Abelian groups that cannot be factored without periodic factor

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Abelian Groups that cannot be Factored Without Periodic Factor.

SAÂNDORSZABOÂ(*)

ABSTRACT- The list of the finite abelian groups that cannot be factored into two of its subsets without one factor being periodic is the same as the list of the finite abelian groups that cannot be factored into any number of its subsets without one factor being periodic.

1. Introduction.

Throughout this paper we will use multiplicative notation for abelian groups. LetGbe a finite abelian group. We denote the identity element of Gbye. LetB;A1;. . .;Anbe subsets ofG. We define the productA1 An

to be

fa₁ a_n:a₁2A₁;. . .;a_n2A_ng:

SupposeBA1 An. We say that the productA1 Anisdirectif eachb inBis uniquely expressible in the form

ba1 an; a12A1;. . .;an2An:

IfBis a direct product ofA₁;. . .;A_n, then the equationBA₁ A_n is said to be afactorizationofB.

A subsetAof a finite abelian groupGwill be callednormalizedife2A.

The factorizationGA1 An is called normalized if each factor is normalized. We say thatAisperiodicif there is an elementa2Gsuch that a6eandaAA. The elementais called aperiodofA. Note that ifaand bare periods ofA, thenabis a period ofAunlessabe. It follows that the

(*) Indirizzo dell'A.: Institute of Mathematics and Informatics,University of PeÂcs, IfjuÂsaÂg u. 6 - 7624 PeÂcs, Hungary.

E-mail: sszabo7@hotmail.com

Mathematics Subject Classification (1991); Primary 20K01; Secondary 52C22

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periods ofAaugmented with the identity element form a subgroupHofG.

Moreover there is a subsetBofGsuch thatABHis a factorization ofA.

If the groupGis a direct product of cyclic groups of orders t₁;. . .;ts re- spectively, then we express this fact shortly saying that G i s of type

(t₁;. . .;t_s). If from the factorizationGA₁ A_nit follows that one of the

factors is periodic for any possible choice of the factorsA1;. . .;An, then we say that G has the Hn-property. (Hn is an abbreviation for the HajoÂs property withnfactors.) Here we do not allow factors with only one element and do not consider groups with only one element.

In 1949, G. HajoÂs [4] called for determining all finite abelian groups with theH₂-property. The complete list of these groups first appeared in 1962 A. D. Sands [8].

(p^a;q);

(p³;2;2);

(p;q;2;2);

(p²;q²);

(p²;2;2;2);

(p;3;3);

(2²;2²);

(p²;q;r);

(p;2²;2);

(3²;3);

(p;p):

(p;q;r;s) (p;2;2;2;2);

(2^a;2);

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Here p;q;r;s are distinct primes the p2 and p3 cases are not ex- cluded anda1 is an integer. Groups whose type is on list (1) and their subgroups have the H₂-property and other groups do not have the H₂- property.

This result has applications in various fields. In geometry [11], combinatorics [5], coding theory [3], number theory [13], Fourier analysis [6].

If a group has theH_n-property for each possible choice ofn, then we say thatGhas theH-property. It is plain that groups with theH-property have theH2-property. We will show that groups with theH2-property are in fact the same as groups with theH-property. We express this fact more formally as a theorem.

THEOREM1. Let G be a finite abelian group. If G has the H₂-property, then G has the H-property.

2. Replacing factors.

We say that in the factorizationGABthe factorBcan bereplacedby B⁰ if GAB⁰ is also a factorization of G. If c is an element of G, then multiplying both sides of the factorization by c we get the factorization GGcA(Bc). In other wordsBcan be replaced byBcfor eachc2G.

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Note thatBis periodic if and only ifBcis periodic. LetGA1 Anbe a factorization. Let a12A1;. . .;an2An multiplying the factorization by a₁¹ a_n¹ we get the normalized factorization G(a₁¹A₁) (a_n¹An).

Thus when we study factorizations with periodic factors we may restrict our attention to normalized factorizations.

Note that GAB is a factorization if and only if jGj jAkBj and AA ¹\BB ¹ feg. From this it is plain thatBcan be replaced byB ¹. Let GAB be a factorization of G. Then each g2G is uniquely expressible in the formgab,a2A,b2B. We callatheA-coordinateofg and we denote it bya(g). Similarly, we callbtheB-coordinateofgand we denote it by b(g). The coordinates of g make sense only relative to the factorizationGAB. IfAis a subset of a finite abelian groupGandqi s an integer, thenA^q will denote the setfa^q:a2Ag.

