推 paggei :想請問 case1為什麼hk∈H => k∈H 呢@@ 10/17 01:26
hk∈H => hk=r ,r∈H => k=h^(-1)r∈H ,h^(-1)∈H (∵H is a subgroup)
=>k∈H
※ 編輯: alasa15 來自: 111.249.11.18 (10/17 01:29)
推 paggei :感謝@@ 10/17 01:33
※ 引述《paggei (XD)》之銘言：
: 以下(G,*,e)為一群，想證 j
: -1 r j -j r
: (1) if a, b∈G, b*a*b = a , then b * a * b = a for some r∈Ｎ
: 這題想不出來啊...想用數學歸納法， k
: k+1 -(k+1) k -k -1 r
: 可是b *a*b 頂多拆b(b * a * b )b 然後中間卡一個a ....
: (2) │G│ is finite, H is a subset of G, prove that H is a subgroup of G
: if and only if e∈H and a*b∈H for all a, b∈G
: 不會證inverse∈H的部分@@
: (3) H, K are subgroup of G, prove that if H∪K is a subgroup of G,
: then H⊆K or K⊆H
第三題我也不確定 打出來參考
Suppose not, ∃h∈H but h not belongs to K
k∈K but k not belongs to H
such that hk∈(H∪K)
consider hk=q∈(H∪K)
case1. q∈(H-K) (difference)
hk∈H => k∈H (a contradiction)
case2. q∈(K-H)
hk∈H => h∈K (a contradiction)
thus q∈(H∩K) only
=>H⊆K or K⊆H
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◆ From: 111.249.11.18