Abstract Algebra- An Interactive Approach, Even Numbered & Old Numbered Solutions. 2nd Edition By Wi

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.

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2

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 à = A − {a1 }, and the function f˜ : à → 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

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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.

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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

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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.

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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

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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.

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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}

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{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.

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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

Answers to Even-Numbered Problems

11

f (H), and so m · a · m−1 ∈ f (H), so f (H) is normal.

20) F1 maps everything to e; F2 maps {e, b, b2 } to e, {a, a · b, a · b2 } to a; F3 maps {e, b, b2 } to e, {a, a·b, a·b2 } to a·b; F4 maps {e, b, b2 } to e, {a, a·b, a·b2 } to a · b2 ; plus 6 other isomorphisms that can be described by x 7→ y · x · y −1 for each y ∈ S3 .

Section 4.3 2) Z12 , Z6 , Z4 , Z3 , Z2 , and the trivial group.

4) D4 , Z8∗ , Z2 , and the trivial group.

6) S3 , Z2 , and the trivial group.

8) The octahedral group, S3 , Z2 , and the trivial group.

10) ϕ1 sends all elements to 1, ϕ2 sends {0, 2} to 1, {1, 3} to 3, ϕ3 sends {0, 2} to 1, {1, 3} to 5, ϕ4 sends {0, 2} to 1, {1, 3} to 7.

12) Since Z3 has no nontrivial subgroups, then the image would have to be all of Z3 , making the kernel to have 2 elements. But S3 has no normal subgroup with 2 elements. |X| |Y | |Z| |X| |Y | |Z| 14) Either |X∩Y | |(X·Y )∩Z| or |X∩(Y ·Z)| |Y ∩Z| .

16) {{{e, a2 }, {a, a3 }}, {{b, a2 · b}, {a · b, a3 · b}}} ≈ {{e, a, a2 , a3 }, {b, a · b, a2 · b, a3 · b}}.

18) The statement is false. A typical element of G/N is gN, whereas a typical element of G/H is gH. So G/H is not a subgroup of G/N, hence (G/N )/(G/H) is meaningless.

20) Kernel must be {e, a · b2 · c, c2 , a · b2 · c3 }. Several solutions, one is to let ϕ(a) = a, ϕ(b) = b, and ϕ(c) = a · b2 .

2) 4)

(12345)

54312 .

(123456)

Section 5.1

315462 .

(1234567) 6) 1 3 7 5 6 2 4 . ( ) 8) 15 24 36 43 52 61 78 87 . ( ) ( ) ( ) ( ) ( ) ( ) 10) 12 23 34 41 , 12 24 31 43 , 13 24 32 41 , 13 21 34 42 , 14 23 31 42 , 14 21 32 43 . ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 12) 12 21 33 44 , 13 22 31 44 , 14 22 33 41 , 11 23 32 44 , 11 24 33 42 , 11 22 34 43 , 12 21 34 43 , 13 24 31 42 , (1234) 4321 . ( ) ( ) ( ) 14) x = 13 22 31 44 , 13 21 34 42 , or 13 24 32 41 . ( ) ( ) 16) Technically, 12 21 34 43 55 ∈ S5 , and 12 21 34 43 ∈ S4 , which are totally different groups. However, there is a natural mapping from S4 to S5 that allows us to consider elements of S4 to also be in S5 .

18) Left·Last·Right.

20) Left·First·Left

12

Answers to Even-Numbered Problems

Section 5.2 2) (1 2 4)(3 7 5 6).

4) (1 9 2 7 6)(4 5 8).

6) 6 and 12.

8) (1 2 3 4 5)(6 7 8) ∈ A8 , since this is an even permutation.

10) If ϕ1 and ϕ2 only move a finite number of integers, then ϕ1 · ϕ−1 2 will move a finite number of integers. Also, if n is the largest integer that ϕ 1 ∪∞ moves, then ϕ1 ∈ Sn in the sense of Problem 16 of section 5.1, so SΩ ⊆ n=1 Sn ⊆ SΩ .

12) If ϕ1 has |ϕ1 (x) − x| < M for all x, and ϕ2 has |ϕ2 (x) − x| < N for −1 all x, then |ϕ−1 (x), and |ϕ1 (ϕ−1 2 (y) − y| < N for all y = ϕ 2 (x)) − x| < M + N for all x. Examples: (1 2)(3 4)(5 6) . . . (2n − 1 2n) . . . ∈ G, but ∈ / SΩ ; (1 2)(4 6)(9 12)(16 20) . . . (n2 n2 + n) . . . ∈ S∞ , but ∈ / G. 14) ϕ is a 13-cycle, so ϕ13 = e, and so ϕ7 · ϕ7 = ϕ. So the “square root” of ϕ is ϕ7 , producing the order 3, 6, 10, 7, 2, 9, Q, A, J, 4, K, 5, 8.

16) If ϕ = (i1 i2 i3 . . . ir ), then x · ϕ · x−1 = (x(i1 ) x(i2 ) x(i3 ) . . . x(ir )).

18) 3-cycle example: (123)(324) = (143); but 4-cycles are odd.

20) a · b = (1 4 3)(2 5 6 7). (a · b)2 is a product of a 3-cycles and two 2-cycles, (a · b)3 is a 4-cycle, and (a · b)4 is a 3-cycle.

2)

{( 1 2 3 4 5 ) ( 1 2 3 4 5 ) ( 1 2 3 4 5 )Section ( 1 2 3 4 5 )5.3 ( 1 2 3 4 5 )} , , , .

12345 23451 34512 45123 , 51234

4) (), (1 2 3 5)(4 8 7 6), (1 3)(2 5)(4 7)(6 8), (1 4 3 7)(2 8 5 6), (1 5 3 2)(4 6 7 8), (1 6)(2 4)(3 8)(5 7), (1 7 3 4)(2 6 5 8), (1 8)(2 7)(3 6)(4 5).

6) −1 must map to a product of two transpositions, like (12)(34). Then ±i, ±j, and ±k map to one of (1324), (1423), (1324)(56), (1324)(57), (1324)(67), (1423)(56), (1423)(57), or (1423)(67). But no combination of these allows i · j = k.

8) f (y) sends z → z · y, so f (y) · f (x) sends z → (z · x) · y = z · (x · y), which is the same as f (x · y).

10) Applying Corollary 5.2: 36 divides 4! · |N |, so 3 divides |N |, hence either H = N, or |N | = 3.

12) Applying Corollary 5.2: 60 divides 4! · |N |, so 5 divides |N |, hence either H = N, or |N | = 5.

14) Applying Corollary 5.2: p · m divides m! · |N |, so p divides |N |, hence H = N, and H is normal.

16) Applying Corollary 5.2: |G| divides p! · |N | = p · (p − 1)! · |N |. But (p − 1)! is coprime to |G|, so |G| divides p · |N |, so |N | ≥ |H|, hence |H| = |N |.

18) Let f (i) represent the ith element in the group. To find a permutation which maps i to j, consider ϕ(f (j) · f (i)−1 ), which maps x to f (j) · f (i)−1 · x.

20) (), (1 2 3 5 9 7)(4 6 12 11 10 8), (1 3 9)(2 5 7)(4 12 10)(6 11 8), (1 4 3 12 9 10)(2 6 5 11 7 8), (1 5)(2 9)(3 7)(4 11)(6 10)(8 12), (1 6 9 8 3 11)(2 12 7 4 5 10), (1 7 9 5 3 2)(4 8 10 11 12 6), (1 8)(2 4)(3 6)(5 12)(7 10)(9 11), (1 9 3)(2 7 5)(4 10 12)(6 8 11), (1 10 9 12 3 4)(2 8 7 11 5 6), (1 11 3 8 9 6)(2 10 5 4 7 12),

Answers to Even-Numbered Problems

13

(1 12)(2 11)(3 10)(4 9)(5 8)(6 7) 22) Using the subgroup {e, a} and the cosets {{e, a}, {b, a·b5 }, {a·b, b5 }, {b2 , a· b2 }, {b3 , a · b7 }, {a · b3 , b7 }, {b4 , a · b4 }, {b6 , a · b6 }}, we get the permutations (), (2 3)(5 6), (1 2 4 5 7 3 8 6), (1 3 8 5 7 2 4 6), (1 4 7 8)(2 5 3 6), (1 4 7 8)(2 6 3 5), (1 5 8 2 7 6 4 3), (1 6 4 2 7 5 8 3), (1 7)(2 3)(4 8)(5 6), (1 7)(4 8), (1 3 4 6 7 2 8 5), (1 2 8 6 7 3 4 5), (1 8 7 4)(2 6 3 5), (1 8 7 4)(2 5 3 6), (1 6 8 3 7 5 4 2), (1 5 4 3 7 6 8 2) Section 5.4 2) 287.

4) 3821.

6) 2668.

8) 38184.

10) P (3, 1, 4, 6, 5, 2).

12) P (1, 7, 4, 3, 2, 5, 6).

14) P (7, 5, 4, 1, 3, 6, 2).

16) P (4, 8, 1, 6, 2, 3, 5, 7).

18) Since the sequence an = the nth permutation contains every element of SΩ , so by definition SΩ is countable.

20) P [7, 6, 4, 1, 2, 5, 3] = (1734)(265) and P [4, 6, 7, 3, 2, 5, 1] = (1437)(265).

Section 6.1 2) {(0, 0), (0, 1), (1, 0), (1, 1), (2, 0), (2, 1)} 7→ {0, 3, 4, 1, 2, 5}.

4) (0, 1) (0, 3) (0, 5) (0, 7) (1, 1) (1, 3) (1, 5) (1, 7) (2, 1) (2, 3) (2, 5) (2, 7) (0, 1) (0, 1) (0, 3) (0, 5) (0, 7) (1, 1) (1, 3) (1, 5) (1, 7) (2, 1) (2, 3) (2, 5) (2, 7) (0, 3) (0, 3) (0, 1) (0, 7) (0, 5) (1, 3) (1, 1) (1, 7) (1, 5) (2, 3) (2, 1) (2, 7) (2, 5) (0, 5) (0, 5) (0, 7) (0, 1) (0, 3) (1, 5) (1, 7) (1, 1) (1, 3) (2, 5) (2, 7) (2, 1) (2, 3) (0, 7) (0, 7) (0, 5) (0, 3) (0, 1) (1, 7) (1, 5) (1, 3) (1, 1) (1, 7) (1, 5) (1, 3) (1, 1) (1, 1) (1, 1) (1, 3) (1, 5) (1, 7) (2, 1) (2, 3) (2, 5) (2, 7) (0, 1) (0, 3) (0, 5) (0, 7) (1, 3) (1, 3) (1, 1) (1, 7) (1, 5) (2, 3) (2, 1) (2, 7) (2, 5) (0, 3) (0, 1) (0, 7) (0, 5) (1, 5) (1, 5) (1, 7) (1, 1) (1, 3) (2, 5) (2, 7) (2, 1) (2, 3) (0, 5) (0, 7) (0, 1) (0, 3) (1, 7) (1, 7) (1, 5) (1, 3) (1, 1) (2, 7) (2, 5) (2, 3) (2, 1) (0, 7) (0, 5) (0, 3) (0, 1) (2, 1) (2, 1) (2, 3) (2, 5) (2, 7) (0, 1) (0, 3) (0, 5) (0, 7) (1, 1) (1, 3) (1, 5) (1, 7) (2, 3) (2, 3) (2, 1) (2, 7) (2, 5) (0, 3) (0, 1) (0, 7) (0, 5) (1, 3) (1, 1) (1, 7) (1, 5) (2, 5) (2, 5) (2, 7) (2, 1) (2, 3) (0, 5) (0, 7) (0, 1) (0, 3) (1, 5) (1, 7) (1, 1) (1, 3) (2, 7) (2, 7) (2, 5) (2, 3) (2, 1) (0, 7) (0, 5) (0, 3) (0, 1) (1, 7) (1, 5) (1, 3) (1, 1) 6) 3 elements of order 2, 2 elements of order 3, no elements of order 4.

