For a positive integer n,we denote byπ(n)the set of all prime divisors of n.For a finite group G,the setπ(G):=π(|G|)is called the prime spectrum of G.Let M<G mean that M is a maximal subgroup of G.We put K(G)=ma...For a positive integer n,we denote byπ(n)the set of all prime divisors of n.For a finite group G,the setπ(G):=π(|G|)is called the prime spectrum of G.Let M<G mean that M is a maximal subgroup of G.We put K(G)=max{|π(G)\π(M)|:M<G}and k(G)=min{|π(G)\π(M)|:M<G}.In this notice,using well-known number-theoretical results,we present a number of examples to show that both K(G)and k(G)are unbounded in general.This implies that the problem"Are k(G)and K(G)bounded by some constant k?",raised by Monakhov and Skiba in 2016,is solved in the negative.展开更多
文摘For a positive integer n,we denote byπ(n)the set of all prime divisors of n.For a finite group G,the setπ(G):=π(|G|)is called the prime spectrum of G.Let M<G mean that M is a maximal subgroup of G.We put K(G)=max{|π(G)\π(M)|:M<G}and k(G)=min{|π(G)\π(M)|:M<G}.In this notice,using well-known number-theoretical results,we present a number of examples to show that both K(G)and k(G)are unbounded in general.This implies that the problem"Are k(G)and K(G)bounded by some constant k?",raised by Monakhov and Skiba in 2016,is solved in the negative.
