factoring abelian groups into uniquely complemented subsets

J. Group Theory 14 (2011), 307–327DOI 10.1515/JGT.2010.046

Journal of Group Theory( de Gruyter 2011

Factoring abelian groups into uniquely complemented subsets

Keresztely Corradi and Sandor Szabo

(Communicated by D. J. S. Robinson)

Abstract. The paper deals with decomposition of a finite abelian group into a direct product ofsubsets. A family of subsets, the so-called uniquely complemented subsets, is singled out. It willbe shown that if a finite abelian group is a direct product of uniquely complemented subsets,then at least one of the factors must be a subgroup. This generalizes Hajos’s factorizationtheorem.

1 Introduction

Let G be a finite abelian group and let A1; . . . ;An be subsets of G. The product of thesubsets A1; . . . ;An is defined to be the set of elements

a1 . . . an; a1 A A1; . . . ; an A An:

Suppose that the product A1 . . .An is equal to B. If each element b A B is uniquelyexpressible in the form

b ¼ a1 . . . an; a1 A A1; . . . ; an A An;

then we say that the product A1 . . .An is direct. If the product A1 . . .An is direct andis equal to B we also will say that the equation B ¼ A1 . . .An is a factorization of thesubset B.

A subset A of G is called a cyclic subset if it is in the form

A ¼ fe; a; a2; . . . ; ar�1g:

Here e is the identity element and a is an element of G, r is an integer such that2c rc jaj. If jaj, the order of a, is equal to r, then A is a cyclic subgroup of orderr. If r < jaj, then A is not a subgroup of G.

The first author was supported by the Hungarian National Science Foundation grant num-ber T-049841.

AUTHOR’S COPY | AUTORENEXEMPLAR


In 1942, Hajos [3] proved that if a finite abelian group is a direct product of cyclicsubsets, then at least one of the factors must be a subgroup.

A subset A of G is called normalized if e A A. A factorization B ¼ A1 . . .An is de-fined to be normalized if each of B;A1; . . . ;An is normalized.

In 1965, Redei showed that if a finite abelian group is a direct product of normal-ized subsets that have prime cardinalities, then at least one of the factors must be asubgroup.

In a restricted version of Hajos’s theorem we assume that each cyclic factor has aprime number of elements. It turns out that the restricted version implies the generalresult. Thus Redei’s theorem can be viewed as a generalization of Hajos’s theorem.Hajos proved his theorem in order to solve a problem of Minkowski from geometry.Redei’s theorem also has interesting applications.

This paper is a contribution to the theory of factorization of abelian groups. Wewill introduce the family of uniquely complemented subsets and show that in somefactorization results cyclic subsets can be replaced by uniquely complemented sub-sets. For instance Hajos’s theorem can be extended replacing the cyclic subsets byuniquely complemented subsets.

2 Complementer factors

Let G be a finite abelian group and let A be a subset of G. A subset B of G is called acomplementer factor of A in G if G ¼ AB is a factorization. In this definition onemay assume that the subsets A and B are normalized. To see why, choose elementsa A A, b A B and multiply the factorization G ¼ AB by a�1b�1 to get the factorizationG ¼ Ga�1b�1 ¼ ðAa�1ÞðBb�1Þ. Here Aa�1 and Bb�1 are normalized subsets. Thusif A has a complementer factor, then so does Aa�1. Further, if B is a complementerfactor of A, then so is Bb�1.

Lemma 1. Let G be a finite abelian group and let A be a normalized subset of G. Set

H ¼ hAi. The subset A has a complementer factor in G if and only if A has a comple-

menter factor in H.

Proof. Suppose that A has a complementer factor B in G, that is, suppose thatG ¼ AB is a normalized factorization of G. Restricting the factorization G ¼ AB toH gives the normalized factorization H ¼ GVH ¼ AðBVHÞ. Therefore A has acomplementer factor BVH in H.

Next assume that A has a complementer factor D in H. In other words assume thatH ¼ AD is a normalized factorization of H. Let C ¼ fc1; . . . ; csg be a complete set ofrepresentatives in G modulo H. Assume that c1 ¼ e. Then the sets Hc1; . . . ;Hcs forma partition of G, that is, G ¼ HC is a factorization of G. Now G ¼ ðADÞC ¼ AðDCÞis a normalized factorization of G. This means that A has a complementer factor DC

in G. r

Let G be a finite abelian group and let A be a normalized subset of G. SetH ¼ hAi. We say that A is uniquely complemented if there is a unique normalized

308 K. Corradi and S. Szabo



complementer factor to A in H. In other words the normalized subset A of G is auniquely complemented subset of G if the following two conditions hold.

(i) There is a normalized subset B of G such that H ¼ AB is a factorization of H.

(ii) The normalized factorizations H ¼ AB, H ¼ AC imply that B ¼ C.

By Lemma 1, if A has a complementer factor in G then it has a complementerfactor in H and conversely if A has a complementer factor in H, then it has a com-plementer factor in G. If A has a unique complementer factor in H, then we call A auniquely complemented factor in G. We would like to point out that it is not assumedthat A has a unique complementer factor in G.

The next examples show that a normalized subset may have more than one com-plementer factor or may not have any complementer factor at all. If a group G is adirect product of cyclic subgroups of orders t1; . . . ; tn, then we say that G is of typeðt1; . . . ; tnÞ.

Let G be a group of type ð4; 4Þ with basis elements x, y. Let

A ¼ fe; xy; x2y3; x3y2g;

B ¼ hxi; C ¼ hyi; H ¼ hAi:

Note that H is equal to G and that H ¼ AB, H ¼ AC are normalized factorizations.In this case A has two complementer factors in H.

Next let G be a group of type ð9Þ with basis element x. Let

A ¼ fe; x; x4g; H ¼ hAi:

Now H ¼ G. We claim that A has no complementer factor in G. In order to verifythe claim assume on the contrary that G ¼ AB is a normalized factorization. Here jAjand jBj are primes and, by the main result of [4], A or B must be a subgroup. Since A

is not a subgroup B must coincide with the unique subgroup K ¼ hx3i of G of order3. The elements e, x, x4 must be pairwise incongruent modulo K . But this is not thecase as x and x4 are in the same coset modulo K .

Lemma 2. Let G be a finite abelian group and let A be a uniquely complemented subset

of G. Let H ¼ hAi. If H ¼ AB is a normalized factorization, then B must be a sub-

group of H.

