FACTORING ABELIAN GROUPS OF. Department of Mathematics, University of Bahrain, P. O. Box

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1 Mathematica Pannonica 7/2 (1996), 197 { 207 FACTORING ABELIAN GROUPS OF ORDER p 4 Sandor Szabo Department of Mathematics, University of Bahrain, P. O. Box Isa Town, State of Bahrain Khalid Amin Department of Mathematics, University of Bahrain, P. O. Box Isa Town, State of Bahrain Received: July 1995 MSC 1991 : 20K01,52C22 Keywords: Factorization of nite abelian groups, Hajos-Redei theory. Abstract: Let G be a nite abelian group of order p 4,where p is a prime. If G is a direct product of three of its subsets A, B, C, where B and C are nonsubgroup cyclic or simulated subsets, then A is a direct product of a subset and a nontrivial subgroup of G. 1. Introduction Let G be a nite abelian group. We will use multiplicative notation in connection with abelian groups. We denote the identity element by e. Let A 1 ::: A n be subsets of G. If the product A 1 :::A n is direct and gives G, then we say that the product A 1 :::A n is a factorization of G. We also call the equation G = A 1 :::A n a factorization of G. Theabove denition is clearly equivalent to the following. Each g in G is uniquely expressible in form g = a 1 :::a n a 1 2 A 1 ::: a n 2 A n : Sometimes another equivalent formulation is useful. The product A 1 ::: :::A n gives G and in addition jgj = ja 1 j :::ja n j holds. Here jaj de-

2 198 S. Szabo and K. Amin notes the cardinality of the subset A of G. We also use the notation jaj to denote the order of the element a of G. The subset A of G is dened to beperiodic if there is an element g 2 G nfeg such that ga = A. Such an element g is called a period of A. All the periods of A together with the identity element e form a subgroup H of G. In fact A is a union of cosets modulo H. Therefore A is a direct product HD, where D is a set of representatives of A modulo H. Clearly D is not necessarily dened uniquely. Beside periodic subsets two other types of subsets play a role in the paper. The subset A of G is called cyclic if it consists of the elements e a a 2 ::: a r;1.inorder to avoid trivial cases we will assume that r 2and jaj r. If jaj = r, then the cyclicsubset A is equal to the cyclic subgroup hai. If jaj > r, then A consists of the \rst" r elements of hai. Let a 2 G such that jaj = r. Thesubset A of G is called simulated if it consists of the elements e a a 2 ::: a r;2 a r;1 u. If u = e, then the simulated subset A is equal to the subgroup hai. If u 6= e, then A diers from the subgroup hai in one element a r;1 u. It is proved in [3] that ifaniteabelian group is a direct product of cyclic subsets, then at least one of the factors must be periodic. By[1] a similar result holds if the factors are simulated instead of being cyclic. Redei [4] proved that if a nite abelian group is a direct product of normed subsets of prime cardinality, then at least one of the factors must be a subgroup. These theorems suggest the following problem. Let G=AB 1 :::B n be a factorization of the niteabelian group G, where each B i is either cyclic or simulated and jaj is a product of two (not necessarily distinct) primes. Does it follow that at least one ofthe factors A B 1 ::: B n is periodic? The main result ofthis paper gives an answer in the armative in a particular case. Namely, let G be a group of order p 4,where p is a prime. If G = ABC is factorization of G, where B and C may be cyclic orsimulated, then at leastoneofthe factors A B C must be periodic. The emphasis is on that jaj is not a prime and nothing is assumed about the structure of A. 2. Preliminaries If A and A 0 are subsets of G such that for every subset B of G, if G = AB is a factorization of G, then G = A 0 B is also a factorization

