Tiling groups with difference sets

We study tilings of groups with mutually disjoint difference sets. Some necessary existence conditions are proved and shown not to be sufficient. In the case of tilings with two difference sets we show the equivalence to skew Hadamard difference sets, and prove that they must be normalized if the group is abelian. Furthermore, we present some constructions of tilings based on cyclotomy and investigate tilings consisting of Singer difference sets.


Introduction
Let G be an additively written group of order v.A (v, k, λ) difference set in G is a ksubset D ⊆ G such that every nonzero element of G can be expressed as a difference x − y with x, y ∈ D in exactly λ ways.Multiplicative notation is sometimes used, in which case the "differences" are written as xy −1 .For basic results on difference sets, see [2,17,22].More recent surveys on difference sets are [14] and [26].
It is not possible to partition the whole group G into disjoint (v, k, λ) difference sets.This follows from the necessary existence condition λ(v − 1) = k(k − 1) when v > k > λ 1, which is assumed throughout the paper.However, if k divides v − 1, it may be possible to partition G \ {0} into difference sets.We introduce the following concept.
Definition 1.Let G be a finite group of order v with identity element 0. A (v, k, λ) tiling of G is a collection {D 1 , . . ., D t } of mutually disjoint (v, k, λ) difference sets such that Tilings of cyclic groups have a nice combinatorial interpretation.We can visualize the group Z v as a necklace of v beads, with the identity element coloured in black.A (v, k, λ) tiling corresponds to a colouring of the remaining beads with t colours, such that there are k beads of every colour.Furthermore, there are exactly λ pairs of equally coloured beads at each possible distance, and for every colour.Here, a pair of beads is thought to have two distances, counted clockwise and counterclockwise.The (31, 6, 1) tiling of Example 2 can be represented as the necklace in Figure 1.Mutually disjoint difference sets have been used to design hopping sequences for multichannel wireless networks [13].Multiple users are assumed to communicate over a number of channels, without synchronization or a common control channel.Every user utilizes the same channel selection strategy, i.e. hopping sequence.Sequences based on difference sets ensure a high rendezvous probability.The difference sets correspond to different channels; they need to be disjoint to ensure that each time slot is assigned a single channel.A (v, k, λ) tiling of G is the extremal case, maximizing the number of channels for a given sequence length v. See also [25] for this application.
A similar concept studied earlier are partitioned difference families (PDFs); see [3,27].A (v, k, λ) tiling with t difference sets can be seen as a (v, [k t , 1 1 ], k − 1)-PDF.A more general concept are disjoint difference families; see [7,10,24].Quite recently, Gnilke, Greferath and Pavčević [11] defined structures called mosaics of combinatorial designs.The simultaneous development of the t difference sets forming a (v, k, λ) tiling of G is a (t + 1)-mosaic of symmetric designs with parameters The layout of our paper is as follows.In Section 2, we prove necessary conditions for the existence of (v, k, λ) tilings and consider small examples.Nonexistence of tilings for some parameter triples satisfying the necessary conditions is proved.In Section 3, tilings consisting of two difference sets are shown to be equivalent to the so-called skew Hadamard difference sets.Constructions of general tilings based on cyclotomy are considered in Section 4, and based on Singer difference sets in Section 5.

Necessary existence conditions and small examples
As already mentioned in the introduction, a necessary condition for the existence of (v, k, λ) difference sets in a group G of order v is λ(v−1) = k(k−1).From this, the number t of difference sets in a (v, k, λ) tiling of G can be expressed as t = (v − 1)/k = (k − 1)/λ and we have the following existence condition for tilings.
An immediate consequence is that v and k are relatively prime (since v and v −1 = t•k are relatively prime).
The condition λ(v − 1) = k(k − 1) is not sufficient for the existence of difference sets; many non-trivial nonexistence results are known.In this paper the existence of (v, k, λ) difference sets is taken for granted, in the sense that only parameter triples satisfying Proposition 3 for which there is at least one difference set are considered admissible for (v, k, λ) tilings.We shall soon see that these two conditions are not sufficient for the existence of (v, k, λ) tilings.
If D is a (v, k, λ) difference set in G, the translates of D (i.e. the sets D + x, x ∈ G) are also difference sets.The set of all translates of D forms a symmetric (v, k, λ) block design with the point set G. Since every two blocks of a symmetric design intersect in λ points, the difference sets forming a (v, k, λ) tiling of G cannot be translates of each other.Using this observation we can prove nonexistence of tilings for some admissible parameters.Proof.From the multiplier theorem [2, Theorem VI.2.11] and [2, Lemma VI.2.5] it follows that there are only two (21,5,1) difference sets up to translation: {3, 6, 7, 12, 14} and {7, 9, 14, 15, 18}.According to the observation above, a (21, 5, 1) tiling of Z 21 would require four difference sets not being translates of each other.Now consider (57, 8, 1) tilings of the group Z 57 .As before, one can see that there are 12 cyclic (57, 8, 1) difference sets up to translation.Seven disjoint difference sets are required for a (57, 8, 1) tiling, so nonexistence of tilings does not follow immediately.We set up a computer search using backtracking and found that there can be at most 5 mutually disjoint cyclic (57, 8, 1) difference sets, thus proving the following result.Proposition 5. A (57, 8, 1) tiling of the cyclic group of order 57 does not exist.
Surprisingly, the non-abelian group of order 57 can be tiled with difference sets.The next example was also found by computer search.Using our backtracking program we examined all admissible groups of order v 50.Tilings exist in 11 of the 18 cases, and in 7 cases there are no tilings.The results are summarized in Table 1.Please note that the table contains only parameters (v, k, λ) satisfying Proposition 3 and only groups G with at least one (v, k, λ) difference set.For example, the parameters (49, 16, 5) and the groups Z 49 and Z 7 × Z 7 are not included because by [17,Theorem 4.38 and Corollary 4.42] there are no (49, 16, 5) difference sets.
A difference set in an abelian group is said to be normalized provided the sum of its elements is 0. The (31, 6, 1) tiling of Example 2 is composed of normalized difference sets, and so are all tilings of abelian groups we found.Therefore we make the following conjecture.
(v, k, λ) In general we have no proof of this conjecture, but in the next section we shall prove it for tilings with two difference sets (t = 2).Proposition 5 follows directly from the conjecture: there is a unique normalized translate of every cyclic (57, 8, 1) difference set and it contains the elements 19 and 38.Hence, normalized (57, 8, 1) difference sets cannot be disjoint.

