推 loribank :謝謝你的回答，可以幫我看看第一題的第二小題和第二 03/14 00:44
→ loribank :題的解法嗎???感謝!!!! 03/14 00:44
→ yusd24 :因為大致上是對的，我就沒有回了XD 03/14 09:34
→ yusd24 :(1) 8個, (2) 168/8=21 03/14 09:36
推 loribank :感謝^^ 03/17 23:02
※ 引述《loribank (小蘿莉銀行)》之銘言：
: 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.
: 第一小題我有算，我算1個或8個
: 第二小題我的想法是，如果只有1個7-Sylow subgroup，那麼P normal in G
: 所以應該是有8個 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.
: 這題我有做，在一個小地方卡住，想請高手指點一下
: (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 (這個情況我不知該如何討論???)
: (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
: #
: 麻煩個位大大了!!!
: 感謝
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※ 編輯: yusd24 來自: 219.71.210.134 (03/13 23:10)