1.

2. (a) Either n is even or it is odd; that is, n = 2k or n = 2k + 1. Then n² = 4k² or n² = 4(k² + k) + 1.

3.

4. (a) If then Hence from which xy = 0; therefore x = 0 or 497 y = 0, contrary to hypothesis.

5. (a) n = 11 is a counterexample because then n² + n + 11 has 11 as a factor.

1. (a) {x x = 5k where

2. (a)

3.

(a) Not equal: −1 A but −1 ∉ B	(c) Equal to {a, l, o, y}
(e) Not equal: 1 A but 1 ∉ B	(g) Equal to { − 1, 0, 1}

4.

5. (a) True. As B ⊆ C, each element of B (in particular, A) is an element of C. (c) False. A = {1}, B = C = {{1}, 2}.

6. Every element of A ∩ B is in both A and B by definition, so A ∩ B ⊆ A and A ∩ B ⊆ B. If X ⊆ A and X ⊆ B, then x X implies that x A and x B; that is, x A ∩ B. Hence, X ⊆ A ∩ B.

11. (a) (x, y) A × (B ∩ C) if and only if x A and y B ∩ C; if and only if x A and y B, and x A and y C; if and only if (x, y) A × B and (x, y) A × C; if and only if (x, y) (A × B) ∩ (A × C). Hence A × (B ∩ C) and (A × B) ∩ (A × C) have the same elements.

1.

2.

3.

7.

9. If βα = 1_A, then α is one-to-one so, as |A| = |B| is finite, α is also onto. Hence α⁻¹ exists so α⁻¹ = 1_Aα⁻¹ = βαα⁻¹ = β1_B = β. Then αβ = αα⁻¹ = 1_B and β⁻¹ = (a⁻¹)⁻¹ = α.

11. ϕ⁻¹(x, y) = α₂ where α₂(1) = x and α₂(2) = y.

14. (b) ⇒ (c). If αγ = αδ, then γ = 1_Aγ = (βα)γ = β(αγ) = β(αδ) = (βα)δ = 1_Aδ = δ.

15. (c) ⇒ (a) If b₀ B − α(A), choose a₀ A, and define β: B → B by:

Deduce b₀ = α(a₀) using (c).

1.

2.

3.

7.

10. (c) |A_≡| = |Q| = n by (c) of the preceding exercise.

2. (e) .

3. (c) If 3^2k+1 + 2^k+2 = 7m, then 3^2k+3 + 2^k+3 = 7(9m − 2^k+2).

7. Clear if n = 1. In general, such a (k + 1) digit number must end in 4, 5, or 6, and those are 3^k of each by induction. We are done since 3 · 3k = 3^k+1.

10. (a) If k ≥ 2 cents can be made up, there must be a 2-cent or a 3-cent stamp. In the first case, replace a 2-cent stamp by a 3-cent stamp; in the second case, replace a 3-cent stamp by two 2-cent stamps.

15. If p₁ is true and p_R ⇒ p_R+1, show X = {n p_nisfalse} is empty.

17. If p_n is “ n has a prime factor”, then p₂ is true. Assume p₂, . . ., p_k are all true. If k + 1 is a prime, we are done. If k + 1 = ab write 2 ≤ a ≤ k and 2 ≤ b ≤ k, then a (and b) has a prime factor by strong induction. Thus, k + 1 has a prime factor.

18.

(a) an = 2(− 1)n

(c)

24.

1. (a)

2. (a) n/d = 134.293 . . ., so q = 134. Then r = 113.

9.

11. (a) If d = xm + yn, where then

15. If d = gcd (m, n) and d₁ = gcd (m₁, n₁), then d m and d n, so d m₁ and d n₁ by hypothesis. Thus d d₁.

19. If d = gcd (m, n) and d₁ = gcd (km, kn), then d|m and d|n, so kd|km and kd|kn. Hence kd|d₁. To show that d₁|kd, write km = qd₁ and kn = pd₁. We have d = xm + yn where x and so kd = xkm + ykn = xqd₁ + ypd₁. Thus d₁|kd.

27. Let d = gcd(m, p^k). Then d p^k so d = p^j, j ≤ k. Show that j > 0 contradicts gcd(m, p^k) = 1.

30.

31.

33. (a) 25, 200 has 90 positive divisors.

41. (a) gcd(28, 665, 22, 869) = 63, and lcm(28, 665, 22, 869) = 10, 405, 395.

1.

2.

3.

8. (a) 7

9. (a) 7

15. One of a, a + 1 is even so 2 a(a + 1)(a + 2); similarly, one of a, a + 1, a + 2 is a multiple of 3. Since gcd(2, 3) = 1, it follows that 2 · 3 = 6 divides a(a + 1)(a + 2).

17. Compute for 0 ≤ a ≤ b.

22.

(a)

(c)

25.

(a)

(c) No solution

(e)

27.

(a)	(c) No solution

31. (1)⇒(2). Let n = p^ka where p is a prime and p6 a. If a > 1 then gcd(n, a) = a > 1, so has no inverse in _n. By (1), let in _n. Then n a^k, so p a^k, so p a, a contradiction.

35. (a) Working modulo p, means . Thus so or by Theorem 7.