Tiling with two difference sets
Suppose that a group G can be tiled with two difference sets {D 1 , D 2 }.Then the parameters (v, k, λ) are of the form (4n − 1, 2n − 1, n − 1) for some n 2, i.e.D 1 and D 2 are Paley-type difference sets.The canonical example are the Paley difference sets, i.e. the nonzero squares F (2) q } in the additive group of the field F q of order q ≡ 3 (mod 4).The nonsquares q are also a difference set, and therefore {F (2) q } the electronic journal of combinatorics 22(2) (2015), #P2.56 is a (q, (q − 1)/2, (q − 3)/4) tiling of the elementary abelian group of order q.
A special class of Paley-type difference sets are the skew Hadamard or antisymmetric difference sets.A difference set D in a group G is skew Hadamard provided D ∪ (−D) = G \ {0} holds.Since −D = {−x | x ∈ D} is clearly also a difference set, {D, −D} is a tiling of G with two difference sets.For a long time it was conjectured that the Paley difference sets were the only examples of skew Hadamard difference sets in abelian groups.Two new series of skew Hadamard difference set in elementary abelian groups of order 3 m , m odd, were constructed in [5,6].Since then there have been other constructions, see [9] and [20].
It is natural to ask whether any tiling of a group G with two difference sets {D 1 , D 2 } necessarily comes from a skew Hadamard difference set, i.e. whether D 2 = −D 1 must hold.We shall prove this by using the group ring Z[G].Now we switch to multiplicative notation, and write D (−1) instead of −D.Also, we identify a subset S ⊆ G with the corresponding group ring element g∈S g.
holds, i.e. the two difference sets are skew Hadamard.
Proof.Since D 1 and D 2 are difference sets, the following relations in the group ring By the definition of a tiling, we have from the right and using (1), we get Similarly, multiplying D + 1 = G by D 2 from the left and using (2), we get Now D 2 = D (−1) 1 follows by subtracting (3) and (4).
By this theorem, all known restrictions for skew Hadamard difference sets also apply to tilings of groups with two difference sets.For example, it is known that the Paley difference sets are indeed the only examples in cyclic groups [15].See the survey [26] for other known restrictions in the abelian case.One known result about skew Hadamard difference sets in elementary abelian groups is that they must be normalized; see [23,Lemma 4.3].We are going to improve this result by proving it without the 'elementary abelian' assumption.Theorem 9. Every skew Hadamard difference set in an abelian group of order |G| > 3 is normalized.
Proof.Let D be a skew Hadamard difference set in an abelian group G and let h = d∈D d.By [4], the order v = |G| = p m is a prime power and the quadratic residues modulo v are multipliers fixing D. If p > 3, we can find a quadratic residue t > 1 such that t − 1 is coprime to p, and then g t = g for all g ∈ G \ {1}.Now from D (t) = D we have Next we look at the case p = 3.According to the fundamental theorem of finitely generated abelian groups, For every g ∈ G, we write g = (g 0 , g 1 , . . ., g l ), where g i ∈ Z 3 N i .Here the identity element is (0, . . ., 0) and we use additive notation on the coordinates.If N i > 1, we can again find a quadratic residue t > 1 such that (t − 1, 3 N i ) = 1 and conclude that h i = 0. Now only the case N i = 1 remains.
Without loss of generality, we assume that N 0 = 1.Let 2 ≡ 1 (mod 3) and λ = v−3 4 ≡ 0 (mod 3), we have Since D is a difference set, D • D (−1) = (k − λ) + λG.Counting elements with g 0 = 0 on the left-hand and right-hand side yields and taking this modulo 3 On the other hand, since D is a skew Hadamard difference set Again, by counting elements with g 0 = 0 we have and since m > 1, modulo 3 we get It is routine to verify that from ( 5), ( 6), (7) follows n 0 ≡ 1 (mod 3) and n 1 ≡ n 2 ≡ 0 (mod 3).From this we have h 0 ≡ n 1 + 2n 2 ≡ 0 (mod 3).Now we see that all components of h are equal to 0, and h is the identity element.Hence, D is normalized.
For p > 3, our proof is essentially the same as in [23], and for p = 3 our proof is simpler and more general.From Theorem 8 and Theorem 9 we have the following result.Here ω = 3 is a primitive element of F 31 .In general, assume we have a (q, k, λ) difference set in F q fixed by the multiplier m, such that the order | m | = r divides k.Then D can be written as a union of cosets The problem of tiling F q by multiples of D is equivalent to tiling the set of integers {0, 1, . . ., n − 1} by cyclic shifts of {c 1 , c 2 , . . ., c k/r } modulo n = (q − 1)/r.This can be expressed as the following theorem.
Theorem 13.Let ω be a primitive element and m a multiplicative subgroup of order r of the finite field F q .Suppose we have a (q, k, λ) difference set in (F q , +) which is a union of cosets for c 1 , c 2 , . . ., c k/r ∈ {0, 1, . . ., n − 1}, n = (q − 1)/r.Then there exists a (q, k, λ) tiling of (F q , +) by multiples of D if and only if there If we take the (31, 6, 1) difference set D = ω 0 5 ∪ω 3 5 , then r = 3, n = 10, C = {0, 3} and the problem of finding the b i 's has two solutions: 0, 2, 4, 6, 8 and 1, 3, 5, 7, 9.The first one gives the tiling of Example 2, and the second one a (31, 6, 1) tiling that corresponds to the mirror image of the necklace in Figure 1.

