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Problem 1. Let G be a group and let H, K be two subgroups of G. Define the set HK = {hk : h ∈ H, k ∈ K}. a) Prove that if both H and K are normal then H ∩ K is also a normal subgroup of G. b) Prove that if H is normal then H ∩ K is a normal subgroup of K. c) Prove that if H is normal then HK = KH and HK is a subgroup of G. d) Prove that if both H and K are normal then HK is a normal subgroup of G. e) What is HK when G = D16 , H = {I, S}, K = {I, T 2 , T 4 , T 6 }? Can you give geometric description of HK? Solution: a) We know that H ∩ K is a subgroup (Problem 3a) of homework 33). In order to prove that it is a normal subgroup let g ∈ G and h ∈ H ∩ K. Thus h ∈ H and h ∈ K. Since both H and K are normal, we have ghg −1 ∈ H and ghg −1 ∈ K. Consequently, ghg −1 ∈ H ∩ K, which proves that H ∩ K is a normal subgroup. b) Suppose that H G. Let K ∈ k and h ∈ H ∩ K. Then khk −1 ∈ H (since H is normal in G) and khk −1 ∈ K (since both h and k are in K), so khk −1 ∈ H ∩ K. This proves that H ∩ K K. c) Let x ∈ HK. Then x = hk for some h ∈ H and k ∈ K. Note that x = hk = k(k −1 hk). Since k ∈ K and k −1 hk ∈ H (here we use the assumption that H G), we see that x ∈ KH. This shows that HK ⊆ KH. To see the opposite inclusion, consider y ∈ KH, so y = kh for some h ∈ H and k ∈ K. Thus y = (khk −1 )k ∈ HK, which proves that KH ⊆ HK and therefoere HK = KH. To prove that HK is a subgroup note that e = e · e ∈ HK. If a, b ∈ HK then a = hk and b = h1 k1 for some h, h1 ∈ H and k, k1 ∈ K. Thus ab = hkh1 k1 . Since HK = KH and kh1 ∈ KH, we have kh1 = h2 k2 for some k2 ∈ K, h2 ∈ H. Consequently, ab = h(kh1 )k1 = h(h2 k2 )k1 = (hh2 )(k2 k1 ) ∈ HK (since hh2 ∈ H and k2 k1 ∈ K). Thus HK is closed under multiplication. Finally, 1 a−1 = (hk)−1 = k −1 h−1 ∈ KH. Since KH = HK, we see that HK is closed under taking inverses so it is a subgroup. Note. We in fact proved that whenever HK = KH then HK is a subgroup. Two subgroups H, K are called permutable if HK = KH. The first part of c) says that a normal subgroup is permutable with any other subgroup. d) Let g ∈ G and x ∈ HK. Then x = hk for some h ∈ H and k ∈ K. Thus gxg −1 = g(hk)g −1 = (ghg −1 )(gkg −1 ) ∈ HK (since ghg −1 ∈ H and gkg −1 ∈ K). This proves that HK is a normal subgroup. e) By the very definition, HK = {I, T, T 2 , T 4 , T 6 , S, ST 2 , ST 4 , ST 6 }. Note that if we think of D16 as of symmetries of the regular 8−gon A1 A2 ...A8 (with T (Ai ) = Ai+1 and S being the reflection in the axis passing through A1 ) then HK consists of those symmetries which take the square A1 A3 A5 A7 into itself. In particular HK is a subgroup of D16 . This also follows from c) since K is a normal subgroup of D16 (prove it!). Note that HK can be identified with D8 (how ?). Problem 2. Let G be the group from Problem 2 of Homework 31. Let H = {Ta,b ∈ G : a is a rational number}. Prove that H is a normal subgroup of G. Solutuion: First note that H is indeed a subgroup. In fact, e = T1,0 ∈ H. Recall that Ta,b Tc,d = Tac,ad+b If Ta,b , Tc,d ∈ H then a, c ∈ Q so ac is rational and −1 −1 therefore Ta,b Tc,d ∈ H. Also, Ta,b = Ta−1 ,−a−1 b so Ta,b ∈ H (since a−1 is rational). This shows that H is a subgroup. Now if Ta,b ∈ G and Tc,d ∈ H then −1 Ta,b Tc,d Ta,b = Tac,ad+b Ta−1 ,−a−1 b = Taca−1 ,ac(−a−1 b)+ad+b = Tc,−cb+ad+b ∈ H since c is rational. Thus H is a normal subgroup. 2