8) 7 elements of order 2, 2 elements of order 3, no elements of order 4.

10) 7 elements of order 2, 2 elements of order 3, 8 elements of order 4.

12) 3 elements of order 2, 2 elements of order 3, 4 elements of order 4.

14) 15 elements of order 2, 26 elements of order 3, 16 elements of order 4.

16) R2 (S3 × Z2 ) = 4 · 2 = 8, whereas R2 (A4 ) = 4.

14

Answers to Even-Numbered Problems

18) If A5 ≈ A × B, then A and B cannot both have an element of order 3, lest |A5 | be a multiple of 9. Likewise, A and B cannot both have an element of order 5. If R3 (A) = 21 and R5 (B) = 25, A × B would have too many elements. But if R3 (A) = 21 and R5 (A) = 25, then A has more than half the elements of A5 . ∗ 20) Put elements of Z21 in the order {1, 17, 19, 35, 13, 5, 31, 23, 25, 29, 7, 11}.

Section 6.2 2) By Lemma 6.1, G ≈ H × K, where |H| = Rm (G) and |K| = Rn (G), so |H| · |K| = mn. For each prime p that divides m, p cannot be a factor of |K|, and likewise any prime dividing n cannot be a factor of |H|. Thus, |H| = m and |K| = n.

4) Z200 ≈ Z8 × Z25 , Z8 × Z5 × Z5 , Z4 × Z2 × Z25 , Z4 × Z2 × Z5 × Z5 , Z2 × Z2 × Z2 × Z25 , and Z2 × Z2 × Z2 × Z5 × Z5 .

6) Z300 ≈ Z4 × Z3 × Z25 , Z4 × Z3 × Z5 × Z5 , Z2 × Z2 × Z3 × Z25 , Z2 × Z2 × Z3 × Z5 × Z5 .

8) Z500 ≈ Z4 × Z125 , Z2 × Z2 × Z125 , Z4 × Z5 × Z25 , Z2 × Z2 × Z5 × Z25 , Z4 × Z5 × Z5 × Z5 , Z2 × Z2 × Z5 × Z5 × Z5 .

10) Z675 ≈ Z27 × Z25 , Z27 × Z5 × Z5 , Z9 × Z3 × Z25 , Z9 × Z3 × Z5 × Z5 , Z3 × Z3 × Z3 × Z25 , and Z3 × Z3 × Z3 × Z5 × Z5 .

12) Four abelian groups of order 36.

14) 100.

16) Z16 × Z4 × Z4 × Z2 .

18) n must have no square factor other than 1.

20) 450.

Section 6.3 2) 8: ϕ(a) = a or a (order 4), ϕ(b) = b, a · b, a2 · b, or a3 · b (order 2). Note ϕ(b) ̸= a2 , since ϕ(a2 ) = a2 .

3

4) 12: ϕ(a) = one of 6 elements of order 6, ϕ(b) = one of the 3 elements of order 2 besides ϕ(a3 ).

6) 20160: ϕ(a) = one of the 15 elements of order 2, ϕ(b) = one of the 14 remaining elements of order 2, determining ϕ(a · b). ϕ(c) = one of the 12 remaining elements of order 2, determining ϕ(a · c), ϕ(b · c), and ϕ(a · b · c). ϕ(c) = one of the 8 remaining elements of order 2.

8) If a · b ̸= b · a, then ϕ(x) = a · x · a−1 will be non-trivial, since ϕ(b) ̸= b.

10) If G is non-abelian, say z · y ̸= y · z, then ϕ(x) = y −1 xy. If G is abelian, but not Z2 × Z2 × · · · × Z2 , then ϕ(x) = x−1 . If G ≈ Z2 × Z2 × · · · × Z2 , then ϕ(x1 , x2 , x3 , . . . xn ) = (x2 , x1 , x3 , . . . xn ).

12) (), (b, b2 )(a · b)(a · b2 ), (a, a · b, a · b2 ), (a, a · b2 )(b, b2 ), (a, a · b2 , a · b), (a, a · b)(b, b2 ).

Answers to Even-Numbered Problems

15

14) · e a a2 a3 a·b a2 · b a3 · b b

e e a a2 a3 a·b a2 · b a3 · b b

a

a2

a3

a·b

a2 · b

a3 · b

b

a a2 a3 e b a·b a2 · b a3 · b

2

3

a·b a2 · b a3 · b b e a a2 a3

a ·b a3 · b b a·b a3 e a a2

a ·b b a·b a2 · b a2 a3 e a

b a·b a2 · b a3 · b a a2 a3 e

a a3 e a a3 · b b a·b a2 · b

a e a a2 a2 · b a3 · b b a·b

2

3

16) (), (b, a2 ·b)(a·b, a3 ·b), (a, a3 )(a·b, a3 ·b), (a, a3 )(b, a2 ·b), (b, a·b, a2 ·b, a3 ·b), (b, a3 · b, a2 · b, a · b), (a, a3 )(b, a · b)(a2 · b, a3 · b), (a, a3 )(b, a3 · b)(a · b, a2 · b).

18) Since Z3∗ and Z4∗ both have two elements, we can pick G = Z3 and M = Z4 .

20) Twelve automorphisms: (), (2, 10)(4, 16)(5, 17)(8, 13)(11, 19), (2, 11)(4, 16)(5, 17)(10, 19), (2, 19)(8, 13)(10, 11), (2, 11)(4, 16)(5, 10)(13, 20)(17, 19), (2, 17, 19, 11, 5, 10)(4, 16)(8, 20, 13), (5, 19)(10, 17)(13, 20), (2, 5, 19)(8, 20, 13)(10, 11, 17), (2, 5)(8, 20)(11, 17), (2, 10, 5, 11, 19, 17)(4, 16)(8, 13, 20), (2, 17)(4, 16)(5, 11)(8, 20)(10, 19), (2, 19, 5)(8, 13, 20)(10, 17, 11).

22) There are 48 automorphisms, generated by f (a) = b, f (b) = a, and g(a) = a, g(b) = a · b.

Section 6.4 2) (1, 7).

4) (3, 7).

6) (7, 1).

8) A nontrivial mapping from Z3 to Aut(Z8∗ ) maps 1 to a 3-cycle, which

by Proposition 6.7 doesn’t matter which, so we can assume ϕ1 = (3 5 7).

Z3 ⋊Z8∗ ≈ A4 .

10) Since Aut(Z) ≈ Z2 , we see that ϕ1 (x) = −x. So (x, 0) · (y, a) = (x + y, a), and (x, 1) · (y, a) = (x − y, 1 − a).

12) If ϕ ∈ Aut(G), then (e1 , ϕ) · (g, e2 ) · (e1 , ϕ)−1 = (ϕ(g), e2 ) for all g ∈ G. Here, e1 is the identity of G, and E2 is the identity of Aut(G).

14) If f (i) is an n-tuple, then ψσ (f ) maps f (i) to f (σ −1 (i)), so ψτ (ψσ (f )) would map f (i) to f (σ ∗ −1(τ −1 (i))). But this is the same as ψτ ·σ .

16

Answers to Even-Numbered Problems

16) τ = (1 2) (0, 0, e) (0, 1, e) (1, 0, e) (1, 1, e) (0, 0, τ ) (0, 1, τ ) (1, 0, τ ) (1, 1, τ ) (0, 0, e) (0, 0, e) (0, 1, e) (1, 0, e) (1, 1, e) (0, 0, τ ) (0, 1, τ ) (1, 0, τ ) (1, 1, τ ) (0, 1, e) (0, 1, e) (0, 0, e) (1, 1, e) (1, 0, e) (0, 1, τ ) (0, 0, τ ) (1, 1, τ ) (1, 0, τ ) (1, 0, e) (1, 0, e) (1, 1, e) (0, 0, e) (0, 1, e) (1, 0, τ ) (1, 1, τ ) (0, 0, τ ) (0, 1, τ ) (1, 1, e) (1, 1, e) (1, 0, e) (0, 1, e) (0, 0, e) (1, 1, τ ) (1, 0, τ ) (0, 1, τ ) (0, 0, τ ) (0, 0, τ ) (0, 0, τ ) (1, 0, τ ) (0, 1, τ ) (1, 1, τ ) (0, 0, e) (1, 0, e) (0, 1, e) (1, 1, e) (0, 1, τ ) (0, 1, τ ) (1, 1, τ ) (0, 0, τ ) (1, 0, τ ) (0, 1, e) (1, 1, e) (0, 0, e) (1, 0, e) (1, 0, τ ) (1.0, τ ) (0, 0, τ ) (1, 1, τ ) (0, 1, τ ) (1, 0, e) (0, 0, e) (1, 1, e) (0, 1, e) (1, 1, τ ) (1, 1, τ ) (0, 1, τ ) (1, 0, τ ) (0, 0, τ ) (1, 1, e) (0, 1, e) (1, 0, e) (0, 0, e) 18) D6 ≈ S3 × Z2 .

20) Z2 Wr A3 ≈ Z2 × A4 .

Section 7.1 2) {e}.

4) {(1, 1), (1, 3), (7, 1), (7, 3)}.

6) If G is abelian, then G/Z(G) ≈ {e} is cyclic. If G/Z(G) ≈ Inn(G) is cyclic, then a generator is x 7→ gxg −1 . For each y ∈ G, yxy −1 = g n xg −n for some n, plugging in x = g yields ygy −1 = g, or gy = yg Since gy = yg for all y, so Inn(G) ≈ {e}, and G is abelian.

8) If a is the only element of order 2, then for any g ∈ G, g · a · g −1 must be of order 2, hence g · a · g −1 = a, so g · a = a · g.

10) If z ∈ Z(G), then (z, z, z . . . , z, ()) will be in the center of G Wr H. We claim there are no other elements in the center. If gi ̸= gj in the element x, then there is some permutation σ that sends i to j, and so (e, e, . . . , e, σ) · x ̸= x · (e, e, . . . , e, σ). Likewise, if the permutation of x sends i to j, then x cannot be in the center. Thus, all elements of the center are of the form (z, z, z . . . , z, ()) for some z ∈ G, and clearly this in not in the center unless z ∈ Z(G). There is an obvious isomorphism between the center and Z(G).

12) Since ϕ(h) ∈ H for all automorphisms, in particular ϕ(h) ∈ H for all inner automorphisms. Thus, g · h · g −1 ∈ H for all g ∈ G, so H is normal.

14) If h ∈ H, and ϕ is any automorphism, then ϕ(h)n = ϕ(hn ) = ϕ(e) = e, so ϕ(h) ∈ H. 16) Let ϕ be an automorphism of G. Since N is characteristic, ϕ(n) ∈ N for all n ∈ N, so ϕ can be restricted to form an automorphism on N. Then ϕ(h) ∈ H for all h ∈ H, since H is a characteristic subgroup of N. Hence, H is a characteristic subgroup of G. 18) Center = {e, a3 }, Quotient group D6 /Z(D6 ) ≈ S3 .

20) Z(Dn ) = {e} if n is odd, Z(Dn ) = {e, an/2 } for n even. Note that the non-identity element corresponds to a 180 degree rotation.

Answers to Even-Numbered Problems

17

Section 7.2 2) ND5 ({e}) = D5 , ND5 ({b}) = {e, b}, ND5 ({a · b}) = {e, a · b}, ND5 ({a2 · b}) = {e, a2 · b}, ND5 ({a3 · b}) = {e, a3 · b}, ND5 ({a4 · b}) = {e, a4 · b}, ND5 ({a}) = ND5 ({a2 }) = ND5 ({a3 }) = ND5 ({a4 }) = {e, a, a2 , a3 , a4 }.