3 Factoring abelian groups of order p G, then we shall say that A is replaceable by A 0. We will need the next three lemmas on replaceable factors. They can be proved using the ideas of the proofs of Lemma 1and 2 in [2]. Lemma 1. Let G be a nite abelian group and let A = fe a a 2 ::: ::: a r;1 g be a cyclic subset of G. Then A can be replaced by A 0 = = fe a i a 2i ::: a (r;1)i g for each i, whenever i is prime to r. Lemma 2. The simulated subset A = fe a a 2 ::: a r;2 a r;1 ug of a nite abelian group can be replaced by A 0 = fe a a 2 ::: a r;2 a r;1 u i g for each integer i. Lemma 3. The simulated subseta = fe a a 2 ::: a r;2 a r;1 ug of a - nite abelian group can be replaced bya 0 = fe a a 2 ::: a i;1 a i u a i+1 ::: ::: a r;1 g for each i, 1 i r ; 1. At some instances it will be convenient towork in the group ring Z(G). If G is a nite abelian group, then Z(G) consists of all the elements P g2g gg, where g is an integer. Addition and multiplication are dened between such sums in the same fashion as between polynomials. To the subset A of G we assign the element A = P a2a a of Z(G). We willusethe next argument several times. Let G = ABC beafactorization of G, where B = e b b 2 ::: b r;1 C = e c c 2 ::: c s;2 c s;1 v : Replace C by hci in the factorization G = ABC to get the factorization G = ABhci. This can be done by Lemma 2 with the choice of i = = 0. The factorizations G = ABC and G = ABhci correspond tothe equations G = ABC and G = ABhci respectivelyinthe group ring Z(G). Subtracting the rstfromthe second we get 0 = AB(c s;1 ; ; c s;1 v)andso0=ab(e ; v). From this by multiplying by e ; b we get the equation 0 = A(e ; b r )(e ; v). Now usingthe ideas in the proof of Th. 2 of [6] we can conclude that there are subsets U V of G such that A = U hb r i[v hvi, where the union is disjoint andthe products are direct. Analogous results hold when both B and C are cyclic or both are simulated. For easier reference we state them as a lemma. Lemma 4. Let G = ABC be a factorization of the nite abelian group G, where the factors B and C are one of the following B = e b b 2 ::: b r;1 C = e c c 2 ::: c s;1 B = e b b 2 ::: b r;2 b r;1 u C = e c c 2 ::: c s;2 c s;1 v B = e b b 2 ::: b r;1 C = e c c 2 ::: c s;2 c s;1 v : Then there are subsets U V of G such that A can be represented inthe following forms respectively

4 200 S. Szabo and K. Amin A = U hb r i[v hc s i A = U hui [V hvi A = U hb r i[v hvi where the unions are disjoint and the products are direct. 3. Result Now we are ready to prove the main result of the paper. the fundamental theorem of the nite abelian groups each nite abelian group is a direct product of cyclic groups of prime power order. If G is the direct product of cyclic groups of prime power order t 1 ::: t n respectively, then we say that G is of type (t 1 ::: t n ). Theorem 1. Let p be a prime and G be anabelian group of order p 4. Let G = ABC be a factorization of G, where jaj = p 2, jbj = jcj = = p. Further the factors B and C are cyclic or simulated. Then one of A B C is periodic. Proof. The type of G be can be (p 4 ) (p 3 p) (p 2 p 2 ) (p 2 p p) (p p p p): We distinguish 5 cases depending on the type of G. Then we distinguish 3subcases depending onboth B and C are cyclic both B and C are simulated B is cyclic and C is simulated. CASE 1. G is of type (p 4 ). This case is settled by Th. 1 of [5]. If G is a nite cyclic group and G = A 1 :::A n is a factorization of G, where each ja i j is a prime power, then one ofthe factors is periodic. CASE 2. G is of type (p 3 p). Let G = hxihyi, where jxj = p 3, jyj = p. Subcase 2(a). Both B and C are cyclic, that is, B = e b b 2 ::: b p;1 C = e c c 2 ::: c p;1 : If b p = e or c p = e, then we are done andsowe assume that jbj p 2, jcj p 2.ByLemma4,A = U hb p i[v hc p i. Let b = x y and c = x y be the basis representations of b and c. Now b p = x p and c p = x p. The subgroups of hx p i form a chain. Hence hx p2 i hx p i\hx p i = = hb p i\hc p i.thus x p2 is a period of A. Subcase 2(b). Both B and C are simulated, that is, B = e b b 2 ::: b p;2 b p;1 u C = e c c 2 ::: c p;2 c p;1 v : Here jbj = jcj = p. If u = e or v = e, then we are done andsoweassume that juj p, jvj p. By Lemma 2wemay assume that juj = jvj = p. The elements of G of order p generate the subgroup K = hx p2 yi of order p 2. Now K = BC is a factorization of K. By Redei's theorem one of the factors B and C is a subgroup of K. By