LEMMA1. Let GAB be a factorization of G and let A fa₁;. . .;a_ng.

For each g2G the elements a(ga₁);. . .;a(ga_n) form a permutation of a1;. . .;an.

PROOF. Clearly, a(ga_i)2A. So we will show that a(ga_i)a(ga_j) i m- pliesa_ia_j. From the equations

ga_ia(ga_i)b(ga_i); ga_ja(ga_j)b(ga_j) we get the equations

ga(ga_i)b(ga_i)a_i¹; ga(ga_j)b(ga_j)a_j¹:

Then b(ga_i)a_i¹b(ga_j)a_j ¹ and a_jb(ga_i)a_ib(ga_j). Now as a_i;a_j2A, b(ga_i),b(ga_j)2B, from the factorizationGABit follows thata_ia_j.

This completes the proof.

LEMMA2. Let GAB be a factorization of G and let q be a prime such that q6 j jAj. Then GA^qB is a factorization of G.

PROOF. Choose ana2A,g2Gand defineTto be the set of allqtuples (x1;x2;. . .;xq); x1;x2;. . .;xq2A

for which a(gx1x2 xq)a. First note thatjTj jAj^q ¹. Indeed, choose x1;x2,. . .;xq 12Aarbitrarily, then by Lemma 1,a[(gx1x2 xq 1)xq]a has a unique solution for x_q. Next note that i f (x₁;x₂;. . .;x_q)2T, then

(x₂;. . .;x_q;x₁)2T. We defi ne a graphG. The verti ces ofGare the elements

of T and we draw an arrow from the node (x₁;x₂;. . .;x_q) to the node

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(x2;. . .;xq;x1). The graphG is a union of disjoint cycles. The cycles are of length 1 or of lengthq. Whenx1x2 xq, then the node (x1;x2;. . .;xq) is on a cycle of length 1. Whenx₁;x₂;. . .;xqare not all equal, then the node (x₁;x₂;. . .;x_q) is on a cycle of lengthq. Asq6 j jAjthere must be a cycle of length 1 inG. In other words there is anx₁2Asuch thata(gx^q₁)a.

Consider the factorization GAB. From gx^q₁a(gx^q₁)b(gx^q₁) using a(gx^q₁)a we getgx^q₁ab(gx^q₁), thengax₁^qb(gx^q₁). Hereax₁^q2aA ^q, b(gx^q₁)2B and the equation holds for each g2G. It follows that G(aA ^q)B. Then GA ^qB. Note that jGj jAkBj, jA ^qj jAj imply jGj jA ^qkBjand soGA ^qBis a factorization. NowA ^qcan be replaced byA^qand so it follows thatGA^qBis a factorization.

LEMMA3. Let GAB be a factorization such that e2A,jAj p is a prime. Then GA⁰B is a factorization of G, where A⁰ fe;a;a²;. . .;a^p ¹g, a2An feg.

PROOF. Note thatGA^tBis a factorization ofGwhenevert2,p6t.

Indeed, astis a product of primes we can apply Lemma 2 several times starting with the factorizationGAB. LetA fe;a₁;a₂;. . .;a_p ₁g. The fact thatGA^tBis a factorization is equivalent to that the sets

eB;a^t₁B;a^t₂B;. . .;a^t_p ₁B

form a partition ofG. SetA⁰ fe;a_k;a²_k;. . .;a^p_k ¹g. The fact thatGA⁰Bis a factorization is equivalent to that the sets

eB;a_kB;a²_kB;. . .;a^p_k ¹B

form a partition of G. Si nce G is finite it is enough to show that aⁱ_kB\a^j_kB ; for each i;j, 0i<jp 1. Assume the contrary that aⁱ_kB\a^j_kB6 ;. Multiplying by a_kⁱ we get eB\a^{j i}_k B6 ;. Set tj i.

Clearly, 1tp 1 and sotis prime top. NoweB\a^t_kB6 ;contradicts the fact thatGA^tBis a factorization ofG.

3. Periodicity forcing factorization types.

IfGA1 Anis a factorization ofGandjA1j q1;. . .;jAnj qn, then we call then tuple (q1;. . .;qn) the typeof the factorization. If from each GA₁ Anfactorization of type (q₁;. . .;qn) it follows that at least one of

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the factors is always periodic we call (q1;. . .;qn) a periodicity forcing factorization type for G. In 1965 L. ReÂdei [7] proved that a factorization type (q₁;. . .;qn) is always periodicity forcing ifq₁;. . .;qnare primes.