4) In each case, the normalizer is the subgroup.

6) x ∈ Z(G) ⇔ x · y = y · x for all y ∈ G ⇔ x ∈ NG ({y}) for all y ∈ G. 8) If x · g · x−1 = g, then (x · g · x−1 )k = x · g k · x−1 = gk , so x ∈ NG ({g k }).

10) If g · x · g −1 = x for all g ∈ G, then g · x = x · g for all g ∈ G, so x ∈ Z(G).

Likewise, if x ∈ Z(G), then g · x · g −1 = x for all g ∈ G. 12) {e, a2 }.

14) {e, a2 , a · b, a3 · b}.

16) D4 .

18) {e, a, a2 , a3 , a4 }.

20) D5 .

22) {e}, {e, a3 }, {e, a2 , a4 }, {e, a, a2 , a3 , a4 , a5 }, {e, a2 , a4 , b, a2 · b, a4 · b}, {e, a2 , a4 , a · b, a3 · b, a5 · b}, and D6 .

Section 7.3 2) {1}, {−1}, {i, −i}, {j, −j}, {k, −k}.

4) {e}, {b2 }, {b, a · b, a2 · b}, {b3 , a · b3 , a2 · b3 }, {a, a2 }, and {a · b2 , a2 · b2 }.

6) If g ∈ Z(G), then g · x · g −1 = x for all x, so the conjugacy class is {x}. If g · x · g −1 = x for all g, then g · x = x · g, so x is in the center.

8) If x and y are in the same conjugacy class, then a · x · a−1 = y for some a.

Then g · x · g −1 = x if and only if a · g · a−1 · y · a · g −1 · a−1 = y, so g is in NG ({x}) if, and only if a · g · a−1 is in NG ({y}).

10) |N | ≥ 22, so |N | = 84, 56, 42, 28, or 24 (divisors of 168). |N | is even, so classes of size 1 and 21 are included, making |N | ≥ 46. Then seven divides |N |, so both classes of size 24 are in N, making |N | ≥ 112.

12) |N | ≥ 56, so |N | = 330, 220, 165, 132, 110, 66, or 60 (divisors of 660). |N | ̸= 60, so 11 divides |N |, hence both classes of size 60 are in N, making |N | ≥ 176. Five divides |N |, so both classes of order 132 are in N, making |N | ≥ 385.

14) |N | ≥ 145, so |N | = 1224, 816, 612,408,306,272,204,or 153 (divisors of 2448). 17 divides |N |, hence both classes of size 144 are in N, making |N | ≥ 442. Six divides |N |, so the classes of size 153 and one of the classes of size 272. This makes |N | ≥ 986. |N | would be a multiple of 9, so we need all of the classes of order 272, forcing |N | ≥ 1534.

16) |N | ≥ 166, so |N | = 3960, 2640, 1980, 1584, 1320, 990, 880, 792, 720, 660, 528, 495, 440, 396, 360, 330, 264, 240, 220, 198, 180, or 176 (divisors of 7920). |N | = ̸ 495, so |N | is even, hence classes of size 1 and 165 are in N, making |N | ≥ 606. |N | ̸= 720, so 11 divides |N |, so both classes of size 720 are in N, making |N | ≥ 2046. Five divides |N |, the the class of size 1584 is in N, making |N | ≥ 3630. Three divides |N |, so the class of size 440 is in N, making |N | ≥ 4620.

18

Answers to Even-Numbered Problems

18) Class of size 1: e; size 40: (123); size 40 : (123)(456); size 45: (12)(34); size 72: (12345); size 72: (12354); size 90: (12)(3456).

20) Nontrivial normal subgroups are {e, a, a2 , a3 , a4 } and {e, b2 , a, a·b2 , a2 , a2 · b2 , a3 , a3 · b2 , a4 , a4 · b2 }.

Section 7.4 2) Not possible, for 8 is not a divisor of the size of the group, 18.

4) Not possible, for the center would have size 2, and this is not a divisor of the size of the group, 33.

6) This is possible. In fact, it is the sizes of the conjugacy classes of S4 .

8) If p divides |Z(G)|, then by Lemma 6.2 there is a subgroup of order p of Z(G). Otherwise, there must be a term in the class equation which is not 1, and not a multiple of p. This means |NG ({g})| is a multiple of p, which by induction has a subgroup of order p.

10) Only one 3-Sylow subgroup and one 11-Sylow subgroup, so both are normal, hence G ≈ Z3 × Z11 ≈ Z33 .

12) Only one p-Sylow subgroup P , and there is a subgroup H of order 2 such that G = P · H. Either G ≈ P × H ≈ Z2p , or G ≈ P ⋊ϕH. By Problem 19, there is only one element of Zp∗ of order 2, so P ⋊H ≈ D2p .

14) Only one 11-Sylow subgroup N, If H is any 3-Sylow subgroup, then H ·N is a subgroup of order 33, which by Problem 10 is a normal subgroup ≈ Z33 . By picking any 2-Sylow subgroup, we have G ≈ Z2 × Z33 ≈ Z66 , or G ≈ Z33 ⋊ϕZ2 for some ϕ. Since Z33 has only 3 elements of order 3, there are at most 4 non-isomorphic groups of order 66, and the 4 listed are non-isomorphic.

16) Factors of |G| are 1, q, p, pq, so only one p-Sylow subgroup.

18) If |G| = 24, let H be a 2-Sylow subgroup H of order 8, and Corollary 5.2 says either H is normal, or a subgroup of order 4 is normal. If q < p, there is only one p-Sylow subgroup, hence normal. Finally, if q > p + 1, and there is more than 1 p-Sylow subgroup, then q ≡ 1 (mod p), or q = np + 1 for n ≥ 2.

Having p2 ≡ 1 (mod q) leads to a contradiction, so there are either 1 or p3 q-Sylow subgroups, the latter producing p3 (q − 1) elements of order q, leaving only p3 for a single p-Sylow subgroup.

20) Five subgroups of order 4, one subgroup of order 5.

Section 8.1 ∗ 2) Z15 ⊇ {1, 2, 4, 8} ⊇ {1, 4} ⊇ {1}. ∗ 4) Z21 ⊇ {1, 8, 13, 20} ⊇ {1, 20} ⊇ {1}.

6) Q ⊆ {1, i, −1, −i} ⊆ {1, −1} ⊆ {1}.

8) D5 ⊆ {e, b, b2 , b3 , b4 } ⊆ {e}.

10) S6 ⊇ A6 ⊇ {e}.

12) Z4 ⊇ {0, 2} ⊇ {0}, and Z8∗ ⊇ {1, 3} ⊇ {1}.

14) A5 × A5 ⊇ {e} × A5 ⊇ {e} × {e}.

16) One example: S3 ⊇ A3 ⊇ {e}.

Answers to Even-Numbered Problems

19

18) Use the exact same argument as the Jordan-Hölder theorem only use Problem 17 instead of the refinement theorem.

20) G ⊇ {1, 6212, 13016, 19853, 24132, 25315, 33108, 38807} ⊇ {1, 6212, 13016, 25315} ⊇ {1, 13016} ⊇ {1}. The middle group could be replaced with {1, 13016, 19853, 24132} or {1, 13016, 33108, 38807}.

Section 8.2 2) If n > 4, Sn contains An as a non-abelian simple subgroup. S1 ≈ {e}, S2 ≈ Z2 , S3 ⊇ A3 ⊇ {e}, and S4 ⊇ A4 ⊇ {(), (12(34), (13)(24), (14)(23)} ⊇ {(), (12)(34)} ⊇ {e}.

4) G′ = {ϕ(x) = x + c | c ∈ R}, both G′ ≈ R and G/G′ ≈ R. −1 −1 −1 ) · k1−1 · 6) For h ∈ H and x = h−1 1 · k1 · h1 · k1 ∈ [H, K], h · x = ((h1 · h −1 (h1 · h−1 ) · k1 ) · (k1 · h · k1 h−1 ) · h ∈ [H, K] · h. Since all such x generate [H, K], h · [H, K] ⊆ [H, K] · h. Likewise, k · [H, K] ⊆ [H, K] · k for all k ∈ K. Thus, [H, K] is a normal subgroup of the group generated by H and K. 8) (D4 )′ = {e, b2 }, (D4 )′′ = {e}.

10) Q′ = {1, −1}, Q′′ = {1}.

12) If ϕ is an automorphism of G, then ϕ(x−1 ·y −1 ·x·y) = (ϕ(x))−1 ·(ϕ(y))−1 · ϕ(x) · ϕ(y) ∈ G′ . Since G′ is generated by all such elements, ϕ(x) ∈ G′ for all x ∈ G′ .

14) By induction assume Gn−1 contains the G(n−1) = (n − 1)th derived subgroup. Gn = [G, Gn−1 ] ⊆ [G, G(n−1) ] ⊆ [G(n−1) , G(n−1) ] = G(n) . If |Gn | = 1, then |G(n) | = 1, and the group is solvable.

16) Define Z0 = {e}, and define Zi for i > 0 to be such that Zi+1 /Zi = Z(G/Zi ). Since each center is nontrivial by Corollary 7.2, Zn = G for some n.

Define Gi as in Problem 14, then by induction, Gj ⊆ Zn−j , since if x ∈ Zn−j+1 and g ∈ G, then (g −1 x−1 gx)Zn−j = (g −1 Zn−j )(x−1 Zn−j )(gZn−j )(xZn−j ) = (g −1 Zn−j )(gZn−j )(x−1 Zn−j )(Zn−j ) = Zn−j .

So Gn ⊆ Z0 = {e}.

18) Q ⊇ {1, −1} ⊇ {1}. For compositions series, add {1, −1, i, −i}.

20) G′ ≈ Q, which is a normal subgroup of G, and there is a 3-Sylow subgroup H for which H · G′ = G. Hence, G is isomorphic to a semi-direct product Q⋊ϕZ3 , and since Aut(Q) ≈ S4 , Z3 must map to a 3-cycle in S4 , but all 3-cycles are conjugate, so there is only one possible semi-direct product G ≈ Q⋊Z3 .

Section 8.3 2) D5 ⊇ {e, a, a , a , a } ⊆ {e}. ∗ 4) Z24 ⊇ {1, 5, 7, 11} ⊇ {1, 5} ⊇ {1}.

6) Z3 ⋊Z4 ⊇ {e, a, a2 } ⊇ {e}.

8) Z2 × Z4 × Z8 ⊇ Z2 × Z4 × {0} ⊇ Z2 × {0} × {0} ⊇ {e}.

10) Z2 × Z2 × Z2 × Z2 .

2

3

4

20

Answers to Even-Numbered Problems

12) Yes. A polycyclic series of maximum length would be a composition series, and so the composition factors will be isomorphic by the Jordan-Hölder theorem.

14) If a = x·y and b = y ·x, then a2 = y 3 , so a is of order 4, and by Problem 17 so is b. In fact, b2 = a2 , and b · a = a3 · b, so a and b generate a subgroup isomorphic to Q. If we let c = x, then c3 = e, c · a = a · b · c, c · b = a · c, so this is a group of order 24.

16) j 4 = e, i2 = j 2 , i−1·j·i = j 3 . Group is listed as {e, j, j 2 , j 3 , i, i·j, i·j 2 , i·j 3 }.

18) B ⊇ {1, L, P, T, I, M, Q, U } ⊇ {1, L, P, T } ⊇ {1}; if a = J, b = I, and c = P , then a2 = c2 , b2 = c2 , c4 = e, a−1 · b · a = b · c2 , a−1 · c · a = c, b−1 · c · b = c.