5 Factoring abelian groups of order p Subcase 2(c). B is cyclic and C is simulated, that is, B = e b b 2 ::: b p;1 C = e c c 2 ::: c p;2 c p;1 v : Here we may assume that jbj p 2 and jcj = jvj = p. By Lemma4, A = U hb p i[v hvi. If U = or V =, then A is periodic and sowe assume that U 6= and V 6=. If jbj = p 3,then ju jjhb p ij p 2 and hence V =. We assume that jbj = p 2. Therefore hb p i = hx p2 i. If v 2hx p2 i, then hvi = hx p2 i and soa is periodic. Thus we assume that v 62 hx p2 i. Replace C by hci in the factorization G = ABC to get the factorization G = ABhci. Let us compute Ahci. U x p2 [ V hvi hci = U x p2 hci [V hvihci: Ahci = The product ABhci is direct so the products hx p2 ihci and hvihci must be direct. They both must be equal to K. Note that x p2 v form a basis for K. By Lemma 3inC c can be replaced by c i for each i, 1 i p ; 1, so we have p ; 1choices for c. Namely, c may be x p2 v i, 1 i p ; 1. Now C = e x p2 v i x 2p2 v 2i ::: x (p;2)p2 v (p;2)i x (p;1)p2 v (p;1)i+1 : There is a j such that 1 j p ; 2and ji (p ; 1)i + 1 (mod p) since (j +1)i 1 (mod p) is solvable. Let t 2 U. The product thx p2 ic is direct since it is part of ABC. Compute hx p2 ic. Note that the sets x p 2 x jp2 v ji = x p2 v ij and cl x p2 x (p;1)p2 v (p;1)i+1 = x p2 v (p;1)i+1 are parts of hx p2 ic and they have the same elements. This is a contradiction. CASE 3. G is of type (p 2 p 2 ). Let G = hxi hyi, where jxj = p 2, jyj = p 2. Subcase 3(a). Both B and C are cyclic, that is, B = e b b 2 ::: b p;1 C = e c c 2 ::: c p;1 : If b p = e or c p = e, then we are done andsowe assume that jbj = p 2, jcj = p 2. Let L = hbi. If c 2 L, then L = BC is a factorization of L and so by Redei's theorem one of the factors is a subgroup of L. Thus we may assume that c 62 L. Wemay choose x y to beb c respectively. Now B = e x x 2 ::: x p;1 C = e y y 2 ::: y p;1 : We show that if G = ABC is a normed factorization, then A x p y or A x y p : To show this let a 0 a 2 A and a 0 a ;1 = x y. If p - and p -, then

6 202 S. Szabo and K. Amin a 0 = ax y contradicts the factorization G = AB 0 C 0 what we get from the factorization G = ABC by replacing B C by B 0 C 0,where B 0 = e x x 2 ::: x (p;1) C 0 = e y y 2 ::: y (p;1) : This replacement is possible by Lemma 1. Let a = x y 2 A. The previous argument with a 0 = e gives that pj or pj. If pj for each a 2 2 A, then A hx p yi. Similarlyifpj for each a 2 A, then A hx y p i. Thus we may assume that there are a = x y a 0 = x 0 y 0 2 A such that pj, p - and p - 0, pj 0,then p - ( 0 ; ), p - ( 0 ; ). So a 0 a ;1 = x 0 ; y 0 ; leads to a contradiction. Let M = hx p yi and N = hx y p i. If A hx p yi = M, then M = = AC is a factorization and soa(e ; y p )=0shows that y p is a period of A. Similarly ifa hx y p i = N, then N = AB is a factorization and so A(e ; x p )=0shows that x p is a period of A. Subcase 3(b). Both B and C are simulated, that is, B = e b b 2 ::: b p;2 b p;1 u C = e c c 2 ::: c p;2 c p;1 v : Here jbj = jcj = p and wemay assume that juj = p, jvj = p. The elements of G of order p generate the subgroup K = hx p y p i of order p 2. Now K = BC is a factorization of K. By Redei's theorem one of the factors B and C is a subgroup of K. Subcase 3(c). B is cyclicand C is simulated, that is, B = e b b 2 ::: b p;1 C = e c c 2 ::: c p;2 c p;1 v : Here we may assume that jbj p 2 and jcj = jvj = p. By Lemma4, A = U hb p i[v hvi. If U = or V =, then A is periodic and sowe assume that U 6= and V 6=. If v 2hb p i,then hvi = hb p i and soa is periodic. Thus we assume that v 62 hb p i. Now b p v form a basis for K = hx p y p i. In the factorization G = ABC replace C by hci to obtain the factorization G = ABhci. Compute Ahci. Ahci = U b p [ V hvi hci = U b p hci [V hvihci: Both hb p ihci and hvihci must be direct and equal to K. Note that b p v form a basis for K. By Lemma 3inCccan be replaced by c i for each i, 1 i p ; 1, so we have p ; 1choices for c. Namely, c may be b p v i, 1 i p ; 1. Now C = e b p v i b 2p v 2i ::: b (p;2)p v (p;2)i b (p;1)p v (p;1)i+1 : There is a j such that 1 j p ; 2and ji (p ; 1)i + 1 (mod p) since (j +1)i 1 (mod p) is solvable. Let t 2 U. The product thb p ic is direct since it is part of ABC. Compute hb p ic. Note that the sets