1.

3.

7. (a) 24

11. (a)

13.

17. (a) (1432)(576)

18. Odd

19.

25. It suffices to show that any pair of transpositions is a product of 3-cycles. If k, l, m, and n are distinct, this follows from (k l)² = ε, (k l)(m n) = (k m l)(k m n), and (k l)(k m) = (k m l).

1.

2. (a)

7. M × N is commutative if and only if both M and N are commutative. (m, n) is a unit if and only if both m and n are units, and then (m, n)⁻¹ = (m⁻¹, n⁻¹).

9. (a) a²⁴a = a²⁵ = (a⁵)⁵ = (b⁵)⁵ = b²⁵ = b²⁴b = a²⁴b. Cancel a 24 times

13. If , show that (vw)u = 1_U.

18. (2)⇒(1). If (ab)⁻¹ = x then (ab)x = 1, so a(bx) = 1. Then (bx)a = 1 by (2). So a is a unit. Similarly for b.

1.

3.

8. (a) Every element σ satisfies σ² = ε.

13. α is onto because α(g⁻¹) = g for all g G; α is one-to-one because g⁻¹ = h⁻¹ implies that g = (g⁻¹)⁻¹ = (h⁻¹)⁻¹ = h.

23. (a) If g = g⁻¹, then g² = gg⁻¹ = 1; if g² = 1, then g⁻¹ = g⁻¹1 = g⁻¹g² = g.

29. (a) We first establish left cancellation: If gx = gy in G, then x = y. In fact, let hg = e. Then gx = gy implies x = ex = hgx = hgy = ey = y. With this, the fact that hg = e = e · e = hge gives g = ge by left cancellation. This shows that e is the unity. Finally, h(gh) = (hg)h = eh = h = he, so gh = e, again by left cancellation. Thus, h is the inverse of g.

1.

6. (a) 1² = 1. If x = g², y = h², then xy = (gh)² and x⁻¹ = (g⁻¹)².

7. (a) 1 = g⁰, g^kg^m = g^k+m, and (g^k)⁻¹ = g^−k; the subgroup test applies.

8. (a) 1 = xx⁻¹ X . Clearly X is closed. If , we obtain . Hence, X is a subgroup; clearly x = x¹ X so X⊆ X .

15.

17. ⇒. If H K, let h H − K. If k K, show kh ∉ K, hence kh H, whence k H.

1.

2.

4.

7.

8. (a) (123)(45)

9.

11. (a) If G = a where o(a) = n, let g = a^k. Then gⁿ = (a^k)ⁿ = a^kn = (aⁿ)^k = 1^k = 1.

16.

(a) H = G

(c)

(e) H = = is even

17. (a) X ⊆ Y ⊆ Y , Y a subgroup, so X ⊆ Y by Theorem 8.

25. Let G = g and H = h where o(g) = m, o(h) = n. As |G × H| = |G| |H| = mn, it suffices to show that o((g, h)) = nm. We have (g, h)^nm = (g^nm, h^nm) = (1, 1). If (g, h)^k = (1, 1), then g^k = 1 and h^k = 1, so m k and n k. But gcd(n, m) = 1, then implies nm k (Theorem 5 §1.2)), so o(g, h) = mn, as required.

27. (a) If A ⊆ B, show that g^a = g^bq, Since |g| =∞, a = qb. Conversely, if a = qb, then g^a B, so A ⊆ B.

8. (a) If let α(1) = m. Then α(k) = α(k · 1) = k[α(1)] = km. Thus, α is multiplication by m, and each such map is a homomorphism →

12.

19. (a) If z Z(G), then σ(z) Z(G₁) because, given g₁ = σ(g) in G₁, σ(z) · g₁ = σ(zg) = σ(gz) = g₁ · σ(z). Hence, σ: Z(G) → Z(G₁) is a mapping. It is one-to-one because σ is, and clearly holds. If z₁ Z(G₁), let z₁ = σ(z), z G. If g G, then σ(gz) = σ(g) · z₁ = z₁ · σ(g) = σ(zg), so gz = zg because σ is one-to-one. Thus, z Z(G), and σ is onto.

23. Write . Suppose is an isomorphism, write . Then so r = 1, so so , a con-tradiction.

25. is infinite cyclic, so is infinite cyclic too, say In particular q² = k₀q, so . Thus, a contradiction.

31. If σ: G → G is an automorphism, then o(σ(a)) = 2, so σ(a) = a. Because σ(1) = 1, σ = 1_G and autG = {1_G}.

33. Let , o(a) =∞. If σ(a) = a^m, , then a = σ(a)^k = a^mk. As o(a) =∞ this gives 1 = mk, whence m = ± 1. If m = 1, then σ = 1_G; if m = − 1 then σ(g) = g⁻¹ for all g G.

1.

5. (a) a ≡ a because a⁻¹a = 1 H. If a ≡ b then b⁻¹a H, and it follows that a⁻¹b = (b⁻¹a)⁻¹ H, so b ≡ a. Finally, if a ≡ b and b ≡ c then b⁻¹a H and c⁻¹b H, so c⁻¹a = (c⁻¹b)(b⁻¹a) H and a ≡ c.

9.