20) D ⊇ {1, L, M, N, O, P, Q, R} ⊇ {1}; if a = S and b = L, then a2 = b4 , b8 = e, a−1 · b · a = b7 .

Section 8.4 2) (1, 1, 0, 1, 0, 0, P (1, 2, 6, 3, 5, 4)).

4) (1, 0, 0, 0, 0, 0, P (4, 2, 3, 5, 1, 6)).

6) (0, 1, 1, 0, 1, 0, (1 6 3 5 2)).

8) (1, 1, 0, 0, 1, 1, (1 3)(2 6)).

10) No, since (Z2 Wr A6 )′ still has a center of order 2, as we saw in the text, whereas (Z2 Wr A6 )/Z would be centerless.

12) The number of 2-Sylow subgroups must be an odd divisor of 11520, which can only be 45, 15, 9, 5, 3, or 1. But there are 4352 elements that would be in a 2-Sylow subgroup, and 15 · 256 = 3840 wouldn’t be enough, so there must be 45 2-Sylow subgroups.

14) |(Z3 Wr S8 )′ | = 88179840.

16) a2 · b−1 · a−1 .

18) (a−1 · b−1 )3 · (a · b)3 · a · b−1 · a−1 · b−2 20) L · R−1 · L · R−1 · L · R−2 .

Section 9.1 2) x · (−y) = x · (−y) + [x · y + −(x · y)] = [x · (−y) + x · y] + −(x · y) = x · [(−y) + y] + −(x · y) = x · 0 + −(x · y) = −(x · y).

4) x · (y − z) = x · (y + (−z)) = x · y + x · (−z) = x · y + −(x · z) = x · y − x · z.

6) (x + y) · (x − y) = x · (x − y) + y · (x − y) = x · x − x · y + y · x − y · y = (x2 − y 2 ) + (y · x − x · y).

8) (a1 − b1 i − c1 j − d1 k) + (a2 − b2 i − c2 j − d2 k) = (a1 + a2 ) − (b1 + b2 )i − (c1 + c2 )j(d1 + d2 )k.

10) (a + bi + cj + dk) · (a − bi − cj − dk) = a2 + b2 + c2 + d2 . Replace x with x to get the other half.

12) (x + 1) · (x −√ 1) = x2 + x − x − √ 1 = x2 − 1. √ 14)√If a = x1 +y1 2 and 2 +y2 2, then a+b = (x1 +x2 )+(y1 +y2 )√2 ∈ √ b = x√ Z[ 2], −a = −x1 − y1 2 ∈ Z[ 2], a · b = (x1 x2 + 2y1 y2 ) + (x1 y2 + x2 y1 ) 2 ∈

Answers to Even-Numbered Problems

21

√ Z[ 2]. Associative laws and distributive laws hold because these are real numbers.

16) If e1 and e2 are both multiplicative identities, then e1 = e1 · e2 = e2 .

18) ⊕ and ⊗ are both closed, and both are clearly commutative. (x ⊕ y) ⊕ z = x + y + z − 2 = x ⊕ (y ⊕ z), x ⊕ 1 = 1 ⊕ x = x so 1 is the additive identity. x⊕(2−x) = 1, so 2−x is the additive inverse. (x⊗y)⊗z = x+y+z −xy−xz − yz + xyz = x ⊗ (y ⊗ z), x ⊗ (y ⊕ z) = 2x + y + z − xy − xz − 1 = (x ⊗ y) ⊕ (x ⊗ z).

20) When n = 5, 7, or 11, Zn is a field. Otherwise, there are zero divisors in Zn .

Section 9.2 2) [[0, a], [0, b]].

4) [[a + b, 0], [0, 0]].

6) [[a + b, a + b], [a + b, a + b]].

8) [[2 · a, 2 · b], [a, b]].

10) [[b, a + b], [a + b, a]].

12) A cyclic additive group must have a generator g. Then all elements can be expressed as ng for some n, and (ng) · (mg) = (nm)g = (mg) · (ng).

14) By induction in n: (m + n)x = (m + n − 1)x + x = mx + (n − 1)x + x = mx + nx.

16) Since n(−x) + nx = n(−x + x) = 0, we have n(−x) = −nx.

18) Define 2a = 0, and a2 = 0.

20) InitRing("a", "b","c") DefineRing([2, 2, 2],[[a + c, b +c, c] [b, b, 0], [c, c, 0]])

Section 9.3 2) (x + y) · (x−1 − x−2 y) = e + x−1 y − x−1 y − x−2 y 2 = e.

4) In T4 , a · c = 0 but c · a = c.

6) x = x2 = (−x)2 = −x.

8) 2x = (2x)3 = 8x3 = 8x, so 6x = 0.

10) (e − a)2 = e(2 − )e · a(− a )· e +(a)2 = e − a − a + a = e − a. n−1 12) First show n−1 = ni . Then i−1 + i (

( ) ( ) ( ) ) n − 1 n−2 n − 1 n−3 2 n − 1 n−1 (x + y) · x + x y+ x y + ··· + y 1 2 n−1 [ ( )] [( ) ( )] n−1 n−1 n−1 = xn + 1 + xn−1 y + + xn−2 y 2 + 1 1 2 [( ( ) ( )] ) n−1 n−1 n−1 n ··· + + xy n−1 + y n−2 n−1 n−1 ( ) ( ) ( ) n n−1 n n−2 2 n n = xn + x y+ x y + ··· + y .

1 2 n n−1

14) a + b and 3a + b.

16) Neither T4 nor T8 have irreducible elements.

18) a and 3a are prime, but not irreducible.

22

Answers to Even-Numbered Problems

20) InitRing("e", "a", "b") T8 = DefineRing([2,2,2], [[e,a,b],[a,a,a],[b,b,b]])

Section 10.1 √ √ 2) Subring. (x1 − x2 ) + (y1 − y2 ) 2 and (x1 x2 + 2y1 y2 ) + (x1 y2 + x2 y1 ) 2 are in the set.

4) Subring. (x1 y2 − x2 y1 )/(y1 y2 ) and (x1 x2 )/(y1 y2 ) are in the set. (Numerators will be even, and denominators will be odd.)

6) Not a subring, since not closed √ under multiplication. √ 8) Subring. (x1 − x2 ) + (y1 − y2 ) 2 and (x1 x2 + 2y1 y2 ) + (x1 y2 + x2 y1 ) 2 are in the set, since (x1 x2 + 2y1 y2 )√is even. √ 10) Subring. (x1 − x2 ) + (y1 − y2 ) 3 and (x1 x2 + 3y1 y2 ) + (x1 y2 + x2 y1 ) 3 are in the set, since (x1 x2 + 3y1 y2 + x1 y2 + x2 y1 ) is even for all combinations.

12) If a, b ∈ B, then there are x, z ∈ R such that a = x · y and b = z · y. Then a − b = (x − z) · y ∈ I, and a · b = (x · y) · (z · y) = (x · y · z) · y ∈ B. 14) If a and b are nilpotent, then( am )= bn = 0 for m and n. By ( some ) m+n−2 Problem 12, (a − b)m+n = am+n − m+n am+n−1 b + m+n a b2 − · · · + 1 2 ( ) m m+n m n m+n m (−1) = 0. So a−b is nilpotient, and (a·b) = am ·bm = m a b +· · · b 0, so a · b is nilpotient.

16) {0}, {0, a, 2a, 3a}, {0, 2a}, {0, b}, {0, a + b, 2a3a + b}, {0, 2a + b, b, 2a}, and the whole ring.

18) {0}, {0, e}, {0, a}, {0, b}, {0, c}, {0, d}, {0, f }, {0, e, c, g}, {0, e, a, d}, {0, e, b, f }, {0, a, b, c}, {0, c, d, f }, and the whole ring.

20) {0}, {0, a, 2a, −a}, {0, 2a}, {0, b, 2a, 2a + b}, {0, a + b, 2a, −a + b}, and the whole ring.

Section 10.2 2) n = 4.

4) n = 192.

6) If a ∈ Ip , then pn a = 0 for some n, and if z ∈ R, then pn (a · z) = 0, so a · z ∈ Ip . Likewise z · a ∈ Ip .

8) {0}, {0, c} and the whole ring.

10) + {0, c} {a, b} · {0, c} {a, b} {0, c} {0, c} {a, b} {0, c} {0, c} {0, c} {a, b} {a, b} {0, c} {a, b} {0, c} {a, b} 12) + {0, b} {a, a + b} {2a, 2a + b} {3a, 3a + b} {0, b} {0, b} {a, a + b} {2a, 2a + b} {3a, 3a + b} {a, a + b} {a, a + b} {2a, 2a + b} {3a, 3a + b} {0, b} {2a, 2a + b} {2a, 2a + b} {3a, 3a + b} {0, b} {a, a + b} {3a, 3a + b} {3a, 3a + b} {0, b} {a, a + b} {2a, 2a + b}

Answers to Even-Numbered Problems · {0, b} {a, a + b} {0, b} {0, b} {0, b} {a, a + b} {0, b} {a, a + b} {2a, 2a + b} {0, b} {2a, 2a + b} {3a, 3a + b} {0, b} {3a, 3a + b} 14) + {0, c} {e, g} {a, b} {d, f } {0, c} {0, c} {e, g} {a, b} {d, f } {e, g} {e, g} {0, c} {d, f } {a, b} {a, b} {a, b} {d, f } {0, c} {e, g} {d, f } {d, f } {a, b} {e, g} {0, c} 16) + ⟨6⟩ 2 + ⟨6⟩ 4 + ⟨6⟩ ⟨6⟩ ⟨6⟩ 2 + ⟨6⟩ 4 + ⟨6⟩ 2 + ⟨6⟩ 2 + ⟨6⟩ 4 + ⟨6⟩ ⟨6⟩ 4 + ⟨6⟩ 4 + ⟨6⟩ ⟨6⟩ 2 + ⟨6⟩

{2a, 2a + b} {0, b} {2a, 2a + b} {0, b} {2a, 2a + b}

23

{3a, 3a + b} {0, b} {3a, 3a + b} {2a, 2a + b} {a, a + b}

· {0, c} {e, g} {a, b} {d, f }

{0, c} {e, g} {a, b} {d, f } {0, c} {0, c} {0, c} {0, c} {0, c} {e, g} {a, b} {d, f } {0, c} {a, b} {a, b} {0, c} {0, c} {d, f } {0, c} {d, f }

· ⟨6⟩ 2 + ⟨6⟩ 4 + ⟨6⟩

⟨6⟩ ⟨6⟩ ⟨6⟩ ⟨6⟩

2 + ⟨6⟩ 4 + ⟨6⟩ ⟨6⟩ ⟨6⟩ 4 + ⟨6⟩ 2 + ⟨6⟩ 2 + ⟨6⟩ 4 + ⟨6⟩

18) If a, b ∈ B, then there are x, z ∈ R such that a = x · y and b = z · y. Then a − b = (x − z) · y ∈ B, and if c ∈ R, then a · c = c · a = (c · x) · y ∈ B. 20) (e + I) · (x + I) = (x + I) · (e + I) = x + I, so e + I is the unity element.

22) {0}, {0, a, 2a, −a}, {0, 2a}, {0, b, 2a, 2a + b}, {0, a + b, 2a, −a + b}, and the whole ring.

Section 10.3 2) Let e′ = ϕ−1 (e). Then ϕ(x · e′ ) = ϕ(x) · ϕ(e′ ) = ϕ(x), and ϕ(e′ · x) = ϕ(e′ ) · ϕ(x) = ϕ(x). Since ϕ is one-to-one, x · e′ = e′ · x = x.

