** Open Journal of Discrete Mathematics** Vol.1 No.1(2011), Article ID:4575,5 pages DOI:10.4236/ojdm.2011.11001

Factoring Elementary p-Groups for p ≤ 7

Institute of Mathematics and Informatics, University of Pécs, Ifjúság, Pécs, Hungary

E-mail: sszabo7@hotmail.com

Received March 10, 2011; revised March 20, 2011; accepted April 5, 2011

**Keywords:** factorization of finite abelian groups, full-rank subset, full-rank factorization, periodic subset, periodic factorization, Rédei’s conjecture, Corrádi’s conjecture

Abstract

It is an open problem if an elementary p-group of rank k ≥ 3 does admit full-rank normalized factorization into two of its subsets such that one of the factors has p elements. The paper provides an answer in the p ≤ 7 special case.

1. Introduction

Let G be a finite abelian group. We use multiplicative notation in connection with abelian groups and e will denote the identity element of G. For two subset A and B of G the product AB is defined to be consisting of all the elements ab, where,. If

imply, , then we say that the product AB is direct. If the product AB is direct and is equal to G, then the equation is called a factorization of G.

We say that a subset A of G is normalized if. The factorization is called a normalized factorization if the factors A and B are normalized. A normalized subset A of G is termed a full-rank subset of G if. Here denotes the span of A in G. In other words is the smallest subgroup of G containing A. A normalized factorization is called a full-rank factorization if the factors A and B are full-rank subsets of G.

Let p be a prime and let k be a positive integer. A group that is a direct product of k cyclic groups of order p is called an elementary p-group of rank k. In 1970 in the open problems section in his book [4] L. Rédei advanced the following conjecture.

Conjecture 1 Let p be a prime. An elementary p-group of rank 3 does not admit any full-rank factorization.

Let be a normalized factorization, where G is an elementary p-group of rank 3. One of the factors A and B must have p elements while the other factor must have elements. A normalized subset of order p in the case cannot contain three generator elements of the group. Thus Rédei’s conjecture holds for and for the remaining part we may restrict our attention to the case.

In 1998 S. Szabó and C. Ward [7] carried out a computer assisted exhaustive search to verify Rédei’s conjecture for. In a private conversation K. Corrádi proposed the following generalization to Rédei’s conjecture.

Conjecture 2 Let p be a prime and let G be an elementary p-group of rank. If is a normalized factorization of G such that and, then at least one of the factors does not span G.

The normalized factor A can contain only generator elements of G and so the generalized conjecture certainly holds for and so it is enough to deal with the case.

In this note we verify Corrádi's conjecture for.

2. The k = 3 Case

At certain points in this paper we rely on some elementary concepts of graph theory. We presents these here. Let Γ be a simple graph, that is, Γ does not have double edges or loops. The set of vertices of Γ is denoted by V. Suppose U is a subset of V. If each two distinct elements of U are always connected in Γ by an edge of Γ, then the subgraph Δ of Γ spanned by U is called a clique of Γ. The set of vertices of Δ is U. If U has k elements, then we say that Δ is a clique of size k of Γ. Sometimes we express this fact simply by saying the Δ is a k-clique of Γ. The following problem is called the listing version of the k-clique problem.

Problem 1 Given a finite simple graph Γ and a positive integer k. List all k-cliques of Γ.

By the complexity theory of computation, Problem 1 belongs to the NP complete complexity class. Loosely speaking Problem 1 is computationally hard. We will solve two instances of the k-clique problem. In these cases the sizes of Γ are not overly large and the existing algorithms presented in [1,3] can handle them.

Let p be a prime and let G be an elementary p-group of rank 3 with basis elements x_{1}, x_{2}, x_{3}, where. Let be a normalized factorization of G such that,.

**Proposition 1** For, implies that B is a subgroup of G.

**Proof**. As, we may choose the basis elements x_{1}, x_{2}, x_{3} such that. We will work with the subset of A, where

For each i, j, we set. Choose an. Multiplying the factorization by gives the normalized factorization. By Lemma 5 of [2], in the factorization the factor can be replaced by to get the normalized factorization. Since the product is direct, by Lemma 2.1 of [6],

Plainly, and so

holds for each i, j,. Set

Clearly,. We define a graph Γ. The nodes of Γ are the elements of G. Two nodes are connected if. We may call T a test set since we use it for testing if a pair is an edge of Γ.

The graph Γ has nodes. We focus our attention on cliques of size p^{2} in Γ. The reason is the following. If the products are direct for each i, j, , then and so the elements of B form the nodes of a clique of size p^{2} in Γ. Conversely, if the elements of B are the nodes of a clique of size p^{2} in Γ, then and hence the products are direct for each i, j,.