10. (a) 6

12. (a) If o(g) = m, we show m = 12. We have m 12 by Lagrange's theorem. If m ≠ 12, then m 4 or m 6, so g⁴ = 1 or g⁶ = 1, contrary to hypothesis.

16. (a) If 1 = xm + yn, where then g = g¹ = (g^m)^x(gⁿ)^y = 1^x1^y = 1.

25. (a) Because a^kba^k = b implies that a^k+1ba^k+1 = aba = b, it holds for k ≥ 0. But aba = b gives b = a⁻¹ba⁻¹, so a^−kba^−k = b follows for k ≥ 1 in the same way.

31. If , let Kh₁, . . ., Kh_n be the distinct cosets of K in H. This means H = Kh₁ ∪ ∪ Kh_n, a disjoint union. Then Hg ⊆ Kh₁g ∪ ∪ Kh_ng is clear, and it is equality because K ⊆ H. Thus, each coset of H in G is the union of n K-cosets. If this gives . Conversely, if is finite, then is clearly finite and is finite by the hint since each H-coset is a union of K-cosets.

3. (a) If σ = (123), the group of motions is

4. (a) If σ = (1234), the group of motions is

6. (a) If σ = (123)(456) and τ = (14)(26)(35), the group G of motions is G = {ε, σ, σ², τ, τσ, τσ²} ≅ D₃.

1.

2. If D₄ = {1, a, a², a³, b, ba, ba², ba³}, where |a| = 4, |b| = 2, and aba = b, the normal subgroups are {1}, D₄, Z = {1, a²} = Z(D₄), K₁ = {1, a², b, a²b} and K₁ = {1, a², ba, ba³}.

5. First aKa⁻¹ is a subgroup, by Theorem 5 §2.3, and aKa⁻¹ ⊆ aHa⁻¹ ⊆ H because H G. If h H, we must show h(aKa⁻¹)h⁻¹ ⊆ aKa⁻¹. We have h⁻¹Kh = K as K H, so h(aKa⁻¹)h⁻¹ = ha(h⁻¹Kh)a⁻¹h⁻¹ = (hah⁻¹)K(hah⁻¹)⁻¹ = K because K G.

11. Let H and K be subgroups of G with and . Then H ∩ K = {1} by Lagrange's theorem. Moreover, H G because it is unique of its order, and similarly K G. Hence, G ≅ H × K by the Corollary to Theorem 6. Since p and q are primes, H and K are cyclic of relatively prime orders. Hence, H × K is cyclic by Exercise 25 §2.4.

15. (a) Conclude that Ka = G  K = Kb⁻¹, so ab K.

17. (a) If H = a^d , d n, let n = md. Since ba^k is self-inverse for all k, we have (ba^k)⁻¹a^dt(ba^k) = ba^ka^dtba^k = b(a^k+dt)ba^k = b · b · a^−k−dt · a^k = b²a^−ka^−dta^k = a^−dt H.

24. (c) Let G = C₂ × C₂, where o(a) = 2, and let H = C × {1}. Then H G because G is abelian. But σ: G → G given by σ(x, y) = (y, x) is an automorphism and σ(H) H. Thus, H is not characteristic in G.

26. (a) Write K = core H = ∩ _aaHa⁻¹. Clearly 1 K. If g, g₁ K, then gg₁ aHa⁻¹ for all a, so gg₁ K. Also, g⁻¹ a⁻¹H(a⁻¹)⁻¹ for all a, so g K. Hence, K is a subgroup. If g G, k K then gkg⁻¹ g[(g⁻¹a)H(g⁻¹a)⁻¹]g⁻¹ = aHa⁻¹ for all a, as required.

1.

3. (a) 6, 4, 3, 12

4. (a) 12

5. (a) 1, 2, 2, 2

7.

13.

19.

25. (a)

4. (a) 1 α⁻¹(X) because α(1) = 1 X; if g and h α⁻¹(X), then α(gh) = α(g)α(h) X and α(g)⁻¹ = α(g⁻¹) X, shows that gh α⁻¹(X) and g⁻¹ α⁻¹(X). If X α(G), let h α⁻¹(X), g G. Then α(ghg⁻¹) = α(g)α(h)α(g)⁻¹ α(g)Xα(g)⁻¹ = X, so ghg⁻¹ α⁻¹(X). Hence, α⁻¹(X) G.

8. (a) If |g| = 6, then the choice of α(g) K₄ determines α: C₆ → K₄. If α(g) = 1, then α is trivial. If x ≠ 1 in K₄, then o(x) = 2 and we define α_x: C₆ → K₄ by α_x(g^k) = x^k. This mapping is well defined because

g^k = g^m ⇒ 6|(k − m) ⇒ 2|(k − m) ⇒ x^k = x^m.

Hence, α_x is a homomorphism and α_x(g) = x. Thus, these are the only nontrivial homomorphisms.