4) If x2 = x and x ̸= 0, then ϕ(x) = ϕ(x2 ) = ϕ(x)2 . Since ϕ is one-to-one, ϕ(x) is non-zero.

6) The additive properties of Rop are inherited from R, and obviously the multiplication is closed. So we only need to prove the associate law and distributive law. a ∗ (b ∗ c) = a ∗ (c · b) = (c · b) · a = c · (b · a) = (b · a) ∗ c = (a ∗ b) ∗ c, a ∗ (b + c) = (b + c) · a = b · a + c · a = a ∗ b + a ∗ c, (b + c) ∗ a = a · (b + c) = a · b + a · c = b ∗ a + c ∗ a.

8) {{0, 2a}, {a, 3a}, {b, 2a + b}, {a + b, −a + b}} 7→ {0, a, c, b} or {0, b, c, a}.

10) Since a non-commutative ring must have a non-cyclic additive group, the smallest such ring would have additive group of Z2 × Z2 . If x2 = y for two nonzero elements x and y, then x·y = y·x, and the whole ring would commute.

Thus, x2 = 0 or x for all x ∈ R. If two nonzero elements have x2 = y 2 = 0, then x · y ̸= x or else (x · y) · y = x ̸= x · (y · y), likewise x · y ̸= y. Also x · y ̸= x + y, or else x · (x · (x + y)) = x + y ̸= (x · x) · (x + y). This means that x · y = 0, and similarly y · x = 0, and the ring would commute. So there are at least two elements for which x2 ̸= 0, call them a and b. Then a2 = a, b2 = b. If (a + b)2 = a + b, then a · b = b · a, so we need (a + b)2 = 0. Then

24

Answers to Even-Numbered Problems

a · b ̸= a + b, or else (a · b) · b = (a + b) · b = a ̸= a · (b · b). Likewise, a · b ̸= 0, otherwise (a + b)2 = 0 would force b · a = a + b. So for a · b ̸= b · a, one must be a, and the other b, yielding T4 and T4op respectively.

12) 2+⟨8⟩ is a generator of the additive group of ⟨2⟩/⟨8⟩, but for every element x of ⟨2⟩/⟨8⟩, x2 = ⟨8⟩ or 4 + ⟨8⟩, so there is no multiplicative identity.

14) Z6 , 2Z12 , 3Z18 , and 6Z36 .

16) Z30 , 2Z60 , 3Z90 , 5Z150 , 6Z180 , 10Z300 , 15Z450 , and 30Z900 .

18) Clearly {ne | n ∈ Z} is in S, and this is a subring, so this is S. Let n be the smallest positive integer for which ne = 0. If no such positive integer exists, ϕ(n · e) = n gives an isomorphism from S to Z, otherwise this gives an isomorphism from S to Zn .

20) R1 : a2 = a · b = b · a = b2 = 0; R2 : a2 = b, b2 = a · b = b · a = 0; R3 : a2 = a, b2 = a · b = b · a = 0; R4 : a2 = a, a · b = b, b2 = b · a = 0; R5 : a2 = a, b · a = b, b2 = a · b = 0; R6 : a2 = b2 = a, a · b = b · a = b; R7 : a2 = a, b2 = a · b = b · a = b; R8 : a2 = a, a · b = b · a = b, b2 = a + b.

Section 10.4 2) {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} 7→ {0, 0, 0, 0, 0, 0, 0, 0, 0, 0}, {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}, {0, 5, 0, 5, 0, 5, 0, 5, 0, 5}, or {0, 6, 2, 8, 4, 0, 6, 2, 8, 4}.

4) Yes. The image is 6Z30 ≈ Z5 .

6) Yes, since clearly ϕ(x + y) = (x + y) mod 5 = ϕ(x) + ϕ(y), and ϕ(x · y) = (x · y) mod 5 = ϕ(x) · ϕ(y).

8) Yes, note that (6x mod 10) mod 5 = x mod 5.

10) Only homomorphisms: f (x) = 0 and ϕ(x) = x. (1 must map to a solution to x2 = x.)

12) If x2 = x, then ϕ(x2 ) = ϕ(x)2 , so ϕ(x) is idempotent.

14) If I is an ideal in Im(ϕ), then K = ϕ−1 (I) is an ideal in R, so it is generated by a single element a. Then I = ϕ(K) is generated by ϕ(a).

16) If k ∈ K and a ∈ I, a · k ∈ K since K is an ideal of R, so K is an ideal of I as well. The mapping ϕ : R/K 7→ R/I given by ϕ(x + K) = x + I is a homomorphism, whose image is all of x + I, and whose kernel is I/K. So I/K is an ideal of R/K, and be the first isomorphism theorem for rings (10.2), (R/K)/(I/K) ≈ I/K. 18) I2 = Ideal(R, 2*a + b) Q = Coset(R, I2) i = RingHomo(R, Q) HomoDef(i, a, a + I2) HomoDef(i, b, b + I2) FinishHomo(i) 'Homomorphism defined'

20) Only two possibilities: a maps to a or b − a, and b maps to 2 · a.

Answers to Even-Numbered Problems

25

Section 11.1 2) bx2 − a.

4) 2ax2 + (a + b)x.

6) bx4 + bx3 − a.

8) 2.

10) Let the identity e have order n in the additive group. Then the characteristic cannot be less than n, but nx = n(x · e) = (ne) · x = 0 for all x ∈ R. 12) Let n be the order of the ring R. Then nx = 0 for all x ∈ R, so the characteristic would be at most n.

14) The ring defined by Tables 9.3 and 9.4 gives an example.

16) (x + y)2 = x2 + 2x · y + y 2 = x2 + y 2 .

18) For the ring defined by Tables 9.3 and 9.4, x = a, y = −a.

20) x2 , x2 + 1, x2 + 2, x2 + x, x2 + x + 1, x2 + x + 2, x2 + 2x, x2 + 2x + 1, x2 + 2x + 2, 2x2 , 2x2 + 1, 2x2 + 2, 2x2 + x, 2x2 + x + 1, 2x2 + x + 2, 2x2 + 2x, 2x2 + 2x + 1, 2x2 + 2x + 2.

22) x3 , x3 +1, x3 +x, x3 +x+1, x3 +x2 , x3 +x2 +1, x3 +x2 +x, x3 +x2 +x+1.

24) All factorizations reveal triple roots. Reason: For real numbers, (x+y)3 = x3 + 3x2 y + 3xy 2 + y 3 , but since we are working mod 3, (x + y)3 = x3 + y 3 .

Section 11.2 ( u ) (( x ) ( z )) ( u ) ( xw+yz ) ( uxw+uyz ) 2) v · = v · = , whereas y + w yw vyw ( u ) ( x ) ( u ) ( z ) ( ux ) ( uz ) ( uxvw+vyuz ) ( uxw+uyz ) = . v · y + v · w = vy + vw = v 2 yw vyw 4) (0, 1) ≡ (0, 2), (1, 1) ≡ (2, 2), (1, 2) ≡ (2, 1). + {(0, 1), (0, 2)} {(1, 1), (2, 2)} {(2, 1), (1, 2)} {(0, 1), (0, 2)} {(0, 1), (0, 2)} {(1, 1), (2, 2)} {(2, 1), (1, 2)} {(1, 1), (2, 2)} {(1, 1), (2, 2)} {(2, 1), (1, 2)} {(0, 1), (0, 2)} {(2, 1), (1, 2)} {(2, 1), (1, 2)} {(0, 1), (0, 2)} {(1, 1), (2, 2)} · {(0, 1), (0, 2)} {(1, 1), (2, 2)} {(2, 1), (1, 2)} {(0, 1), (0, 2)} {(0, 1), (0, 2)} {(0, 1), (0, 2)} {(0, 1), (0, 2)} {(1, 1), (2, 2)} {(0, 1), (0, 2)} {(1, 1), (2, 2)} {(2, 1), (1, 2)} {(2, 1), (1, 2)} {(0, 1), (0, 2)} {(2, 1), (1, 2)} {(1, 1), (2, 2)} ( ) 6) Consider ϕ ab = a · b−1 , which ( ) is defined since K is a field. Clearly ϕ is a homomorphism, the kernel is z0 , and the image is all of K. √ 8) {x + y 2 | x, y ∈ Z}.

10) (x3 + x + 1)/(x2 + x).

12) 1/(x2 + x).

14) (x3 + x2 + x)/(x2 + 1).

16) (x3 + 1)/x2 .

18) (x2 + x + 1)/(x + 1).

20) (ix2 + ix + 1 + i)/(x3 + (1 + i)x2 + 2ix + 1).

26

Answers to Even-Numbered Problems

Section 11.3 2) If u′ (x) + iv ′ (x) = i(u(x) + iv(x)), then u′ (x) = −v(x) and v ′ (x) = u(x).

Hence u′′ (x) = −u(x), and v ′′ (x) = −v(x). But since e0 = 1, u(0) = 1 and v(0) = 0. This means that u′ (0) = −v(0) = 0 and v ′ (0) = u(0) = 1. Then u(x) = cos x and v(x) = sin(x) are the unique solutions to the initial value problems.

4) πi + 2kπi, where k ∈ Z. √ √ 6) ±1, 1/2 ±√i 3/2, −1/2 ± i 3/2.

8) −2, 1 ± i 3.

10) . . . , e−9π/2 , e−5π/2 , e−π/2 , e3π/2 , e7π/2 , . . . .

12) Let z be one solution to ez = x. Then xy = e(z+2kπi)y = ezy · (e2yπi )k .

Let a = ezy and r = e2yπi . Note that |r| ̸= 1 since y is complex. So we get an infinite number of solutions ark for k ∈ Z. 14) If x and y are nth roots of 1, then (x · y −1 )n = xn · (y n )−1 = 1, so x · y −1 is an nth root of 1.

16) If either x or y are 0, then both sides are 0 or both undefined, so we may assume that x ̸= 0 and y ̸= 0. Let a and b be values such that x = ea and y = eb . Then xz = e(a+2kπi)z = e(az+2kzπi) , and y z = e(bz+2jzπi) .

Since ea+b = x · y, we see that (xy)z = e(az+bz+2kzπi) . So (xz ) · (y z ) = e(az+bz+2kzπi+2jzπi) = e(az+bz+2kzπi) = (x · y)z .

18) False: 4(1/2) · 4(1/2) = (±2) · (±2) = ±4, yet 4(1/2+1/2) = 41 = 4.

20) parametric plot3d([r*cos(t), r*sin(t), log(r)], (t, -pi, pi), (r, .01, 3))

The surface is single valued, and looks like a tornado funnel.

Section 11.4 2) Let x be any nonzero element, so that x2 > 0. Then b = x2 + a > a.

4) b · d − a · c = b · d − b · c + b · c − a · c = b · (d − c) + c · (b − a) > 0.

6) Since (x − y)2 ≥ 0, x2 − 2xy + y 2 ≥ 0, so x2 + y 2 ≥ 2xy.

8) Let f (x) ∈ Z[x]+ have leading term axm and g(x) ∈ Z[x]+ have leading term bxn . Then f (x) · g(x) has a leading term abxm+n which is in Z[x]+ since ab > 0. f (x) + g(x) will have leading term of either axm , bxn , or (a + b)xm , depending on whether m > n, n > m, or n = m. In any case f (x) + g(x) ∈ Z[x]+ . Finally, either the polynomial is 0, or the leading term is either positive or negative, so the law of trichotomy holds. √ 10) For x > 0 in the standard ordering, then x = ( x)2 > 0 in any ordering, so there is no non-standard ordering of R. Hence if there were a non-trivial automorphism ϕ, then ϕ(P ) = P . Also, ϕ(1) = 1 since the identity must map to the identity. Then ϕ(2) = ϕ(1) + ϕ(1) = 2, and likewise ϕ(n) = n for all integers n. Then ϕ(p/q) = ϕ(p)/ϕ(q) = p/q for all rationals. If ϕ(x) = y ̸= x, then there is a rational p/q between x and y, but then ϕ(x − p/q) = y − p/q, which contradicts ϕ(P ) = P .