Proof. Choose an element b A B. As the factorization H ¼ AB is normalized, it fol-lows that b A H. Multiplying the factorization H ¼ AB by b�1 we get the normalizedfactorization H ¼ Hb�1 ¼ AðBb�1Þ. As A is uniquely complemented, it follows thatB ¼ Bb�1. Thus

B ¼ 7b AB

Bb�1:

This means that b 0b�1 A B for each b 0; b A B and so B is a subgroup of H. r

Factoring abelian groups into uniquely complemented subsets 309



https://www.researchgate.net/publication/226590357_Die_neue_Theorie_der_endlichen_abelschen_Gruppen_und_Verallgemeinerung_des_Hauptsatzes_von_Hajs?el=1_x_8&enrichId=rgreq-86f91954-eef0-4270-8336-ecbb3f742a6e&enrichSource=Y292ZXJQYWdlOzI0MzEwNjQ2ODtBUzoxOTc5MjA0MjgxNzEyNjVAMTQyNDE5OTMxNTE3Mg==

Let H be a subgroup and let A be a subset of the finite abelian group G. The set ofcosets faH : a A Ag is denoted by ðAHÞ=H.

Lemma 3. Let G be a finite abelian group and let H be a subgroup of G. Let A, B be

subsets of G such that

G=H ¼ ½ðAHÞ=H�½ðBHÞ=H� ð1Þ

is a factorization of the factor group G=H, and suppose further that the elements of

A are pairwise incongruent modulo H and the elements of B are pairwise incongruent

modulo H. Then G ¼ ABH is a factorization of G.

Proof. As the elements of A are pairwise incongruent modulo H, the cosets aH, a A A

are distinct. In other words, if a1H ¼ a2H and a1; a2 A A then a1 ¼ a2. Similarly, ifb1H ¼ b2H and b1; b2 A B then b1 ¼ b2.

In order to show that G ¼ ABH is a factorization of G choose an element g A G.Since (1) is a factorization of G=H, the coset gH can be represented in the form

gH ¼ ðaHÞðbHÞ; a A A; b A B:

Therefore g A abH and so g ¼ abh for some h A H.Suppose that

a1b1h1 ¼ a2b2h2; with a1; a2 A A; b1; b2 A B; h1; h2 A H: ð2Þ

It follows that a1b1H ¼ a2b2H, that is, ða1HÞðb1HÞ ¼ ða2HÞðb2HÞ. By the factoriza-tion (1), we get a1H ¼ a2H, b1H ¼ b2H. Then it follows that a1 ¼ a2, b1 ¼ b2 asrequired. The equation (2) now gives h1 ¼ h2. This completes the proof. r

Lemma 4. Let G be a finite abelian group and let A be a uniquely complemented

subset of G. Let H ¼ hAi and let K be a subgroup of H. If the elements of A are

pairwise incongruent modulo K , then ðAKÞ=K is a uniquely complemented subset of

G=K .

Proof. As A is a uniquely complemented subset in G, it has a complementer factor inG. By Lemma 1, there is a subset B of H such that H ¼ AB is a normalized factori-zation of H. Considering the factor group H=K we get the factorization

H=K ¼ ½ðAKÞ=K�½ðBKÞ=K �

of H=K . This means that ðAKÞ=K has a complementer factor in H=K .In order to make a more detailed analysis we set

A ¼ fa1; . . . ; amg; B ¼ fb1; . . . ; bng:




As the elements of A are pairwise incongruent modulo K the cosets a1K ; . . . ; amK aredistinct. On the other hand the cosets

b1K ; . . . ; bnK ð3Þ

are not necessarily distinct. We may assume that b1K ; . . . ; brK are all the distinctcosets among (3) since this is only a matter of indexing the elements of B. SetB1 ¼ fb1; . . . ; brg. Note that BK ¼ B1K .

Suppose that there is a subset C of G such that

H=K ¼ ½ðAKÞ=K �½ðCKÞ=K �

is a factorization. Let C ¼ fc1; . . . ; csg. Here we may choose C such that the cosetsc1K ; . . . ; csK are distinct. Namely, from each coset of ðCKÞ=K we simply choose onlyone element.

Lemma 3 is applicable to the factorizations

H=K ¼ ½ðAKÞ=K �½ðB1KÞ=K �; H=K ¼ ½ðAKÞ=K �½ðCKÞ=K �

and gives that H ¼ AB1K , H ¼ ACK are factorizations of H. As A is a uniquelycomplemented subset of G, it follows that B1K ¼ CK . Using BK ¼ B1K we getBK ¼ CK . Then it follows that ðBKÞ=K ¼ ðCKÞ=K as required. r

The next lemma is a generalization of the previous one.

Lemma 5. Let G be a finite abelian group and let A be a uniquely complemented subset

of G. If K is a subgroup of G such that the elements of A are pairwise incongruent mod-

ulo K , then ðAKÞ=K is a uniquely complemented subset of G=K .

Proof. Let H ¼ hAi. We claim that ðAKÞ=K spans ðHKÞ=K in G=K , that ishðAKÞ=Ki ¼ ðHKÞ=K . To prove the claim note that as H ¼ hAi, each h A H canbe represented in the form

h ¼ aað1Þ1 . . . aaðsÞ

s ; a1; . . . ; as A A;

where að1Þ; . . . ; aðsÞ are integers. The computation

hK ¼ ðaað1Þ1 . . . aaðsÞ

s ÞK ¼ ðaað1Þ1 ÞK . . . ðaaðsÞ

s ÞK ¼ ða1KÞað1Þ . . . ðasKÞaðsÞ

shows that each hK A ðHKÞ=K can be represented in the form

hK ¼ ða1KÞað1Þ . . . ðasKÞaðsÞ; a1; . . . ; as A A;

where að1Þ; . . . ; aðsÞ are integers.




Next we claim that ðAKÞ=K has a complementer factor in ðHKÞ=K. In order toverify the claim set L ¼ H VK . Note that L is a subgroup of K. As the elements ofA are pairwise incongruent modulo K , it follows that the elements of A are pairwiseincongruent modulo L. Note that L is a subgroup of H. By Lemma 4, ðALÞ=L has acomplementer factor in H=L. In other words there is a normalized subset B of G suchthat

H=L ¼ ½ðALÞ=L�½ðBLÞ=L� ð4Þ

is a factorization of H=L. We may assume that the elements of B are pairwise incon-gruent modulo L. By Lemma 3, H ¼ ABL is a factorization of H. Therefore for eachh A H there are a A A, b A B, l A L such that h ¼ abl. It follows that hK ¼ ðaKÞðbKÞ.Thus ðHKÞ=L ¼ ½ðAKÞ=L�½ðBKÞ=L�.

