K. Corr´adi - S. Szab´o PERIODIC FACTORIZATION OF A FINITE ABELIAN 2-GROUP

K. Corr´adi - S. Szab´o

PERIODIC FACTORIZATION OF A FINITE ABELIAN 2-GROUP

Abstract.

Let G be a finite abelian 2-group that is a direct product of a cyclic group and an elementary group. Suppose that G is a direct product of its subsets A₁, . . . ,An of cardinality two or four. Then one of the subsets A₁, . . . ,Anis periodic. The subset A_iis periodic if A_ig= A_iholds with a nonidentity element g of G. This is a generalization of an earlier result of A. D. Sands and S. Szab´o.

1. Introduction

Throughout the paper G will be a finite abelian group. We use multiplicative notation. The identity element is denoted by e. The symbol “⊂” denotes a not necessarily strict inclusion,|a|

denotes the order of the element a∈ A,|A|denotes the cardinality of the subset A of G. If G is a direct product of its subsets A₁, . . . ,An, then we express this fact saying that the equality G= A₁· · ·Anis a factorization of G. If e∈A_i, then we say that the subset A_iis normed. We call the factorization G = A₁· · ·A_nnormed if each A_i is normed. A subset A of G is called periodic if there is a g∈G\{e}such that Ag=A. The element g is a period of A. If G is a direct product of cyclic groups of orders t₁, . . . ,t_s respectively, then we say G is of type(t₁, . . . ,t_s).

A. D. Sands and S. Szab´o [2] proved that if G is of type(2, . . . ,2)and G = A₁· · ·An is a factorization, where|A₁| = · · · = |A_n| = 4, then one of the factors A₁, . . . ,A_n is periodic.

We will prove the following generalization of this theorem. Let G be a finite abelian 2-group and let G = A₁· · ·A_n be a factorization of G, where each|A_i|is either 2 or 4. If G is of type(2^λ,2, . . . ,2), then one of the factors A₁, . . . ,A_nis periodic. We accomplish this using characters of G.

Ifχis a character and A is a subset of G, then we denote the sum

X

a∈A χ (a)

byχ (A). Ifχ (A)=0, thenχannihilates A. We denote by Ann(A)the set of characters of G that annihilates A.

If A and A⁰are subsets of G such that given any subset B of G, if G=A B is a factorization of G, then G= A⁰B is also a factorization of G, then we say that A is replaceable by A⁰. There is a character test for replaceability due to L. R´edei [1] which reads as follows. If|A| = |A⁰|and Ann(A)⊂Ann(A⁰), then A can be replaced by A⁰

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2. The result

Let G be a finite abelian group and let A= {e,a,b,c}be a subset of G. We define a subset A⁰ by A⁰= {e,a}{e,b}. Since the equation c=abd is solvable for d, A can be written in the form A= {e,a,b,abd}. We need the next lemma.

LEMMA1. If|a| =2, then (a) Ann(A)⊂Ann(A⁰),

(b) A is periodic if and only if d=e, (c)χ (A)=0 impliesχ (d)=1.

Proof. (a) Letχ be a character of G for which 0= χ (A) = 1+χ (a)+χ (b)+χ (c). As

|a| = 2, it follows thatχ (a) = −1 orχ (a) = 1. Ifχ (a) = 1, then χ (A) = 0 gives that χ (b)=χ (c)= −1. Using this we have

χ (A⁰)=1+χ (a)+χ (b)+χ (a)χ (b)=1+1−1−1=0.

Ifχ (a)= −1, thenχ (A)=0 gives thatχ (b)=ρandχ (c)= −ρ, whereρis a root of unity.

Using this we have

χ (A⁰)=1+χ (a)+χ (b)+χ (a)χ (b)=1−1+ρ−ρ=0.

(b) If d=e, then A= A⁰and so A is periodic with period a. Conversely, assume that A is periodic with period g. Note that g²is also a period of A if g²6=e. Using this observation we may assume that|g| =2. From e∈A it follows that g∈ A.

If g=a, then

Aa= {a,e,ab,bd} = {e,a,b,abd} =A

gives that{ab,bd} = {b,abd}. Here either ab = b or bd = b. The first one leads to the contradiction a=e. The second one gives d =e.

If g=b, then

Ab= {b,ab,e,ad} = {e,a,b,abd} = A

gives that{ab,ad} = {a,abd}. Hence either ab = a or ad = a. The first one leads to the contradiction b=e. the second one gives d=e.