Answers to Even-Numbered Problems

27

12) T1 · T2 sends y to (x3 + x)y ′′′ + 2x2 y ′′ + (x2 − 2x + 1)y ′ − 2xy; T2 · T1 sends y to (x3 + x)y ′′′ + (1 − x2 )y ′′ + (x2 + 1)y ′ − 2xy.

14) If S and T are the same degree and in P , then ϕ(S + T ) = ϕ(S) + ϕ(T ), so S + T is in P . If S has a higher degree than T , ϕ(S + T ) = ϕ(S), so in either case S + T is in P . Since ϕ(S · T ) = ϕ(S) · ϕ(T ), S · T will also be in P . Clearly the √ trichotomy condition is satisfied, so W is an ordered ring. √ 16) ( 3 2 − 4 3 4 − 11)/43.

18) Assume a, b and c are integers with no prime factors in common. The only way for b3 + 3bc2 − c3 to be even is if both b and c are even, forcing a to be odd. Then replacing b = 2x and c = 2y shows a3 + (x3 − 3ax2 ) + (3xy 2 − 3axy) + (6x2 y) − (3ay 2 + y 3 ) = 0. But each expression in parenthesis can be shown to be even, so we have a contradiction.

20) ϕ(x + y cos(π/7) + z cos(2π/7)) = x − z/2 + z cos(π/7) − (y + z) cos(2π/7).

Section 12.1 2) q(x) = x/2 + 7/4, r(x) = 3x/4 + 3/4.

4) q(x) = x3 + x2 + x, r(x) = x2 + x + 1.

6) q(x) = 3x, r(x) = 2.

8) 2x2 + x + 2.

10) (x + 1)(x2 + x + 2).

12) f (0) = 4(mod 13), f (1) = f (3) = f (9) = 5(mod 13), f (2) = f (5) = f (6) = 12(mod 13), f (4) = f (10) = f (12) = 3(mod 13), f (7) = f (8) = f (11) = 9(mod 13), so Proposition 12.3 applies.

14) x2 + x − 3.

16) x2 + x + 1.

18) x2 + x + 2.

20) s(x) = x, t(x) = −1.

22) s(x) = −(3x + 4)/64, t(x) = (3x + 13)/64.

Section 12.2 2) (x + 1)(x2 + x + 2).

4) (x + 3)(x2 + 6x + 3).

6) (x + 4)2 (x2 + 4x + 2).

8) (x2 + 1)(x2 + 2x + 2).

10) 3 and 15 are irreducible.

12) Yes. The only non-units are 3 and 6, which are associates, so these are irreducible.

14) f (x) mod 3 = x3 +2x+2, which is irreducible in Z3 [x]. (f (0) = 2, f (1) = 5, f (2) = 14.)

16) Suppose f (x) = g(x) · h(x), with g(x) of degree m and h(x) of degree n.

The only way for f (ai ) to be prime is if either g(ai ) = ±1, or h(ai ) = ±1.

By Corollary 12.2, g(ai ) = ±1 for at most 2m values, and h(ai ) = ±1 for at most 2n values, so f (x) would be prime for at most 2n + 2m = 2d values of

28

Answers to Even-Numbered Problems

a. But we found 2d + 1 values, so f (x) is irreducible. Additional note: values for which f (a) = ±1 can be counted twice!

18) f (0) = 2, f (±1) = 13, f (±3) = 173, f (±5) = 877, f (±13) = 30253. Since the polynomial is irreducible in Z[x], by Problem 15 it is also irreducible in Q[x].

20) (5 + 2a)−1 = 5 − 2a, so (5 + 2a)2 = 49 + 20a is also a unit.

Section 12.3 2) {0, 2, 4, 6, 8, 10}, {0, 3, 6, 9}.

4) {0, 2, 4, 6, 8, 10, 12, 14, 16}, {0, 3, 6, 9, 12, 15}.

6) No, every non-zero would be a unit.

8) It suffices to show that b + I has a multiplicative inverse whenever b ∈ R, b∈ / I. But {br + a|r ∈ R, a ∈ I} is an ideal that is strictly contains I, so it must be R. Hence, e = bc + d for some c ∈ R and d ∈ I, so (b + I)(c + I) = bc + I = e + I. 10) If I is a maximal ideal, by Problem 8, R/I is a field, hence has no zero divisors, so by Proposition 12.7, I is a prime ideal.

12) If g(x) · h(x) ∈ I, then g(0) · h(0) = 0, so either g(0) = 0 or h(0) = 0, hence one of the polynomials is in I, so I is prime. Since the non-trivial ideal in Problem 11 properly contains I, I is not maximal.

14) If I is a prime ideal, since R is a PID, it is generated by some element p, which would be prime. By Lemma 12.1, p is irreducible. Then by Lemma 12.6, so R/⟨p⟩ will be a field. So by Problem 9, I = ⟨p⟩ is a maximal ideal.

16) Let x ̸= 0. Since ⟨x2 ⟩ is a prime ideal, and x · x ∈ ⟨x2 ⟩, then x ∈ ⟨x2 ⟩, so there is some y for which x = x2 · y, and canceling we get x · y = 1, so x is invertible.

18) A PID is a UFD, so every nonzero, non-unit x can be uniquely factored into irreducible elements, so x has an irreducible factor. But in a PID, irreducible elements are prime. √ √ √ √ 20) 11 and 13 are prime in Z[ −5]. Also, −5, 2 −5 + 3, and −5 + 6 are prime.

Section 12.4 2) −4 + i = −1 · (5 + 3i) + (1 + 4i) = (−i + i)(5 + 3i) + (4 − i).

4) a = b · u for some unit u, so µ(a) = µ(b · u) ≥ µ(b), and µ(b) = µ(a · u−1 ) ≥ µ(a).

2 6) x2 can be 0, 1, 2, or 4 (mod 7), and likewise for −6y √ √ . So the sum is 0 (mod 7) only if x = y = 0. Now if (x + y 6) · (a + b 6) is a multiple of 7, then (x2 − 6y 2 ) · (a2 − 6b2 ) is a multiple of 7, so one of these factors, say x2 − 6y 2 , is a multiple √ of 7. But then both x and y are multiples of 7, so the original factor (x + y 6) is a multiple of 7.

8) Since N (a + bi) = a2 + b2 is prime, Proposition 12.8 shows that a + bi is irreducible, hence prime.

10) If a+bi is a factor of p, then a−bi will also be a factor, so (a+bi)·(a−bi) = a2 +b2 will be a factor of p. But p is prime in the ordinary sense, so a2 +b2 = p.

Problem 9 does the other direction.

Answers to Even-Numbered Problems

29

√ 12) Let q = (1 − 4n + 1)/2, and x + yq = x + yq. If N (a) = ±1, then b = a is such that a · b = ±1, so a has an inverse. Likewise, if a has an inverse a−1 , then 1 = N (1) = N (a · a−1 ) = N (a) · N (a−1 ), so N (a) = ±1. If N (a) = p, and a = b · c, then N (b) · N (c) = p, and so either b or c is a unit. √ 14) Let t = x · y −1 = u + vq ∈ Q( 5), and round u and v to the nearest integers i and j. If p = i + jq, then N (p − t) = a2 + ab − b2 , where a and b are both less than 1/2, so |N (p−t)| ≤ 3/4. Hence µ(r) = |N (r)| = |N (p·y −x)| = |N (p − t) · N (y)| < |N (y)| = µ(y).

16) If a · b is a multiple of 2, then N (a) · N (b) is a multiple of N (2) = 4, so either N (a) or N (b) is even, say N (a). x2 + xy + 5y 2 can only be even if both x and y are, so a is a multiple of 2, hence 2 is prime. To show 3 is prime, repeat the argument, but we need to show x2 + xy + 5y 2 is a multiple of 3 only if both x and y are. This can be done via a small table for x, y ∈ 0, 1, 2. √ 18) Let b√be the greatest integer not exceeding √ 2Im(z)/ 19. Then Im(z−bq) = Im(z − 19b/2) will be between 0 and 19/2. Let a be the closest √ integer to Re(z − bq), and let x = a + bq. Then 0 ≤ Im(z − x) < 19/2, and −1/2 ≤ Re(z − x) ≤ 1/2.

20) Letting z = m−1 x, we let y be as Problem 19 so that either |z − y| < 1 or |2z − y| < 1. We can extend N (x) to Q(q) by N (a + bq) = a2 + ab + 5b2 , a, b ∈ Q. In fact, N (z) = zz = |z|2 . So |m−1 x − y| < 1 or |2m−1 x − y| < 1, or |x − my| < |m| or |2x − my| < |m|. But x − my and 2x − my are in I, and we chose m to have minimum nonzero absolute value, so either x = my or 2x = my. In the first case, we can double y to get 2x = my.

22) If I is an ideal that is not a principle ideal, we can let m be the nonzero element of I with least N (m), and let x ∈ I, x ̸∈ ⟨m⟩. From Problem 21 we can find a y (not a multiple of 2) such that x = (m/2)y. Then xy = myy/2 ∈ I, and yy is some odd number, say 2n + 1. Since m(2n + 1)/2 = nm + m/2 ∈ I, and m ∈ I, then m/2 ∈ I, but this contradicts the fact that m was chosen to have minimum N (m). √ 24) 3, 5, 11, 13, 19, 29, and 37 are prime in Z[ 2]. In general, primes for which p ≡ 3 or 5 (mod 8).

Section 13.1 2) w + 3.

4) 3w + 4.

6) 2.

8) 4w + 1.

10) Let y be a root of x2 + x + 1 in the extension field. + 0 1 y y+1 · 0 1 y y+1 0 0 1 y y+1 0 0 0 0 0 1 1 0 y+1 y 1 0 1 y y+1 y y y+1 0 1 y 0 y y+1 1 y+1 y+1 y 1 0 y+1 0 y+1 1 y

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Answers to Even-Numbered Problems

12) Let y be a root of x2 + x + 2 in the extension field. + 0 1 2 y y+1 y+2 2y 2y + 1 2y + 2 0 0 1 2 y y+1 y+2 2y 2y + 1 2y + 2 1 1 2 0 y+1 y+2 y 2y + 1 2y + 2 2y 2 2 0 1 y+2 y y + 1 2y + 2 2y 2y + 1 y y y+1 y+2 2y 2y + 1 2y + 2 0 1 2 y+1 y+1 y+2 y 2y + 1 2y + 2 2y 1 2 0 y+2 y+2 y y + 1 2y + 2 2y 2y + 1 1 0 1 2y 2y 2y + 1 2y + 2 0 1 2 y y+1 y+2 2y + 1 2y + 1 2y + 2 2y 1 2 0 y+1 y+2 y 2y + 2 2y + 2 2y 2y + 1 2 0 1 y+2 y y+1 · 0 1 2 y y+1 y+2 2y 2y + 1 2y + 2

0 1 2 y y+1 y+2 2y 2y + 1 2y + 2 0 0 0 0 0 0 0 0 0 0 1 2 y y+1 y+2 2y 2y + 1 2y + 2 0 2 1 2y 2y + 2 2y + 1 y y+2 y+1 0 y 2y 2y + 1 1 y + 1 y + 2 2y + 2 2 0 y + 1 2y + 2 1 y+2 2y 2 y 2y + 1 0 y + 2 2y + 1 y + 1 2y 2 2y + 2 1 y 0 2y y y+2 2 2y + 2 2y + 1 y + 1 1 0 2y + 1 y + 2 2y + 2 y 1 y+1 2 2y 0 2y + 2 y + 1 2 2y + 1 y 1 2y y+2

14) All we need is at irreducible polynomial of degree 3 in Z3 [x], such as f (x) = x3 + 2x + 1. Then Z2 [x]/⟨f (x)⟩ is a field of order 27.

