{"title":"Some New Results on Purely Singular Splittings","authors":"P. Yuan","doi":"10.4208/cmr.2020-0048","DOIUrl":null,"url":null,"abstract":"Let G be a finite abelian group, M a set of integers and S a subset of G. We say that M and S form a splitting of G if every nonzero element g of G has a unique representation of the form g=ms with m∈M and s∈S, while 0 has no such representation. The splitting is called purely singular if for each prime divisor p of |G|, there is at least one element of M is divisible by p. In this paper, we continue the study of purely singular splittings of cyclic groups. We prove that if k≥2 is a positive integer such that [−2k+1,2k+2]∗ splits a cyclic group Zm, then m=4k+2. We prove also that if M=[−k1,k2] ∗ splits Zm purely singularly, and 15 ≤ k1+k2 ≤ 30, then m = 1, or m = k1+k2+1, or k1 = 0 and m=2k2+1. AMS subject classifications: 20D60, 20K01, 94A17","PeriodicalId":66427,"journal":{"name":"数学研究通讯","volume":null,"pages":null},"PeriodicalIF":0.0000,"publicationDate":"2022-01-01","publicationTypes":"Journal Article","fieldsOfStudy":null,"isOpenAccess":false,"openAccessPdf":"","citationCount":"0","resultStr":null,"platform":"Semanticscholar","paperid":null,"PeriodicalName":"数学研究通讯","FirstCategoryId":"1085","ListUrlMain":"https://doi.org/10.4208/cmr.2020-0048","RegionNum":0,"RegionCategory":null,"ArticlePicture":[],"TitleCN":null,"AbstractTextCN":null,"PMCID":null,"EPubDate":"","PubModel":"","JCR":"","JCRName":"","Score":null,"Total":0}
引用次数: 0
Abstract
Let G be a finite abelian group, M a set of integers and S a subset of G. We say that M and S form a splitting of G if every nonzero element g of G has a unique representation of the form g=ms with m∈M and s∈S, while 0 has no such representation. The splitting is called purely singular if for each prime divisor p of |G|, there is at least one element of M is divisible by p. In this paper, we continue the study of purely singular splittings of cyclic groups. We prove that if k≥2 is a positive integer such that [−2k+1,2k+2]∗ splits a cyclic group Zm, then m=4k+2. We prove also that if M=[−k1,k2] ∗ splits Zm purely singularly, and 15 ≤ k1+k2 ≤ 30, then m = 1, or m = k1+k2+1, or k1 = 0 and m=2k2+1. AMS subject classifications: 20D60, 20K01, 94A17