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SOLUTION MANUAL for Abstract Algebra: An Interactive Approach, Even Numbered & Old Numbered Solutions. 2nd Edition by William Paulsen

Note: Odd Numbered Questions Answers are the end of Text Book.

Answers to Even-Numbered Problems Section 0.1 2) q = 15, r = 12 4) q = −21, r = 17 6) q = 87, r = 67 8) q = −1, r = 215 10) 1 + n < 1 + (n − 1)2 = n2 + 2(1 − n) < n2 12) If (n − 1)2 + 3(n − 1) + 4 = 2k, then n2 + 3n + 4 = 2(k + n + 1). 14) If 4n−1 − 1 = 3k, then 4n − 1 = 3(4k + 1). 16) (1 + x)n = (1 + x)(1 + x)n−1 ≥ (1 + x)(1 + (n − 1)x) = 1 + nx + x2 (n − 1) ≥ 1 + nx 18) (n − 1)2 + (2n − 1) = n2 . 20) (n − 1)2 ((n − 1) + 1)2 /4 + n3 = n2 (n + 1)2 /4. 22) (n − 1)/((n − 1) + 1) + 1/(n(n + 1)) = n/(n + 1). 24) 4 · 100 + (−11) · 36 = 4. 26) (−6) · 464 + 5 · 560 = 16. 28) (−2) · 465 + 9 · 105 = 15. 30) (−54) · (487) + (−221) · (−119) = 1. 32) Let c = gcd(a, b). Then c is the smallest positive element of the set A = all integers of the form au + bv. If we multiply all element of A by d, we get the set of all integers of the form dau + dbv, and the smallest positive element of this set would be dc. Thus, gcd(da, db) = dc. 34) Since both x/gcd(x, y) and y/gcd(x, y) are both integers, we see that (x · y)/gcd(x, y) is a multiple of both x and y. If lcm(x, y) = ax = by is smaller then (x · y)/gcd(x, y), then (x · y)/lcm(x, y) would be greater than gcd(x, y). Yet (x · y)/lcm(x, y) = y/a = x/b would be a divisor of both x and y. 36) 2 · 3 · 23 · 29. 38) 7 · 29 · 31. 40) 3 · 132 · 101. 42) u = −222222223, v = 1777777788. 44) 34 · 372 · 3336672 . Section 0.2 2) If a/b = c/d, so that ad = bc, then ab(c2 +d2 ) = abc2 +abd2 = a2 cd+b2 cd = cd(a2 + b2 ). Thus, ab/(a2 + b2 ) = cd/(c2 + d2 ). 4) a) One-to-one, 3x + 5 = 3y + 5 ⇒ x = y. b) Onto, f ((y − 5)/3) = y. 6) a) One-to-one, x/3 − 2/5 = y/3 − 2/5 ⇒ x = y. b) Onto, f (3y + 6/5) = y.