(c) Let , where o(a) = 3, o(b) = 2, and aba = b, and let , o(c) = 4. If α: D₃ → C₄ is a homomorphism, write K = ker α. Then K D₃, so K = {1}, or K = D₃. Now K = {1} is impossible as α is not one-to-one (|D₃| = 6 does not divide |C₄| = 4). If K = D₃, then α is trivial. So assume that α is not trivial. Then α(G) ≅ G/K = {K, bK}, so α(G) is the unique subgroup of C₄ of order 2: α(G) = {1, c²}. If ϕ: G → G/K is the coset map, there is an isomorphism σ: G/K → α(G) such that α = σϕ. Clearly, σ(K) = 1 and σ(bK) = c². Hence

This is the only nontrivial homomorphism.

10. (a) No. If α: S₃ → K₄ were onto, then K₄ ≅ S₃/ker α, and |K₄| = 4 would divide |S₃| = 6.

(c) Yes. S₃/A₃ ≅ C₂, say σ: S₃/A₃ → C₂ is an isomorphism. If ϕ: S₃ → S₃/A₃ is the (onto) coset map, then σϕ: S₃ → C₂ is an onto homomorphism.

13. Let G be simple. If α: G → G₁ is nontrivial, ker α ≠ G, so ker α = {1} by sim-plicity. So α is one-to-one and G ≅ α(G) ⊆ G₁. Conversely, if G₁ has a subgroup G₀ and σ: G → G₀ is an isomorphism, then σ: G → G₁ is a (one-to-one) homo-morphism, which is nontrivial because G₀ ≠ {1} (it is simple).

17. (a) If g G′, write g = [a₁, b₁][a₂, b₂] [a_n, b_n], where [a, b] = a⁻¹b⁻¹ab. Then .

21. (a) Define because z ≠ 0). Then α is a homomorphism because , and ker . Thus, the isomorphism theorem gives

33. (a) has subgroups H = {0}, {0, 2}, and . Hence {1}, so these are the possible images.

1.

2.

7. (a) Detects 3, corrects 1.

11. (a) As k = 3 and t = 2, n must satisfy . If n = 3, 4, . . ., 8, this expression reads 7 ≤ 1, 11 ≤ 2, 16 ≤ 4, 22 ≤ 8, 29 ≤ 16, and 37 ≤ 32. Hence n ≥ 9. [Note: For n = 9, it reads 46 ≤ 64.]

15. (a) If C is a (4, 2)-code that corrects one error, the weight of C must be at least 3 so that the nonzero words in C are contained in {1111, 1110, 1101, 0111}. But the sum of any two of these words is not in the set.

20.

21.

1. (a) Not an additive group.

(c) h(f + g) ≠ hf + hg can happen (try h(x) = x²).

3. (a) because the column sums are

The rest of the subring test is routine.

7.

14. Compute (1 + sr) [1 − s(1 + rs)⁻¹r].

15. (3) ⇒ (1). Let e be the unique right unity. Given b R, show that r(e + eb − b) = r for all r R. Now use the uniqueness.

16. (a) is a subring by Theorem 5. It is centered because s · (k1_n) = ks = (k1_n)_s for all s R by Theorem 2.

18.

21. (a) (1 − 2e)² = 1 − 4e + 4e² = 1

22. (a) If a = (1 − e)re then the fact that e² = e gives ea = 0 and ae = a. It follows that a² = (ae)a = a(ea) = a · 0 = 0.

23. (4) ⇒ (1). If r R, a = (1 − e)re is nilpotent so u = 1 + a is a unit and so commutes with e by (4). Conclude that re = ere.

29.

36. (a) If is an isomorphism, then a = σ(i) satisfies a² = − 1, a contradiction. (c) If then is a division ring, a contradiction.

1.

3. Idempotents ={0, 1}; nilpotents = {0}

7. If ab = 0 show that (ba)² = 0.

9. Try for various primes p.

15. If z Z(R) and za = 1, showing that a Z(R) is sufficient. Given r R, (ra − ar)z = raz − arz = r · 1 − 1 · r = 0, so (as za = 1) ra − ar = 0.

16. (a) If o ≠ a K and ab = 1 with b K, conclude that b = a⁻¹ where a⁻¹ is the inverse in F.

19. is a subfield of R by Example 4. If F is any subfield of R then (because 1 R), and hence (because for all n, m ≠ 0 in ). If also F, this means for all . Thus .

23. If R = {r₁, r₂, . . ., r_n} and 0 ≠ a R, then ar₁, ar₂, . . ., ar_n are distinct (ar_i = ar_j implies r_i = r_j as a ≠ 0). Hence {ar₁, ar₂, . . ., ar_n} has n elements, and so equals R. Hence, 1 = ar_i for some i.

26. (a) If and show

29. (a) If r = i and s = 1 in , consider a = r + sω in . Then aa^* = r² + s² = 0, but a ≠ 0 and a^* ≠ 0 in . Thus, is not a field. In let a = 1 + 2ω. Then aa^* = 1² + 2² = 0, and a ≠ 0 ≠ a^*. So is not a field. However, is a field. If a = r + si ≠ 0 in then aa^* = r² + s² and it suffices to show r² + s² ≠ 0 in . Suppose r² + s² = 0. If r = 0 or s = 0 then a = 0, contrary to hypothesis. Thus r ≠ 0 ≠ s. Then 0 = s⁻¹(r² + s²) = (s⁻¹r)² + 1 so (s⁻¹r)² = − 1 in . This is not the case because 0² = 0, 1² = 1 = 6², 2² = 4 = 5², 3² = 2 = 4² in .

