標題:
另一條 Group theory
發問:
If H and K are subgroup of G with K normal in G, show that (a) HK is a subgroup of G and (H∩K)
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最佳解答:
I will still refer to the book, Algebra , by Michael Artin. In fact, part(a) is just the results of p63(8.6)(b) and p60(7.1) , part(f) is just a direct application of First Isomorphism Theorem, p68(10.9). (a) The trick is to show HK=KH first. ?h ∈H and for ?k ∈K, hk = (hkh-1)h ∈KH since hkh-1 ∈K as K is normal. Similarly, kh= h(h-1kh) ∈HK and this shows that HK=KH Now, h1k1h2k2 = h1h'2k'1k2 ∈HK (Since HK=KH, there exists h'2 and k'1 such that k1h2 = h'2k'1 ) Since e ∈H and e ∈K, e= ee ∈HK Besides, (hk)-1 = k-1h-1 ∈KH =HK This has shown that HK is a subgroup of G. Since intersection of subgroup is still a subgroup and (H∩K) ? H , (H∩K)
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