Answers to Even-Numbered Problems

8) a) One-to-one, if x > 0, y < 0 then y = 3x > 0. b) Onto, if y ≥ 0, f (y/3) = y. If y < 0, f (y) = y. 10) a) Not one-to-one f (1) = f (2) = 1. b) Onto, f (2y − 1) = y. 12) a) One-to-one, if x even, y odd, then y = 2x + 2 is even. b) Not onto, f (x) ̸= 3. 14) a) Not one-to-one f (5) = f (8) = 24. b) Not onto, f (x) ̸= 1. 16) Suppose f were one-to-one, and let B̃ = f (A), so that f˜ : A → B̃ would be a bijection. By lemma 0.5, |A| = |B̃|, but |B̃| ≤ |B| < |A|. 18) Suppose f were not one-to-one. Then there is a case where f (a1 ) = f (a2 ), and we can consider the set Ã = A − {a1 }, and the function f˜ : Ã → B would still be onto. But |Ã| < |B| so by Problem 17 f˜ cannot be onto. Hence, f is one-to-one. 20) x4 + 2x2 . 22) x3 − 3x{+ 2. 3x + 14 if x is even, 24) f (x) = 6x + 2 if x is odd. 26) If f (g(x)) = f (g(y)), then since f is one-to-one, g(x) = g(y). Since g is onto, x = y. 28) There is some c ∈ C such that f (y) ̸= c for all y ∈ B. Then f (g(x)) ̸= c since g(x) ∈ B. 30) If x even and y odd, f (x) = f{ (y) means y = x + 8 is even. Onto is proven x + 3 if x is even, −1 by finding the inverse: f (x) = x − 5 if x is odd. 32) Associative, (x ∗ y) ∗ z = x ∗ (y ∗ z) = x + y + z − 2. 34) Not associative, (x ∗ y) ∗ z = x − y − z, x ∗ (y ∗ z) = x − y + z. 36) Yes. 38) Yes. 40) Yes. 42) f (x) is both one-to-one and onto. Section 0.3 2) 55 4) 25 6) 36 8) 7 10) 10 12) 91 14) 43 16) 223 18) 73 20) 1498 22) 3617 24) 3875 26) First find 0 ≤ q ≤ u · v such that q ≡ x(mod u) and q ≡ y(mod v). Then find k so that k ≡ q(mod u · v) and k ≡ z(mod w). 28) 12

Answers to Even-Numbered Problems

30) 4 32) 35 34) 17 36) 30 38) 51 40) 3684623194282304903214 42) 21827156424272739145155343596495185185220332 44) 1334817563332517248 Section 0.4 2) Since 1+2⌊an ⌋ is an integer, 1+2⌊an ⌋−an will have the same denominator as an . Thus, the numerator an+1 is the denominator of an . Note that the fractions will already be in lowest terms. 4) Since the sequence begins b0 = 0, b1 = 1, b2 = 1, b3 = 2,. . . we see that the equations are true for n = 1. Assume both equations are true for the previous n, that is, b2n−2 = bn−1 and b2n−1 = bn−1 + bn . Then by the recursion formula, b2n = bn−1 + (bn−1 + bn ) − 2(bn−1 mod (bn−1 + bn )). But (bn−1 mod (bn−1 + bn )) = bn−1 , since bn−1 + bn > bn−1 . So b2n = bn . Then we can compute b2n+1 = bn−1 + bn + bn − 2(bn−1 + bn mod bn ). But (bn−1 + bn mod bn ) = (bn−1 mod bn ), and bn+1 = bn−1 + bn − 2(bn−1 mod bn ). Thus, b2n+1 = bn + bn+1 . 6) a2n+1 = b2n+1 /b2n+2 = (bn + bn+1 )/bn+1 = (bn /bn+1 ) + 1 = an + 1. 8) If ai = aj for i > j, then because an+1 is determined solely on an , a2i−j = ai . In fact, the sequence will repeat forever, so there would be only a finite of rational numbers in the sequence. But this contradicts that every rational is in the sequence, which is an infinite set. 10) In computing the long division of p/q, the remainders at each stage is given by the sequence in Problem 9. Since this sequence eventually repeats, the digits produced by the long division algorithm will eventually repeat. 12) If p3 /q 3 = 2 with p and q coprime, then 2|p, but replacing p = 2r shows 2|q too. 14) If p2 /q 2 = 5 with p and q coprime, then 5|p, but replacing p = 5r shows 5|q too. 16) If p3 /q 3 = 3 with p and q coprime, then 3|p, but replacing p = 3r shows 3|q too. 18) If 1/a were rational, then a = 1/a−1 would be rational. 20) Given x and y, choose any irrational z, and find a rational q between x − z and y − z. Then q + z is irrational by Problem 19. √ √ 22) x2 = 5 + 2 6, and 6 is irrational, so x2 is too. If x were rational, then x2 would be rational. √ √ 24) 2 − 2 and 2 are both irrational, but the sum is 2. √ √ 26) a6 = 2 + 4, a102 = 2 + 8.

