d H (u, x) = 3 and u and x have the same r-coordinate. Let o be the point opposite to x on C.
Case 1. s ≥ 3 (hence r ≥ 5). Case 1.1. If u ∈ S, then one of the other 3 points of distance 3 from x is in S. (These are the ones that differ from x in their r-coordinate.) Since r ≥ 5, modifying only the rcoordinate of o we may build up 3 more 4D cubes having vertices x, y, z, u. Applying the previous argument using y, z, u ∈ S we obtain that S contains a pair of points of Hamming distance 3 from x differing only in their r-coordinate, which contradicts the fact that r ∤ |S|.
Case 1.2. If u ∈ S, then let u ′ be a point which only differs in its s-coordinate from u and have different s-coordinate from x. Such u ′ exists since s ≥ 3. Then u ′ ∈ S since s ∤ |S|. Hence we may apply the previous argument by exchanging the role of u and u ′ .
Note that similar argument works if p ∤ |S| (resp. q ∤ |S|) by changing the role of p (resp. q) and s. Therefore we can assume that pq | |S|.

Case 2. s = 2 (and pq | |S|).
Note that the projection S pqs is a set in Z pqs as r ∤ |S|. Therefore |S| can be either pq or spq = 2pq. The latter case is not possible since s ∤ |S|, thus |S| = |Λ| = pq. Now we project Λ on Z 2pq and so at least half of the elements of Λ project on the same Z pq -coset. Easy calculation shows if min{p, q} ≥ 3 (which clearly holds), then there are three pairs of elements of Λ 2pq having only different p-, q-, pq-coordinates, respectively.
Since (S, Λ) is a spectral pair, these conditions imply that Φ p Φ q Φ pq | m S over F r [x]. Applying Lemma 2.17 and rs ∤ |S| which gives Φ r , Φ s ∤ m S , m Λ , we conclude that (ii) holds, i.e., Φ p Φ q Φ pq | m S . This means that S pq = cZ pq , for some c ∈ N. As |S| = pq, we have that c = 1 and plainly, S tiles Z N .
Lemma 3.4. Let N = pqrs and let (Λ, S) be a spectral pair. Assume that r, s ∤ |S| = |Λ|, φ N ∤ m S and φ N | m Λ . Then S is a tile.
We distinguish two cases according to the conditions of the previous two Lemmata.
Case 1. Assume that for every x ∈ Λ we have x + Z p ⊆ Λ or x + Z q ⊆ Λ. If q ∤ |Λ| (i.e. |Λ| = p), then there is no point x in Λ such that x + Z q ⊂ Λ. Thus Λ is a Z p -coset, hence Λ − Λ ∈ Z p . Therefore, by spectrality, S intersects every Z qrs -coset once, when S is a tile. A similar argument works if |Λ| = q.
It remains to handle the case |S| = |Λ| = pq. Assume first that (x + Z p ) ∪ (x + Z q ) ⊂ Λ. Then Φ p Φ q Φ pq | m S . Thus S pq = Z pq and since |S| = pq we obtain that S is a tile.
Assume now that there are x and y such that x + Z p and y + Z q are disjoint subsets of Λ. Then Λ pq = Z pq , a contraction.
Finally, if Λ is the union of Z p -cosets or Z q -cosets only, then S is a tile by Summary 1.

Case 2.
There exists an x ∈ Λ such that x + Z p ⊆ Λ and x + Z q ⊆ Λ Note that the proof of Lemma 3.3 shows that there is no such spectral set except if |Λ| = pq and s = 2. Since Λ is tile we have Λ pq = Z pq . Since p ≥ 3, there are at least 3 elements of Λ pq in each Z p -cosets. Two of them have the same s-coordinate as well since s = 2. Thus Φ p | m S or Φ pr | m S . Similarly we obtain Φ q | m S or Φ qr | m S . There are 3 elements of Λ pq such that the Hamming distance of any two of them is 2. Thus again two of them have the same s-coordinate.
By applying Lemma 2.17 we obtain Φ p Φ q Φ pq | m S since r ∤ |S| and s ∤ |S| (so cases (i) and (iii) are excluded). Thus S pq = Z pq and since |S| = pq, we obtain that S is a tile.
3.2 Proof of Case (II): Φ pqrs ∤ m S and Φ pqrs ∤ m Λ We start with a technical lemma as follows.
Lemma 3.5. Let N = p ′ p ′′ q ′ q ′′ and let B ⊆ Z N be primitive. Then, for every pair p ′ , p ′′ of distinct primes with p ′ p ′′ | N we have belongs to the aforementioned union, completing the proof.
