本文選題：有限群　切入點：子群的個數　出處：《西南大學》2017年碩士論文　論文類型：學位論文

【摘要】：本文研究同階子群個數之集合對群結構的影響.設G是一個有限群.n(G)表示群G中所有同階子群的個數組成的集合.本文得出了當n(G)= {1,3,4}時G的所有Sylow子群的結構,以及當G為內冪零群時群G的結構.主要得到了下述定理:引理2.6設P是一個非交換p群,H=h是P的循環(huán)極大子群且|H| =pn.假設H在P中有補A =a,則其p階子群的個數有如下結果:(1)當p = 2 且 P =a,b|a2n = b2 = 1,b-1 ab = a-1時,其 2 階子群有 2n + 1個;(2)當p = 2,n ≥ 3 且 P =a,b|a2n = b2 = 1,b-1 ab=a-1+2n-1時,其 2 階子群有2n-1+ 1個;(3)當 p = 2,n ≥ 3 且 P=a,b|a2n = b2 = 1,b-1ab=a1+2n-1時,其 2 階子群有3個;(4)當 p = 3 且 P=a,b|a3n = b3 = 1,b-1ab = a1+3n-1時,其 3 階子群有 4個.定理3.1設G為有限群,且|G| = 2α3βq1α1q2α2 …qnαn,其中qi為大于3的素數,α,β為非負整數,αi,n均為正整數.如果n(G)= {1,3,4},則α0,β0且G的Sylow-2子群P2和Sylow-3子群P3不能都正規(guī)且具有如下性質:(a)當P2循環(huán)時,G的Sylow 2-子群只有3個;當P2不循環(huán)時,P2(?)G且具有如下結構:如果 α=2,則 P2 = C2× C2;如果 α = 3,則 P2 = C4 × C2 或 P2 =a,b|a4 = 1,b2=a2,b-1ab=a-1;如果α ≥ 4,則P2=C2α-1×C2 或P2=a,b|a2α-1 =b2=1,b-1ab = a1+2α-2.(b)當P3循環(huán)時,G的Sylow 3-子群有4個;當P3不循環(huán)時,P3(?)G且具有如下結構: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循環(huán)且Qi(?)G,i=1,2,…,n.定理4.1若G為內冪零群,且n(G)= {1,3,4},則G必與下列群同構:(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，

本文編號：1634363