Abstract Algebra Theory and Applications - Computer Science ...

HINTS AND SOLUTIONS 393Chapter 2. Groups1. (a) {. . . , −4, 3, 10, . . .}. (c) {. . . , −8, 18, 44, . . .}. (e) {. . . , −1, 5, 11, . . .}.2. (a) Not a group. (c) A group.6. · 1 5 7 111 1 5 7 115 5 1 11 77 7 11 1 511 11 7 5 18. Pick two matrices. Almost any pair will work.15. There is a group of order 6 that is nonabelian.16. Look at the symmetry group of an equilateral triangle or a square.17. There are actually five different groups of order 8.18. Letσ =( )1 2 · · · na 1 a 2 · · · a nbe in S n . All of the a i ’s must be distinct. There are n ways to choose a 1 ,n − 1 ways to choose a 2 , . . ., 2 ways to choose a n−1 , and only one way tochoose a n . Therefore, we can form σ in n(n − 1) · · · 2 · 1 = n! ways.24. (aba −1 ) n = (aba −1 )(aba −1 ) · · · (aba −1 ) = ab(aa −1 )b(aa −1 )b · · · (aa −1 )ba −1 =ab n a −1 .29. abab = (ab) 2 = e = a 2 b 2 = aabb ⇒ ba = ab.33. H 1 = {id}, H 2 = {id, ρ 1 , ρ 2 }, H 3 = {id, µ 1 }, H 4 = {id, µ 2 }, H 5 = {id, µ 3 },S 3 .39. id = 1 = 1 + 0 √ 2, (a + b √ 2 )(c + d √ 2 ) = (ac + 2bd) + (ad + bc) √ 2, and(a + b √ 2 ) −1 = a/(a 2 − 2b 2 ) − b √ 2/(a 2 − 2b 2 ).44. Not a subgroup. Look at S 3 .47. a 4 b = ba ⇒ b = a 6 b = a 2 ba ⇒ ab = a 3 ba = ba.Chapter 3. Cyclic Groups1. (a) False. (c) False. (e) True.2. (a) 12. (c) Infinite. (e) 10.3. (a) 7Z = {. . . , −7, 0, 7, 14, . . .}. (b) {0, 3, 6, 9, 12, 15, 18, 21}.(c) {0}, {0, 6}, {0, 4, 8}, {0, 3, 6, 9}, {0, 2, 4, 6, 8, 10}.(g) {1, 3, 7, 9}. (j) {1, −1, i, −i}.