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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)<| H (b) φ(hk)=(H∩K)h is a homomorphism HK to H/(H∩K) (c) φ is onto H/(H∩K) (d) ker(φ) is K (e) φ is not multiple-valued [ if h1k1=h2k2, show that (H∩K)h1=(H∩K)h2 ] (f) HK/K ≈ H/(H∩K)
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) < H. It remains to show that (H∩K) is normal to H. For any g ∈(H∩K) and h ∈H, hgh-1 ∈K since K is normal. Besides, hgh-1 ∈H since g ∈(H∩K) ? H . This means that hgh-1 ∈(H∩K) and hence (H∩K) is normal to H. (b) φ(h1k1h2k2) = φ(h1h2h2-1k1h2k2) = φ(h1h2k3k2) ...since K is normal, ?k3 =h2-1k1h2 = (H∩K)h1h2 = (H∩K)h1 (H∩K)h2 = φ(h1k1) φ(h2k2) (c) ?s ∈H/(H∩K) , ?h ∈H such that s= (H∩K)h Now, h= he ∈HK Hence, φ is onto. (d) ?k ∈K , φ(k)= φ(ek)= (H∩K)e ? K ?Ker(φ) On the other hand, ?g ∈Ker(φ) ?HK, we have φ(g)= (H∩K)e and g=hk for some h ∈H and k ∈K . Now, φ(hk)= φ(g)= (H∩K)e ? h ∈(H∩K) ?K Hence, g=hk ∈K (e) If h1k1=h2k2 , then (H∩K)h1=φ(h1k1) =φ(h2k2) =(H∩K)h2 (f) By First Isomorphism Theorem, HK/Ker(φ) ≈ Im(φ) i.e. HK/K ≈ H/(H∩K) by part (c), (d) and (e). 圖片參考：http://i175.photobucket.com/albums/w130/bjoechan2003/My%20Cat%2020070530/DSCN2698.jpg
其他解答:
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)<| H (b) φ(hk)=(H∩K)h is a homomorphism HK to H/(H∩K) (c) φ is onto H/(H∩K) (d) ker(φ) is K (e) φ is not multiple-valued [ if h1k1=h2k2, show that (H∩K)h1=(H∩K)h2 ] (f) HK/K ≈ H/(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) < H. It remains to show that (H∩K) is normal to H. For any g ∈(H∩K) and h ∈H, hgh-1 ∈K since K is normal. Besides, hgh-1 ∈H since g ∈(H∩K) ? H . This means that hgh-1 ∈(H∩K) and hence (H∩K) is normal to H. (b) φ(h1k1h2k2) = φ(h1h2h2-1k1h2k2) = φ(h1h2k3k2) ...since K is normal, ?k3 =h2-1k1h2 = (H∩K)h1h2 = (H∩K)h1 (H∩K)h2 = φ(h1k1) φ(h2k2) (c) ?s ∈H/(H∩K) , ?h ∈H such that s= (H∩K)h Now, h= he ∈HK Hence, φ is onto. (d) ?k ∈K , φ(k)= φ(ek)= (H∩K)e ? K ?Ker(φ) On the other hand, ?g ∈Ker(φ) ?HK, we have φ(g)= (H∩K)e and g=hk for some h ∈H and k ∈K . Now, φ(hk)= φ(g)= (H∩K)e ? h ∈(H∩K) ?K Hence, g=hk ∈K (e) If h1k1=h2k2 , then (H∩K)h1=φ(h1k1) =φ(h2k2) =(H∩K)h2 (f) By First Isomorphism Theorem, HK/Ker(φ) ≈ Im(φ) i.e. HK/K ≈ H/(H∩K) by part (c), (d) and (e). 圖片參考：http://i175.photobucket.com/albums/w130/bjoechan2003/My%20Cat%2020070530/DSCN2698.jpg
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