作者yusd24 (阿乡)
看板Math
标题Re: [代数] 请教两题simple group
时间Sun Mar 13 23:09:26 2011
※ 引述《loribank (小萝莉银行)》之铭言：
: 1.Let G be a simple group of order 168.
: (1) How many Sylow 7-subgroups of G?
: (2) Let P be a Sylow 7-subgroup. Show that │N(P)│= 21,
: where N(P) is the normalizer of P in G, ie
: N(P)={g in G│gPg^-1=P}
: (3) Prove or disprove that G has a subgroup of order 14.
No,
Let H be a such subgroup.
H contains an element of orfer 7, H contains a Sylow 7 of G.
Denote this Sylow 7 by K. Then K is normal in H, ([H:K]=2)
Therefore, H is contained as a subgroup in N(K), but |N(K)|=21.
Contradiction.
: 第一小题我有算，我算1个或8个
: 第二小题我的想法是，如果只有1个7-Sylow subgroup，那麽P normal in G
: 所以应该是有8个 7-Sylow subgroup
: 2. Show that a group of order 72 is not simple.
If there are only 1 Sylow 3, then we are done.
If there are 4 Sylow 3, say P1, P2, P3, P4. Let G act on X={P1,P2,P3,P4}
by conjugation. (g → gPig^{-1}). This induces a map f from G to S4,
Since |G|=72, f can't be injective. Hence Ker(f) is non-identity.
G is not simple.
: 这题我有做，在一个小地方卡住，想请高手指点一下
: (pf.) 72=2^3 * 3^2
: Consider 3-sylow subgroups
: (1) If 3-sylow subgroups is normal ,then done.
: (2) Otherwise, there exist 4 3-sylow subgroups
: Let H,K be two of them
: => │H∩K│=(│H││K│)/│HK│= 9*9/72 =1.... & │H∩K│││K│=9
: => │H∩K│= 3
: ∵ H、K are abelian
: => H∩K is normal in H & H∩K is normal in K.
: i.e N_G(H∩K)＞H and N_G(H∩K)＞K
: => 9 =│H│││N_G(H∩K)│││G│=72
: =>│N_G(H∩K)│= 18 or 36 or 72
: (i) │N_G(H∩K)│= 18 (这个情况我不知该如何讨论???)
: (ii) │N_G(H∩K)│= 36
: [G:N_G(H∩K)]= 72/36 = 2 => N_G(H∩K) is normal in G
: => G is not simple
: (iii) │N_G(H∩K)│= 72
: =>│N_G(H∩K)│= │G│= 72
: => N_G(H∩K)=G
: => H∩K is normal in G
: #
: 麻烦个位大大了!!!
: 感谢
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◆ From: 219.71.210.134
※ 编辑: yusd24 来自: 219.71.210.134 (03/13 23:10)
1F：推 loribank :谢谢你的回答，可以帮我看看第一题的第二小题和第二 03/14 00:44
2F：→ loribank :题的解法吗???感谢!!!! 03/14 00:44
3F：→ yusd24 :因为大致上是对的，我就没有回了XD 03/14 09:34
4F：→ yusd24 :(1) 8个, (2) 168/8=21 03/14 09:36
5F：推 loribank :感谢^^ 03/17 23:02