32. (a) If q is a unit in then 1 = N(1) = N(qq⁻¹) = N(q)N(q⁻¹), so N(q) is a unit in R. Conversely, qq^* = N(q) shows q⁻¹ = N(q)⁻¹q^* if N(q) R^*.

1.

3. (c) If R is commutative then (r + A)(s + A) = rs + A = sr + A = (s + A)(r + A).

7. Let where a = (n, m), b = (k, l). Show that ml = 0.

9. (a) A = R because i is a unit in R. So |R/A| = 1.

(c) R/A = {0 + A, 1 + A, 2 + A, 3 + A, 4 + A} and |R/A| = 5

10. If 0 ≠ z Z(R), then Rz is a nonzero ideal of R (it contains z ≠ 0) and so Rz = R. Hence 1 Rz, say 1 = az. It is enough to show that a Z(R). If r R, then

so ra = ar, as required.

11. (a) Show that A = {r R nr = 0} is an ideal of R.

14. (a) X + Y is a subgroup because (1) 0 = 0+ 0 X + Y; (2) if r = x + y is in X + Y then −r = (− x) + (− y) X + Y; and (3) if also r′ = x′ + y′ is in X + Y then r + r′ = (x + x′) + (y + y′) X + Y. We have X ⊆ X + Y because x = x + 0 X + Y for all x X. Similarly Y ⊆ X + Y.

15. A ∩ S is clearly an additive subgroup of S. If a A ∩ S and s S, then sa A because A is an ideal and sa S because a S. Thus, sa A ∩ S; similarly, as A ∩ S.

21. (a) (r + A)² = r² + A = r + A for all r R because r² = r by hypothesis.

22. (a) If e² = e in R, show that (e + A)² = e + A in R/A. By hypothesis, e A or 1 − e A. But 0 is the only nilpotent idempotent.

23.

25. (c) If u is a unit then u Ru implies Ru = R by Theorem 2. Conversely, if Ru = R then 1 Ru, say 1 = vu, . Hence, u is a unit (R is commutative).

33. (c) In is nilpotent. If then is not nilpotent.

34. (c) Write If is an ideal of then so k pⁿ. Hence, k = p^t for t ≤ n, so A ⊆ M where It follows that M is the unique maximal ideal of so is local and

1.

3. If is a general ring homomorphism, let θ(1) = e. Show that e² = e so either e = 1or e = 0. In the last case θ(k) = θ(k · 1) = θ(k) · θ(1).

9. 0 and R up to isomorphism

10. (4) Clearly, θ(r⁰) = θ(1) = 1 = θ(r)⁰. If θ(rⁿ) = θ(r)ⁿ for some n ≥ 0, then θ(rⁿ⁺¹) = θ(rⁿ · r) = θ(rⁿ) · θ(r) = θ(r)ⁿ · θ(r) = θ(r)ⁿ⁺¹. (5) Note first that θ(u) · θ(u⁻¹) = θ(uu⁻¹) = θ(1) = 1 and, similarly, that θ(u⁻¹) · θ(u) = 1. So θ(u⁻¹) = θ(u)⁻¹. If k ≥ 0, then (4) gives (5): If k = − m, m > 0, then θ(u^k) = θ[(u⁻¹)^m] = θ(u⁻¹)^m = [θ(u)⁻¹]^m = θ(u)^k.

13. In ₇ this is 4n² = 2 and this has a solution (n = 2) in ₇. In ₁₁ it is 7m² = 9, or m² = 8 · 9 = 72 = 6. But m² = 0, 1, 3, 4, 5, 9 in ₁₁, so there is no solution.

17. R ≅ R for any ring R because 1_R: R → R is an isomorphism. If R ≅ S, say σ: R → S is an isomorphism, then σ⁻¹: S → R is also an isomorphism, so S ≅ R. If also S ≅ T, where τ: S → T is an isomorphism, then τσ: R → T is an isomorphism and R ≅ T.

21. If is a ring homomorphism, then because 1 ∉ ker θ (θ(1) = 1 ≠ 0). Thus ker θ = 0 because is a field, from which If θ(i) = a, then a² = θ(i)² = θ(i²) = θ(− 1) = − 1, a contradiction.

29. (a) The map given by is an onto homomorphism with kernal .

37. Define by . Show that θ is a ring homomorphism and .

41. If , show that e² = e. If is defined by σ(r) = re, show that σ is a ring isomorphism. Hence

3.

3. (a) 500

4. (a) In : 4, 5, 1, 2; in : 4, 5

5. (a) In : 0, 1; in all four elements are roots; in : 0, 1, 3, 4

7. (a) Let uxⁿ and bx^m be the leading terms of f and g, where u is a unit. The leading term of f g is ubx^n+m because ub ≠ 0 (otherwise b = u⁻¹(ub) = 0). Hence, f g ≠ 0 and deg (f g) = n + m = deg f + deg g.

14. (a) q = x³ + 3x² − 3x + 5, r = − x − 3 = 5x + 3

16. (a) 3, 5

17.

23.

25.

