本文選題：有限群　切入點(diǎn)：子群的個(gè)數(shù)　出處：《西南大學(xué)》2017年碩士論文　論文類型：學(xué)位論文

【摘要】：本文研究同階子群個(gè)數(shù)之集合對(duì)群結(jié)構(gòu)的影響.設(shè)G是一個(gè)有限群.n(G)表示群G中所有同階子群的個(gè)數(shù)組成的集合.本文得出了當(dāng)n(G)= {1,3,4}時(shí)G的所有Sylow子群的結(jié)構(gòu),以及當(dāng)G為內(nèi)冪零群時(shí)群G的結(jié)構(gòu).主要得到了下述定理:引理2.6設(shè)P是一個(gè)非交換p群,H=h是P的循環(huán)極大子群且|H| =pn.假設(shè)H在P中有補(bǔ)A =a,則其p階子群的個(gè)數(shù)有如下結(jié)果:(1)當(dāng)p = 2 且 P =a,b|a2n = b2 = 1,b-1 ab = a-1時(shí),其 2 階子群有 2n + 1個(gè);(2)當(dāng)p = 2,n ≥ 3 且 P =a,b|a2n = b2 = 1,b-1 ab=a-1+2n-1時(shí),其 2 階子群有2n-1+ 1個(gè);(3)當(dāng) p = 2,n ≥ 3 且 P=a,b|a2n = b2 = 1,b-1ab=a1+2n-1時(shí),其 2 階子群有3個(gè);(4)當(dāng) p = 3 且 P=a,b|a3n = b3 = 1,b-1ab = a1+3n-1時(shí),其 3 階子群有 4個(gè).定理3.1設(shè)G為有限群,且|G| = 2α3βq1α1q2α2 …qnαn,其中qi為大于3的素?cái)?shù),α,β為非負(fù)整數(shù),αi,n均為正整數(shù).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(zhì):(a)當(dāng)P2循環(huán)時(shí),G的Sylow 2-子群只有3個(gè);當(dāng)P2不循環(huán)時(shí),P2(?)G且具有如下結(jié)構(gòu):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(dāng)P3循環(huán)時(shí),G的Sylow 3-子群有4個(gè);當(dāng)P3不循環(huán)時(shí),P3(?)G且具有如下結(jié)構(gòu):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(nèi)冪零群,且n(G)= {1,3,4},則G必與下列群同構(gòu):(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(hào)：1634363