Answers to Even-Numbered Problems

Section 1.1 2) 12 steps. 4) y = y · e = y · (x · y ′ ) = (y · x) · y ′ = e · y ′ = y ′ , so y = y ′ . 6) x = a−1 · b. 8) After 2 flips, Terry will be facing towards the audience again, so it would be a rotation. 10) FlipRt·Spin = Spin·FlipLft. Other answers are possible. 12) (FlipRt·Spin)2 ̸= Stay·Stay. Other answers are possible. 14) FlipRt, FlipRt, and Spin cannot be expressed in terms of RotLft or RotRt. Section 1.2

2) 0 1 2 3 4 5 6) 0 3 6 9 12 15 18 21

0 0 1 2 3 4 5

1 1 2 3 4 5 0

2 2 3 4 5 0 1

3 3 4 5 0 1 2

4 4 5 0 1 2 3

5 5 0 1 2 3 4

4) 0 2 4 6

0 3 6 9 12 15 18 21 0 3 6 9 12 15 18 21 3 6 9 12 15 18 21 0 6 9 12 15 18 21 0 3 9 12 15 18 21 0 3 6 12 15 18 21 0 3 6 9 15 18 21 0 3 6 9 12 18 21 0 3 6 9 12 15 21 0 3 6 9 12 15 18

0 0 2 4 6

2 2 4 6 0

4 4 6 0 2

6 6 0 2 4

1 1 2 4 5 7 8

2 2 4 8 1 5 7

4 4 8 7 2 1 5

8) 1 2 4 5 7 8

5 5 1 2 7 8 4

7 7 5 1 8 4 2

8 8 7 5 4 2 1

12) 10) 1 3 5 9 11 13

1 1 3 5 9 11 13

3 3 9 1 13 5 11

5 5 1 11 3 13 9

9 9 13 3 11 1 5

11 11 5 13 1 9 3

13 13 11 9 5 3 1

1 5 7 11 13 17 19 23

1 5 7 11 13 17 19 23 1 5 7 11 13 17 19 23 5 1 11 7 17 13 23 19 7 11 1 5 19 23 13 17 11 7 5 1 23 19 17 13 13 17 19 23 1 5 7 11 17 13 23 19 5 1 11 7 19 23 13 17 7 11 1 5 23 19 17 13 11 7 5 1

14) Since f (x) = f (x), x ∼ x. If f (x) = f (y), then f (y) = f (x), so y ∼ x. Finally, if f (x) = f (y) and f (y) = f (z), then f (x) = f (z), so x ∼ z. 16) 15. 18) 11. 20) 5. 22) 97.

Answers to Even-Numbered Problems

24) 491. 26) 353. Section 1.3 2) Yes, this is a group. 4) Not closed, no identity, hence no inverses. 6) Yes, this is a group. 8) No additive inverses. 10) Not closed, no identity, hence no inverses. 12) 2 has no inverse. 14) Yes, this is a group, e = −3. 16) If e1 and e2 are two identity elements, then e1 = e1 · e2 = e2 . 18) Note that (x · y) · (y −1 · x−1 ) = x · x−1 = e, so y −1 · x−1 is the inverse of x·y 20) Consider the set {a · x | x ∈ S} which is a subset of S, yet must contain the same number of elements. 22) To show x · y = y · x, start with (x · y)2 = x · y · x · y = e. 24) If a3 = e then (a−1 )3 = e. Furthermore, if a ̸= e, then a−1 ̸= a. So the non-identity solutions pair oﬀ, and with the identity we have an odd number of solutions. 26) · a b c d a b c d

b a d c a b c d d c a b c d b a

28) 9 → 6, 27 → 18, 81 → 54, 243 → 162, 5 → 4, 25 → 20, 125 → 100. Conjecture (p − 1)n/p. ∗ ∗ | · |Zn∗ |. | = |Zm 30) If m and n are coprime, |Zmn

Section 2.1 2) 1, 3, 5, 9, 11, and 13. 4) 1, 5, 7, 11, 13, 17, 19, and 23. 6) 2, 6, 7, and 8. 8) 3 and 5. 10) No generators 12) No generators 14) 96. 16) 320. 18) 1210. 20) 1680.