16) By Proposition 13.1, Q[x]/⟨x2 − 2⟩ is a field with a root to the√equation 2 x2 = 2. Thus, there √ is an isomorphism sending Q[x]/⟨x − 2⟩ to Q[ 2] which sends this root to 2.

18) {0, 1, y 2 + y, y 2 + y + 1} is a subfield of order 4, where y is the root of x4 + x + 1 in the field extension. There is no subfield of order 8.

20) {0, 1, 2, y 3 + y 2 + 2y, y 3 + y 2 + 2y + 1, y 3 + y 2 + 2y + 2, 2y 3 + 2y 2 + y, 2y 3 + 2y 2 + y + 1, 2y 3 + 2y 2 + y + 2} is a subfield of order 9, where y is the root of x4 + x + 2 in the field extension. There is no subfield of order 27.

Section 13.2 2) {3, 5, 6, 7, 10, 11, 12, 14}.

4) {3, 11, 12, 13, 17, 21, 22, 24}.

6) There will be ϕ(p − 1) primitive roots.

8) Note that a = (p − 1)/4 is an element of Zp−1 for which a2 is the only element of Zp−1 of order 2. Since Zp∗ ≈ Zp−1 , there is a corresponding element x in Zp∗ . Since −1 is of order 2 in Zp∗ , it must be the only element of order 2, so x2 = −1 in Zp∗ .

10) Since p is not prime in the UFD Z[i], it is not irreducible. Obviously there is not a real factor, and if a + bi is a factor, then by symmetry a − bi is a factor, so (a + bi)(a − bi) = a2 + b2 is a factor of p. But since p is prime in the ordinary sense, we have a2 + b2 = p.

Answers to Even-Numbered Problems

31

12) F (x) is obviously infinite, since every polynomial in F [x] is in F (x). Also, p · f (x) = 0 for all rational functions in F (x), so the characteristic is p.

14) Use Corollary 13.7.

16) If f (x) factored in K, then Zp [x]/⟨f (x)⟩ would be a subfield of order p2 of K. But this contradicts Problem 14.

18) Let F be the field of order pn . Since F ≈ Zp [x]/⟨f (x)⟩ whenever f (x) is n an irreducible polynomial of degree n, and x(p ) − x factors completely in F , n we have that the roots of f (x) are also roots of x(p ) − x. Since f (x) has no n repeated roots in Zp [x]/⟨f (x)⟩, we have that f (x) is a divisor of x(p ) − x in F , and hence in Zp [x].

20) x27 − x = x(x + 1)(x + 2)(x3 + 2x + 1)(x3 + 2x + 2)(x3 + x2 + 2)(x3 + x2 + x + 2)(x3 + x2 + 2x + 1)(x3 + 2x2 + 1)(x3 + 2x2 + x + 1)(x3 + 2x2 + 2x + 2), giving the 8 irreducible monic cubics.

Section 13.3 2) x6 + x3 + 1.

4) x12 + x11 + x10 + x9 + x8 + x7 + x6 + x5 + x4 + x3 + x2 + x + 1. n(n+1)/2 n(n+1)/2 = (ωnn )(n+1)/2 = . If n is odd, ωn 6) ωn · ωn2 · ωn3 · ωnn = ωn n/2 n(n+1)/2 = (ωn )n+1 = (−1)n+1 = −1.

1(n+1)/2 = 1. If n is even, ωn 8) GF(243).

10) Replacing x with 1/y in a d degree polynomial, and then multiplying this by y d , produces a polynomial with the coefficients reversed. Whenever x is a primitive root, then x−1 will be a primitive root of ϕn (x), but also a root of this reversed polynomial. Thus, the reversed polynomial has exactly the same roots, and by Problem 9 will be monic, it must be the same as Φn (x). n n−1 12) Since x(p ) −1 = Φ1 (x)·Φp (x)·Φp2 (x) · · · Φpn (x), and x(p ) −1 = Φ1 (x)· n n−1 Φp (x)·Φp2 (x) · · · Φpn−1 (x), it is clear that Φpn (x) = (x(p ) −1)/(x(p ) −1) = n−1 (Y p − 1)/(Y − 1), where Y = x(p ) . Since p is prime, this is Φp (Y ) = n−1 Φp (x(p ) ).

14) By Proposition 13.8, Φ(pn −1) , which has degree ϕ(pn − 1), has all irreducible factors of degree n, so n|ϕ(pn − 1).

16) Φ7 (x) = (x3 + x + 1)(x3 + x2 + 1).

18) Φ31 (x) = (x5 + x2 + 1)(x5 + x3 + 1)(x5 + x3 + x2 + x + 1)(x5 + x4 + x2 + x + 1)(x5 + x4 + x3 + x + 1)(x5 + x4 + x3 + x2 + 1). All irreducible polynomial of degree 5 are here.

20) The 6th degree Conway polynomial over Z2 is x6 + x4 + x3 + x + 1, the 3rd degree Conway polynomial is x3 + x + 1, and the 2nd degree is x2 + x + 1. If a is a root of the 6th degree, then a9 will be a root to the third degree, since (a9 )3 + (a9 ) + 1 = (a6 + a4 + a3 + a + 1) · 21

(a

+ a19 + a18 + a17 + a16 + a15 + a14 + a11 + a9 + a7 + a2 + a + 1).

a21 will be a root to the second degree polynomial, since (a21 )2 + a21 + 1 = (a6 + a4 + a3 + a + 1) ·

32

Answers to Even-Numbered Problems (a36 + a34 + a33 + a32 + a31 + a30 + a29 + a26 +a24 + a22 + a18 + a17 + a15 + a14 + a10 + a7 + a2 + a + 1).

This suggests a way to find the Conway polynomial: Since gcd((x9 )3 + (x9 ) + 1, (x21 )2 + x21 + 1) = (x6 + x4 + x3 + x + 1) · (x6 + x5 + 1) · (x6 + x5 + x2 + x + 1),

there are only 3 choices for the 6th degree polynomial, and only one is missing the x5 term.

Section 13.4 2) Since a2 + b2 + c2 + d2 is even, either all the numbers are even, or all the numbers are odd, or that two are even, and two are odd. In any case we can rearrange the numbers so that a and b have the same parity, and so do c and d. Then ((a + b)/2)2 + ((a − b)/2)2 + ((c + d)/2)2 + ((c − d)/2)2 = (a2 + b2 + c2 + d2 )/2 = mp/2.

4) Note that xy = xx + mz = mp + mz.

6) From Problem 1 there is some positive m such that a2 + b2 + c2 + d2 = mp, so consider the smallest such m. By Problem 2, m cannot be even. If m is odd, letting x = a + bi + cj + dk, we can use Problems 4 and 5 to find an integer quaternion y such that x = y + mz, |y|2 = mr for some r < m and xy is a multiple of m. By Problem 3, |xy| = m2 pr, so |(xy)/m| = pr. This contradicts minimality unless m = 1.

8) Clearly the difference of two Hurwitz integers is a Hurwitz integer, so we must show these are closed under multiplication. We must show that if x and y are integer quaternions, then x · y, x · (y + z), (x + z) · y and (x + z) · (y + z) are Hurwitz integers. Since the product of two integer quaterions is an integer quaternion, by the distribute law it is sufficient to show x · z, z · y and z 2 are Hurwitz integers. But x·z =

a−b−c−d a+b+c−d a−b+c+d a+b−c+d + i+ j+ k.

2 2 2 2

The numerators are either all even, or all odd. Likewise, z ·y will be a Hurwitz integer. Finally, z 2 = −1 + i + j + k.

10) If t = x · y −1 = a + bi + cj + dk, round a to the nearest integer or half-integer, producing e. If e is an integer, round b, c, and d to the nearest integer, otherwise round these to the nearest half-integer, producing q. Then |t − q|2 ≤ (1/4)2 + (1/2)2 + (1/2)2 + (1/2)2 < 1. Thus, N (r) = N (x − q · y) = N ((t − q) · y) = N (t − q)N (y) < N (y).

12) If x = y · z, then N (x) = N (y)N (z) = p. Thus, either N (y) or N (z) must be 1, and so either y or z is a unit.

Answers to Even-Numbered Problems

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14) Clearly x is an associate of x, since we can pick u1 = u2 = 1. If x = −1 u1 · y · u2 , then u−1 1 · x · u2 = y. Finally, if x = u1 · y · u2 and y = u3 · z · u4 , then x = (u1 · u3 ) · z · (u4 · u2 ).

16) R = 2(C13 −3C1 C22 +C23 +6C12 C3 −3C1 C2 C3 +9C1 C32 −3C2 C32 +C33 +2C43 − 6C4 C52 + 2C52 + 12C42 C6 − 6C4 C5 C6 + 18C4 C62 − 6C5 C62 + 2C63 − 6C1 C4 C7 − 12C3 C4 C7 +6C2 C5 C7 +12C3 C5 C7 −12C1 C6 C7 −6C2 C6 C7 −18C3 C6 C7 +4C73 + 6C2 C4 C8 −6C3 C4 C8 +6C1 C5 C8 −6C2 C5 C8 +12C1 C6 C8 +6C3 C6 C8 −12C7 C82 + 4C83 −12C1 C4 C9 +12C2 C4 C9 −18C3 C4 C9 −6C1 C5 C9 +6C3 C5 C9 −18C1 C6 C9 + 6C2 C6 C9 −6C3 C6 C9 +24C72 C9 −12C7 C8 C9 +36C7 C92 −12C8 C92 +4C93 ). Since this is real, w−1 = v/R. To show R ̸= 0, suppose R = 0 for some rational C1 through C9 . But multiplying by the common denominator, we can get an integer solution to R = 0 , and by dividing by any common factors, we can get an integer solution for which C1 through C9 have no common factors.

Then C13 + C1 C22 + C23 + C1 C2 C3 + C1 C33 + C2 C32 + C33 ≡ 0(mod 2). The only combination for this to be true is if C1 , C2 , and C3 are all even. Substituting C1 = 2B1 , C2 = 2B2 , and C3 = 2B3 into R, and factoring out 2 reveals that C43 + C4 C52 + C53 + C4 C5 C6 + C4 C62 + C5 C62 + C63 ≡ 0(mod 2). This forces C4 , C5 , and C6 to be even, so further replacing C4 = 2B4 , C5 = 2B5 , and C6 = 2B6 into R, and dividing by 2, reveals C72 + C7 C82 + C83 + C7 C8 C9 + C7 C92 + C8 C92 + C93 ≡ 0(mod 2), which once again forces C7 , C8 , and C9 to be even. But this contradicts that C1 through C9 have no common factors, so R ̸= 0.

2 18) c2 = c3 = c4 = c√ 6 = c9 = 0, and c7 = c8 . Since (a · b) = 2, the normalizer 3 is isomorphic to Q( 2).

20) 13 = (3 + 2i)(3 − 2i) = (1 + 2i + 2j + 2k)(1 − 2i − 2j − 2k). These two factorizations are not related by associates.

Section 14.1 √ 2) {1, 5}. √ √ √ √ √ √ √ 4) {1, 2, 3, 5, 6, 10, 15, 30}.

6) Since ω9 satisfies a polynomial equation of degree 6, Q(ω9 ) is a 6-dimensional extension of Q. Hence, {1, ω9 , ω92 , ω93 , ω94 , ω95 } is a basis.