Tilings using Singer difference sets
Let q be the power of a prime, n 3 an integer, and G the cyclic group of order q n −1 q−1 .Singer difference sets are difference sets in G with parameters ( q n −1 q−1 , q n−1 −1 q−1 , q n−2 −1 q−1 ).The classical construction comes from projective geometry P G(n − 1, q).Let ω be a primitive element of F q n and Tr(x) = x + x q + • • • + x q n−1 the trace mapping from F q n to F q .Let β be a nonzero element in F q n and r an integer coprime to q n −1 q−1 .Then the set of integers forms a Singer difference set with addition modulo q n −1 q−1 .There are also other constructions of cyclic difference sets with these parameters, see [12,19].We are going to consider using classical Singer difference sets to tile the group G; such tilings will be called classical Singer tilings.Several small examples fall into this category.
Example 14.Take q = 2, n = 3 and the irreducible polynomial f (x) = x 3 + x 2 + 1 over F 2 to construct the field F 2 3 .Take the primitive element ω = x and β = 1.Then by substituting r = 3 in (8) we get the squares in F * 7 , and from r = 1 the non-squares.Hence, the Paley tiling of F 7 is a classical Singer tiling.
To prove our main result in this section, we need the following lemma which can be found in [18,Chapter 7].
the electronic journal of combinatorics 22(2) (2015), #P2.56 Lemma 17.If ϕ : F q → F q is an arbitrary function, then there exists a unique polynomial g ∈ F q [x] with deg(g) < q representing ϕ, in the sense that g(c) = ϕ(c) for all c ∈ F q .To be precise, g Furthermore, for f, g ∈ F q [x] we have f (c) = g(c) for all c ∈ F q if and only if f (x) ≡ g(x) (mod x q − x).
Theorem 18.When q n −1 q−1 > q • n+q−2 n−1 + 1, there are no classical Singer tilings of the cyclic group of order q n −1 q−1 .Proof.Assume that there are β i and r i where i = 1, 2, . . ., q, such that the classical Singer difference sets determined by (β j , r j ) form a tiling.That means Tr(β i ) = 0 for each i and there is a unique i such that Tr(β i a r i ) = 0 for each a ∈ F q n \ F q .Hence τ (a) := q i=1 Tr(β i a r i ) q−1 = 0, a ∈ F q ; q − 1, otherwise.
The last equality follows from the fact that a∈Fq a i = −1, i ≡ 0 (mod q − 1) and i = 0; 0, otherwise.
the electronic journal of combinatorics 22(2) (2015), #P2.56 On the other hand, as the polynomial Tr(β i x r i ) q−1 has at most n+q−2 n−1 terms, τ (x) has at most q • n+q−2 n−1 terms.Note that reducing modulo x q n − x cannot increase the number of terms of the polynomial.By Lemma 17, we have a contradiction when q n −1 q−1 > q • n+q−2 n−1 + 1.
According to Theorem 18, for each given q there is an integer n q such that, when n n q there are no classical Singer tilings.We list several q's and the corresponding n q 's in the following table.q 2, 3, 4 5, . . ., 17

Proposition 4 .
A (21, 5, 1) tiling of the cyclic group of order 21 does not exist.

Table 1 :
Tilings of admissible groups of order v 50.
a, b | a 5 = b 8 = 1, a 4 b = ba Conjecture 7. The difference sets in a tiling of an abelian group are necessarily normalized.