Answers to Even-Numbered Problems (r −1)

22) For ϕ(n) = 14, either pi − 1 or pi i must be a multiple of 7 for some prime pi . In the first case, pi ≥ 29, so ϕ(n) ≥ 28. In the latter case, pi = 7 and ri ≥ 2, so ϕ(n) ≥ 42. 24) Yes, 8 elements are generators: 2, 3, 8, 12, 13, 17, 22, and 23. 26) Zn∗ is cyclic if n is twice the power of an odd prime. Section 2.2 2) b2 · a = b · (a · b2 ) = (a · b2 ) · b2 = a · b · b3 = a · b. 4) Answers will vary depending on how the elements are labelled. The group will be isomorphic to A4 . 6) a · b · c3 . 8) a · b2 · c3 . 10) a · b2 · c2 . 12) a · c. 14) a · b · c. 16) a · c3 . 18) The group has 20 elements. 20) InitGroup("e") AddGroupVar("a", "b") Define(a^3, e) Define(b^5, e) Define((a*b)^2, e) G = Group(a, b) len(G) 60

Section 2.3 2) {0}, {0, 2, 4, 6, 8, 10, 12, 14, 16, 18}, {0, 4, 8, 12, 16}, {0, 5, 10, 15}, {0, 10}, and the whole group. 4) {1}, {1, 8}, {1, 4, 7}, and the whole group. 6) {1}, {1, 2, 4, 8}, {1, 4}, {1, 4, 7, 13}, {1, 11}, {1, 14}, {1, 4, 11, 14}, and the whole group. 8) R2 (G) = 10, R3 (G) = 9, R4 (G) = 16, and R6 (G) = 18. For these examples, Rk (G) is a multiple of k. 10) R9 (G) = 9, and R3 (G) = 3, so six elements of order 9. 12) When n = k, an element is of order k if, and only if, it is a generator. If k is a divisor of n, and m is a divisor of k, then Rm (Zk ) = Rm (Zn ). Thus, computing the elements of order k in both Zk and Zn will give the same results. 14) If g is a generator, than only g and g −1 have finite order. 16) If a and b are of finite order, then am = bn = e for some m > 0 and n > 0. Then (a · b−1 )mn = e, so a · b−1 is of finite order. 18) (y · x · y −1 )2 = e, but y · x · y −1 ̸= e, so y · x · y −1 = x. 20) x2 ̸= e if, and only if, x−1 ̸= x, so these elements pair oﬀ, leaving an even number of elements. Thus, there is an even number of solutions to x2 = e.

Answers to Even-Numbered Problems

22) The order of the inverse element is always the same as the order of the element. 24) b · f has order 15, b · f · r · f 2 has order 6, f · b · r has order 24.

