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MATH 103A Homework 5 - Solutions Due February 15, 2013 Version February 4, 2013 Assigned reading: Chapters 4, 6,7 of Gallian. Recommended practice questions: Chapter 4 of Gallian, exercises Chapter 6 of Gallian, exercises 2, 6, 11, 12, 15, 16, 18 Chapter 7 of Gallian, exercises 3, 4, 5, 6, 7, 10, 15, 17, 26, 33 Assigned questions to hand in: (1) (Gallian Chapter 6 # 10) Let G be a group. Prove that the mapping α g g 1 for all g G is an automorphism if and only if G is Abelian. Solutions: Suppose α is an automorphism. Let x, y G. We want to show that xy Applying the property of automorphism to xy 1 : α xy 1 definition of α xy 1 1 yx. xy, but also autom. α xy 1 α y 1x1 α y 1α x1 y 1 1 x1 1 yx. Thus, xy yx. Conversely, suppose G is abelian. To prove that α is an automorphism, we need two facts: (1) WTS α is a bijection. It is sufficient to exhibit an inverse for α. In fact, we will show that α is its own inverse. Let x G, then α α x α x1 x1 1 x. Thus, α α identity and α has an inverse so is a bijection. (2) WTS α preserves the operation. Let x, y G. Then α xy xy 1 y 1x1 abelian x1 y 1 α xα y , as required. (2) (Gallian Chapter 6 # 22) Let φ be an automorphism of a group G. Prove that H G : φ x x is a subgroup of G. x Solution: We use the one-step subgroup test. Nonempty? Consider e G. By Theorem 6.2.1, φ e e so e H. Thus, H . Closed? Let a, b H. WTS that ab1 H. By definition of H, φ a a and φ b b. 6.2.2 φ ab1 φ aφ b1 φ aφ b1 ab1 . Thus, ab1 satisfies the condition for membership in H. (3) (Gallian Chapter 6 # 27) Let r U n. Prove that the mapping α : Zn α s sr mod n for all s Zn is an automorphism of Zn . Zn defined by Solution: Fix n and r. By question 3 on homework 1 (Ch 0 # 11), there is x such that rx mod n 1; in fact, x mod n r 1 in U n. (1) Bijection? We will prove that the function β s sr 1 mod n (for r 1 U n) is the inverse of α. Note that the domain and codomain of α and β is Zn , as required. Let s Zn . α β s α sr 1 mod n sr 1 mod n s r 1 r mod n s, mod nr by associativity of multiplication mod n. β α s β sr mod n sr mod nr 1 mod n s rr 1 mod n s, by associativity of multiplication mod n. Thus, α and β are inverse functions and are bijections. (2) Preserves operation? Let s, t Zn . Then α s t s tr mod n sr mod n tr mod n α s α t, by distributivity of multiplication mod n over addition mod n. (4) (Gallian Chapter 6 # 28) The group group? What if Z is replaced by R? 1 a 0 1 :aZ is isomorphic to what familiar Solution: It is isomorphic to Z or R under addition (respectively). To prove this, consider 1 a the function φ a. 0 1 1 a works. (1) Bijection? The inverse ψ a 0 1 1 a 1 b and be given. Then (2) Preserves operation? Let 0 1 0 1 φ 1 a 0 1 1 b 0 1 Matrix Mult φ 1 ba 0 1 baabφ 1 a 0 1 φ 1 b 0 1 . (5) (Gallian Chapter 6 # 34) Prove or disprove that U 20 and U 24 are isomorphic. Solution: These groups are not isomorphic. We prove this by demonstrating that every element in U 24 has order 1 or 2 whereas there is an element of U 20 with order 4. U 24 1, 5, 7, 11, 13, 17, 19, 23. To prove that x 2 for x U 24, it is sufficient to show that x2 12 72 13 2 19 2 49 1, 52 25 mod 24 1, 112 mod 24 1, 169 mod 24 1, 361 mod 24 1, 2 121 mod 24 1, 17 2 289 mod 24 1, 23 2 529 mod 24 1. 1 for each x U 24. On the other hand, U 20 1, 3, 7, 9, 11, 13, 17, 19 and 3 1, 3, 9, 7 so 3 4 2. By Theorem 6.2.7, two finite isomorphic groups have exactly the same number of elements of every order. Since this condition fails for U 20 and U 24, they cannot be isomorphic. (6) (Gallian Chapter 6 # 48) Let φ be an isomorphism from a group G to a group Ḡ and let a G. Prove that φ C a C φ a. Solution: Fix G and a. We prove subset inclusion in two directions. Let y φ C a. Then there is x C a such that y φ x. WTS that y namely that y commutes with φ a: Because x yφ a φ xφ a φ xa C a C φ a, φ ax φ aφ x φ ay, as required. Let y C φ a. Then yφ a φ ay. WTS that y φ C a, namely that there is x C a with y φ a. Since φ is an isomorphism, it has an inverse and φ1 y exists. We claim that φ1 y C a. Recall that by Theorem 6.3.1, φ1 is an isomorphism. So: C φa 1 φ φ ay φ1 φ aφ1 y aφ1 y , as required. Thus, φ1 y C a and φ φ1 y y so y φ C a. φ1 y a φ1 y φ1 φ a φ1 yφ a Because y (7) (Gallian Chapter 7 # 8) Suppose that a has order 15. Find all the left cosets of a5 in a. Solution: The subgroup H a e, a 5 5 , a5 2 , a5 3 , . . . e, a5 , a10 . The left cosets of H are of the form xH for x G, namely ak a5 since each x G is ak for some k eH a2 H e, a 5 a 2 Z. In particular, ,a 10 , aH a, a 6 , a11 , a7 , a12 , a3 H a3 , a8 , a13 a4 H a4 , a8 , a14 but a5 H a5 , a10 , a15 a5 , a10 , e eH. Thus, there are 5 distinct left cosets of this subgroup. (8) (Gallian Chapter 7 # 24) Suppose that H and K are subgroups of G and there are elements a, b G such that aH bK. Prove that H K. Solution: Let a, b G such that aH bK. In particular, since H is a subgroup of G, e G and a ae bK so there is k K such that a bk. Note that a1 bk 1 k 1 b1 and that k 1 K (since K is a subgroup of G). 3 Suppose x H. We will show that x K. By definition ax ax bK. Namely, there is y K such that ax by. x a1 ax a1 ax a1 by k 1 b1 by k 1 y aH so by assumption, K since k 1 , y K and K is a subgroup of G so is closed under the operation. (9) (Gallian Chapter 7 # 26) Suppose that G is a group with more than one element and G has no proper, nontrivial subgroups. Prove that G is prime. (Do not assume at the outset that G is finite.) Note that by Corollary 3 to Lagrange’s theorem, this exercise proves that G is cyclic. Solution: G contains at least the elements e, x for x some non-identity element. We have proved that x G. Since x e, x e. By assumption that G contains no proper nontrivial subgroups, it must be the case that x G. We consider two cases. Case 1: x . By Theorem 4.1, for each i, j Z, xi xj iff i j. In particular, e x2n : n Z xn : n Z. But, x2n : n Z x2 is a subgroup of G. In this case, we have a proper nontrivial subgroup of G, which contradicts the assumption on G. Case 2: x . By the Fundamental Theorem of Cyclic Groups (Theorem 4.3), each positive divisor of x corresponds to a subgroup of G x of that order. To guarantee that G has no proper nontrivial subgroups, x can have no positive divisors other than 1 and x. That is, x is prime. But, x x G so we have shown that G is prime. 4