Corollary 3.6. Let N = pqrs and let (S, Λ) be a spectral pair in Z N such that S is primitive. Then there exist x and y in S such that they have different p-and q-coordinates. With other Note that if S is a spectral set, then the previous corollary can be applied to both S and Λ since Λ is also spectral in this case. Now, we handle a special case which will be crucial in the sequel.
Proposition 3.7. Let S ⊂ Z pqrs be a spectral set that is primitive and suppose Φ p Φ q | m S and Φ N ∤ m S . Then S is a tile.
Proof. We prove this statement in two steps.
Lemma 3.8. Under the assumptions of Proposition 3.7 we have S pq = cZ pq for some c ∈ N.
Proof. By Corollary 3.6, if Φ N ∤ m S , then either Φ pq | m S , or Φ pqr | m S or Φ pqs | m S . If Φ p Φ q Φ pq | m S , then we are done. Thus, by contradiction, let us assume Φ pq ∤ m S . By the symmetry of the role of r and s we can assume that Φ pqr | m S . Thus We may apply Lemma 2.17. Since r ∤ |S| = |Λ| and s ∤ |S| = |Λ| we obtain Φ p Φ q Φ pq | m S . Proof. We proceed by contradiction. Take two elements t 1 , t 2 ∈ S with the same p-and q-coordinates. Their r-or s-coordinates must be different since otherwise we have Φ s | m Λ or Φ r | m Λ , which was excluded. So let t 1 = (a 1 , a 2 , a 3 , a 4 ) and t 2 = (a 1 , a 2 , a ′ 3 , a ′ 4 ) ∈ S, where a 3 = a ′ 3 and a 4 = a ′ 4 . As p, q ≥ 2 and c > 1, clearly there are elements t 3 , t 4 having a ′ 1 in the first and a ′ 2 in their second coordinate, where a 1 = a ′ 1 and a 2 = a ′ 2 . Since Φ N ∤ m Λ , it implies that there are no points of Hamming-distance 4 in S. Thus, we may assume that t 3 = (a ′ 1 , a ′ 2 , a 3 , a ′ 4 ), t 4 = (a ′ 1 , a ′ 2 , a ′ 3 , a 4 ), since t 3 and t 4 cannot differ only in their r-or s-coordinate. If p, q ≥ 3, then there are a ′′ 1 ∈ {a 1 , a ′ 1 } and a ′′ 2 ∈ {a 2 , a ′ 2 }. Then in the Z rs -coset (a ′′ 1 , a ′′ 2 , * , * ) every element is of Hamming distance 4 from one of t 1 , t 2 , t 3 , t 4 , a contradiction. If c ≥ 3, then we also get a contradiction, as there is an element t = (a 1 , a 2 , a ′′ 3 , a ′′ 4 ) with a ′′ 3 ∈ {a 3 , a ′ 3 } and a ′′ 4 ∈ {a 4 , a ′ 4 } and so d H (t, t 4 ) = 4. Hence we can assume without loss of generality that c = 2 and p = 2. So every element has either a 1 or a ′ 1 in its first coordinate. Iterating the argument as above, we get that for every element having a 1 in its first coordinate is of the form (a 1 , * , a 3 , a 4 ) or (a 1 , * , a ′ 3 , a ′ 4 ). Similarly, every element having a ′ 1 in its first coordinate is of the form (a ′ 1 , * , a 3 , a ′ 4 ) or (a ′ 1 , * , a ′ 3 , a 4 ). Moreover all of these elements are in S. This implies that |S| = 4q. Thus S is not a tile. Therefore we have to show that S is not a spectral set.
By contradiction, assume that (S, Λ) is a spectral pair for some |S| = |Λ| = 4q. Then there is a Z pq -coset which contains at least 2 elements of Λ, i.e., these elements have the same p-and q-coordinates. Then their r-and s-coordinate should be different, otherwise Φ r | m S or Φ s | m S , which is excluded. This implies that Φ rs | m S . Hence the 2D cube rule holds for S rs on Z rs , i.e., S rs is the sum of Z r -and Z s -cosets. All the elements of S can have only 2 different r-coordinates and 2 different s-coordinates, thus S rs does not contain either a Z r -coset or a Z s -coset. This contradiction shows that S is not a spectral set.
By Lemma 3.8, S pq = cZ pq for some 0 < c ∈ N. By Lemma 3.9, c = 1, and then S is a tile, finishing the proof of Proposition 3.7.
Proof. The proof is the same for the two statements and by duality it is enough to show that if Φ pqr | m S , then there is a pair of points in S pqr of Hamming distance 3. This implies that the preimage of these points are of distance 3 in S as Φ N ∤ m S , which implies Φ pqr | m Λ .
