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# all2 - Math 4124 Wednesday March 16 Fifth Homework...

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Math 4124 Wednesday, March 16 Fifth Homework Solutions 1. Let G be a group of order 10 with a normal subgroup H of order 2. (a) Prove that H is contained in the center of G (hint: If 1 = x H and g G , then 1 = gxg - 1 H ). (b) Prove that G is abelian. (c) Let a G \ H . Prove that aH has order 5 in G / H . (d) Prove that G has an element y of order 5. (e) Let 1 = x H . Prove that the order of xy is neither 2 nor 5. (f) Prove that G is cyclic. (g) Prove that G = Z / 10 Z . (a) Since | H | = 2, it has two elements, one of which is the identity 1; we’ll call the other element x . Let g G . The g 1 g - 1 = 1. Also gxg - 1 H \ 1 (because H G ), hence gxg - 1 = x . It follows that g commutes with every element of H and we deduce that H is contained in the center of G . (b) Note that G / H is cyclic because it has order 5, which is a prime, and all groups of prime order are cyclic. Since H is contained in the center of G , we conclude that G is abelian. (c) By Lagrange’s theorem, | aH | divides | G / H | = 5. Since aH = 1 because a / H , we see that | aH | = 5. (d) Since | aH | = 5, we see that | a | = 1 or 2. Also | a | divides 10 by Lagrange’s theorem. Therefore | a | = 5 or 10. If | a | = 5, we’re finished. On the other hand if | a | = 10, then | a 2 | = 5 as required. (e) Note that | x | = 2. Since G is abelian, ( xy ) 2 = x 2 y 2 = y 2 = 1 and ( xy ) 5 = x 5 y 5 = x = 1. Therefore | xy | = 2 or 5. (f) Note that xy = 1. Therefore | xy | = 1, 2 or 5. It follows from Lagrange’s theorem that | xy | = 10 and we conclude that G is cyclic. (g) All cyclic groups of order 10 are isomorphic to Z / 10 Z . The result follows, be- cause G is cyclic of order 10. 2. 3.3.8 on page 101. Let p be a prime and let G be the group of p -power roots of 1 in C . Prove that the map z z p is a surjective homomorphism. Deduce that G is isomorphic to a proper quotient of itself. Define θ : G G by θ g = g p . Then for x , y G , we have ( xy ) p = x p y p because G is abelian, consequently θ ( xy ) = ( xy ) p = x p y p = θ x θ y ,

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so θ is a homomorphism. Suppose g G , then g = e 2 k π i / p n for some k , n Z . Set h = e 2 k π i / p n + 1 . Then h G and θ h = g . Therefore θ is surjective. Finally e 2 π i / p ker θ , so ker θ = 1. The result now follows from the first isomorphism theorem. 3. Let G be a group, and suppose there is a homomorphism of G onto S 3 (the symmetric group of degree 3) with kernel K . Determine the number of subgroups of G which contain K , and show that exactly three of these subgroups are normal. By the subgroup correspondence theorem, the subgroups containing K are in a one- to-one correspondence with the subgroups of G / K , and this correspondence preserves normality. This means we have to determine the subgroups of S 3 . Any subgroup of S 3 has order dividing 6, so any subgroup not equal to 1 or S 3 has prime order and is therefore cyclic. It follows easily that the subgroups of S 3 are 1, S 3 , ( 12 ) , ( 23 ) , ( 31 ) , ( 123 ) , so there are six subgroups of G which contain K . Finally the normal subgroups of S 3 are 1, S 3 and ( 123 ) . We conclude that the number of normal subgroups containing K is 3, as required.
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