We need to show that

ða1KÞðb1KÞ ¼ ða2KÞðb2KÞ; a1; a2 A A; b1; b2 A B ð5Þ

implies a1K ¼ a2K , b1K ¼ b2K . Using (5) we get that there is a k A K for whicha1b1 ¼ a2b2k. By rearranging we get a1b1a

�12 b�1

2 ¼ k. The left-hand side is an elementof H. The right-hand side is an element of K . It follows that there is an l A L suchthat a1b1 ¼ a2b2l. Multiplying by L gives

ða1LÞðb1LÞ ¼ ða2LÞðb2LÞ:

From the factorization (4) it follows that a1L ¼ a2L, b1L ¼ b2L. There are ele-ments l1; l2 A L such that a1 ¼ a2l1, b1 ¼ b2l2. Multiplying by K gives a1K ¼ a2K ,b1K ¼ b2K , as required.

Let ðCKÞ=K be another complementer factor of ðAKÞ=K in ðHKÞ=K , that is, as-sume that

ðHKÞ=K ¼ ½ðAKÞ=K �½ðCKÞ=K �

is a normalized factorization of ðHKÞ=K . We may assume that the elements of C arepairwise incongruent modulo K .

We claim that ðBKÞ=K ¼ ðCKÞ=K . In order to prove the claim let d1; . . . ; dt A K bea complete set of representatives in KH modulo H. Lemma 3 is applicable to thefactorizations

ðHKÞ=K ¼ ½ðAKÞ=K �½ðBKÞ=K �; ðHKÞ=K ¼ ½ðAKÞ=K �½ðCKÞ=K �

and it shows that

HK ¼ ABK ¼ AðBKd�1i Þ; HK ¼ ACK ¼ AðCKd�1

i Þ




are factorizations of HK . Restricting these factorizations to H we get the factoriza-tions

H ¼ HK VH ¼ AðBKd�1i VHÞ; H ¼ HK VH ¼ AðCKd�1

i VHÞ

of H. As A is a uniquely complemented subset of G, it follows that

BKd�1i VH ¼ CKd�1

i VH:

Multiplying by di we get BK VHdi ¼ CK VHdi. Using this the routine computation

BK ¼ BK VHK ¼ BK V ½Hd1 U � � �UHdt�

¼ ðBK VHd1ÞU � � �U ðBK VHdtÞ

¼ ðCK VHd1ÞU � � �U ðCK VHdtÞ

¼ CK V ½Hd1 U � � �UHdt� ¼ CK VHK ¼ CK

shows that BK ¼ CK . It follows that ðBKÞ=K ¼ ðCKÞ=K as we claimed. r

3 Distorted cyclic subsets

Let G be a finite abelian group and let L be a subgroup of G such that jLjd 3.Choose an element l A Lnfeg and an element d A ðGnLÞU feg. The subset

A ¼ ðLnflgÞU fldg

constructed from the subgroup L by using l and d is called a simulated subset.Clearly jLj ¼ jAj.

In addition to the simulated subsets and cyclic subsets defined in Section 1 we willconsider distorted cyclic subsets. Let A ¼ fe; a; a2; . . . ; ar�1g be a cyclic subset of afinite abelian group G such that rd 3. If i is an integer with 1c ic r� 1 and d isan element of G such that aid B Anfaig, then we call the set

B ¼ ðAnfaigÞU faidg

a distorted cyclic subset of G. The condition aid B Anfaig implies that jAj ¼ jBj. Ofcourse we could add the condition jAj ¼ jBj to the definition of a distorted cyclic sub-set, and then this new condition would imply aid B Anfaig.

A subset A of G is defined to be periodic if there is an element g A Gnfeg such thatAg ¼ A. The element g is called a period of A. If g and h are periods of A and gh0 e,then gh is also a period of A. The periods of A augmented with the identity elementform a subgroup H of G. We call H the subgroup of periods of A. One can verifythat there is a subset B of G such that the product HB is direct and is equal to A.The subset B is far from uniquely determined. If A is normalized, then B can bechosen to be a subset of A.




Lemma 6. Let G be a finite abelian group and let A be a cyclic subset of G. If A has a

complementer factor in G, then A is a uniquely complemented subset of G.

Proof. Let A ¼ fe; a; a2; . . . ; ar�1g be a cyclic subset of G. Set H ¼ hAi ¼ hai. SinceA has a complementer factor in G, by Lemma 1, A has a complementer factor in H

too. Let H ¼ AB be a normalized factorization of H. This means that the subsets

eB; aB; a2B; . . . ; ar�1B

form a partition of H. Multiplying the factorization H ¼ AB by a gives the factoriza-tion H ¼ ðAaÞB and so the subsets

aB; a2B; . . . ; ar�1B; arB

form a partition of H. Comparing the partitions gives that B ¼ arB. Thereforear A B.

Let jBj ¼ s. The equation jHj ¼ jAj jBj ¼ rs shows that jaj ¼ rs. It follows thatjarj ¼ s ¼ jBj. From ar A B it follows that B ¼ hari, that is, B is uniquely de-termined. r

Lemma 7. Let G be a finite abelian group and let A be a simulated subset of G. If A has

a complementer factor in G, then A is a uniquely complemented subset of G.

Proof. Let A ¼ ðLnflgÞV fldg be a simulated subset of G. (Here L is a subgroup ofG, jLjd 3, l A Lnfeg, d A ðGnLÞU feg.) Set H ¼ hAi ¼ hL; di. Since A has a com-plementer factor in G, it has a complementer factor in H too. Let H ¼ AB be a nor-malized factorization of H. By [7, Lemma 3], in the factorization H ¼ AB the factorA can be replaced by L to give the factorization H ¼ LB. Further it follows thatB ¼ dB.

Let K ¼ hdi. There is a normalized subset C of B such that the product KC

is direct and is equal to B. From the factorization H ¼ LB ¼ LCK it followsthat LKHH. On the other hand H ¼ hL; diHLK . Thus H ¼ LK . In particularC ¼ feg and B ¼ K . Therefore B is uniquely determined. r

Lemma 8. Let G be a finite abelian group and let B be a distorted cyclic subset of G. If

B has a complementer factor in G, then B is a uniquely complemented subset of G.

