The Influence On The Structure Of The Finite Group By The Number Of Subgroups With The Same Oder In A Finite Group Under A Certain Condition

The number question of p-group is about the number of each type subgroup ,element or subset in a finite group.In,turn,the structure and properties of p-group determined by the number of each type subgroups are also the impertant topics of the number question of p-group.This article starts studying from these two aspects. First,we calculate the number of subgroups of each order of p-group with circle maximal subgroup,or(pⁿ,p^m)type,then base on these,we get the complete clssfication of p-group with p + 1 nontrival subgroups of each order and a new characteriation of (p^m,pⁿ)type p-group with the number of circle subgroups of each order.In addition ,we also give out the complete classfication of the finite group with the number of the same order subgroups≤3.Worthy of note,this discussion is not only limited in p-group.We get the following theorems:Theorem 2.1 Let G be a finite p- group with maximal circle subgroup.Then(1)If G is(I)type group,then s_k(G)= 1 with k = 0,1,...,n.(2)If G is(â…¡),(â…¢),or(â…¥)type group,then s_k(G)=(?)(3)If G is(â…£),(â…¤),or(â…¦)type group,then s_k(G)=(?)especially,s_l(G)= 1 for(â…£)type group;s_l(G)= 2^n-1+ 1 for(â…¤)type group;s_l(G)= 2^n-2+ 1 for(â…¦)type group.Theorem 2.2 Let G be a finite p- group.If the number of notrival subgroups of each order of G is p+1,then G is only(â…¡),(â…¢)or(â…¥)type group.Note:1.A finite p- group with maximal circle subgroup have seven types, say them by turn(â… ),(â…¡),(â…¢),(â…£),(â…¤),(â…¥),(â…¦)type in[18].2.We denote the set of the number of subgrous with the same order of G by N(G).Theorem 3.1 Let G be(pⁿ,p^m)type Abel p- group with m≤n.Then the number of subgroups of order p^k of G is:Theorem 3.2 Let G be a Abel group,and∣G∣ = p^n+m,m≤n.If then G is(pⁿ,p^m)type Abel p- group.Theorem 4.1 Let G be a finite group.Then N(G)= {1} if only if G is a circle group.Theorem 4.2 Let G be a finite group.If N(G)= {1,2},then G is not exist.Theorem 4.3 Let G be a finite group,and N(G)= {1,3}.(1)If G is a p- group,∣G∣ = pⁿ,n≥2,then G is Q₈,or one of two types fllowing:â… )(2^n-1,2)type 2-group;â…¡)Half generalized quaternion group:G =＜a,b＞,n≥4, a^2n-1= 1,b² = 1,b^-1ab = a^1+2n-2.(2)If G is not a p- group,then â… )If all Sylow subgroups of G are normal,then G is one of three types fllowing:(â…°)Q₈×,2(?)o(a);(â…±)p×< u >,2(?)o(u),p is a(2^n-1,2)type Abel 2- group;(â…²)P×< v >,2(?)o(v),P is a Half generalized quaternion group.â…¡)If G has non-normal Sylow subgroup,then G =(P(?)P₁)×< u >,P =< a > ,P₁ =< b >,o(a)= 2ⁿ,n≥1,o(b)= 3,2(?)o(u),3(?)o(u).Corollary 1 Let G be a finite p- group.If N(G)= {1,p + 1},then G is only Q₈,(â…¡),(â…¢),or(â…¥).Theorem 4.4 Let G be a finite group.If N(G)= {1,2,3},thenG is not exist.