If g=abd, then

Aabd= {abd,bd,ab²d,e} = {e,a,b,abd} = A

gives that{ab²d,bd} = {a,b}. Now either ab²d= b or bd = b. The first equality gives the contradiction abd=e, the second one provides d=e.

(c) Ifχ (A)=0, then by part (a),χ (A⁰)=0 and so

0=χ (A)−χ (A⁰)=χ (ab)χ (d)−χ (ab)=χ (ab)

χ (d)−1 . This completes the proof.

After this preparation we are ready to prove the main result of the paper.

THEOREM1. Let G be a finite group of type(2^λ,2, . . . ,2). If G=A₁· · ·Anis a normed factorization of G, where|A_i|is either 2 or 4 for each i , 1 ≤i ≤n, then at least one of the factors A₁, . . . ,Anis periodic.

Proof. The|G| =2 case is trivial. So we assume that|G| ≥4 and proceed by induction on|G|.

Clearly G is a direct product of its subgroups H and K of types(2^λ)and(2, . . . ,2)respectively.

Ifλ=1, then G is of type(2, . . . ,2). This special case is covered by [2] Theorem 9. So for the remaining part of the proof we may assume thatλ ≥2. Let H = hxiand K = hy₁, . . . ,ysi, where|x| =2^λand|y₁| = · · · = |y_s| =2. Consider a characterχof G that is faithful on H or equivalently for whichχ (x)=ρ, whereρis a primitive(2^λ)th root of unity.

Let A_i = {e,a_i}be a factor of order 2. If 0=χ (A_i)= 1+χ (a_i), thenχ (a_i)= −1 or χ (a²_i)=1 and so a²_i =e. Therefore A_iis periodic with period a_i. So in the remaining part of the proof we may assume thatχ (A_i)6=0 whenχis faithful on H and|A_i| = 2. Asχis not the principal character of G, it follows that 0=χ (G)=χ (A₁)· · ·χ (An)and soχ (A_i)=0 for some i , 1≤i≤n. Thus we may assume that|A_i| =4 for some i .

Let A_i = {e,a_i,b_i,c_i}be a factor of order 4. If

0=χ (A_i)=1+χ (a_i)+χ (b_i)+χ (c_i),

then one ofχ (a_i),χ (b_i), χ (c_i)must be −1 and so one of a_i², b²_i, c²_i must be e. Thus there is at least one factor of order 4 that contains at least one second order element. We choose the notation such that A₁, . . . ,Amare all the factors of order 4 containing at least one second order element. If m=1, thenχ (A₁)=0 for eachχthat is faithful on H . Now, by [2] Theorem 1, A₁ is periodic. So we may assume that m≥2.

Let us consider an A_i = {e,a_i,b_i,c_i}with 1≤i≤m. We choose the notation such that

|a_i| =2. Further c_ican be written in the form c_i =a_ib_id_iwith a suitable d_i ∈G. By Lemma 1 A_i is periodic if and only if d_i = e. Thus we may assume that d_i 6=e. Also by Lemma 1 χ (A_i)=0 impliesχ (d_i)=1. From this it follows that A_ican be replaced by

{e,a_i,b_i,a_ib_id_i^k}

for each integer k. In particular, we may assume that|d_i| =2 for each i , 1≤i ≤m. Also A_i can be replaced by

{e,a_i,b_i,a_ib_i} = {e,a_i}{e,b_i}.

If b²_i = e, then each element of A_i\ {e}is of order 2. We will say that A_i is a type 1 factor.

Now A_i can be replaced by H_iB_i, where H_i = ha_i,b_iiand B_i = {e}. If b_i²6=e, then a_iis the only second order element in A_i. We will say that A_iis a type 2 factor. In this case A_i can be replaced by H_iB_i, where H_i = {e,a_i} = ha_iiand B_i= {e,b_i}.

The subgroup H has a unique subgroup L = hx^2λ−1iof order 2. From the factorization G= H₁B₁· · ·H_mB_mA_m+1· · ·A_n

it follows that the product H₁· · ·H_mis direct. So there can be only one subgroup H_ifor which L ⊂H_i. Such an H_idoes not necessarily exists. But if it does, then we choose the notation such that L⊂ H₁. We claim that L6⊂H₁may be assumed.