LEMMA4. Let q1p1 ps, where p1;. . .;psare primes. If(q1;. . .;qn) is a periodicity forcing factorization type for an abelian group, then so is (p1;. . .;ps;q2;. . .;qn).

PROOF. Suppose that (q1;. . .;qn) is a periodicity forcing factorization type for the finite abelian groupGand consider a normalized factorization

GB1 BsA2 An; (2)

where

jB1j p1;. . .;jBsj ps;jA2j q2;. . .;jAnj qn:

We would like to show that at least one of the factorsB₁;. . .;B_s,A₂;. . .;A_n is periodic. If one of the factors is periodic then there is nothing to prove so we assume that none of the factors is periodic.

If the order of each element inBiis a power ofpi, then we defineCito beB_i. Assume that there are elements inB_iwhose order is not a power of p_i. In factorization (2), by Lemma 2,B_ican be replaced by a normalized subsetC_isuch that the orders of each element inC_iis a product of at most two distinct primes andC_i is not a subgroup ofG. Note that asjC_ij i s a prime ande2Ci,Ciis periodic if and only ifCiis a subgroup ofG. Setting CC1 Cs we get a factorizationGCA2 An. The type of this factorization is (q1;q2;. . .;qn). So by our assumption one of the factors

C;A₂;. . .;A_nis periodic. If one ofA₂. . .;A_nis periodic, then we are done

so we assume thatCis periodic. Now a periodic subsetCofGis factored into subsets C₁;. . .;C_s. In addition, there is a subset A of G such that GCA is a factorization. Simply set AA2 An. The hypotheses of Theorem 2 of [2] are satisfied therefore this theorem gives that at least one of the factorsC₁;. . .;C_s must be periodic.

This contradiction completes the proof.

IfAis a subset andxis a character ofG, then we will use the notation x(A) to denote the sum X

a2A

x(a):

In this paper character ofGalways means irreducible character ofG. The set of all charactersxofGwithx(A)0 is called theannihilatorset ofA and will be denoted by Ann(A).

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LEMMA5. Let G be a finite abelian group with the H2-property and let q1p be a prime. In the p3case we assume that the p-component of G is cyclic. (In the p2 case nothing is assumed about the p-component of G.) Then(q₁;q₂;q₃)is a periodicity forcing factorization type for G.

PROOF. LetGA1A2A3be a normalized factorization ofGsuch that jA1j q1,jA2j q2,jA3j q3. We would like to show that one of the factors A₁,A₂,A₃is periodic. If one ofA₁,A₂,A₃is periodic, then there is nothing to prove. So we assume that none of the factors is periodic. Assume thatp3.

In this case thep-component ofGis cyclic. By Lemma 3,A₁can be replaced byA⁰₁ fe;a;a²;. . .;a^p ¹gfor eacha2A1n feg. IfA⁰₁is a subgroup ofG for eacha2A1n feg, then as thep-component ofGis cyclic,A1is equal to the unique subgroup ofGthat haspelements. ButA₁ is not a subgroup.

This contradiction gives that for somea2A₁n fegthe subsetA⁰₁ is not a subgroup. This is equivalent toa^p6e. For the remaining part of the proof we choose an A⁰₁ which is not a subgroup. In the p2 case we set A⁰₁A1 fe;ag. Herea²6easA1is not a subgroup. Thusa^p6eholds in both of the casesp2 orp3.

From the factorizationG(A⁰₁A₂)A₃, by theH₂-property ofG, it follows thatA⁰₁A₂orA₃is periodic. SinceA₃is not periodic,A⁰₁A₂is periodic, say with periodg, that is,A⁰₁A2gA⁰₁A2. We claim thatx(A2)0 holds for all charactersxofGwithx(g)61 andx(a^p)61. In order to prove the claim assume that x(g)61 and x(a^p)61. As x(g)61 from x(A⁰₁A₂)x(g)

x(A⁰₁A₂) we get 0x(A⁰₁A₂)x(A⁰₁)x(A₂). Fromx(a^p)61 it follows that x(A⁰₁)60. Thereforex(A₂)0. The fact thatx(a^p)61 andx(g)61 imply x(A2)0 is equivalent to

Ann ha^pi

\Ann hgi

Ann(A₂):

By Theorem 2 of [10], there are subsetsX,YofGsuch that A₂Xha^pi [Yhgi;

where the products are direct and the union is disjoint. Similarly, from the factorizationGA2(A⁰₁A3) it follows thatA⁰₁A3is periodic, say with period h. Then there are subsetsU,V ofGsuch that

A₃Uha^pi [Vhhi;

where the products are direct and the union is disjoint.

