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=m s with m∈M and s∈S,while ...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=m s 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[−2 k+1,2 k+2]^(∗)splits a cyclic group Z m,then m=4 k+2.We prove also that if M=[−k_(1),k_(2)]^(∗)splits Z m purely singularly,and 15≤k_(1)+k_(2)≤30,then m=1,or m=k_(1)+k_(2)+1,or k_(1)=0 and m=2 k_(2)+1.展开更多
文摘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=m s 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[−2 k+1,2 k+2]^(∗)splits a cyclic group Z m,then m=4 k+2.We prove also that if M=[−k_(1),k_(2)]^(∗)splits Z m purely singularly,and 15≤k_(1)+k_(2)≤30,then m=1,or m=k_(1)+k_(2)+1,or k_(1)=0 and m=2 k_(2)+1.