Section 3.1 2) {{0, 4, 8}, {1, 5, 9}, {2, 6, 10}, {3, 7, 11}}. 4) {{1, 4}, {2, 8}, {7, 13}, {11, 14}}. 6) {{1, 7}, {3, 5}, {9, 15}, {11, 13}}. 8) {{1, 5}, {7, 11}, {13, 17}, {19, 23}}. 10) Left cosets: {e, a · b}, {a, b}, {b2 , a · b2 }. Right cosets: {e, a · b}, {a, b2 }, {b, a · b2 }. 12) 11. 14) 11. 16) 9. 18) 3. 20) 2. 22) 3. 24) Since xH is a subgroup, e ∈ xH, so xh = e for some h ∈ H, and x = h−1 ∈ H. 26) Possible orders are 1, 3, 11, and 33. An element g of order 33 would make g 11 of order 3. Each element of order 11 generates a subgroup containing 10 such elements, so there cannot be 32 elements of order 11. 28) All elments have order 1, p, q, or pq, If there were only 1 subgroup of order p and one subgroup of order q, there would be p − 1 elements of order p and q − 1 elements of order q. But 1 + (p − 1) + (q − 1) < pq, so there is an element of order pq. 30) Each element of order p generates a subgroup with p elements. Suppose there are n subgroups of order p. Each will contain the identity, but diﬀerent subgroups cannot share any other elements. Thus, there are n(p − 1) elements of order p. 32) Left and Right cosets: {{e, a·b2 ·c, c2 , a·b2 ·c3 }, {a, b2 ·c, a·c2 , b2 ·c3 }, {b, a·b· c, b·c2 , a·b·c3 }, {a·b, b·c, a·b·c2 , b·c3 }, {b2 , a·c, b2 ·c2 , a·c3 }, {a·b2 , c, a·b2 ·c2 , c3 }}. Section 3.2 2) 14, 24, 21, 28, 14, 0, 5, 9, 0, 9, 5, 26 4) 31, 9, 5, 14, 0, 4, 1, 5, 3, 27 6) ALL SYSTEMS GO 8) REVERSE POLARITY 10) If n = p2 , we can let x = p, and if rs ≥ 2, (xr )s ≡ 0 (mod n) ̸= x (mod n). 12) f −1 (x) = x23 mod 55. 14) f −1 (x) = x103 mod 143. 16) f −1 (x) = x61 mod 437. 18) f −1 (x) = x1571 mod 2717.

Answers to Even-Numbered Problems

20) xk = b if, and only if, (x · y −1 )k = e. Since there are n solutions to this equation, there are n solutions to xk = b. 22) Answers will vary. 24) “If p and q are close together, n can be factored by taking the square root.” Section 3.3 2) {Stay}, {Stay, RotLft, RotRt}, and the whole group. 4) If h is in the intersection of the two normal subgroups, then g · h · g −1 would be in both subgroups, hence in the intersection. 6) Let g1 = x1 · y1 and g2 = x2 · y2 be two elements of X · Y . Then g1 g2−1 = −1 (x1 · x−1 2 ) · (y1 · y2 ) ∈ X · Y . 8) If x, y ∈ Z, g · y = y · g for all g ∈ G, so y −1 · g = g · y −1 , hence g · (x · y −1 ) = x · (g · y −1 ) = (x · y −1 ) · g for all g ∈ G, so x · y −1 ∈ Z. 10) Let H = {e, h} be a normal subgroup. Then g · h · g −1 ∈ H for all g ∈ G. But g · h · g −1 ̸= e, least h = e. So g · h · g −1 = h, indicating g · h = h · g for all g ∈ G, thus h ∈ Z. 12) Let f (x) = mx + b ∈ G, and n(x) = x + c ∈ N, so f −1 (x) = (x − b)/m. Then (f · n · f −1 )(x) = f (n(f −1 (x))) = x + mc ∈ N. 14) Since K is normal, hK = Kh for all h ∈ H, so the union of all such cosets is H · K = K · H. 16) Let h ∈ H ∩ K. If g ∈ H, then g · h · g −1 ∈ K, since K is normal. But g · h · g −1 is also in H, so g · h · g −1 ∈ H ∩ K. 18) By Lagrange’s theorem, H has order 2p, p, 2, or 1. But if H has order 2p, p or 1, then H would be normal. 20) {e, b, a · c, b2 , c2 , a · b · c, b · c2 , a · b2 · c, a · c3 , b2 · c2 , a · b · c3 , a · b2 · c3 }. Section 3.4 2) {0, 4, 8} {1, 5, 9} {2, 6, 10} {3, 7, 11} 4) {0, 6} {1, 7} {2, 8} {3, 9} {4, 10} {5, 11}

{0, 4, 8} {0, 4, 8} {1, 5, 9} {2, 6, 10} {3, 7, 11} {0, 6} {0, 6} {1, 7} {2, 8} {3, 9} {4, 10} {5, 11}