Proof. Let A ¼ fe; a; a2; . . . ; ar�1g be a cyclic subset of G. Let B ¼ ðAnfaigÞU faidgbe a distorted cyclic subset. Here 1c ic r� 1 and

aid B fe; a; a2; . . . ; ai�1gU faiþ1; . . . ; ar�1g:

Consequently, jAj ¼ jBj ¼ r. If d ¼ e, then B is a cyclic subset of G. By Lemma 6, Bis a uniquely complemented subset of G. So for the rest of the proof we may assumethat d0 e.




Set H ¼ hBi ¼ ha; di. As B has a complementer factor in G, by Lemma 1, it alsohas a complement factor in H. Let H ¼ BC be a normalized factorization of H. Herewe distinguish two cases depending on rd 4 or r ¼ 3.

Let us deal first with the case when rd 4. By [2, Lemma 1], in the factorizationH ¼ BC the factor B can be replaced by A to give the factorization H ¼ AC. FurtherC ¼ Cd. As d0 e, it follows that d is a period of C. Set L ¼ hdi. There is a normal-ized subset D of C such that the product LD is direct and is equal to C.

From the factorization H ¼ AC ¼ ADL it follows that the product AL is direct.In particular the elements of A are pairwise incongruent modulo L. Considering thefactor group H=L and the factorization H ¼ ADL we get the factorization

H=L ¼ ½ðALÞ=L�½ðDLÞ=L�

of H=L. There is a subset D1 of D such that the elements of D1 are pairwise incon-gruent modulo L and

H=L ¼ ½ðALÞ=L�½ðD1LÞ=L�

is a factorization of H=L. Here D1L ¼ DL.Note that

ðALÞ=L ¼ feL; aL; . . . ; ar�1Lg ¼ fðaLÞ0; ðaLÞ1; . . . ; ðaLÞr�1g

is a cyclic subset of H=L and so it follows that

ðaLÞr½ðD1LÞ=L� ¼ ðD1LÞ=L:

Set M ¼ hari. Now ðMLÞ=L is a subgroup of H=L. There is a subset E of D1

such that the product ½ðMLÞ=L�½ðELÞ=L� is direct and is equal to ðD1LÞ=L. We maychoose E such that the elements of E are pairwise incongruent modulo L.

From the factorization

H=L ¼ ½ðALÞ=L�½ðD1LÞ=L� ¼ ½ðALÞ=L�½ðELÞ=L�½ðMLÞ=L�;

by Lemma 3, it follows that H ¼ AMEL is a factorization. In particular AMLHH.On the other hand

H ¼ ha; di ¼ hAM; diHAML:

Therefore H ¼ AML and consequently E ¼ feg. This gives D1 ¼ AM ¼ hai. ThusD1 is uniquely determined. Therefore D1L is uniquely determined. Using D1L ¼ DL

we see that DL is uniquely determined. The factorization H ¼ BC ¼ BðDLÞ showsthat B is a uniquely complemented subset in H.

Let us turn to the case when r ¼ 3. In this case [2, Lemma 1] is not applicable sincerd 4 is an assumption of the lemma. We deduce the equation C ¼ dC in a di¤erent




way. First note that each normalized subset B of G with jBj ¼ 3 is a distorted cyclicsubset of the form B ¼ fe; a; a2dg. To see why, let B ¼ fe; a; bg. Plainly there is anelement d in G such that b ¼ a2d.

The normalized factorization H ¼ BC means that the sets

C; aC; a2dC ð6Þ

form a partition of H. Multiplying the factorization H ¼ BC by a we get the factori-zation H ¼ aH ¼ ðaBÞC and so the sets

aC; a2C; a3dC

form a partition of H. Comparing the two partitions shows that

C U a2dC ¼ a2C U a3dC:

If a2dC V a3dC0q, then C V aC0q. This violates the partition (6). Thusa2dCH a2C and so dCHC. Consideration of cardinalities reveals that dC ¼ C.Using this from (6) it follows that the sets

C; aC; a2C

form a partition of H. In other words H ¼ AC is a factorization of H. From thispoint on we can argue in the same way as in the case when rd 4. r

In the proof of [2, Lemma 1] characters were used at one point. Sands [6] presenteda character-free proof for this result.

4 Replacing factors

A normalized subset A of a finite abelian group G is called a full-rank subset ifhAi ¼ G. A normalized factorization G ¼ AB is called a full-rank factorization if Aand B are full-rank subsets of G.

Lemma 9. Suppose that the normalized factorization G ¼ AB is not full-rank, say

H ¼ hAi and jG : Hj ¼ sd 2.

(i) Then there are elements b1; . . . ; bs A B such that the factorization G ¼ AB forks

into simultaneous normalized factorizations

H ¼ AðBb�11 VHÞ; . . . ;H ¼ AðBb�1

s VHÞ

of H. In addition b1 ¼ e may be assumed.

(ii) If in each of the forked factorizations above the factor A can be replaced by the

normalized subset D, then in the factorization G ¼ AB the factor A can be replaced

by D.




https://www.researchgate.net/publication/267186811_A_note_on_distorted_cyclic_subsets?el=1_x_8&enrichId=rgreq-86f91954-eef0-4270-8336-ecbb3f742a6e&enrichSource=Y292ZXJQYWdlOzI0MzEwNjQ2ODtBUzoxOTc5MjA0MjgxNzEyNjVAMTQyNDE5OTMxNTE3Mg==

Proof. (i) Let B ¼ fb1; . . . ; bng with b1 ¼ e. The factorization G ¼ AB meansthat the sets Ab1; . . . ;Abn form a partition of G. As Abi HHbi, it follows thatGHHb1 U � � �UHbn. There are elements c1; . . . ; ct A B such that Hc1; . . . ;Hct arethe distinct cosets in fHb1; . . . ;Hbng. We may assume that c1 ¼ b1. Further, wemay assume that c2 ¼ b2; . . . ; ct ¼ bt, since this is only a matter of renaming theelements b2; . . . ; bn. The sets BVHb1; . . . ;BVHbt form a partition of B.

Multiplying the normalized factorization G ¼ AB by b�1i we get the normalized

factorization G ¼ Gb�1i ¼ AðBb�1

i Þ. Restricting this factorization to H gives the nor-malized factorization H ¼ AðBb�1

i VHÞ of H for 1c ic t. It is clear that the cardi-nality of Bbi VH is independent of the choice of i, say u. Using this the equation

B ¼ ðBVHb1ÞU � � �U ðBVHbtÞ

shows that jBj ¼ tu. The factorization G ¼ AB gives that jGj ¼ jAj jBj. NowjGj ¼ jAjtu and jHj ¼ jAju imply that t ¼ jGj=jHj ¼ s. This completes the proof ofpart (i).