7 Factoring abelian groups of order p b p b jp v ji = b p v jp and b p b (p;1)p v (p;1)i+1 = b p v (p;1)i+1 are parts of hb p ic and they have the same elements. Thisisacontradiction. CASE 4. G is of type (p 2 p p). Let G = hxihyihzi, where jxj = p 2, jyj = jzj = p. Subcase 4(a). Both B and C are cyclic, that is, B = e b b 2 ::: b p;1 C = e c c 2 ::: c p;1 : Here we may assume that jbj = p 2, jcj = p 2. By Lemma 4,A = U hb p i[ [ V hc p i. Let b = x y z and c = x y z be the basis representations of b and c. Now b p = x p and c p = x p. Thesubgroups of hx p i form a chain. So hx p i = hb p i = hc p i.thus x p is a period of A. Subcase 4(b). Both B and C are simulated, that is, B = e b b 2 ::: b p;2 b p;1 u C = e c c 2 ::: c p;2 c p;1 v : Here jbj = jcj = p and by Lemma 2wemay assume that juj = p, jvj = = p. Replace B C by hbi hci in the factorization G = ABC to get the factorization G = Ahbihci. This gives that the product hbihci is direct. If u v 2hb ci, then BC = hb ci is a factorization and sobyredei's theorem we are done. We assume that u 62 hb ci and v 2hb c ui, that is, v = b c u Now B = e b b 2 ::: b p;2 b p;1 u C = e c c 2 ::: c p;2 c p;1 b c u A = U hui [V b c u : Here we assume that U 6= and V 6= since otherwise A is periodic. Assume rstthat =0. If =0,then 6= 0. By Lemma 2we may assume that = 1. Let t 2 V. The product thbib is direct since the product ABC is direct. On the other hand b 2hbi and b 2 B. This is a contradiction. If 6= 0,then by Lemma 2wemay assume that =1. If =0, then u is a period of A. Thus we assume that 6= 0. Let t 2 V. The product thb uib is direct. The elements b (p;1)+p;1 e b b 2 ::: p p;2 belong to hb uib. So(p ; 1) + p ; ::: p; 2 (mod p) and hence (p ; 1) + p ; 1 p ; 1(mod p). Thus =0. Butweknowthis is not the case. Secondly assume that 6= 0. By Lemma 2 we may assume that =1. Now C = e c c 2 ::: c p;2 c p;1 b cu = e c c 2 ::: c p;2 b u Let t 2 U. Theproduct thuibc is direct. Clearly huibc = hb uic. But this a contradiction since b u 2hb ui and b u 2 C.

