Solutions to A First Course in Abstract Algebra John B. Fraleigh sixth edition ISBN 0-201-33596-4 Addison Wesley Longman by Ben Hekster PO Box 391852 Mountain View, CA 94039-1852 heksterb@acm.org http://www.hekster.org/Academic/Mathematics/ These are completely unofficial and unverified worked solutions by me. Corrections welcome. Typeset in Galliard and Gill Sans using AppleWorks 5 on Apple Macintosh. Copyright © 1999-2005 All Rights Reserved by: Ben Hekster. 1 Abstract Algebras set + binary operation binary algebraic structure + associative binary operation semigroup + identity monoid + inverse group + finite order + commutative operation finite group commutative group + prime order + generator symmetric group cyclic group + prime order alternating group prime order group Glossary : reads as “so that” + , ⋅ summation, multiplication over i i i ( { ordered, unordered set over i i i ∧,∨,<,> scalar operators ∩,∪,⊂,⊃ set operators = congruent modulo n n < is normal to, is ideal to ( ) fx function application f x commutative group abelian group maximal p-group Sylow p-group 1 §0.1 Preliminaries 1. proving theorems 2. set 3. precision? 4. definition 5. A triangle with vertices P, Q, R is the collection of points X such that • X is in the line segment PQ, or • X is in the line segment QR, or • X is in the line segment RP. 6. An equilateral triangle is a triangle with vertices P, Q, R such that the length of the line segment PQ equals both the length of the line segment QR and the length of the line segment RP. 7. A right triangle is a triangle with vertices P, Q, R in which the two line segments through one of its vertices (say PQ and PR) are such, that for any point X on PQ there is no point Y on PR such that the length of the line segment XY is less than the length of the line segment XP. 8. The interior of a triangle is the collection of points X such that the line segments XP, XQ, XR from X to its vertices P, Q, R have only the vertices in common with the triangle. 9. A circle with center C and radius r is the collection of points X such that the length of the line segment XC equals r. 10. A disk with center C and radius r is the collection of points X such that the length of the line segment XC is less than or equal to r. 11. Define the relationship between PQ and PR in 7. to be a right angle. Then, a rectangle with vertices P, Q, R, S is the collection of points formed by the four line segments PQ, QR, RS, SP, where PQ is at a right angle to QR, QR to RS, RS to SP, and SP to PQ. 12. Let n and m be even integers. Then by (2), there are integers p, q such that n = 2p, m = 2q. Then n + m = 2p + 2q = 2(p + q), so n + m is even. 13. Let n, m, p, q as in 12. Then nm = 2p · 2q = 4pq. Since pq is an integer, 4pq is an integral multiple of 4. 14. Define an odd integer m to be an integer such that there exists another integer n such that m = 2n + 1. Let r be an even integer and s an odd integer. Then there are integers p, q such that r = 2p, s = 2q + 1. So r + s = 2p + 2q + 1 = 2(p + q) + 1, so r + s is odd. 15. counterexample 16. A B F G M, C D J, E H K N, I, L, O. 17. 1, 2, 4, 8, 16, 31 (the conjecture is false). 