Solution (incorrect!) (i) Group G: Modular multiplication is commutative, so G is abelian. Hence, by Theorem B35, G is cyclic. In (G, X21) we have: 2² = 4, 2³ = 8, 24 = 16, 25 16 ×21 2 = 11, 26 11 X21 2 = 1. Thus (2) = {1, 2, 4, 8, 11, 16} = G, and hence 2 is a generator for G. Group H: We can see from the table that (e) = {e, r, s, t, u, v}, (s) = {s, e,r, u, v, t}, (u) = {u, v, t, s, e, r}, Thus (H, *) is cyclic, because every element of H is a generator. (ii) (G, X21) and (Z6, +6) are both cyclic groups of order 6, so they are isomorphic and we can find an isomorphism by using Strategy B6. From part (i), 2 is a generator for G. We choose the generator 1 for Z6. This gives the following isomorphism: : (G, X21) → (Z6, +6) 110 2 1 (r) = {r, s, e, v, t, u}, (t) = {t, u, v, e, r, s}, (v) = {v, t, u, r, s, e}. 412 8 3 114 165. (b) Write out a correct solution to the exercise.
Solution (incorrect!) (i) Group G: Modular multiplication is commutative, so G is abelian. Hence, by Theorem B35, G is cyclic. In (G, X21) we have: 2² = 4, 2³ = 8, 24 = 16, 25 16 ×21 2 = 11, 26 11 X21 2 = 1. Thus (2) = {1, 2, 4, 8, 11, 16} = G, and hence 2 is a generator for G. Group H: We can see from the table that (e) = {e, r, s, t, u, v}, (s) = {s, e,r, u, v, t}, (u) = {u, v, t, s, e, r}, Thus (H, *) is cyclic, because every element of H is a generator. (ii) (G, X21) and (Z6, +6) are both cyclic groups of order 6, so they are isomorphic and we can find an isomorphism by using Strategy B6. From part (i), 2 is a generator for G. We choose the generator 1 for Z6. This gives the following isomorphism: : (G, X21) → (Z6, +6) 110 2 1 (r) = {r, s, e, v, t, u}, (t) = {t, u, v, e, r, s}, (v) = {v, t, u, r, s, e}. 412 8 3 114 165. (b) Write out a correct solution to the exercise.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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