8 204 S. Szabo and K. Amin Subcase 4(c). B is cyclic and C is simulated, that is, B = e b b 2 ::: b p;1 C = e c c 2 ::: c p;2 c p;1 v : We may assume that jbj = p 2 and jcj = jvj = p. By Lemma 4,A = = U hb p i[v hvi. Weassume that U 6= and V 6= since otherwise A is periodic. Let t 2 U. The product thb p ic is direct since it is part of the product ABC. If c 2 hb p i, then c = b pi for some i, 1 i p ; 1. This leads to the contradiction c 2hb p i and c 2 C. Thus we assume that c 62 hb p i. Wedistinguish two cases depending on v 2 hb p ci or v 62 hb p ci. If v 2hb p ci, then v = b p c. If =0,then A is periodic by b p. If 6= 0,then by Lemma2wemay assume that =1. Now C = e c c 2 ::: c p;2 c p;1 b p c = e c c 2 ::: c p;2 b p : Let t 2 U. Theproduct thb p ic is direct since it is part of the product ABC. But b p 2hb p i and b p 2 C is a contradiction. Turn to the case when hb p c vi is of type (p p p). From the factorization G = ABC it follows that 0 = AC(e ; b p ) and so AC is periodic by b p. Consequently V hvic is periodic by b p. Note that hvic = hc vi. If t 2 V,then e 2 t ;1 V hvic = t ;1 V hc vi. Now hb p i t ;1 V hc vi. As t ;1 V hc vi is a union cosets modulo hc vi and the elements of hb p i are incongruent modulo hc vi, it follows that p 3 = = hb p ihc vi t ;1 V hc vi = jv jp 2. This gives that jv jp and sowe get the contradiction that U =. CASE 5. G is of type (p p p p). Each element ofg nfeg is of order p and so a cyclic subset of G is a subgroup of G. Thus the only case we should consider is when both B and C are simulated, that is, B = e b b 2 ::: b p;2 b p;1 u C = e c c 2 ::: c p;2 c p;1 v : Here jbj = jcj = p and wemay assume that juj = p, jvj = p. Replace B C by hbi hci in the factorization G = ABC to get the factorization G = Ahbihci. This gives that the product hbihci is direct. So the group hb c u vi is one ofthe types (p p), (p p p), (p p p p). Turn rst to the case when hb c u vi is of type (p p). Now u v 2 2hb ci. Further BC = hb ci is a factorization andsobyredei's theorem we are done. Secondly consider the case when hb c u vi is of type (p p p). We assume that u 62 hb ci and v 2hb c ui, that is,v = b c u Now B = e b b 2 ::: b p;2 b p;1 u C = e c c 2 ::: c p;2 c p;1 b c u

9 Factoring abelian groups of order p A = U hui [V b c u : Here we assume that U 6= and V 6= otherwise A is periodic. Assume rstthat =0. If =0,then 6= 0. By Lemma 2we may assume that = 1. Let t 2 V. The product thbib is direct since the product ABC is direct. On the other hand b 2hbi and b 2 B. This is a contradiction. If 6= 0,then by Lemma 2wemay assume that =1. If =0, then u is a period of A. Thus we assume that 6= 0. Let t 2 V. The product thb uib is direct. The elements b (p;1)+p;1 e b b 2 ::: p p;2 belong to hb uib. So(p ; 1) + p ; ::: p; 2 (mod p) and hence (p ; 1) + p ; 1 p ; 1(mod p). Thus =0. Butweknowthis is not the case. Secondly assume that 6= 0. By Lemma 2 we may assume that =1. Now C = e c c 2 ::: c p;2 c p;1 b cu = e c c 2 ::: c p;2 b u Let t 2 U. Theproduct thuibc is direct. Clearly huibc = hb uic. But this a contradiction since b u 2hb ui and b u 2 C. Finallyturn to the case when hb c u vi is of type (p p p p). From the factorization G = Ahbihci it follows that A is a complete set of representatives modulo hb ci and so A = u i v j a ij :0 i j p ; 1 a ij 2hb ci : We may assume that e 2 A, that is, a 00 = e. We also know that A = U hui [V hvi. Here we assume that U 6= and V 6= since otherwise A is periodic. As the rolesofu and v are symmetric, wemay assume that e 2 U. Then hui A. Thismeans that a i0 = e for each i, 0 i p ;1. Let u i v j a ij 2 V hvi. Now u i v j a ij hvi A. This gives that a ij 's are equal for each j, 0 j p ; 1. As a i0 = e, wehave a ij = e for each j, 0 j p ; 1. Then u i v j a ij hvi = u i hvi. Therefore u i 2 V hvi. On the other hand u i 2 U hui. This is a contradiction since V hvi and U hui are disjoint. This completes the proof. 4. Examples In this section we exhibit examples to show that the conditions of the problem proposed in the introduction cannot be relaxed in general. Let G = Q 5 i=1 hx ii, where jx i j = r i 3. Set