In order to prove this claim let us consider A₁= {e,a₁,b₁,c₁}and distinguish two cases depending on whether A₁is of type 1 or type 2.

If A₁is of type 1, then it can be written in the forms

A₁= {e,a₁,b₁,a₁b₁d₁}, A₁= {e,b₁,c₁,b₁c₁d₁⁰}, A₁= {e,a₁,c₁,a₁c₁d₁⁰⁰} and can be replaced by the subgroups

H₁= ha₁,b₁i, H₁⁰= hb₁,c₁i, H₁⁰⁰= ha₁,c₁i

respectively. If L ⊂ H₁, then one of a₁, b₁, a₁b₁is equal to x^2λ−1. In the a₁ = x^2λ−1 case a₁6∈H₁⁰since obviously a₁6=e, a₁6=b₁, a₁6=c₁and a₁=b₁c₁combined with c₁=a₁b₁d₁ leads to the d₁ = e contradiction. In the b₁ = x^2λ−1 case b₁ 6∈ H₁⁰⁰ since clearly b₁ 6= e, b₁ 6=a₁, b₁6=c₁and b₁ =a₁c₁leads to the d₁=e contradiction. In the a₁b₁= x^2λ−1 case a₁b₁6∈H₁⁰since a₁b₁=e, a₁b₁=b₁, a₁b₁=c₁, a₁b₁=b₁c₁leads in order to the a₁=b₁, a₁=e, d₁=e, a₁=c₁contradictions.

If A₁ is of type 2, then a₁² = e, b²₁ 6= e and A₁ can be written in the form A₁ = {e,a₁,b₁,a₁b₁d₁}and can be replaced by H₁B₁, where H₁= {e,a₁}, B₁= {e,b₁}. If L⊂H₁, then a₁=x^2λ−1. Now replace A₁by

A⁰₁=b⁻¹₁ A₁= {b₁⁻¹,b⁻¹₁ a₁,e,a₁d₁} = {e,a₁⁰,b⁰₁,a₁⁰b⁰₁d₁⁰},

where a₁⁰ =a₁d₁, b⁰₁ =b⁻¹₁ a₁, d₁⁰ =d₁. The only second order element in A⁰₁is a⁰₁=a₁d₁ which is not equal to x^2λ−1. Here A⁰₁is replaceable by H₁⁰B₁⁰, where

H₁⁰= ha₁⁰i = ha₁d₁i, B₁⁰ = {e,b⁰₁}.

Thus in each case we may assume that L6⊂H₁.

Replace A_iby H_iB_iin the factorization G= A₁· · ·A_nto get the factorization G=A₁· · ·A_i−1(H_iB_i)A_i+1· · ·A_n,

where 1≤i≤m. This leads to the factorization

G= A₁· · ·A_i−1B_iA_i+1· · ·A_n of the factor group G=G/H_i. Here

A_j = {H_i,a_jH_i,b_jH_i,c_jH_i} or A_j = {H_i,a_jH_i}, B_i= {H_i,b_iH_i} or B_i= {H_i}.

As|G|<|G|, by the inductive assumption it follows that either B_i or A_j is periodic for some j , 1≤ j≤n, j6=i .

If B_iis periodic, then|B_i| = |B_i|must be 2 and consequently A_imust be of type 2. Since B_i is periodic, it follows that(b_iH_i)²=b_i²H_i = H_iand so b²_i ∈ H_i = {e,a_i}. We know that b_i²6=e and hence b_i²=a_i. Let

b_i =x^βy₁^β1· · ·y_s^βs and a_i=x^αy^α₁¹· · ·y_s^αs, whereα=2^λ−1, 0≤β≤2^λ−1, 0≤α₁, β₁, . . . , α_s, β_s≤1. Now

b²_i =(x^βy₁^β1· · ·y_s^βs)²=x^2β =x^αy₁^α1· · ·y^α_s^s =a_i gives thatα₁= · · · =αs =0 and so L⊂H_iand this is a contradiction.

If A_jis periodic and|A_j| = |A_j| =2, then in a similar way a²_j ∈ H_i and a²_j 6=e lead to the contradiction L ⊂ H_i. Therefore if A_j is periodic, then|A_j| = |A_j| =4. The periodicity of A_j implies that A_j contains a second order element, say(a_jH_i)² = a²_jH_i = H_i. Hence a²_j ∈ H_i. As a²_j 6=e in the known way leads to the contradiction L⊂H_i, it follows that a²_j =e.