We call a clique normalized if e is one of its nodes. A computer assisted inspection reveals that each normalized cliques of size p^{2} in Γ is a subgroup of G.

One can draw the following conclusion. If is a normalized factorization of G, where, then B must be a subgroup of G.

This completes the proof.

For p = 5 the graph Γ has 5^{3} = 125 nodes. The search found 30 cliques of size 5^{2} = 25. Each of them was a coset modulo some subgroup of order 25 of G. (The subgroup that plays the role of the modulus of course may vary from case to case.) In particular the normalized cliques of size 25 in Γ correspond to subgroups of G.

For p = 7 the graph Γ has 7^{3} = 343 nodes. The inspection gave 140 cliques of size 7^{2} = 49. Each of them turned out to be a coset modulo some subgroup of order 49.

For p = 11 the graph Γ does contain normalized cliques of size 11^{2} that are not subgroups of G. So our approach to verify Rédei’s conjecture (or Corrádi’s conjecture) breaks down for .

The above mentioned computer searches are not particularly demanding in terms of the time of computation. However, one cannot be cautious enough in connection with computer aided proofs. Therefore, in order to be on the safe side we used the algorithms described in [1,3] respectively as these algorithms have well tested implementations.

One can view the elements of G as points of the 3-dimensional affine space. Using geometrical terminology one can say that for a clique of size p^{2} in Γ is a 2-dimensional linear complex in. A 2-dimensional linear complex is a translated copy of some 2-dimensional subspace of.

3. The k = 4 Case

Let p be a prime and let G be an elementary p-group of rank 4 with basis elements, where. Let be a normalized factorization of G such that,.

**Proposition 2** For, implies that B is a subgroup of G.

**Proof.** We may assume that since this is only a matter of choosing the basis elements suitably. We set, where

We know that. For each i, j, we set. By Lemma 5 of [2], in the factorization the factor A can be replaced by to get the normalized factorization. As the product is direct, by Lemma 2.1 of [6], it follows that

(1)

for each i, j,.

We partition B into subsets. Each can be represented uniquely in the form

(2)

The set consists of each for which. Note that

In particular it follows that. We use now equations (1) only for i, j, to conclude that

(3)

holds for each i, j, k, ,.

Set. (The index 4 intends to indicate that x_{4} is missing from the basis in the definition of.) Set

Obviously. We define a graph. The nodes of are the elements of. Two nodes are connected if.

The graph has nodes. In addition and. Note that is isomorphic to the graph Γ used in the proof of Proposition 1. Consequently the nodes of a clique of size p^{2} in form a 2-dimensional linear complex in.

From (3) one can see that and consequently the elements of form the nodes of a clique of size p^{2} in. Using geometrical terminology one may say that is a 2-dimensional linear complex in.

We set out now to prove that the union of the p disjoint 2-dimensional linear complexes forms a 3-dimensional linear complex. This will show that B is in fact a subgroup of G of order p^{3}.

We partition B into. Each can be represented uniquely in the form (9). The set contains each for which. Note that

Therefore in particular holds. We use now equations (1) only for i, j, to conclude that

(4)

holds for each i, j, k, ,.

Set. (The meaning of the index 1 is that x_{1} is missing from the basis in the definition of the subgroup.) Set

Plainly. We define a graph. The nodes of are the elements of. Two nodes are connected if.

The graph has nodes. In addition and. Therefore, in fact is isomorphic to the graph Γ we defined in the proof of Proposition 1. From (4) it follows that and so is a 2-dimensional linear complex in.

Let us observe that is a subgroup of G of order p. Using geometrical terminology is a 1-dimensional linear complex in. We may view as a union of p disjoint 1-dimensional linear complexes. Similarly, we may view as a union of p disjoint 1-dimensional complexes. In addition each of these linear complexes is a translated copy of. Using the 1-dimensional linear complexes

analogously we can conclude that B is a union of p^{2} disjoint 1-dimensional complexes each of which is a translated copy of. The translation vectors form a 2-dimensional linear complex. Therefore B is a 3-dimensional linear complex in.

This completes the proof.

For the k = 6 case we need a corollary of Proposition 2. Set

and define a graph Γ. The nodes of Γ are the elements of G. Two nodes are connected if.

**Corollary 1 **Each clique of size p^{3} in Γ corresponds to a 3-dimensional linear complex in.

4. The k ≥ 5 Case

Let p be a prime and let G be an elementary p -group of rank 5 with basis elements, where. Let be a normalized factorization of G such that,.

**Proposition 3** For, implies that B is a subgroup of G.

**Proof. **The proof is similar to the proof of Proposition 2 and we just outline the argument. It may be assumed that. We set, where

Clearly. For each i, j, we set. From the factorization we get the normalized factorization. From the directness of the product it follows that

(5)

for each i, j,.