(a)

(c) 2, −1

(e) None

34. (c) If u let denote the conjugate of u. Define by This is a homomorphism (it is evaluation at 0 followed by conjugation) but it is not evaluation at a for any . Indeed, if θ = ϕ_a then while ϕ_a(i) = i.

38. (a) Show that R[x]/P[x] ≅ (R/A)[x] as rings (Exercise 37). Use Theorem 2.

4.

5.

8. (a) As f is monic, we may assume that both factors are monic (Exercise 6). Hence Now equate coefficients.

15.

18.

22. (a) Eisenstein Criterion, with p = 3

23. (a) f(x + 1) = x⁴ + 4x³ + 6x² + 6x + 2, so use the Eisenstein criterion with p = 2.

31. (a) If f is irreducible in K[x] it cannot factor properly in F[x].

35.

39.

42. (a) Let 1 = mf + kg with m and k in F[x]. If h = pf and h = qg, then

h = hmf + hkg = (qg)mf + (pf)hg = (qm + pk)fg.

2

3

5

4. (a) Here t² = t. Idempotents: 0, t, 1, 1− t; nilpotents: 0; units: a + bt, where a ≠ 0 ≠ a + b

7. (a) 5(− 1 + t + t²)

14.

21. (a) If x² + ax + b is not irreducible over F, it must have a root u F. Thus, u² + au + b = 0. Take c = 2u + a. Then c² = 4(u² + ua) + a² = − 4b + a², con-trary to hypothesis.

25. (a) Write polynomials as f = f(x). Then d f + g because d = uf + vg for some . On the other hand, f d and g d because d is a common divisor of f and g. Hence f ⊆ d and g ⊆ d , so f + g ⊆ d .

2.

2. (a) (y²z²) + (x³ + xyz + x²z) + (x² + xz − yz) + (3x − 3y)

7. (a)

3.

(a)

(c)

11.

12. (a)

13. (a) x³ − 17x² − 14x − 9

7. ±1

10.

12.

14.

16. (a) If p q, suppose p is irreducible. If q = ab in R then p ab so p a or p b. Thus q a or q b, so q is irreducible. The converse is the same.

23. No.

27. Write d = gcd [a, gcd (b, c)] and d₁ = gcd [gcd (a, b), c]. Then d divides a and gcd (b, c), so it divides all a, b, and c. Thus d divides gcd (a, b) and c, which gives d|d₁. Similarly d₁|d, so d d₁. Moreover, this result shows that d divides a, b, and c and that every common divisor of a, b, and c divides d. Hence, gcd (a, b, c) exists and d gcd (a, b, c).

31. If m lcm(a₁, . . ., a_n) exists in R then a_i|m for each i shows m ⊆ a_i for each i, and hence that m ⊆ A where we write A = a_i ∩ ∩ a_n . But r A means a_i|r for each i, so m|r by definition. Thus, r m and we have m = A. Conversely, if A = m thena_i|m for each i (because m a_i ); and, if a_i|r for each i, then r A = m so m|r. Thus, m is a least common multiple of the a_i.

35. Use Gauss' lemma.

37. If f = ug, u is a unit, write . Since f and g are primitive, show a b.

39. (a) Show that R is a subring of

40. (a) Show

1. No, in .

5. Let A = a , a ≠ 0. If a is a unit then |R/A| = 1. Otherwise, by Theorem 4 §3.3, let B/A be any ideal of R/A, say B = b . Show that there are at most finitely many such divisors b of a up to associates.

8. (a) Write R is a subring of R because , and and p does not divide nn′. Thus, R is an integral domain. Given in R, if p does not divide m then is a unit in R (with inverse . Conversely, if is a unit, say , then mm′ = nn′ so p does not divide m (it does not divide n or n′), so does not divide m}.

13. (b) where

15. (b) a = 5b + (− 1), where δ(− 1) = 1 < 2 = δ(b)

24. (a)

26. (a) (1) ⇒ (2). Given a ≠ 0, b ≠ 0, let a, b = d . Then d = ra + sb for some r, s R, so if k|a and k|b in R then k|d. But d|a and d|b because a, b d .  (2) ⇒ (1). Given A = a, b , clearly A is principal if a = 0 or b = 0. Otherwise let g = gcd(a, b) d where d = ra + sb for r, s R. Then d A so d ⊆ A. On the other hand, d|a and d|b so a d and b d . Hence a, b ⊆ d .

35. No. If so, and , then . But 2 = 1 + 1 > 0.

1.

2.

7.

11. {(1, − 1, 0), (1, 1, 1), (a, 0, 0)} for any a ≠ 0 in

15. I, A, A², A³, A⁴ cannot be independent.

19. (a) {1, r, . . ., rⁿ} is not independent because dim _F(R) = n. So a₀ + a₁r + + a_nrⁿ = 0 where the a_i F are not all zero.

22. (b) If {u₁, . . ., u_m} ⊆ {u₁, . . ., u_m, . . ., u_n} then implies that where a_m+1 = = a_n = 0. So a_i = 0 for 1 ≤ i ≤ m.

25. If is in then so, writing a_i = − 1, with a_i ≠ 0. Hence is dependent. Conversely, if where some a_i ≠ 0, then is in

31. (a) They are additive subgroups by group theory. If kerϕ then for all a F. If im ϕ, say , then

1.