8) Q(ω9 , ω6 ) = Q(ω18 ), and ω18 satisfies a polynomial equation of degree 6, 5 2 3 4 , ω18 , ω18 } is , ω18 Q(ω18 ) is a 6-dimensional extension of Q. Hence, {1, ω18 , ω18 a basis. √ 10) {1, 3}.

12) If every element of V can be expressed as a linear combination of the xi , we have a basis. Otherwise, we can add a vector so that the new set is still linearly independent. Since V is finite-dimensional, repeating the process will eventually give us a basis.

14) For n ∈ Z and r ∈ R, we define nr as in Definition 9.8, and properties 2, 3, and 4 come from Problems 13, 14 and 15.

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Answers to Even-Numbered Problems

16) Clearly R/I is an abelian group under addition, r((a + I) + (b + I)) = (r·a+I)+(r·b+I), (r+s)(a+I) = (r·a+I)+(s·a+I), (rs)(a+I) = r(s·a+I), and e(a + I) = a + I. 18) If x ∈ A + (B ∩ C), then x ∈ A + B, and since A ⊆ C, x ∈ C. Thus, x ∈ (A + B) ∩ C. Now suppose c ∈ (A + B) ∩ C. Then c = a + b, where a ∈ A, b ∈ B, and c ∈ C. Then b = c − a, so b ∈ C, hence b ∈ B ∩ C. So c ∈ A + (B ∩ C), and we have inclusion in both directions.

20) ⟨41/36, −1/18, 1/4⟩.

Section 14.2 2) x3 − 5.

4) x6 − 6x4 − 4x3 + 12x2 − 24x − 4.

6) x6 + 2x3 − 4.

8) x8 − 4x6 + 8x4 − 8x2 + 2. √ √ √ √ √ √ 10) Six roots: 3 2 ± 2, ω3 3 2 ± 2, ω32 3 2 ± 2. √ √ √ √ √ √ 3 3 3 12) Six roots: ± 5 − 1, ω3 ± 5 − 1, ω32 ± 5 − 1. √ √ √ 14) Eight roots: ± ± ± 2 − 1 + 1, where each ± can be either + or − independently of the other ± symbols.

16) F (u)(v) is the smallest subfield containing both v and F (u), and F (u) is the smallest subfield containing both u and F . Hence F (u)(v) = F (u, v), the smallest field containing u, v, and F . By symmetry, F (v)(u) = F (u, v), too. √ √ √ √ √ √ √ √ 18) √ ϕ0 (x) = √ x, ϕ1 ( √ 2) = 2,√ϕ1 ( 3) = − 3, ϕ2 ( 2) = − 2, ϕ2 ( 3) = 3, ϕ3 ( 2) = − 2, ϕ3 ( 3) = − 3. √ √ 20) x2 + 4x − 1 = (x + 2 + 5)(x + 2 − 5).

Section 14.3 √ √ 2) 2 + 5. √ 4) 3 2 + i.

6) ω15 . . . . .

8) r1 = 1.25992, r2 = −0.62996 + 1.09112i, r3 = −0.62996 − 1.09112i. r22 = 2 .

2 . −0.7937 − 1.37473i, r3 = −0.7937 + 1.37473i, r2 r3 = r1 = 1.5874. √ 10) Factors into (x2 − 7)(x2 + 1), so splitting field is Q( 7, i). √ √ √ √ 12) The roots are ± 1 ± 2, and all four roots are in Q( 1 + 2). √ √ 14) Letting ϕ(f ( t)) = f (t) creates a homomorphism from F ( t) to F (t), and the kernel is just 0, so we have an isomorphism. √ √ √ √ 3 16) If a = m + n, then m = (a −√ (3m + n)a)/(2n − 2m), and n = √ (a3 −√ (3n + √ m)a)/(2m − 2n). So Q( m, n) is in Q(a), and clearly Q(a) is in Q( m, n).

18) Splitting field = Q(a), where a5 = −a4 + 4a3 + 3a2 − 3a − 1; 5-dimensional extension.

20) Splitting field = Q(a, b, c), where a5 = −3a2 − 5a − 10, b4 = −ab3 − a2 b2 − a3 b − 3b2 − 3ab − 3a2 − a4 − 5, c3 = −bc2 − ac2 − b2 c − abc − a2 c − 3c − b3 − ab2 − a2 b − a3 − 3b − 3a; 60-dimensional extension.

Answers to Even-Numbered Problems

35

Section √ √ √ 15.1 √ √ 2) fix(ϕ0 ) = Q( 2, 3), fix(ϕ1 ) = Q( 2), fix(ϕ2 ) = Q(√3), fix(ϕ3 ) = Q( 6) 4) The fixed field of the identity automorphism is Q( 6 2), the fixed field of √ 3 the other automorphism is Q( 2).

6) D4 .√ √ √ √ √ √ √ √ √ 8)√ϕ0 ( 2)√= 2,√ ϕ0 ( 5) ϕ1 ( 5) = − 5; ϕ2 ( 2) = √ = 5;√ϕ1 ( 2)√= 2, √ − 2, ϕ2 ( 5) = 5; ϕ3 ( 2) = − 2, ϕ3 ( 5) = − 5.

10) The only non-trivial automorphism is the Frobenius automorphism f (x) = x2 .

12) The Frobenius automorphism f (x) = x3 is of order 4, so the Galois group is {i(x), f (x), f (f (x)), f (f (f (x)))}, where i(x) is the identity mapping.

14) The first extension is of order 5, so the Galois group must contain a 5cycle. Also, the complex conjugate automorphism switches two roots, so is a single 2-cycle. Now any 5-cycle and 2-cycle in S5 generate all of S5 , so the Galois group is isomorphic to S5 .

16) Z1 , Z2 , Z3 , Z4 , Z2 × Z2 , S3 , D8 , A4 , S4 . (Subgroups of S4 .)

18) 6 automorphisms, the most possible.

20) 4 automorphisms, sending a to a, −a, a − a3 , or a3 − a, where a is a root.

Group is isomorphic to Z8∗ .

22) 5 automorphisms, sending a to a, 2 − a2 , a3 − 3a, a4 − a3 − 3a2 + 2a + 1 or 4a2 − a4 − 2, where a is a root. Group is isomorphic to Z5 .

Section 15.2 2) Let ϕ be an automorphism that generates the Galois group. For an element of S4 to have order 4, it must be a 4-cycle, so ϕ is a 4-cycle of the four roots, ϕ : r1 → r2 → r3 → r4 → r1 . Then ϕ(k) = k, where k = r12 r2 + r22 r3 + r32 r4 + r42 r1 .

So k is in the fixed field of ϕ, and since ϕ generates the Galois group, k ∈ Q. 4) If the Galois group is Z5 , the roots of the polynomial can be rearranged such that r12 r2 + r22 r3 + r32 r4 + r42 r5 + r52 r1 is rational.

6) One solution: r1 = 1.827090915, r2 = 1.338261213, r3 = −0.209056927, r4 = −1.956295201, r12 r2 + r22 r3 + r32 r4 + r42 r1 = 11.

8) If a is a root, then all roots are in Q(a), hence |GalQ (F )| ≤ 4. There is an automorphism that sends a to a2 − 2, and this would send a2 − 2 to (a2 − 2)2 − 2, which can’t be a or else a would satisfy x4 − 4x2 − x + 2 = 0.

So there is an automorphism that is not of order 2, hence GalQ (F ) ≈ Z4 .

10) S3 .

12) Z2 .

14) Z2 .

16) GalQ (K) ≈ D4 , with 8 elements.

18) GalQ (K) ≈ Z5 ⋊Z4 , with 20 elements.

20) GalQ (K) ≈ Z2 × Z2 , with 4 elements.

22) GalQ (K) ≈ A5 , with 60 elements.

Section √ √ √ √ √ 15.3 2) Q, Q( 2), Q( 5), Q( 10), Q( 2, 5).

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Answers to Even-Numbered Problems

4) Since GalQ (F ) ≈ Z5∗ ≈ Z4 , and Z4 has only one nontrivial subgroup of order 2, then GalQ (F ) has only one nontrivial subfield of dimension 2, fixed by the mapping of order 2 sending every element to its complex conjugate.

Since √ −1 ω5 + ω5−1 is real, it is in this subfield. ω + ω = 2 cos(2π/5) = ( 5 − 1)/2, 5 5 √ so the nontrivial subfield is Q( 5). ∗ 6) Since Z15 has 3 elements of order 2, namely 4, 11 and 14, the 3 automor4 11 14 phisms send ω15 to either ω15 , ω15 , or ω15 .

6 6 36 6 8) ϕ1 (ω7 + ω7 ) = ω7 + ω7 = ω7 + ω7 . ϕ2 (ω73 + ω75 + ω76 ) = ω76 + ω710 + ω712 = ω73 + ω75 + ω76 .

10) Since ϕ fixes F , and also u, then ϕ fixes F (u), and hence is in GalF (u) (E).

12) Since |GalF (E)| = p, and there is only one group of order p, GalF (E) ≈ Zp . √ √ 14) F = Q, K = Q( 3 2), E = Q( 3 2, ω3 ). √ 16) For ϕ(ω8 ) = ω83√ , fix(ϕ) = Q(ω8 + ω83 ) = Q( −2). For ϕ(ω8 ) √ = ω85 , 2 5 7 fix(ϕ) = Q(ω8 ) = Q( −1). For ϕ(ω8 ) = ω8 , fix(ϕ) = Q(ω√8 + ω8 ) = Q( 2).

6 2 11 3 , fix(ϕ) or ω16 18) For ϕ(ω16 ) = ω16 √ = Q(ω16 + ω16 ) = Q( −2). For ϕ(ω16 ) = 4 13 5 ω16 or ω16 , fix(ϕ) = Q(ω16 ) = Q( −1). √ 9 7 3 7 3 ) = Q( −5). + ω20 + ω20 , fix(ϕ) = Q(ω20 +√ω20 or ω20 20) For ϕ(ω20 ) = ω20 5 17 13 ) = Q( −1). , fix(ϕ) = Q(ω20 or ω20 For ϕ(ω20 ) = ω20 Section 15.4 √ √ √ 2) Q( −1, 4 3).

4) Q(ω5 , 5 6). √ √ √ √ √ √ √ 3 3 8) Q( 29, ω3 , −14 + 4 29, 14 + 4 29).

6) Q( 2, 3).

10) Since △ABC is similar to △ADE, BC/AB = DE/AD Thus, DE = a · b.

12) Simply move the point D to between A and B, and construct a new perpendicular so that E will be between A and C. 14) The identity is proven by squaring both sides. Thus, we can construct the real and complex parts of the square root of a constructible complex number. By induction, any number √ in Fn of the form a + bi would have a and b constructible. Also, since i = −1 is constructible in the new sense, any a + bi would also be constructible in the new sense if a and b are constructible.

Thus, the two definitions are equivalent.

16) The pentagon can be constructed if ω5 is a constructible complex number.

But Q(ω5 ) is a Galois extension of Q of dimension 4 = 22 , so by Problem 15, ω5 is constructible.

18) The 257-gon can be constructed if ω257 is a constructible complex number. But Q(ω257 ) is a Galois extension of Q of dimension 256 = 28 , so by Problem 15, ω257 is constructible.

2 4 8 16 32 64 128 4 16 20) Q(ω17 + ω17 + ω17 + ω17 + ω17 + √ ω17 + ω17 + ω17√, ω17 + ω17 + ω17 + 64 16 2 ω √17 , ω17 + ω17 , ω17 ) = Q(a = (−1 √+ 17)/2, b = (a + a + 4)/2), c = (b + b2 − 2ab − 2b + 2a + 6)/2, (c + c2 − 4)/2).