10 206 S. Szabo and K. Amin T 0 = e x 3 x 2 3 ::: xr 3;2 3 x r 3;1 3 x 4 T 1 = e x 2 x 2 2 ::: xr 2;2 2 x r 2;1 2 x 5 T 2 = = T r 1;1 = hx 2 i: Now wedene A B C by A = T 0 hx 2 i[x 1 T 1 hx 3 i[x 2 1 T 2hx 3 i[:::[ x r 1;1 1 T r 1;1hx 3 i B = e x 4 x 2 4 ::: xr 4;2 4 x r 4;1 4 x 2 C = e x 5 x 2 5 ::: xr 5;2 5 x r 5;1 5 x 3 : We claim that G = ABC is a factorization of G. In order to verify this rst we show that T 0 hx 2 ibc = T 1 hx 3 ibc = = T r 1;1hx 3 ibc = hx 2 x 3 x 4 x 5 i: Indeed, T 0 hx 2 ibc = T 0 hx 2 x 4 ic = hx 2 x 3 x 4 ic = hx 2 x 3 x 4 x 5 i and T 1 hx 3 icb = T 1 hx 3 x 5 ib = hx 2 x 3 x 5 ib = hx 2 x 3 x 4 x 5 i: The remaining cases can be veried in a similar way. Now ABC = ; T 0 hx 2 i[x 1 T 1 hx 3 i[x 2 1 T 2hx 3 i[:::[ x r 1;1 1 T r 1;1hx 3 i BC = = T 0 hx 2 ibc [ x 1 T 1 hx 3 ibc [ x 2 1 T 2hx 3 ibc [ :::[ x r 1;1 1 T r 1;1hx 3 ibc = = e x 1 x 2 1 ::: xr 1;1 1 x2 x 3 x 4 x 5 = x1 x 2 x 3 x 4 x 5 = G: It is clear that B C are not periodic. The subset A is not periodic since it is a disjoint union of periodic subsets which have noperiodin common. To makethe example more concrete let us choose r 1 ::: r 5 to be 3. In this case G is of type ( ), B C are simulated subsets jaj =3 3 and none ofthe factors is periodic. In the special case when r 1 ::: r 5 are pairwise relatively primes G is a cyclic group. In the next example the factors B and C are cyclic. Let G = hxi hyi, where jxj = rst, jyj = uv, and r s t u v 2. Set T 0 = e y u y 2u ::: y (v;2)u y (v;1)u x r T 1 = e x rs x 2rs ::: x (t;2)rs x (t;1)rs y T 2 = = T r;1 = hx rs i: Now dene A B C by A = T 0 hx rs i[xt 1 hy u i[x 2 T 2 hy u i[:::[ x r;1 T r;1 hy u i B = e x r x 2r ::: x (s;1)r C = e y y 2 ::: y u;1 : Clearly B C are not periodic and A is not periodic since it is a disjoint union of periodic subsets without common period. We claim that G = = ABC is a factoring of G. In order to verify this claim we rst show that

11 Factoring abelian groups of order p T 0 hx rs ibc = T 1 hy u ibc = T 2 hy u ibc = = T r;1 hy u ibc = hx r yi: Indeed, T 0 hx rs ibc = T 0 hx r ic = hx r yi and T 1 hy u icb = T 1 hyib = hx r yi: The remaining cases can be veried in a similar way. Using these facts ABC = ; T 0 hx rs i[xt 1 hy u i[x 2 T 2 hy u i[:::[ x r;1 T r;1 hy u i BC = = T 0 hx rs ibc [ xt 1 hy u ibc [ x 2 T 2 hy u ibc [ :::[ x r;1 T r;1 hy u ibc = = e x x 2 ::: x r;1 hx r yi = hx yi = G: If we choose r s t u v to be 2, then G is of type ( ), jaj =2 3, jbj = jcj =2. Ifrst, and uv are relatively primes, then G is a cyclic group. References [1] CORR ADI, K. and SANDS, A. U. and SZAB O, S.: Simulated factorizations, Journal of Alg., 151 (1992), 12{25. [2] CORR ADI, K. and SZAB O, S: Solution to a problem of A. D. Sands, Communications in Algebra, 23 (1995), 1503{1510. [3] HAJ OS, G.: Uber einfache undmehrfache Bedeckung des n-dimensionalen Raumes mit einem Wurfelgitter, Math. Zeitschr. 47 (1941), 427{467. [4] REDEI, L.: Die neue Theorie der endlichen Abelschen Gruppen undverallge- meinerung des Hauptsatzes von Hajos, Acta Math. Acad. Sci. Hungar. 16 (1965), 329{373. [5] SANDS, A. D.: Factorization of cyclic groups, Colloquium on abelian groups, Tihany, Hungary, Sept. 1963, 139{146. [6] SANDS, A. D. and SZAB O, S.: Factorization of periodic subsets, Acta Math. Hungar. 57 (1991), 159{167.