(ii) Suppose that in the normalized factorization H ¼ AðBb�1i VHÞ the factor A

can be replaced by D to give the normalized factorization H ¼ DðBb�1i VHÞ for

i A f1; . . . ; sg. Multiplying by bi we get

Hbi ¼ AðBVHbiÞ; Hbi ¼ DðBVHbiÞ

for each i A f1; . . . ; sg. From the proof of part (i) the sets BVHb1; . . . ;BVHbs form apartition of B. The computation

G ¼ AB ¼ A½ðBVHb1ÞU � � �U ðBVHbsÞ�

¼ AðBVHb1ÞU � � �UAðBVHbsÞ

¼ DðBVHb1ÞU � � �UDðBVHbsÞ

¼ D½ðBVHb1ÞU � � �U ðBVHbsÞ� ¼ DB

shows that G ¼ DB. Consideration of cardinalities gives that G ¼ DB is a factoriza-tion of G. This completes the proof of part (ii). r

Lemma 10. Let G be a finite abelian group and H a non-trivial subgroup.

(i) If G is an elementary 2-group, the index jG : Hj is equal to 2, G ¼ AH is a normal-

ized factorization, then A must be a subgroup.

(ii) If G is not an elementary 2-group or jG : Hjd 3, then there is a normalized

factorization G ¼ A1 . . .AnH, where each Ai is a non-subgroup cyclic or simulated

subset.

Proof. (i) From the factorization G ¼ AH, it follows that jGj ¼ jAj jHj and sojAj ¼ jGj=jHj ¼ jG : Hj ¼ 2. Each element of Gnfeg has order 2. This implies thatA is a subgroup of G.




(ii) Assume first that G is an elementary 2-group. By the hypotheses of the lemma,jG : Hjd 4 must hold. There is a subgroup K of G such that the product HK is directand is equal to G. Choose elements h A Hnfeg, k A Knfeg and set

A1 ¼ ðKnfkgÞU fkhg:

Plainly A1 is a non-subgroup simulated subset. The computation

HA1 ¼ H½ðKnfkgÞU fkhg� ¼ HðKnfkgÞUHkh

¼ ðHKnHkÞUHhk ¼ ðHKnHkÞUHk

¼ HK ¼ G

gives that G ¼ A1H is a factorization of G.Consider the case when G is not an elementary 2-group. We claim that there

is an element c A GnH such that jcjd 3. Suppose that this is not the case. If H

is an elementary 2-group, then each element of Gnfeg has order 2, and a contradic-tion follows. Choose h A H with jhjd 3 and choose x A GnH. Now hx B H and soe ¼ ðhxÞ2 ¼ h2. This is a contradiction. Therefore there is an element c A GnH suchthat jcjd 3.

Set jcj ¼ n. Let t be the least positive integer satisfying ct A H. Note that t is adivisor of n. In particular tc n. Set

C ¼ fe; c; c2; . . . ; ct�1g; K ¼ hH; ci:

The elements of C form a complete set of representatives in K modulo H and soK ¼ CH is a factorization of K .

We claim that in the factorization K ¼ CH the factor C can be replaced by a non-subgroup cyclic or simulated subset. Indeed if t0 n, then C is a non-subgroup cyclicsubset and we may choose A1 to be C. If t ¼ n, then C is a subgroup. Now choose anelement h A Hnfeg and set

A1 ¼ ðCnfcn�1gÞU fcn�1hg:

As nd 3 and h B C, A1 is a non-subgroup simulated subset. The computation

HA1 ¼ H½ðCnfcn�1gÞU fcn�1hg� ¼ ðHCnHcn�1ÞUHcn�1h

¼ ðHCnHcn�1ÞUHhcn�1 ¼ ðHCnHcn�1ÞUHcn�1

¼ HC ¼ K

proves that K ¼ A1H is a factorization of K .If K ¼ G, then we have a desired factorization for G and there is nothing to do. If

K0G, we replace the pair ðH;GÞ by ðK ;GÞ and repeat the whole argument. Con-tinuing in this way finally we end up with a required factorization of G. r




Lemma 11. Let G be a finite abelian group and let G ¼ AB be a normalized factoriza-

tion, where A is a uniquely complemented subset. If A is not a subgroup, then there are

subsets A1; . . . ;An such that G ¼ A1 . . .AnB is a factorization and each Ai is a non-

subgroup cyclic or simulated subset.

Proof. Set H ¼ hAi. Restricting the factorization G ¼ AB to H we get the facto-rization H ¼ AðBVHÞ. By Lemma 2, BVH is a subgroup of H. Let K ¼ BVH.If K ¼ feg, then H ¼ A. This contradicts the fact that A is not a subgroup. ThusK0 feg.

If H is an elementary 2-group and jH : K j ¼ 2, then by Lemma 10, A is again asubgroup. So if H is an elementary 2-group, then jH : K jd 4. By Lemma 10, thereare subsets A1; . . . ;An such that H ¼ A1 . . .AnK is a factorization and each Ai is anon-subgroup cyclic or simulated subset. Set D ¼ A1 . . .An. Now

H ¼ A1 . . .AnK ¼ A1 . . .AnðBVHÞ ¼ DðBVHÞ:

In the factorization H ¼ AðBVHÞ the factor A can be replaced by D. Note thatas A is a uniquely complemented factor in G, it follows that B is a subgroup of G.Therefore B ¼ Bb�1

i for each bi A B. We conclude that H ¼ DðBb�1i VHÞ is a fac-

torization of H for each bi A B. Consequently Lemma 9 is applicable and in thefactorization G ¼ AB the factor A can be replaced by D to give the factorizationG ¼ DB ¼ A1 . . .AnB. This completes the proof. r

5 Extensions of Hajos’ theorem

We are ready to prove the main results of the paper.

Theorem 1. Let G be a finite abelian group and let G ¼ A1 . . .An be a normalized

factorization of G, where each Ai is a uniquely complemented subset. Then there is a

permutation B1; . . . ;Bn of the factors A1; . . . ;An such that

B1 HB1B2 H � � �HB1B2 . . .Bn

is an ascending chain of subgroups of G.

