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MATH 103A Homework 4 - Solutions Due February 4, 2013 Version February 5, 2013 Assigned reading: Chapters 3-4 of Gallian. Recommended practice questions: Chapter 3 of Gallian, exercises 11, 12, 13, 14, 15, 16, 17, 20, 22, 23, 28, 31, 33, 34, 37, 50, 53, 78 Chapter 4 of Gallian, exercises 4, 5, 7, 13, 24, 30, 63, 82 Assigned questions to hand in: (1) (Gallian Chapter 3 # 10) How many subgroups of order 4 does D4 have? Solution: Recall that D4 e, R90 , R180 , R270 , F, F R90, F R180 , F R270. There are three subgroups of order 4, one cyclic and two not: R90 R270 e, R90 , R180 , R270 e, F, R180 , F R180 e, R180 , F R90, F R270 . (2) (Gallian Chapter 3 # 26) Prove that a group with two elements of order 2 that commute must have a subgroup of order 4. Solution: Let G be such a group and let a, b G be elements of order 2 with ab Consider the subgroup of G containing a and b. Then it must contain ba. e, a, a2 , a3 , . . . , a1 , . . . , b, b2 , b3 , . . . , b1 , . . . , ab, ba, abb, baba, . . . But, since a b 2, a1 a and b1 b. Moreover, all powers of a are either a or e and all powers of b are either b or e. Thus, we’ve reduced the subgroup to e, a, b, Since ab and products containing both a, b. ba, i1 j1 i2 j2 a b a b a in jn b i1 a in j1 b jn ab, a, or b depending on the parity of i1 . . . in and j1 jn . Thus, the subgroup of G containing a, b is e, a, b, ab. It remains to argue that ab a and ab b. Suppose, towards a contradiction, that ab a. By cancellation (left multiplication by a1 ), this gives b e so b 1, contradicting b 2. The symmetric argument shows that ab b. (3) (Gallian Chapter 3 # 45) Let G be an Abelian group with identity e and let n be some fixed integer. Prove that the set of all elements of G that satisfy the equation xn e is a subgroup of G. Give an example of a group G in which the set of all elements of G that satisfy the equation x2 e does not form a subgroup of G. x G : xn e. To show this is a Solution: Let G be an abelian group and H subgroup, we use the one-step subgroup test: Is H nonempty? Take x e. Then em e for all m so in particular en e and e G. For a, b H, is ab1 H? Let a, b H. Then an e and bn e. We check if ab1 H by considering the criterion for membership in H: ab1 n abelian anbn an bn 1 ee1 ee e. Thus, ab1 H. For the example, consider the group D4 . Then x D4 : x2 e e, R180 , F, F R90, F R180 , F R270. But, this set is not a group because it is not closed under multiplication. For example, F R90 F R180 F F R270 R180 R90 , which is not in the set. (4) (Gallian Chapter 3 # 69) Let H be a subgroup of R under addition. Let K Prove that K is a subgroup of R under multiplication. 2a : a H . Solution: We use the one-step subgroup test: Is K nonempty? Since H is nonempty, let a H and then 2a K. Let x, y K. Is xy1 K? By definition, there are a, b H such that x 2a and y 2b . Then y 1 2b and xy 1 2a 2b 2ab . But, H is a subgroup of R under addition so, since a, b H, a b H as well. Thus, 2ab K and xy 1 K. (5) (Gallian Chapter 4 # 28) Let a be a group element that has infinite order. Prove that ai aj if and only if i j. Solution: Suppose i j. Then either ai aj or ai is the inverse of aj . In either case, they generate the same group. For the converse, suppose ai aj . Then ai aj and aj ai . By definition, there are m, n Z such that ai amj and aj ani . Therefore, aimj e and aj ni e. But, since a , the only way this can happen is if i mj 0 and j ni 0. In other words, ij and j i. This implies that i j. (6) (Gallian Chapter 4 # 38) Consider the set 4, 8, 12, 16. Show that this set is a group under multiplication modulo 20 by constructing its Cayley (multiplication) table. What is the identity element? Is the group cyclic? If so, find all of its generators. Solution: 2 4 8 12 16 4 16 12 8 4 8 12 4 16 8 12 8 16 4 12 16 4 8 12 16 The Cayley table shows closure under the operation (which is associative) and e 41 16, 4, 81 12, 121 8, 161 16. This group is cyclic. 8 80 , 81, 82, 83, . . . 16, 8, 4, 12 12 120, 121, 122, 123, . . . 16, 12, 4, 8. Note that 16 16 and 4 16, 4 so the only generators of the group are 8 and 12. (7) (Gallian Chapter 4 # 64) Let a and b belong to a group. If a and b are relatively prime, show that a b e. Solution: Recall that the intersection of two subgroups is itself a subgroup, and is a subgroup of each group. That is, a b a and a b b. By the Fundamental Theorem of Cyclic Groups (Theorem 4.3), every subgroup of a cyclic group is cyclic. Therefore, a b is cyclic. Moreover, the same theorem says that the order of this group divides both a and b. But, by Corollary 1 of Theorem 4.1 (page 79), a a and b b. We assume that these two numbers are relatively prime. Therefore, the only common divisor is 1. Thus, 1 a b. The only group of order 1 is the trivial group, so a b e. 3

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