2.

3.

4. (a) x² − 2x + 2

7. (a) so . We claim that the minimal poly-nomial is . Its roots in are and neither is in , so it is irreducible in , as required.

12. (a) {1, u, u²}, where (c) where (e)

13.

17. If F(u) ⊇ L ⊇ F write p = [F(u): F] = [F(u): L][L: F]. Thus [L: F] = 1 or p; so L = F or [L: F] = [F(u): F], whence L = F(u) by Theorem 8 §6.1.

19. Let . Show that with degree 2 exists such that f(u) = 0, say f = ax² + bx + c. Conclude that so . We may as-sume a, b, and c are integers, so . If d = p²e, , p a prime, then . Continue until where m is square free.

21. (a) Write L = F(u), so Thus and [L: F] = m by hypothesis, and Hence, we simply show that If p and m are the minimal polynomials of over L and F, respectively, then p|m by Theorem 3, so

23. If then , gcd(f, g) = 1. Then h(π) = 0, where h(x) = 2g²(x) − f²(x), and this is a contradiction if h ≠ 0 in . But h = 0 means 2g² = f² so, since gcd(f, g) = 1, f|2. Thus f = ± 1, ±2, g²(x) = ± 1, . This forces deg g = 0; , g = ± 1, . Thus g = ± 1, , a contradiction. Thus h ≠ 0 and has led to a contradiction. So

26. (a) Let f(u²) = 0, 0 ≠ f F[x]. Use g where g(x) = f(x²) ≠ 0.

31. (a) We show F(u) = Q, where Q = {f(u)g(u)⁻¹| f, g F[x]; g(u) ≠ 0}. Since u is transcendental over F, f(u) ≠ 0 whenever f(x) ≠ 0. Thus Q is a subfield of E containing F and u, so F(u) ⊆ Q. But any subfield of E containing F and u must contain Q, so F(u) ⊇ Q. Thus F(u) = Q.

33. Show that if u is algebraic over A then u A, contrary to hypothesis. If f(u) = 0 where f ≠ 0 in A[x], let , . Show that u L(u) where is a finite extension of F.

1. (a) and [E: Q] = 2. (c) and

2. (a)

4.

6. (a) No. If were the splitting field of then . Thus would be algebraic, contradicting the fact that π or e is transcendental.

9. If gcd(f, g) = 1 let 1 = fh + gk; h, k in F[x]. Suppose E ⊇ F is an extension containing a common root u, that is f(u) = 0 = g(u). Then substitution gives 1 = f(u)h(u) + g(u)k(u) = 0, a contradiction. So no such extension E exists. Conversely, let d = gcd(f, g). If d ≠ 1 then degd ≥ 1 so let E ⊇ F be a field containing a root u of d. Then d|f and d|g means f(u) = 0 = g(u), contrary to hypothesis. So d = 1.

13. Show that the roots of x^p − 1 are so is the splitting field. By Theorem 6, Appendix A, x^p − 1 = (x − 1)Φ_p, where Φ_p = x^p−1 + x^p−2 + + x + 1 is irreducible over by Example 13 §4.2.

20. (a) We show Clearly is algebraic. We must show that if u E is algebraic over then u A. Since u is algebraic over A, we show that u ∉ A implies u is transcendental over A. We have E = A(π) so this follows from Exercise 31 §6.2 if we can show that π is transcendental over A. But if π were algebraic over A it would be algebraic over , contrary to the preceding exercise.

1.

4.

5. If GF(16) = {a + bt + ct² + dt³ a, b, c, d in , then t is primitive. The subfields are , and

8.

9. If G ⊆ C^*, |G| = n, then where u = e^2πi/n.

17. Let d = gcd(f, f′) and write d = fg + f′h where g and h are in F[x]. If d = 1, suppose f has a repeated root a in E ⊇ F. Then x − a divides both f and f′ in E[x], and so divides d = 1, a contradiction. Conversely, if d ≠ 1, let E be a splitting field of f over F. Then d|f implies d has a root a in E, so x − a divides f and f′, a contradiction by Theorem 3.

22. (a) If is a field containing a root u of f, conclude that f is the minimal polynomial of u over . If then and so |E| = pⁿ. Then u is a root of so f|h in E[x], say h = qf. But h = q₀f + r in by the division algorithm, so this holds in E[x]. By the uniqueness in E[x], we get and

3. Yes. Bisect 30, after constructing that from a 30–60–90-triangle.

5. No. A sphere of radius 1 has volume and, if this is the volume of a cube with side a, then . But a is not constructible since it is not even algebraic over . For if a is a root of . Then π is a root of . This is impossible as π is transcendental over

5. (a) In B₄: 1 + t, t + t² = t(1 + t), t² + t³ = t²(1 + t) and 1 + t³ = t³(1 + t). The other members 0, 1 + t², t + t² and 1 + t + t² + t³ all lie in smaller ideals. (See Example 3.)

7. (a) 1 + x⁷ = (1 + x)(1 + x + x³)(1 + x² + x³) so there are 2 · 2 · 2 = 8 divisors in all. Thus, there are 7 codes (excluding 1 + x⁷ = 0).

