Abstract Algebra Theory and Applications - Computer Science ...

398 HINTS AND SOLUTIONS9. For any homomorphism φ : Z 24 → Z 18 , the kernel of φ must be a subgroupof Z 24 and the image of φ must be a subgroup of Z 18 .14. Let a, b ∈ G. Then φ(a)φ(b) = φ(ab) = φ(ba) = φ(b)φ(a).18. False.19. If a ∈ G is a generator for G, then aH is a generator for G/H.25. Since eg = ge for all g ∈ G, the identity is in C(g). If x, y ∈ C(g), then xyg =xgy = gxy ⇒ xy ∈ C(g). If xg = gx, then x −1 g = gx −1 ⇒ x −1 ∈ C(g) ⇒C(g) is a subgroup of G. If 〈g〉 is normal in G, then g 1 xg1 −1 g = gg 1xg1 −1 forall g 1 ∈ G.28. (a) Let g ∈ G and h ∈ G ′ . If h = aba −1 b −1 , then ghg −1 = gaba −1 b −1 g −1 =(gag −1 )(gbg −1 )(ga −1 g −1 )(gb −1 g −1 ) = (gag −1 )(gbg −1 )(gag −1 ) −1 (gbg −1 ) −1 .We also need to show that if h = h 1 · · · h n with h i = a i b i a −1i b −1i , then ghg −1is a product of elements of the same type. However, ghg −1 = gh 1 · · · h n g −1 =(gh 1 g −1 )(gh 2 g −1 ) · · · (gh n g −1 ).Chapter 10. Matrix Groups and Symmetry1.1 [‖x + y‖ 2 + ‖x‖ 2 − ‖y‖ 2] 2= 1 [〈x + y, x + y〉 − ‖x‖ 2 − ‖y‖ 2]2= 1 [‖x‖ 2 + 2〈x, y〉 + ‖y‖ 2 − ‖x‖ 2 − ‖y‖ 2]2= 〈x, y〉.3. (a) An element of SO(2). (c) Not in O(3).5. (a) 〈x, y〉 = x 1 y 1 + · · · + x n y n = y 1 x 1 + · · · + y n x n = 〈y, x〉.7. Use the unimodular matrix ( 5 22 110. Show that the kernel of the map det : O(n) → R ∗ is SO(n).13. True.17. p6m.).Chapter 11. The Structure of Groups1. Since 40 = 2 3 · 5, the possible abelian groups of order 40 are Z 40∼ = Z8 × Z 5 ,Z 5 × Z 4 × Z 2 , and Z 5 × Z 2 × Z 2 × Z 2 .4. (a) {0} ⊂ 〈6〉 ⊂ 〈3〉 ⊂ Z 12 .(e) {((1), 0)} ⊂ {(1), (123), (132)} × {0} ⊂ S 3 × {0} ⊂ S 3 × 〈2〉 ⊂ S 3 × Z 4 .