OJDMOpen Journal of Discrete Mathematics2161-7635Scientific Research Publishing10.4236/ojdm.2017.72005OJDM-75432ArticlesPhysics&Mathematics Non-Full Rank Factorization of Finite Abelian Groups KhalidAmin1Department of Mathematics, College of Science, University of Bahrain, Bahrain, Kingdom of Bahrain* E-mail:170420170702515323, November 201614, April 2017 17, April 2017© Copyright 2014 by authors and Scientific Research Publishing Inc. 2014This work is licensed under the Creative Commons Attribution International License (CC BY). http://creativecommons.org/licenses/by/4.0/

Tilings of p-groups are closely associated with error-correcting codes. In , M. Dinitz, attempting to generalize full-rank tilings of <span style="font-family: Euclid Math Two">Z</span><sup>n</sup><sub style="margin-left:-6px;">2</sub> to arbitrary finite abelian groups, was able to show that if p &ge;5, then <span style="font-family: Euclid Math Two">Z</span><sup>n</sup><sub style="margin-left:-6px;">p</sub> admits full-rank tiling and left the case p=3, as an open question. The result proved in this paper the settles of the question for the case p=3.

Factorization of Abelian Groups Error-Correcting Codes
1. Introduction

A factorization of a finite abelian group G is a collection of subsets

A 1 , ⋯ , A i , ⋯ , A k of G such that each element g ∈ G can be represented in the form g = a 1 ⋯ a i ⋯ a k . In this case, we write G = A 1 , ⋯ , A i , ⋯ , A k and if each A i contains the identity element e of G , we say we have a normalized factori- zation of G .

The notion of factorization of abelian groups arose when G. Hajós  found the answer to “Minkowski’s conjecture” about lattice tiling of ℝ n by unit cubes or clusters of unit cubes. The geometric version of “Minkowski’s conjecture” can be explained as follows:

A lattice tiling of ℝ n is a collection { T i : i ∈ I } of subsets of ℝ n such that ∪ i ∈ I T i = ℝ n and int ( T i ) ∩ int ( T j ) = ∅ , if i ≠ j , i , j ∈ I . Two unit cubes are called twins if they share a complete ( n − 1 ) -dimensional face. Minkowski was wondering if there exists a tiling of ℝ n by unit cubes such that there are no twins! Minkowski’s conjecture is usually expressed as follows:

Each lattice tiling of ℝ n by unit cubes contains twins.

As mentioned above, it was G. Hajós  who solved Minkowski’ conjecture. His answer was in the affirmative, after translating the conjecture into an equivalent conjecture about finite abelian groups. Its group―theoretic equivalence 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 ”.

Rėdei  generalized Hajos’s theorem to read as follows:

“If G is a finite abelian group and G = A 1 ⋯ A i ⋯ A k is a normalized factori- zation of G , where each of the subsets A i contains a prime number of elements, then at least one of the subsets A i is a subgroup of G ”.

2. Preliminaries

A tiling is a special case of normalized factorization in which there are only two subsets, say A and B of a finite abelian groups G , such that G = A B is a factorization of G .

A tiling of a finite abelian group G is called a full-rank tiling if G = A B implies that 〈 A 〉 = 〈 B 〉 = G , where 〈 A 〉 denotes the subgroup generated by A . In this case, A and B are called full-rank factors of G . Otherwise, it is called a non-full-rank tiling of G . As suggested by M. Dinitz  and also in that of O. Fraser and B. Gordon  , finding answers to certain questions is sometimes easier in one context than in others. In this connection consider the group, ℤ p n viewed as a vector space of n -tuples ( x 1 , x 2 , ⋯ , x n ) over ℤ p . Then subspaces correspond to subgroups. Moreover, ℤ p n is equipped with a metric, called Hamming distance d H , which is defined as follows:

For x = ( x 1 , x 2 , ⋯ , x n ) and y = ( y 1 , y 2 , ⋯ , y n ) ,

d H ( x , y ) = | { i : 1 ≤ i ≤ n , x i ≠ y i } | .

With respect to this metric, the sphere S ( x , e ) with center at x and radius e is the set S ( x , e ) = { y : d H ( x , y ) ≤ e } .

A perfect error-correcting code is a subset C of ℤ p n such that

∪ x ∈ C S ( x , e ) = ℤ p n and S ( x , e ) ∩ S ( y , e ) = ∅ , if x ≠ y .

Observe that in the language of tiling, this says that ℤ p n = C S ( 0 , e ) is a factorization of ℤ p n  .

Factorization and Partition

Let G = A B be a factorization of a finite Abelian group G . Then the sets

{ a B : a ∈ A } form a partition of G . Also, | G | = | A | | B | , where | A | as before denotes the number of elements of A .

Definition

Let A and A ′ be subsets of G . We say that A is replaceable by A ′ , if whenever G = A B is a factorization of G , then so is G = A ′ B .

Redei  showed that if G = A B is a factorization of G , where

A = { e , a 1 , a 2 , ⋯ , a p − 1 } , and p is a prime, then A is replaceable by 〈 a i 〉 , for each i , 1 ≤ i ≤ p − 1 .

Definition

A subset A of G is periodic, if there exists g ∈ G , g ≠ e such that

g A = A . It is easy to see that if A is periodic, then A = H C , where H is a proper subgroup of G  .

Before we show the aim of this paper, we mention the following observation. If G = A B is a factorization of G , then for any a ∈ A , and b ∈ B , then so is G = a − 1 A b − 1 B , so we may assume all factorizations G = A B are normalized.

Theorem

Let G = ℤ 3 n and assume G = A B is a factorization of G , where | A | = 3 , then either A or B is a non-full-rank factor of G .

Proof:

Note that | G | = 3 n . We induct on n .

If n = 1 , then | B | = 1 . Thus, B is a non-full-rank factor of G .

Let n > 1 and assume the result is true for all such groups of order less than 3 n .

Let A = { e , a , b } . Then in G = A B , by Rédei  , A can be replace by

A ′ = { e , a , a 2 } .

If a 3 = e , then A is a subgroup of G . Thus, 〈 A 〉 ≠ G , so A is a non-full- rank factor of G .

If a 3 ≠ e , then from G = { e , a , a 2 } B , we get the following partition of G :

G = e B ∪ a B ∪ a 2 B ⋯ ( ∗ )

from which we get

G = a B ∪ a 2 B ∪ a 3 B ⋯ ( ∗ ∗ ) .

Comparing ( ∗ ) with ( ∗ ∗ ) , we obtain B = a 3 B . Thus, B is periodic, from which it follows that B = H C , where H is a a proper subgroup of G . Now, from G = A B , we obtain the factorization G / H = A B / H = ( A / H ) ( B / H ) of the quotient group G / H , which is of order less than 3 n . So, by inductive assumption, either 〈 A H / H 〉 ≠ G / H or 〈 B H / H 〉 ≠ G / H from which it follows that either 〈 A 〉 ≠ G or 〈 B 〉 ≠ G . That is either A or B is a non-full-rank factor of G QED.

Cite this paper

Amin, K. (2017) Non-Full Rank Factorization of Finite Abelian Groups. Open Journal of Discrete Mathematics, 7, 51-53. https://doi.org/10.4236/ojdm.2017.72005

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