We prove it by contradiction. If there is no pair of points of Hamming distance 3 in S pqr , then either all of the points are of distance 1 from each other, but then one can easily see that S is not primitive or there is a pair of points x, y ∈ S pqr of Hamming distance 2. Since the 3D cube rule holds on S pqr and S pqr is a set (as Φ s ∤ m S ), there are at least two other points in any 3D cube containing both x and y of odd distance from x. There are 4 points of odd Hamming distance from x on each of these 3D cubes. Two of them are excluded, since each of them are of Hamming distance 3 from either x or y. The remaining two points that are on the 2D cube spanned by x and y must be in the set S pqr (by 3D cube rule). Then all of the points of Z pqr which is not in the plane determined by x, y is of Hamming distance 3 from one of these 4 points, so none of them can be in S pqr . By Summary 1, S is primitive, thus S pqr should also be primitive, a contradiction.
, then every pair of points in Λ having different p-and qcoordinates must have different r-coordinate ( resp. s-coordinate) as well or S is a tile.
Proof. Suppose there are points x, y ∈ Λ with different p-and q-coordinates and with the same r-coordinate. Then their projection x ′ , y ′ ∈ Λ pqr is of Hamming distance 2. By assumption the 3D cube rule holds for Λ pqr , thus there are at least two more points z ′ , w ′ in every 3D cube containing x ′ and y ′ .
We have two cases. If one of these additional points are in 2D cube determined by x ′ , y ′ , then this implies that Φ p Φ q Φ pq | m S over F s . Thus by Lemma 2.17 we have that either Φ p Φ q | m S and Φ pq | m S , then S is a tile and we are done by Proposition 3.7; or Φ r | m S or Φ s | m Λ , which cases were excluded. Otherwise both of the additional points z ′ , w ′ differ in their r-coordinates from x ′ , y ′ , respectively. Without loss of generality we can assume that d H (x ′ , w ′ ) = d H (y ′ , z ′ ) = 3. Hence, the preimage z of z ′ has the same s-coordinate as y; and the preimage w of w ′ have the s-coordinate as x, because their distance in Z pqrs is at least 3 and it cannot be 4, since Φ s ∤ m S . Thus d H (x, w) = d H (y, z) = 3.
If r ≥ 3, then taking another cube containing x ′ and y ′ we get z ′′ and w ′′ similarly as above and then z ′ and z ′′ differ only in their r-coordinates, thus Φ r | m S , which was excluded and we are done.
If r = 2, then we have two cases whether x and y have the same or different s-coordinates. If they have the same s-coordinates, then so do z and w. By the first part of the proof, there is no more points of the 3D cube C(x ′ , w ′ ) in Λ pqr . On the other hand, using the fact that r = 2, one can see that every point of Z pqr \ C(x ′ , w ′ ) is of Hamming distance 3 from at least one of x ′ , y ′ , z ′ , w ′ . Hence, every point of S has the same s-coordinate thus Λ is not primitive so S is a tile, by Summary 1. Assume now that x ′ and y ′ have different s-coordinates. This implies that Φ pqs | m S . By Lemma 3.10, it implies that Φ pqs | m Λ and the projection of x and w in Λ pqs is of distance 2. Repeating the argument above for these case and using s > 2 we get a contradiction. Proposition 3.12. Let (S, Λ) be a spectral pair. Assume that Φ N ∤ m S , Φ N ∤ m Λ and Λ is a tile. Then S is a tile.
If |Λ| = pq, then Λ pq = Z pq by [19]. Since Φ N ∤ m S we have that for every pair of elements x, y of Λ if the p-coordinate and q-coordinate of x and y are different, then their ror s-coordinate is the same.
If Λ pq = Z pq , then there is a pair of points having different p-and q-that have different r-coordinate as well. Otherwise, Λ is not primitive, and then S is a tile by Summary 1. Let x 1 and y 1 be such a pair in Λ, so their p-, q-and r-coordinates are different. So the s-coordinate of x 1 and y 1 is the same since Φ N ∤ m S . Thus Φ pqr | m S , which implies Φ pqr | m Λ by Lemma 3.10. Thus by Lemma 3.11 if the p-and q-coordinates of a pair of points of Λ are different, then so do their r-coordinates. Again, we obtain that the s-coordinate of these points is the same. Then it is easy to see from Λ pq = Z pq that the s-coordinate of every point of Λ is the same. Thus Λ is not primitive, and then S is a tile.
Finally, assume |Λ| = p. Suppose by contradiction that S is not a tile. Then there is a pair x and y in S such that p | x − y. Then by our assumptions we have Φ qr | m Λ or Φ qs | m Λ or Φ rs | m Λ or Φ qrs | m Λ . We may assume without loss of generality that Φ rs | m Λ or Φ qrs | m Λ .
