# ahw5 - 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 6 = x H and g G , then 1 6 = 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 6 = 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 6 = 1 because a / H , we see that | aH | = 5. (d) Since | aH | = 5, we see that | a | 6 = 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

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## This note was uploaded on 01/02/2012 for the course MATH 4124 taught by Professor Staff during the Spring '08 term at Virginia Tech.

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ahw5 - Math 4124 Wednesday, March 16 Fifth Homework...

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