Proof. First we show that in the factorization G ¼ A1 . . .An at least one of the factorsmust be a subgroup. Suppose that this is not the case. Set Ci ¼ A1 . . .Ai�1Aiþ1 . . .An.By Lemma 11, there are subsets Di;1; . . . ;Di;mðiÞ such that G ¼ Di;1 . . .Di;mðiÞCi is afactorization of G and each Di; j is a non-subgroup cyclic or simulated subset. Then

G ¼Yni¼1

YmðiÞ

j¼1

Di; j

is a factorization of G, where none of the factors is a subgroup. This contradicts [7,Theorem 1].




Therefore at least one of the factors, say A1 is a subgroup. Let H1 ¼ A1. Consider-ing the factor group G=H1 we see that

G=H1 ¼ ½ðA2H1Þ=H1� . . . ½ðAnH1Þ=H1� ð7Þ

is a normalized factorization of G=H1. By Lemma 5, each ðAiH1Þ=H1 is a uniquelycomplemented subset of G=H1. Again one of the factors in (7), say ðA2H1Þ=H1, isa subgroup of G=H1. Thus H2 ¼ A1A2 is a subgroup of G. Continuing in this wayfinally we may assume that

A1 HA1A2 H � � �HA1A2 � � �An

is an ascending chain of subgroups. r

The next result extends [5, Theorem 5] which itself is a generalization of Hajos’stheorem.

Theorem 2. Let G be a finite abelian group whose 2-component is cyclic and let

A1; . . . ;An be uniquely complemented subsets of G. Let G ¼ BA1 . . .An be a normal-

ized factorization of G, where jBj is a product of two distinct primes. Then either B is

periodic or at least one of the factors A1; . . . ;An is a subgroup of G.

Proof. Assume on the contrary that B is not periodic and none of the factorsA1; . . . ;An is a subgroup of G. By Lemma 10, there are non-subgroup cyclic or simu-lated subsets Ci;1; . . . ;Ci;mðiÞ such that in the factorization G ¼ BA1 . . .An the factorAi can be replaced by the product Ci;1 . . .Ci;mðiÞ. Therefore

G ¼ BYn

i¼1

YmðiÞ

j¼1

Ci; j

is a normalized factorization of G. Here B is not periodic and none of the factors Ci; j

is a subgroup of G. This contradicts [5, Theorem 5]. r

6 A localization result

We start with two lemmas.

Lemma 12. Let G be a finite abelian group and let G ¼ AB be a normalized factoriza-

tion. If A is a uniquely complemented subset and A is not a subgroup, then B is periodic.

Proof. Choose an element b A B. Multiplying the factorization G ¼ AB by b�1

we get the factorization G ¼ AðBb�1Þ. Set H ¼ hAi. Restricting the factorizationG ¼ AðBb�1Þ to H gives the factorization H ¼ AðBb�1 VHÞ. By Lemma 2, Bb�1 VH

is a subgroup of H. This subgroup does not depend on the choice of b. Let




K ¼ Bb�1 VH. If K ¼ feg, then H ¼ A and so A is a subgroup. This is not the case.Thus K0 feg. Now

KH 7b AB

Bb�1:

By [7, Lemma 6], each k A Knfeg is a period of B. r

The next lemma is a generalization of [5, Theorem 1].

Lemma 13. Let G be a finite abelian group and let A1; . . . ;An be uniquely comple-

mented non-subgroup subsets of G. Assume that the product A1 . . .An is direct and

that jGj ¼ pjA1j . . . jAnj, where p is a prime. If there is a normalized subset B of G

such that G ¼ BA1 . . .An is a factorization, then B is uniquely determined.

Proof. Let G ¼ BA1 . . .An be a normalized factorization of G. We would like to showthat B is uniquely determined. We proceed by induction on n.

Suppose first that n ¼ 1. Let H ¼ hA1i. The cosets modulo H form a partition ofG. Thus there are elements b1; . . . ; bs A B such that b1 ¼ e and the sets

BVHb1; . . . ;BVHbs

form a partition of B. Multiplying the factorization G ¼ A1B by b�1i we get

the factorization G ¼ A1ðBb�1i Þ. Restricting this to H gives the factorization

H ¼ A1ðBb�1i VHÞ. By Lemma 2, Bb�1

i VH is a subgroup of H and this subgroupdoes not depend on the choices of bi. Let K ¼ Bb�1

i VH. If K ¼ feg, then A1 ¼ H,that is, A1 is a subgroup of G. This contradicts to the hypotheses of the lemma. ThusK0 feg. In particular jK jd 2. Now Kbi ¼ BVHbi and consequently

G ¼ A1B ¼ A1ðKb1 U � � �UKbsÞ

is a factorization of G. Considering cardinalities yields that jGj ¼ jA1j jK js. Compar-ing this with jGj ¼ jA1jp one can see that jK js ¼ p. It follows that jK j ¼ p and s ¼ 1.Thus B ¼ BVHb1 ¼ K and so B is uniquely determined.

Now suppose that nd 2. In order to prove that B is uniquely determined assumeon the contrary that there is a normalized factorization G ¼ B 0A1 . . .An such thatB0B 0. There are elements b A B, b 0 A B 0 such that b B B 0, b 0 B B.

Set C ¼ fe; b; b2; . . . ; bp�1g. In the factorization G ¼ BA1 . . .An, by [8, Lemma 3],the factor B can be replaced by C to give the factorization G ¼ CA1 . . .An. ByLemma 6, C is a uniquely complemented subset. By Theorem 1, at least one of thefactors C;A1; . . . ;An is a subgroup of G. However only C can be a subgroup of G.Again by Theorem 1, there is a permutation B1; . . . ;Bn of the factors A1; . . . ;An suchthat

CHCB1 HCB1B2 H � � �HCB1 . . .Bn




is a chain of subgroups of G. We may assume that B1 ¼ A1; . . . ;Bn ¼ An since this isonly a matter of indexing the factors in the factorization G ¼ CA1 . . .An. Thus wemay assume that

CHCA1 HCA1A2 H � � �HCA1 . . .An

is a chain of subgroups of G. Set

Li ¼ CA1 . . .Ai; Hi ¼ hAii;

for each i with 1c ic n. Note that Ai HLi implies Hi HLi.Set C 0 ¼ fe; b 0; ðb 0Þ2; . . . ; ðb 0Þp�1g. Repeating the above argument with C 0 in place

of C gives that in the factorization G ¼ C 0A1 . . .An, the factor C 0 must be a subgroupand there is an Ai for which C 0Ai is a subgroup of G. Set L ¼ C 0Ai.