If Φ rs | m Λ , then Λ rs is the sum of Z r -cosets and Z s -cosets by Lemma 2.16. Since r ∤ |Λ| and s ∤ |Λ| both types appear in the sum. Thus we have Φ pq | m S or Φ q | m S or Φ p | m S .
The last option would imply that S is equidistributed mod p and since |S| = p, it is a tile, which contradicts our assumption. Φ q | m S is excluded since q ∤ |S|. Finally, one can see that Φ pq | m S and |S| = p also implies that S pq is a Z p -coset by Lemma 2.16, and hence S is a tile, a contradiction.
Assume Φ qrs | m Λ . Then the 3D cube rule holds for Λ qrs . We claim that in this case there is a pair of points in Λ qrs of Hamming distance 1. This follows simply from the fact that if z ∈ Λ qrs such that none of the point of Hamming distance 1 from z is in Λ qrs , then every point z ′ with d H (z, z ′ ) = 3 is in Λ qrs by the 3D cube rule.
We obtain that Φ up | m S or Φ u | m S , where u ∈ {q, r, s}. The latter is excluded by u ∤ |S|. If Φ up | m S and |S| = p, then by Lemma 2.16, S up is a Z p -coset so S is a tile, which is a contradiction.
Corollary 3.13. Let (S, Λ) be a spectral pair of Z N . Assume Φ N ∤ m S and Φ N ∤ m Λ . If Φ pqr | m S ( resp. Φ pqs | m S ), then every pair of points in Λ having different p-and q-coordinates must have different r-coordinate ( resp. s-coordinate) as well or S is a tile.
Proof. Applying Lemma 3.10, either the statement follows directly from Lemma 3.11 or Λ is a tile. Then by Proposition 3.12, S is a tile. Lemma 3.14. If Φ pqr | m S , then Φ pqs ∤ m S , and also if Φ pqs | m S , then Φ pqr ∤ m S . The same holds for Λ.
Proof. We take x and y obtained from Corollary 3.6, i.e., they have different p-and qcoordinates. By Corollary 3.13, if Φ pqr | m S holds, then their r-coordinates are different, as well. Similarly, if Φ pqs | m S holds, then so their s-coordinates. If both Φ pqr Φ pqs | m S holds, then d H (x, y) = 4, which is a contradiction. The statement for Λ directly follows by applying Lemma 3.10.
By Lemma 3.14, without loss of generality, we can assume that Φ pqs ∤ m S , and thus Φ pqs ∤ m Λ by Lemma 3.10.
Lemma 3.15. If Φ pqs ∤ m S , then S is not primitive. that Φ p | m Λ which was excluded. Similar argument shows that z 4 = (a 3 , b 2 , c 2 ) ∈ S pqr since y ′ ∈ S pqr , by changing the role of x ′ and y ′ .
As (a 2 , b 2 , c 1 ) and (a 1 , b 1 , c 2 ) are not in S pqr by Corollary 3.13, applying the 3D cube rule for C(x ′ , z 4 ) and C(y ′ , z 3 ) we get that z 5 = (a 1 , b 2 , c 1 ) and z 6 = (a 2 , b 1 , c 2 ) are in S pqr . Then the projection of these 4 points {x ′ , y ′ , z 5 , z 6 } in S pq is a 2D cube, although it is known that S pq is in (x + Z p ) ∪ (y + Z q ). This contradiction shows that Φ pqr ∤ m S . By Corollary 3.6, if Φ N ∤ m S and Φ pqr ∤ m S and Φ pqs ∤ m S , then Φ pq | m S . In this case, the projection S pq is the sum of Z p -and Z q -cosets. As S pqs ⊂ (x + Z p ) ∪ (y + Z q ) ∪ (z + Z s ), it follows that S pq is the sum of a Z p -and a Z q -coset. Let x ′′ , y ′′ , and z ′′ denote the projection of x, y and z on Z pq , respectively. Clearly, z ′′ = (x ′′ + Z p ) ∩ (y ′′ + Z q ). Since all elements of (x ′′ + Z p ) \ {z ′′ } have different p-and q-coordinates from any elements of (y ′′ + Z q ) \ {z ′′ } and Φ pqr , Φ pqs ∤ m S , the preimages of the elements (x ′′ + Z p ) \ {z ′′ } has the same r-and s-coordinate. Since the elements of (x ′′ + Z p ) have also the same q-coordinate, it follows that their preimages differ only in their p-coordinates. As p ≥ 3 (by our assumption), we have |(x ′′ + Z p ) \ {z ′′ }| ≥ 2, and hence Φ p | m Λ , which was excluded. This contradiction shows the statement.