18. Suppose that i is the square of an odd integer k. Then ∃ i∈(cid:2):k =2l +1⇒i =k2 =(2l +1)2 =4l2+4l +1 Since i is also even, ∃j ∈(cid:2): i =2j ⇒ 4l2+4l +1=2j ⇒ 2l2+2l + 1 = j ∉(cid:2) 2 which is a contradiction, so k cannot be odd. Since k must be even, ∃l∈(cid:3): k =2l⇒ i =k2 =(2l)2 = 4l2 so i is indeed an integral multiple of 4. 19. Let n = 0, then (n+3)2 =32 =9>/ 9. 20. Let n2+2=3⇒ n2 =1⇒ n=−1∨n=+1, so n is not unique. 21. Let n=2⇒ n2+4=22+4=8. 22. Let n=3⇒ n2+5=32+5=14. 23. Let n=−3⇒ n2+5=(−3)2+5=14. With 22., n is not unique. 24. Let n=0: n2 >n⇐ 02 >0, which is a contradiction. 25. Let n ∈(cid:3),n <0⇒ n2 > 0⇒ n2 >0 >n. ( ) 2 26. Let x = 1 ⇒ x2 n⇐ 22 >2⇐ 4>2. n=0: n2 =n⇐ 02 =0⇐ 0=0 28. Let , so x is not unique. n=1: n2 =n⇐ 12 =1⇐ 1=1 29. Let j be an odd integer, so ∃k∈(cid:2): j =2k+1 ⇒ j 2 =(2k+1)2 =4k2 +4k+1 = k2 +k∈(cid:3) ( ) 30. ∃m∈(cid:3): n =3m+1, so n2 =(3m+1)2 =9m2+6m+1=33m2+2m +1, and 3 m2+2m is integral. 31. Let n=−2: n3 A . ( ) Let A be such that A =ℵ. Then the power set of A has P(A) >ℵ, and P P(A) > P(A), ad infinitum. 20. a. It is possible to define addition in (cid:3) in terms of the union of disjoint sets, so 2+3=5⇐ A =2, B =3, A∪B =5. i. 3+ℵ = {0}∪ (cid:2)+ (=*) (cid:2)+ =ℵ , where (*): φ:(cid:2)+ →{0}∪ (cid:2)+ :ma m−1. 0 0 ( ) (*) ( )  m odd: 1(m−1)+ 1 ii. ℵ +ℵ = (cid:2)+ − 1 ∪ (cid:2)+ = (cid:2)+ =ℵ , where (*) φ: (cid:2)+ → (cid:2)+ − 1 ∪ (cid:2)+ :m a  2 2. 0 0 2 0 2  m even: 1m 2 b. It is possible to define multiplication in (cid:3) in terms of a Cartesian product: 2⋅3={1,2}×{1,2,3}=6, so fig 14 ℵ ⋅ℵ = (cid:2)+ × (cid:2)+ = (cid:2)+ =ℵ . 0 0 0 21. 102 digits, 105 digits. By extrapolation, 1 0ℵ0 would equal the number of digits of the form 0.###…, where ‘#’ is { } repeated ℵ times— name this set R. Since any number in R ′ = x ∈(cid:4)|0 ≤x <1 can be expressed arbitrarily 0 precise by an element of R, R ′⊆R. Since R ⊆R′, R =R′. By Exercise 15, R′ =ℵ, so R =ℵ and 10ℵ0 =ℵ. Similar arguments can be made in terms of duodecimal and binary expansions of numbers of R′, so 1 2ℵ0 =2ℵ0 =ℵ. 22. Since P((cid:2))(1=7)2(cid:2) = 2ℵ0 (1=8)ℵ; P((cid:2))(1=9){0,1}(cid:2) The next higher cardinals after ℵ are 0 { }(cid:2) { } ℵ= 0,1 = 0,1 exp (cid:2), {0,1}{0, 1}(cid:2) ={0,1}exp {0,1} exp (cid:2) et cetera. 28. xRy ⇒∃i:x,y ∈P ⇒y,x ∈P ⇒y Rx (symmetric) i i xRx ⇔∃i:x ∈P ⇔x ∈S (reflexive) i xRy ∧y Rz ⇒∃i:x,y ∈P ∧∃j:y,z ∈P i j ⇒(P disjoint⇒i = j)∃i:x,y,z ∈P (transitive). i i ⇒y Rz 29. not reflexive because 0⋅0>/ 0⇒0R/ 0 30. not symmetric because 2≥1,1≥/ 2⇒2R1,1R/ 2. 5 31. R is a relation, because y x = x xRx R   x = y ⇒ y = x ⇒xRy ⇒y Rx   x x = y ∧ y = x ⇒ x = z xRy ∧y Rz ⇒xRz  . 0−3 =3≤3, 3−6 =3≤3, 0−6 =6≤/ 3⇒ 32. so R is not transitive. 0R3, 3R6, 0R/ 6 33. The number of digits of n ∈(cid:2)+ is base 10 notation is 1+10logn. Obviously R is reflexive and symmetric, and transitive. R 0 10 100 34. R is congruence modulo 10 on (cid:2)*. 35. a. {1, 3, 5, …}, {2, 4, 6, …} b. {1, 4, 7, …}, {2, 5, 8, …}, {3, 6, 9, …} c. {1, 6, 11, …}, {2, 7, 12, …}, {3, 8, 13, …}, {4, 9, 14, …}, {5, 10, 15, …} 36a. ∀r ∈(cid:2): r −r =0 =0⋅n ⇒r ~r ( ) ( ) ∀r,s,r ~s : ∃q∈(cid:2): r −s =qn ⇒ s −r =− qn = −q n⇒ s ~r ∀r,s,t,r ~ s,s ~t: ∃p,q ∈(cid:2): r −s =pn, s −t=qn r −s +s −t =pn +qn r −t =(p +q)n⇒ r ~t r −s r s b. ∀r,s ∈(cid:2)+,r ~s : ∃q∈(cid:2): r −s =qn ⇒(n∈(cid:2)) = − =q n n n ∃r ′ ,s ′ ∈(cid:2),r ′′ , s ′′ ∈(cid:3): r =r ′ n +r ′′ , s = s ′ n +s ′′ ,0 ≤r ′′ , s ′′

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