SOLUTION MANUAL First Course in Abstract Algebra A 8th Edition by John B. Fraleigh All Chapters Full Complete

CONTENTS 1. Sets and Relations 1 I. Groups and Subgroups 2. Introduction and Examples 4 3. Binary Operations 7 4. Isomorphic Binary Structures 9 5. Groups 13 6. Subgroups 17 7. Cyclic Groups 21 8. Generators and Cayley Digraphs 24 II. Permutations, Cosets, and Direct Products 9. Groups of Permutations 26 10. Orbits, Cycles, and the Alternating Groups 30 11. Cosets and the Theorem of Lagrange 34 12. Direct Products and Finitely Generated Abelian Groups 37 13. Plane Isometries 42 III. Homomorphisms and Factor Groups 14. Homomorphisms 44 15. Factor Groups 49 16. Factor-Group Computations and Simple Groups 53 17. Group Action on a Set 58 18. Applications of G-Setsto Counting 61 IV. Rings and Fields 19. Rings and Fields 63 20. Integral Domains 68 21. Fermat’s and Euler’s Theorems 72 22. The Field of Quotients of an Integral Domain 74 23. Rings of Polynomials 76 24. Factorization of Polynomials over a Field 79 25. Noncommutative Examples 85 26. Ordered Rings and Fields 87 V. Ideals and Factor Rings 27. Homomorphisms and Factor Rings 89 28. Prime and Maximal Ideals 94 29. Gro¨bner Bases for Ideals 99 VI. Extension Fields 30. Introduction to Extension Fields 103 31. Vector Spaces 107 32. Algebraic Extensions 111 33. Geometric Constructions 115 34. Finite Fields 116 VII. Advanced Group Theory 35. Isomorphism Theorems 117 36. Series of Groups 119 37. Sylow Theorems 122 38. Applications of the Sylow Theory 124 39. Free Abelian Groups 128 40. Free Groups 130 41. Group Presentations 133 VIII. Groups in Topology 42. Simplicial Complexes and Homology Groups 136 43. Computations of Homology Groups 138 44. More Homology Computations and Applications 140 45. Homological Algebra 144 IX. Factorization 46. Unique Factorization Domains 148 47. Euclidean Domains 151 48. Gaussian Integers and Multiplicative Norms 154 X. Automorphisms and Galois Theory 49. Automorphisms of Fields 159 50. The Isomorphism Extension Theorem 164 51. Splitting Fields 165 52. Separable Extensions 167 53. Totally Inseparable Extensions 171 54. Galois Theory 173 55. Illustrations of Galois Theory 176 56. CyclotomicExtensions 183 57. Insolvability of the Quintic 185 APPENDIX Matrix Algebra 187 iv − 0. Sets and Relations 1 1. Sets and Relations √ √ 1. { 3, − 3} 2. The set is empty. 3. {1, −1,2,−2,3, −3,4, −4,5,−5, 6, −6,10, −10,12, −12,15,−15,20,−20,30,−30, 60, −60} 4. {−10,−9,−8,−7,−6,−5,−4,−3, −2,−1,0, 1, 2, 3,4, 5,6,7, 8,9, 10,11} 5. It is not a well-defined set. (Some may argue that no element of Z + is large, because every element exceeds only a finite number of other elements but is exceeded by an infinite number of other elements. Such people might claim the answer should be ∅.) 6. ∅ 7. The set is ∅ because 3 3 = 27 and 4 3 = 64. 8. It is not a well-defined set. 9. Q 10. The set containing all numbers that are (positive, negative, or zero) integer multiples of 1, 1/2, or 1/3. 11. {(a, 1), (a, 2), (a, c), (b, 1), (b, 2), (b, c), (c, 1), (c, 2), (c, c)} 12. a. It is a function. It is not one-to-one since there are two pairs with second member 4. It is not onto B because there is no pair with second member 2. b. (Same answer as Part(a).) c. It is not a function because there are two pairs with first member 1. d. It is a function. It is one-to-one. It is onto B because every element of B appears as second member ofsome pair. e. Itis a function. It is not one-to-one because there are two pairs with second member 6. It is not onto B because there is no pair with second member 2. f. It is not a function because there are two pairs with first member 2. 13. Draw the line through P and x, and let y be its point of intersection with the line segment CD. 14. a. φ : [0, 1] → [0, 2] where φ(x) = 2x b. φ : [1, 3] → [5, 25] where φ(x) = 5 + 10(x − 1) c. φ : [a, b] → [c, d] where φ(x) = c + d− c (x − a) b a 15. Let φ : S → R be defined by φ(x) = tan(π(x − ) 2 ). 16. a. ∅; cardinality 1 b. ∅, {a}; cardinality 2 c. ∅,{a},{b},{a, b}; cardinality 4 d. ∅,{a},{b},{c},{a, b},{a, c},{b, c},{a, b, c}; cardinality 8 17. Conjecture: |P(A)| = 2 s = 2| A| . Proof The number of subsets of a set A depends only on the cardinality of A, not on what the elements of A actually are. Suppose B = {1, 2, 3, · · · ,s − 1} and A = {1, 2, 3, ,s}. Then A has all the elements of B plus the one additional element s. All subsets of B are also subsets of A; these are precisely the subsets of A that do not contain s, so the number of subsets of A not containing s is |P(B)|. Any other subset of A must contain s, and removal of the s would produce a subset of B. Thus the number of subsets of A containing s is also |P(B)|. Because every subset of A either contains s or does not contain s (but not both), we see that the number of subsets of A is 2|P(B)