{1, 5, 9} {1, 5, 9} {2, 6, 10} {3, 7, 11} {0, 4, 8}

{1, 7} {1, 7} {2, 8} {3, 9} {4, 10} {5, 11} {0, 6}

{2, 6, 10} {2, 6, 10} {3, 7, 11} {0, 4, 8} {1, 5, 9}

{2, 8} {2, 8} {3, 9} {4, 10} {5, 11} {0, 6} {1, 7}

{3, 9} {3, 9} {4, 10} {5, 11} {0, 6} {1, 7} {2, 8}

{3, 7, 11} {3, 7, 11} {0, 4, 8} {1, 5, 9} {2, 6, 10} {4, 10} {4, 10} {5, 11} {0, 6} {1, 7} {2, 8} {3, 9}

{5, 11} {5, 11} {0, 6} {1, 7} {2, 8} {3, 9} {4, 10}

Answers to Even-Numbered Problems 6) {1, 14} {2, 13} {4, 11} {7, 8}

{1, 14} {1, 14} {2, 13} {4, 11} {7, 8}

8) {1, 3, 9} {2, 5, 6} {4, 10, 12} {7, 8, 11}

{2, 13} {2, 13} {4, 11} {7, 8} {1, 14}

{1, 3, 9} {1, 3, 9} {2, 5, 6} {4, 10, 12} {7, 8, 11}

{4, 11} {4, 11} {7, 8} {1, 14} {2, 13}

{2, 5, 6} {2, 5, 6} {4, 10, 12} {7, 8, 11} {1, 3, 9}

{7, 8} {7, 8} {1, 14} {2, 13} {4, 11} {4, 10, 12} {4, 10, 12} {7, 8, 11} {1, 3, 9} {2, 5, 6}

{7, 8, 11} {7, 8, 11} {1, 3, 9} {2, 5, 6} {4, 10, 12}

10)

{Stay, RotRt, RotLft} {FlipRt, FlipLft, Spin} {Stay, RotRt, RotLft} {Stay, RotRt, RotLft} {FlipRt, FlipLft, Spin} {FlipRt, FlipLft, Spin} {FlipRt, FlipLft, Spin} {Stay, RotRt, RotLft}

12) Since Q is abelian, Z is a normal subgroup. If g ∈ Q/Z, then g = (p/q)Z for some rational number p/q, so g q = pZ = Z. 14) Let g be a generator of G, then gN will be a generator of G/N. 16) If g is of order m, then (gH)m = g m · H = H, so gH must have an order that divides m. 18) If hN is an element of H/N, and gN is an element of G/N, then (gN ) · (hN ) · (gN )−1 = (g · h · g −1 ) · N ∈ H/N, since (g · h · g −1 ) ∈ H. 20) {e, a, a2 , a3 , a4 } is a normal subgroup of order 5.

Section 4.1 2) ϕ(f (x · y)) = ϕ(f (x) · f (y)) = ϕ(f (x)) · ϕ(f (y)). 4) 1 7→ 0, −1 7→ 2, ±i can go to either 1 or 3. 6) Z6 = {0, 1, 2, 3, 4, 5} ≈ Z9∗ with order {1, 2, 4, 8, 7, 5}. ∗ with order {1, 5, 7, 17, 13, 11}. 8) Z6 = {0, 1, 2, 3, 4, 5} ≈ Z18 ∗ 10) Z10 = {0, 1, 2, 3, . . . , 9} ≈ Z22 with order {1, 7, 5, 13, 3, 21, 15, 17, 9, 19}. ∗ 12) Z12 = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11} ≈ Z26 , using the arrangement given by {1, 7, 23, 31, 9, 11, 25, 19, 3, 21, 17, 15}.

14) Not true if G is not abelian. 16) ϕ(x · y) = ϕ(x + y) = ex+y = ex · ey = ϕ(x) · ϕ(y). Also note that ϕ is one-to-one and onto from R to G.