IfX ;, thenA2is periodic with periodg. IfU ;, thenA3is periodic with periodh. So we may assume that X6 ; andU6 ;. Choose x2X,

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u2U. Multiply the factorization GA1A2A3 by gx ¹u ¹ to get the factorization

GGgA₁(A₂x ¹)(A₃u ¹):

Now ha^pi A₂x ¹, ha^pi A₃u ¹ contradict the definition of the factorization.

4. Proof of Theorem 1.

We consider a finite abelian groupGwith theH₂-property. ThusGi s a subgroup of a group whose type is on list (1) and we try to establish thatG has theH_n-property for each possible choice ofn. Then1 case is trivial so we assume thatn2.

LetGbe a subgroup of a group of type (p^a;q) and letGA1 Anbe a factorization ofG. IfGi s ap-group, thenjA₁j;. . .;jA_njare powers ofp.

IfGi s not ap-group, thenqdivides one ofjA₁j;. . .;jA_njandqcan divide only one ofjA₁j;. . .;jA_nj. SayqjA₁jand plainly each ofjA₂j;. . .;jA_njmust be a power of p. Now the p-component of G i s cycli c and each of

jA2j;. . .;jAnji s a power ofp. The conditions of Theorem 2 of [9] are met.

By this theorem, one of the factorsA1;. . .;Ani s peri odi c and soGhas the H_n-property. Sincen2 was arbitraryGhas theH-property.

By Theorem 1 of [12] (or by Theorem 10 of [1]), a group of type (2^a;2) has theH-property.

We inspect the remaining groupsGin a case by case manner and sort out the arising cases using ReÂdei's theorem or Lemma 4 or Lemma 5.

(1) Suppose the type ofGis one of the following (p³;2;2); (p²;2;2;2); (p;2;2;2;2):

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Note thatjGjis a product of 5 primes. This is why we treat these cases together. Let GA1 An be a factorization of G. If the type of the factorization is (q₁;q₂;q₃;q₄;q₅), then eachq_iis a prime and ReÂdei's theorem gives that one of the factors is periodic. If the factorization type is (q1;q2;q3;q4), then only oneqican be composite, sayq4. TheH2-property ofGgives that (q1q2q3;q4) is a periodicity forcing factorization type forG.

Now by Lemma 4, (q₁;q₂;q₃;q₄) is a periodicity forcing factorization type forG. If the factorization type is (q₁;q₂;q₃), then at least one of theq_i's is a

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prime, sayq1i s a pri me, and Lemma 5 appli es. IfGis a proper subgroup of a group whose type i s on li st (3), then we can use a si mi lar but to some extent simpler argument.

(2) Let us assume that the type ofGis one of the next (p²;q²)

(p;4;2)

(p²;q;r) (p;q;2;2)

(p;q;r;s) (2²;2²):

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Note thatjGjis a product of 4 primes. LetGA₁ A_nbe a factorization of G. If the factorization type is (q₁;q₂;q₃;q₄), then ReÂdei's theorem implies that this factorization type is periodicity forcing. If the factorization type is (q1;q2;q3), then Lemma 4 is applicable. IfGis a proper subgroup of a group whose type is on list (4), then an analogous but slightly simpler reasoning can be used.

(3) The case whenGis of type (p;3;3), (3²;3), (p;p), that is, whenjGji s a product of at most 3 primes is rather trivial.

This inspection completes the proof.

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[12] S. SzaboÂ, Factoring a certain type of 2-groups by subsets, Rivista di Matematica,6(1997), pp. 25±29.

[13] R. TIJDEMAN,Decomposition of the integers as a direct sum of two subsets, Number Theory (Paris 1992-1993) London Math. Soc. Lecture Note Ser. 215 Cambridge Univ. Press, Cambridge 1995, pp. 261±276.

Manoscritto pervenuto in redazione il 17 aprile 2004 e in forma finale il 19 ottobre 2005

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