Let us settle first the case when i ¼ 1. Now L ¼ C 0A1 and L1 ¼ CA1. Re-stricting the factorization L1 ¼ CA1 to H1 gives the factorization H1 ¼ ðC VH1ÞA1.If C VH1 ¼ feg, then A1 ¼ H1, that is, A1 is a subgroup. This is a contradiction.Thus C VH1 0 feg which implies CHH1. Now C VH1 ¼ C and so

H1 ¼ ðC VH1ÞA1 ¼ CA1

is a factorization of H1. A similar argument shows that H1 ¼ C 0A1 is a factorizationof H1. The normalized factorizations

H1 ¼ C 0A1; H1 ¼ CA1

and the special case of the lemma when n ¼ 1 lead to the contradiction that C ¼ C 0.For the rest of the proof we assume that id 2. From L ¼ C 0Ai it follows

that Ai HL. Restricting the factorization Li ¼ Li�1Ai to L gives the factorizationL ¼ ðLi�1 VLÞAi. The factorizations

L ¼ C 0Ai; L ¼ ðLi�1 VLÞAi;

imply Li�1 VL ¼ C 0 by the case of the lemma when n ¼ 1. In particular C 0 HLi�1.Note that C 0;A1; . . . ;Ai�1 HLi�1 and since G ¼ C 0A1 . . .An is a factorization of

G, the product C 0A1 . . .Ai�1 is direct. Therefore Li�1 ¼ C 0A1 . . .Ai�1 is a factoriza-tion of Li�1. The factorizations

Li�1 ¼ C 0A1 . . .Ai�1; Li�1 ¼ CA1 . . .Ai�1;

and the inductive assumption yield the contradiction that C ¼ C 0. This completes theproof. r

Theorem 3. Let G be a finite abelian group and let A1; . . . ;An be uniquely comple-

mented subsets of G. Let G ¼ A1 . . .An be a normalized factorization of G and let




kð1Þ; . . . ; kðsÞ A f1; . . . ; ng. If the product Akð1Þ . . .AkðsÞ is periodic, then at least one of

the factors Akð1Þ; . . . ;AkðsÞ is a subgroup of G.

Proof. By Theorem 1, one of the factors in the factorization G ¼ A1 . . .An is a sub-group of G. If there is a subgroup among Akð1Þ; . . . ;AkðsÞ, then we are done. So weassume that none of the factors Akð1Þ; . . . ;AkðsÞ is a subgroup of G. We may renamethe factors A1; . . . ;An such that A1; . . . ;Am are subgroups of G and Amþ1; . . . ;An arenot subgroups of G. We may assume that md 1 and mþ 1c kð1Þ < � � � < kðsÞc n.Set

H ¼ A1 . . .Am; B ¼ Akð1Þ . . .AkðsÞ; C ¼ Amþ1 . . .An:

Clearly H is a subgroup of G. Choose a period g of B such that jgj is a prime. SetK ¼ hgi. There is a normalized subset U of B such that the product KU is directand is equal to B. The sets Akð1Þ; . . . ;AkðsÞ occur among the sets Amþ1; . . . ;An. There-fore there is a normalized subset V of C such that the product BV is direct and isequal to C. From the equation

C ¼ BV ¼ ðKUÞV ¼ KðUVÞ

one can read o¤ that g is a period of C.There is a normalized subset D of C such that the product DK is direct and is equal

to C. From the factorization G ¼ HC ¼ HDK one can conclude that the productHK is direct. In particular H VK ¼ feg. It follows that g B H. Choose an elementh A Hnfeg and set

A ¼ ðHnfhgÞU fhgg:

If jHjd 3, then since g B H, it follows that A is a non-subgroup simulated subsetof G. The computation

CA ¼ C½ðHnfhgÞU fhgg� ¼ ðCHnChÞUChg

¼ ðCHnChÞUCh ¼ CH ¼ G

shows that G ¼ AC ¼ AAmþ1 . . .An is a factorization of G. In the factorizationG ¼ AAmþ1 . . .An each factor is a uniquely complemented subset and none of themis a subgroup. This contradicts Theorem 1. This settles the case when jHjd 3.

For the rest of the proof we may assume that jHj ¼ 2. Now A ¼ fe; ghg is a cyclicsubset. If jgjd 3, then A is not a subgroup and we can finish the proof as before. Weare left with the jgj ¼ 2 case. Now from the factorizations

G ¼ AAmþ1 . . .An; G ¼ HAmþ1 . . .An;

by Lemma 13, it follows that H ¼ A and so h ¼ hg. This leads to g ¼ e, a contradic-tion. r




7 Characters

In order to formulate the result of this section we need to introduce some termi-nology. A normalized subset A of a finite abelian group G is called a lacunary cyclicsubset of G if A is in the form

A ¼ 6k

i¼1

gife; ai; a2i ; . . . ; a

ri�1i g;

where k is less than the least prime divisor of jGj and g1; . . . ; gk are given distinctelements of G such that g1 ¼ e and

jAj ¼Xki¼1

ri ¼ r:

If hg1; . . . ; gkiH hai then A is called a lacunary cyclic subset in the restrictedsense.

The normalized subset A of the finite abelian group G is called a T type subset ifthe following conditions hold:

(i) there are subgroups H, K of G such that K ¼ AH, where the product AH isdirect;

(ii) wðAÞ ¼ 0 implies wðhÞ ¼ 1 for each character w of G and each h A H.

The theorems below are the main results of this section.

Theorem 4. Let G ¼ A1 . . .An be a normalized factorization of the finite abelian

group G, where each Ai is either a lacunary cyclic subset in the restricted sense or a

T type subset. Then there is a permutation B1; . . . ;Bn of the factors A1; . . . ;An such

that

B1 HB1B2 H � � �HB1B2 . . .Bn

is an ascending chain of subgroups of G.

Theorem 5. Let G ¼ A1 . . .An be a normalized factorization of the finite abelian group

G, where each Ai is either a lacunary cyclic subset in the restricted sense or a T type

subset and let fkð1Þ; . . . ; kðsÞgH f1; . . . ; ng. If the product Akð1Þ . . .AkðsÞ is periodic,then at least one of the factors Akð1Þ; . . . ;AkðsÞ is a subgroup of G.