Thus A_j contains a second order element, that is 1≤ j≤m. By Lemma 1 the periodicity of A_j implies that d_j ∈H_i.

The summary of the above argument is that for each i , 1≤i≤m there is a j , 1≤ j ≤m such that d_j ∈ H_i and i6= j . We define a bipartite graph0whose nodes are H₁, . . . ,Hm and d₁, . . . ,dmand if d_j ∈H_i, then(H_i,d_j)is a directed edge of0. If(H_i,d_j),(H_k,d_j)are edges of0with i6=k, then d_j ∈H_i∩H_k= {e}which is a contradiction. Thus for each d_jthere is at most one H_isuch that(H_i,d_j)is an edge of0. Further, for each H_ithere is at least one d_j for which(H_i,d_j)is an edge of0. Therefore there is a ono-to-one map f from{H₁, . . . ,Hm}into {d₁, . . . ,d_m}such that H_i, f(H_i)

, 1≤i≤m are all the edges of0.

Let us consider

Am= {e,am,bm,cm}.

(Remember m≥2.) If Amis of type 1, then it can be written in the forms

A_m= {e,a_m,b_m,a_mb_md_m}, A_m = {e,a_m,c_m,a_mc_md_m⁰}, A_m = {e,b_m,c_m,b_mc_md_m⁰⁰} and it can be replaced by the subgroups

Hm = ham,bmi, H_m⁰ = ham,cmi, H_m⁰⁰ = hbm,cmi

respectively. These replacements give rise to the graphs0, 0⁰,0⁰⁰ and the maps f , f⁰, f⁰⁰ respectively. The nodes H₁, . . . ,H_m−1and d₁, . . . ,d_m−1are common in these graphs. After removing the edges joining to Hm,H_m⁰,H_m⁰⁰ and dm,d_m⁰ ,d_m⁰⁰ the remaining parts of the graphs are identical. From this it follows that f(H_m)= f⁰(H_m⁰)= f⁰⁰(H_m⁰⁰). Let d_j be this common value. This leads to the contradiction d_j ∈Hm∩H_m⁰ ∩H_m⁰⁰ = {e}.

If Amis of type 2, then a_m² =e, b²_m6=e and Amcan be written in the form Am= {e,am,bm,ambmdm}

and can be replaced by H_mB_m, where

Hm= {e,am}, Bm= {e,bm}.

The factor Amcan be replaced by

A⁰_m=b⁻¹_m Am= {b⁻¹_m ,b⁻¹_m am,e,amdm} = {e,a_m⁰ ,b_m⁰ ,a_m⁰ b⁰_md_m⁰ },

where a⁰_m=amdm, b_m⁰ =b⁻¹_m am, d_m⁰ =dm. Then A⁰_mcan be replaced by H_m⁰ B_m⁰, where H_m⁰ = {e,a⁰_m}, B⁰_m= {e,b⁰_m}.

The A_m → H_mB_m and A⁰_m → H_m⁰B_m⁰ replacements give rise to the graphs0, 0⁰ and the maps f , f⁰respectively. The nodes H₁, . . . ,H_m−1and d₁, . . . ,d_m−1,dmare common in these graphs. After removing the edges joining to Hm,H_m⁰ the remaining parts of the graphs are identical. From this it follows that f(H_m)= f⁰(H_m⁰). Let d_jbe this common value. This gives the contradiction d_j ∈Hm∩H_m⁰ = {e}.

This completes the proof.

References

[1] R ´EDEIL., Die neue Theorie der endlichen Abelschen Gruppen und Verallgemeinerung des Hauptsatzes von Haj´os, Acta Math. Acad. Sci. Hungar. 16 (1965), 329–373.

[2] SANDS A.D. AND SZABO´ S., Factorization of periodic subsets, Acta Math. Hung. 57 (1991), 159–167.

AMS Subject Classification: 20K01, 52C22.

Kereszt´ely CORR ´ADI

Department of Computer Sciences E¨otv¨os University Budapest 1088 Budapest, HUNGARY S´andor SZAB ´O

Department of Mathematics University of Bahrain P.O. Box: 32038, BAHRAIN

Lavoro pervenuto in redazione il 31.08.1998.