We partition B into subsets, Each can be represented uniquely in terms of the basis in the form

(6)

The set consists of each for which. It follows that. The equations (5) for i, j, give that

(7)

for each i, j, k, ,.

Set and

We define a graph. The nodes of are the elements of. Two nodes are connected if. Note that is isomorphic to the graph Γ in Corollary 1. From (7) one can see that the elements of form the nodes of a clique of size p^{3} in and so

is a 3-dimensional linear complex in.

Next we partition B into, where contains each for which in the representation (6). A routine computation shows that The equations (5) for i, j, imply that

(8)

for each i, j, k, ,.

Set and

We define a graph. The nodes of are the elements of. Two nodes are connected if. Let us observe that is isomorphic to the graph Γ in Corollary 1. From (8) it follows that is a 3-dimensional linear complex in.

Using the fact that is a subgroup of G of order p^{2} one can show that B is a 4-dimensional linear complex in.

This completes the proof.

For the k = 6 case we need a corollary of Proposition 3. Set

and define a graph Γ. The nodes of Γ are the elements of G. Two nodes are connected if.

**Corollary 2** Each clique of size p^{4} in Γ corresponds to a 4-dimensional linear complex in.

Let p be a prime and let G be an elementary p-group of rank 6 with basis elements, where. Let be a normalized factorization of G such that,.

**Proposition 4** For, implies that B is a subgroup of G.

**Proof.** The proof is similar to the proof of Proposition 3 and we do not detail it.

We spell out the main result of this note formally as a theorem.

**Theorem 1** Let G be a finite elementary p-group, where p is a prime and let be a normalized factorization such that. If, then either or.

5. An Application

Let G be a finite abelian group and let A be a subset of G. We say that the subset A is periodic if there is an element such that and. A factorization is called periodic if the factors A and B are both periodic. A. D. Sands has proved the following lemma. (See Lemma 3 of [5].)

**Lemma 1** Let be a factorization of a finite abelian group G such that,. If, then the factorization is periodic.

Motivated by this result we prove the next theorem.

**Theorem 2** Let be a normalized factorization of a finite abelian group G such that is a prime and. If, then either or B is periodic.

**Proof**. Let be a prime and consider a normalized factorization of a finite abelian group G such that, and. We claim that B is periodic.

If or, then by Sands’ lemma it follows that either A or B is periodic. Thus for the remaining part of the proof we may assume that or.

Choose an element. By Lemma 5 of [2], in the factorization the factor A can be replaced by to get the normalized factorization. This factorization is equivalent to the fact that the sets

(9)

form a partition of G. Multiplying the factorization by the element a we get the normalized factorization. This factorization is equivalent to the fact that the sets

(10)

form a partition of G. Comparing the partitions (9) and (10) provides that. Therefore, if, then B is periodic. Thus for the remaining part of the proof we may assume that for each. As, it follows that G is an elementary p-group. From the factorization, by Theorem 1, it follows that either or. Using we get that.

The reader can check that in the course of the proof of Theorem 1 we obtained the following side result. Let G be a finite elementary p-group where is a prime. If is a normalized factorization such that and, then B is a subgroup of G. Clearly, a subgroup B of G is a periodic subset unless. But in our case, by the hypotheses of the theorem, holds.

This completes the proof.

6. References

[1] R. Carraghan and P. M. Pardalos, “An exact algorithm for the maximum clique problem,” Operation Research Letters 9 (1990), 375-382. doi:10.1016/0167-6377(90)90057-C

[2] K. Corrádi, S. Szabó and P. Hermann, “A character free proof for Rédei's theorem,” Mathematica Pannonica 20 (2009), 3-15.

[3] P. R. J. Östergå rd, “A fast algorithm for the maximum clique problem,” Discrete Applied Mathematics 120 (2002), 195-205.

[4] L. Rédei, Lückenhafte Polynome über endlichen Körpern, Birkhäuser Verlag, Basel 1970, (English translation: Lacunary Polynomials over Finite Fields, North-Holland, Amsterdam, 1973.)

[5] A. D. Sands, “On the factorisation of finite abelian groups,” Acta Math. Acad. Sci. Hung. 8 (1957), 65-86. doi:10.1007/BF02025232

[6] S. Szabó and A. D. Sands, “Factoring Groups into Subsets,” CRC Press, Taylor and Francis Group, Boca Raton, 2009.

[7] S. Szabó and C. Ward, “Factoring elementary groups of prime cube order into subsets,” Mathematics of Computation 67 (1998), 1199-1206. doi:10.1090/S0025-5718-98-00929-6