We will show that a lacunary cyclic subset is a uniquely complemented subset.Similarly, we will show that a T type subset is a uniquely complemented subset. Afterthis, Theorem 4 and Theorem 5 become corollaries to Theorem 1 and Theorem 3respectively.




Lemma 14. Let G ¼ AB be a normalized factorization of the finite abelian group G,where

A ¼ 6k

i¼1

gife; ai; a2i ; . . . ; a

ri�1i g

is a lacunary cyclic subset in the restricted sense. Then A is a uniquely complemented

subset of G.

Proof. First we prove that in the factorization G ¼ AB the lacunary cyclic factor A

can be replaced by the cyclic subset

A� ¼ fe; a; a2; . . . ; ar�1g

to give the factorization G ¼ A�B. (Here we do not assume that A is a lacunary cyclicsubset in the restricted sense.) We will use Redei’s character test of replacement toverify this claim. We have to establish that wðAÞ ¼ 0 implies wðarÞ ¼ 1 for each char-acter w of G. One may use the ideas of the proof of [9, Theorem 5.2.3]. So it remainsto show that wðAÞ ¼ 0 implies wðaÞ0 1 for each character w of G. Assume on the con-trary that there is a character w of G with wðAÞ ¼ 0 and wðaÞ ¼ 1. By [9, Lemma3.2.5], wðAÞ ¼ 0 implies wðAlÞ ¼ 0 for each character w of G and for each integer l

that is relatively prime to jGj. Here Al stands for the set fal : a A Ag. So for the char-acter w in the indirect assumption we see that wðAlÞ ¼ 0 for each l, 1c lc k. (Werecall that k is less than the least prime divisor of jGj.)

Note that

0 ¼ wðAlÞ ¼Xb AA

wðblÞ ¼Xk

i¼1

riwðgli Þ ¼

Xk

i¼1

ri½wðgiÞ� l :

Let t1; . . . ; ts be all the distinct values among wðg1Þ; . . . ; wðgkÞ. As g1 ¼ e, we maychoose t1; . . . ; ts such that t1 ¼ 1. We obtain the system of homogeneous linear equa-tions

Xs

i¼1

tli xi ¼ 0; 1c lc s; ð8Þ

where the each xi is a positive integer. The matrix of the system (8) is the Vander-monde matrix

1 1 � � � 1

1 t2 � � � ts

..

. ... . .

. ...

1 ts�12 � � � ts�1

s

266664

377775




whose determinant is not zero as t2; . . . ; ts are distinct. It follows that the system (8)has only the trivial solution x1 ¼ � � � ¼ xs ¼ 0 in the field of complex numbers. Thiscontradiction proves our claim that the factor A can be replaced by A�.

Next we prove that A is a uniquely complemented subset of G. From the nor-malized factorization G ¼ AB we get that G ¼ A�B is also a normalized factoriza-tion of G. Since A is a lacunary cyclic subset in the restricted sense it follows thathAi ¼ hA�i ¼ hai. Restricting the factorizations G ¼ AB, G ¼ A�B to hai we getthe factorizations

hai ¼ AðBVhaiÞ; hai ¼ A�ðBVhaiÞ:

By Lemma 6, BVhai is the unique complement of A� in hai. We conclude thatBVhai is the unique complement of A in hai. r

Lemma 15. Let G ¼ AB be a normalized factorization of the finite abelian group G,where A is a T type subset. Then A is a uniquely complemented subset of G.

Proof. Let H, K be subgroups of G such that K ¼ AH, where the product AH isdirect and wðAÞ ¼ 0 implies wðhÞ ¼ 1 for each character w of G and for each h A H.Set L ¼ hAi. As AHK, it follows that LHK . Let M ¼ H VL.

Restricting the factorization K ¼ AH to L we get the factorization

L ¼ K VL ¼ AðH VLÞ ¼ AM:

Suppose that there is a normalized subset D of L such that L ¼ AD is also a factori-zation. Let w be a character of H for which wðMÞ ¼ 0. Plainly, w is not the principalcharacter on H. Similarly, w is not principal on L.

Let us examine wðAÞ. If wðAÞ ¼ 0, then by the definition of A, wðhÞ ¼ 1 for eachh A H. In other words w is the principal character on H. This is not the case, sowðAÞ0 0 must hold. Applying w to the factorization L ¼ AD gives 0 ¼ wðAÞwðDÞ. Itfollows that wðDÞ ¼ 0. By [9, Lemma 3.2.6], there is a normalized subset C of L suchthat the product CM is direct and is equal to D. From the factorizations

L ¼ AD ¼ AðCMÞ; L ¼ AM

by considering cardinalities we deduce that

jLj ¼ jAj jCj jMj; jLj ¼ jAj jMj:

Dividing the first equation by the second gives that jCj ¼ 1. Therefore C ¼ feg andso D ¼ M, as required. r

References

[1] K. Corradi and S. Szabo. A generalized form of Hajos’ theorem. Comm. Algebra 21 (1993),4119–4125.




[2] K. Corradi and S. Szabo. Direct products of subsets in a finite abelian group. Preprint.[3] G. Hajos. Uber einfache und mehrfache Bedeckung des n-dimensionalen Raumes mit

einem Wurfelgitter. Math. Z. 47 (1942), 427–467.[4] L. Redei. Die neue Theorie der endlichen Abelschen Gruppen und Verallgemeinerung des

Hauptsatzes von Hajos. Acta Math. Acad. Sci. Hungar. 16 (1965), 329–373.[5] A. D. Sands. A generalization of Hajos’ theorem. Int. Electron. J. Algebra 3 (2008), 83–95.[6] A. D. Sands. A note on distorted cyclic subsets. Math. Pannon. 20 (2009), 123–127.[7] S. Szabo. An elementary proof of Hajos’ theorem through a generalization. Math. Japon.

40 (1994), 99–107.[8] S. Szabo. Factoring an infinite abelian group by subsets. Period. Math. Hungar. 40 (2000),

135–140.[9] S. Szabo. Topics in factorization of abelian groups (Birkhauser Verlag, 2004).

Received 13 January, 2010; revised 22 April, 2010

Keresztely Corradi, Department of General Computer Technics, Eotvos L. University,Pazmany P. setany 1/c, 1117 Budapest, Hungary

Sandor Szabo, Institute of Mathematics and Informatics, University of Pecs, Ifjusag u. 6, 7624Pecs, HungaryE-mail: [email protected]




factoring abelian groups into uniquely complemented subsets

Documents