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.
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.
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
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麻煩個位大大了!!!
感謝
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