Answers to Even-Numbered Problems

18) The groups are Z10 : 0 1 2 3 4 5 6 7 8 9 0 0 1 2 3 4 5 6 7 8 9 1 1 2 3 4 5 6 7 8 9 0 2 2 3 4 5 6 7 8 9 0 1 3 3 4 5 6 7 8 9 0 1 2 4 4 5 6 7 8 9 0 1 2 3 5 5 6 7 8 9 0 1 2 3 4 6 6 7 8 9 0 1 2 3 4 5 7 7 8 9 0 1 2 3 4 5 6 8 8 9 0 1 2 3 4 5 6 7 9 9 0 1 2 3 4 5 6 7 8 and the group: e a a2 a3 a4 b a·b 2 3 4 e e a a a a b a·b a a a2 a3 a4 e a · b a2 · b a2 a2 a3 a4 e a a2 · b a3 · b 3 3 4 2 a a a e a a a3 · b a4 · b 4 4 2 3 a a e a a a a4 · b b 4 3 2 b b a ·b a ·b a ·b a·b e a4 a·b a·b b a 4 · b a3 · b a2 · b a e a2 · b a2 · b a · b b a4 · b a 3 · b a 2 a a3 · b a3 · b a2 · b a · b b a4 · b a 3 a2 a4 · b a4 · b a3 · b a2 · b a · b b a4 a3

a2 · b a2 · b a3 · b a4 · b b a·b a3 a4 e a a2

a3 · b a3 · b a4 · b b a·b a2 · b a2 a3 a4 e a

a4 · b a4 · b b a·b a2 · b a3 · b a a2 a3 a4 e

∗ ∗ with order {1, 2, 8, 4, 11, 7, 13, 14}. = {1, 3, 7, 9, 11, 13, 17, 19} ≈ Z15 20) Z20

Section 4.2 2) If g is a generator of G, and x ∈ Im(ϕ), then x = ϕ(g n ) = (ϕ(g))n for some n, and hence ϕ(g) generates Im(ϕ). 4) ϕ(x · y) = ϕ(x + y) = −(x + y) = −x + (−y) = ϕ(x) + ϕ(y) = ϕ(x) · ϕ(y). Since ϕ(ϕ(x)) = x for all x, this must be one-to-one and onto. 6) ϕ(x · y) = (x · y)6 = x6 · y 6 = ϕ(x) · ϕ(y). Kernel is ±1, Image is the positive real numbers. 8) ϕ(x · y) = ϕ(x × y) = ln |x × y| = ln |x| + ln |y| = ϕ(x) + ϕ(y) = ϕ(x) · ϕ(y). Kernel is ±1. 10) ϕ(f · g) = ϕ(f (t) + g(t)) = f ′ (t) + g ′ (t) = ϕ(f ) + ϕ(g) = ϕ(f ) · ϕ(g). Kernel is all constant polynomials. 12) ϕ(1) = 1, ϕ(2) = 7, ϕ(4) = 4, ϕ(7) = 7, ϕ(8) = 13, ϕ(11) = 1, ϕ(13) = 13, ϕ(14) = 4. 14) ϕ(1) = ϕ(7) = ϕ(17) = ϕ(23) = 1, ϕ(11) = ϕ(13) = ϕ(27) = ϕ(29) = 9, ϕ(9) = ϕ(15) = ϕ(25) = ϕ(31) = 17, ϕ(3) = ϕ(5) = ϕ(19) = ϕ(21) = 25. 16) ϕ(x · y) = [x · y (mod n)] mod k = x · y mod k = ϕ(x) · ϕ(y). The kernel is the multiples of k, so there are n/k elements in the kernel. 18) By Problem 4.5, f (H) is a subgroup of M . If a ∈ f (H), and m ∈ M then f (h) = a and f (g) = m for some h ∈ H and g ∈ G. Then f (g · h · g −1 ) is in