同阶子群个数之集为{1,3,4}的有限群

发布时间：2018-03-19 12:54

  本文选题：有限群　切入点：子群的个数　出处：《西南大学》2017年硕士论文　论文类型：学位论文

【摘要】：本文研究同阶子群个数之集合对群结构的影响.设G是一个有限群.n(G)表示群G中所有同阶子群的个数组成的集合.本文得出了当n(G)= {1,3,4}时G的所有Sylow子群的结构,以及当G为内幂零群时群G的结构.主要得到了下述定理:引理2.6设P是一个非交换p群,H=h是P的循环极大子群且|H| =pn.假设H在P中有补A =a,则其p阶子群的个数有如下结果:(1)当p = 2 且 P =a,b|a2n = b2 = 1,b-1 ab = a-1时,其 2 阶子群有 2n + 1个;(2)当p = 2,n ≥ 3 且 P =a,b|a2n = b2 = 1,b-1 ab=a-1+2n-1时,其 2 阶子群有2n-1+ 1个;(3)当 p = 2,n ≥ 3 且 P=a,b|a2n = b2 = 1,b-1ab=a1+2n-1时,其 2 阶子群有3个;(4)当 p = 3 且 P=a,b|a3n = b3 = 1,b-1ab = a1+3n-1时,其 3 阶子群有 4个.定理3.1设G为有限群,且|G| = 2α3βq1α1q2α2 …qnαn,其中qi为大于3的素数,α,β为非负整数,αi,n均为正整数.如果n(G)= {1,3,4},则α0,β0且G的Sylow-2子群P2和Sylow-3子群P3不能都正规且具有如下性质:(a)当P2循环时,G的Sylow 2-子群只有3个;当P2不循环时,P2(?)G且具有如下结构:如果 α=2,则 P2 = C2× C2;如果 α = 3,则 P2 = C4 × C2 或 P2 =a,b|a4 = 1,b2=a2,b-1ab=a-1;如果α ≥ 4,则P2=C2α-1×C2 或P2=a,b|a2α-1 =b2=1,b-1ab = a1+2α-2.(b)当P3循环时,G的Sylow 3-子群有4个;当P3不循环时,P3(?)G且具有如下结构:P3= C3β-1 × C3或P3=a,b|a3β-1 = b3 = 1,b-1ab = a1+3β-2).(c)G 的 Sylow qi-子群Qi循环且Qi(?)G,i=1,2,…,n.定理4.1若G为内幂零群,且n(G)= {1,3,4},则G必与下列群同构:(1)G ≌a,c1,c2|a3β= 1,c12=c22=1,c1·c2=c2·c1,c1a=c2,c2a=c1·c2;(2)G ≌a,b,c|a3β= 1,b4 = 1,b2 = c2,b-1cb = c-1,a-1ba = c,a-1ca = bc.
[Abstract]:In this paper, we study the effect of the set of the number of subgroups of the same order on the structure of the group. Let G be a finite group. Let G be a set of the numbers of all subgroups of the same order in G. In this paper, we obtain the structure of all Sylow subgroups of G when the number of subgroups of the same order G = {1 / 3 / 4}. And the structure of group G when G is an inner nilpotent group. The following theorems are obtained: Lemma 2.6 Let P be a noncommutative p-group and H ~ h be a cyclic maximal subgroup of P and H = pn.If H has a complement A in P, then the number of its p-order subgroup is obtained. The number has the following result: 1) when p = 2 and p = 2 b _ 2n = b _ 2 = 1o b ~ (-1) ab = a-1, Its subgroup of order 2 has 2n 1 / 2) when p = 2n 鈮，

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