Factoring by hereditary periodicity forcing subsets (original) (raw)

Acta Mathematica Hungarica, 1991

The paper deals with decomposition of a finite abelian group into a direct product of subsets. A family of subsets, the so-called uniquely complemented subsets, is singled out. It will be 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 Hajó s's factorization theorem.

NONQUASI-PERIODIC FACTORIZATIONS FOR CERTAIN FINITE ABELIAN GROUPS

International Journal of Algebra and Computation, 2007

Hajós [2] asked if each factorization of a finite abelian group is quasi-periodic. Sands [5] exhibited a counter-example. We will show that a sizeable family of finite abelian groups admits nonquasi-periodic factorizations. We also describe a small family whose members have only quasi-periodic factorizations.

Groups admitting only quasi-periodic factorizations

We will show that if a finite abelian group is a direct product of two subgroups of relatively prime orders and one of the subgroups has prime order, then each normalized factorization of the group must be quasiperiodic.

Factoring Finite Abelian Groups by Subsets with Maximal Span

SIAM Journal on Discrete Mathematics, 2006

Nonperiodic product of subsets and Hajós’ theorem

Note di Matematica

G. Hajós proved that if a finite abelian group is a direct product of its cyclic subsets, then at least one of the factors must be a subgroup. We give a new elementary proof of this theorem based on the special case for p-groups.

Extension of Hajós' factorization theorem to some non-abelian groups

Communications on Pure and Applied Mathematics, 1968

This paper represents the author's doctoral dissertation presented to New York University, and results from work done under National Science Foundation Fellowships. Reproduction in whole or in part is permitted for any purpose of the United States Government. LEMMA 3.1. r f 1 is a non-zero integer relatively prime to a natural number m, then (nl I 0 5 n < m) is a complete system of residues modulo m. Using this result we prove the following variant on [14], Hilfssatz 3, p. 33. LEMMA 3.2. Let w , x , andy be elements of ring R, a a unit of R, and r an integer relatively prime to the natural number k. Then, zf for all integers v the equations (3.2.l.v) w = xa"[a; k ] y

Groups Factorized by Pairwise Permutable Abelian Subgroups of Finite Rank

Advances in Group Theory and Applications, 2016

It is proved that a group which is the product of pairwise permutable abelian subgroups of finite Prüfer rank is hyperabelian with finite Prüfer rank; in the periodic case the Sylow subgroups of such a product are described. Furthermore, if G=ABCG = ABCG=ABC is such a non-periodic product with locally cyclic subgroups A, B and C, then the Prüfer rank of GGG is at most 888. Moreover, GGG is soluble of derived length at most 444 and has Prüfer rank at most 6, if AcapBcapC=1A\cap B\cap C = 1AcapBcapC=1, and GGG has a torsion subgroup TTT such that the factor group G/TG/TG/T is locally cyclic and the Sylow ppp-subgroups of TTT are of Prüfer rank at most 222 for odd ppp and at most 666 for p=2p = 2p=2, otherwise.

On factorizations of finite groups

2021

Let G be a finite group and let {A1, . . . , Ak} be a collection of subsets of G such that G = A1 . . . Ak is the product of all the Ai and card(G) = card(A1) . . . card(Ak). We shall write G = A1 · . . . · Ak, and call this a kfold factorization of the form (card(A1), . . . , card(Ak)). We prove that for any integer k ≥ 3 there exist a finite group G of order n and a factorization of n = a1 . . . ak into k factors other than one such that G has no k-fold factorization of the form (a1, . . . , ak).