SOME EXPLICIT CONSTRUCTIONS OF TERNARY NON-FULL-RANK TILINGS OF ABELIAN GROUPS

A tiling of a finite abelian group G is a pair (A,B) of subsets of G, such that both A and B contain the identity element e of G and every g ∈ G can be uniquely written in the form g = ab, where a ∈ A and b ∈ B. A tiling (A,B) of G is called full-rank if 〈A〉= 〈B >= G, Otherwise, it is called a non-full rank tiling. In this paper, we show some explicit constructions of non-full rank tilings of 3−groups of order 34.


Introduction
A tiling of a finite abelian group G is a pair (A, B) of subsets of G containing the identity e of G and every g ∈ G can be uniquely written in the form g = ab, where a ∈ A and b ∈ B.
Tilings are a special case of normalized factorizations of a finite abelian group G, where by a normalized factorization of G is meant a collection of subsets A 1 , A 2 ,...,A n of G, such that e ∈ A i for each i = 1, 2, ..., n and every g ∈ G can be uniquely written in the form g = a 1 a 2 ...a n , E-mail address: kamin35@hotmail.comReceived February 27, 2017 942 KHALID AMIN a i ∈ A i .The notion of factorization of an abelian group into subsets was introduced by G. Hajos [1], when he found the answer to a conjecture by H. Minkowski [4], about lattice tiling of R n by unit cubes or clusters of unit cubes.Hajos first translated Minkowski's conjecture into a question about finite abelian groups and then he solved the question.
The group-theoretic version of Minkowski's conjecture reads as follows: If G is a finite abelian group and G = A 1 ...A i...A k is a normalized factorization of G, where each of the subsets A i is of the form {e, a, a 2 , ..., a k }, where k <| a |; (here | a |denotes order of a).then at least one of the subsets A i is a subgroup of G.

Preliminaries
Hajos made use of the integral group ring Z (G).Corresponding to each subset A of G, we Redei [4] made use of group characters; viz homomorphisms χ from G to the multiplicative group of complex numbers C.These extend to ring homomorphisms χ from Z (G) to the multiplicative group of complex numbers C, where χ (∑ n i g i ) = ∑ n i κ(g i ).He also defined the We will use( * ) to show that our constructions constitute factorizations of a given group G.

Main results
M. Dinitz [1], showed that if p 5, then groups of order p n admit full-rank tiling and left the case p = 3, as an open question.We answer this question by showing some explicit constructions of non-full rank tiliings of groups of order 3 4 .

SOME EXPLICIT CONSTRUCTIONS OF TERNARY NON-FULL-RANK TILINGS OF ABELIAN GROUPS 943
We recall that a finite abelian group G is said to be of type (p α 1 1 , p α 2 2 , ..., p α r r ) if it is a product of cyclic groups of orders p α 1 1 , p α 2 2 , ..., p α r r .If p i = p, for each i, G is called a p−group.We construct non-full-rank tilings of 3−groups of order 3 4 of the following types: We observe that no element of B has order greater than 9.
Note that the possible orders of κ (x) are 1, 3 and 9 and similarly for orders of κ (y).

SOME EXPLICIT CONSTRUCTIONS OF TERNARY NON-FULL-RANK TILINGS OF ABELIAN GROUPS 945
Note that the possible orders of κ (x) are 1, 3 and 9, while the possible orders of κ (y) and κ (z) are 1 and 3 only.So, altogether we have 12 different cases to consider.The result is summarized below: Case Order of χ(x) Order of χ(y) Order of χ(z) χ(A) χ(B) In this case, by construction, we get that, A = G.
Note that the possible orders of κ (x) are 1, 3 only.Similarly with κ(y), κ (u) and κ (v).So, altogether we have 16 different cases to consider. Let where α, β , γ,and δ are primitive 3rd roots of unity.Then This takes care of 4 cases.
The result is summarized below.
Case Order of χ(x) Order of χ(y) Order of χ(u) Order of χ(v) χ(A) χ(B) then by < b > is meant the subgroup of G generated by the support of b; viz.those elements g i such that n i = 0. We will also, use A to mean the subgroup generated by a subset A of G and b 1 , b 2 , ..., b m will denote the subgroup generated by the support of b | and for each non-identity character κ there exists A i such that κ(A I ) = 0....................................( * )

1 A
3 3 , 3 , 3 2 , 3 2 , 3 2 , 3, 3 and (3, 3, 3, 3).Construction non-full rank tiling of 3−groups of type 3 3 , 3 .Let G = x × y , where | x |= 27 and | y |= 3. Let A = x 9 y ∪x x 9 y ∪ x 2 x 9 y and B = x 9 ∪ x 3 y ∪ x 6 y .We will use ( * ) to show that AB is a tiling of G. First note that the possible orders of κ (x) are 1, 3, 9 and 27 and the possible orders of κ (y) are 1 and 3.So, altogether we have 8 different cases to consider.The result is summarized below: Case Order of κ (x) Order of κ (y) κ (A) κ (B) All the other cases, except this one, which we will detail.Let κ (x) = ξ and κ (y) = η, where ξ and η are primitive 9−th roots of unity.Then ) All the other cases, except these, in which case, we get the result by using a similar argument as in the previous case.
In this case, by construction, we get that in fact, neither A = G nor B = G .