11. Since g(x) = 1 + x + x⁴ has no root in , if it factorizes at all, it must do so as g(x) = (a + bx + cx²)(a′ + b′x + c′x²). Thus aa′ = 1 = cc′ so a = a′ = 1 = c = c′. Thus g(x) = (1 + bx + x²)(1 + b′x + x²) so (coefficient of x³) b + b′ = 0 and (co-efficient of x) b + b′ = 1. This is impossible.

17. (c) Write g(x) = 1 + x + x³. We have 1 + x⁷ = (1 + x)(1 + x² + x³)g(x), a product of irreducibles. Hence, 1 + x⁷ = h(x)g(x), where h(x) = 1 + x + x² + x⁴. We have 1 = xg(x) + 1 · h(x), so take e(x) = xg(x) = x + x² + x⁴. Note that e(t)² = e(t²) = t² + t⁴ + t⁸ = t² + t⁴ + t = e(t). So e(t) = t + t² + t⁴ is the idempo-tent generator.

1. (c) Using (a), x + (− x) = 0 = 0x = (1 + (− 1))x = 1x + (− 1)x = x + (− 1)x, so −x = (− 1)x.

2. (a) If α: M → N is onto and R-linear, and if M = Rx₁ + + Rx_n, then we have N = Rα(x₁) + + Rα(x_n). Since some of the α(x_i) may be zero, the result follows.

5. (a) If x = Σa_ik_i then rx = Σ(ra_i)k_i AK

6. Let A = Σ_iRa_i, a_i A, and M = Σ_jRx, x_j M. Use Exercise 5 to show that AM = Σ_i,jRa_ix_j.

7. (a) Define α: N → K + NK by α(n) = n + K for all n N. Show that α is R -linear and kerα = N ∩ K. Every coset in K + NK has the form (k + n) + K, where k K and n N. But (k + n) + K = n + K = α(n), which proves that α is onto. Now the isomorphism theorem applies.

11. (a) Yes. (m, n) = (n, n) + (m − n, 0) shows that M = K+ X; clearly K ∩ X = 0.

16. (a) We have M = π(M) + kerπ because m = π(m) + (m − π(m)) for each m M and π[m − π(m)] = π(m) − π²(m) = 0. If m π(M) ∩ kerπ, let m = π(m₁) with m₁ M. Then 0 = π(m) = π²(m₁) = π(m₁) = m, so π(M) ∩ kerπ = 0.

21. (a) If we must show that Show that where a_i A, and apply the linearity of α.

1. (c) ,

2. (a) The types are (4), (3, 1), (2, 2), (2, 1, 1) and (1, 1, 1, 1). Hence, representative groups are , , , and

3. (a) ,

4. (a) The types are: p-component (2), (1, 1); the q-component (3), (2, 1), (1, 1, 1); and the r-component (4), (3, 1), (2, 2), (2, 1, 1), (1, 1, 1, 1). Hence 2 · 3 · 5 = 30 in all.

7. (a) Thus G(2) has type (2, 2); G(3) has type (1, 1, 1) and G(5) has type (2, 1).

9. (a) The types are (2, 2, 2), (2, 2, 1, 1), (2, 1, 1, 1, 1), (1, 1, 1, 1, 1, 1).

12. (a) T(K) = {k K o(k) ≠ 0} = K ∩ {m M o(m) ≠ 0} = K ∩ T(M).

16. (a) Define σ: K → M/T(M) by σ(k) = k + T(M). This is a group homomor-phism and ker σ = {k K k T(M)} = K ∩ T(M) = T(K). Use the isomor-phism theorem.

20. (c) If m = Σx_i then dm = 0 implies dx_i = 0 for all i. Hence L_d(M) ⊆ Σ_iL_d(M_i).

22. (a) Here L_p(Rx) = {rx p(rx) = 0}. If prx = 0 then pr ann say pr = sp^m. Since m ≥ 1 and R is a domain, this gives r = sp^m−1, and we have shown that (Rx)^p ⊆ R(p^m−1x). The other inclusion is because p^mx = 0. Finally, p(Rx) = Rpx is a routine verification, and px = 0 if m = 1 because o(x) = p^m.

27. (a) L_p(G) consists of 0 and the elements of order p. We have L_p(G) = L_p(G₁) ⊕ ⊕ L_p(G_p) by Exercise 20, and |L_p(G_i)| = p for each i by Exercise 26.

1. (a) XY = {ε, σ, σ²}. Note: XY is a subgroup here, but X and Y are not.

3. as is abelian. Hence H G by the correspondence theorem.

4.

5.

9. (a) Clearly . If let Kg = Kh, h H. Then gh⁻¹ K ⊆ H, so g H. Similarly g H₁, so .

12. (a) H² ⊆ H because H is closed; H ⊆ H² because 1 H.

15. KA = AK and KB = BK are subgroups by Theorem 5 §2.8. Given kb in KB, Ab = bA and Kb = bK by hypothesis, so KA(kb) = AKkb = AKb = AbK = bAK = bKA = KbA = kKbA = (kb)KA. Thus KA KB.

22. (a) Let a ≠ b have order 2, show that H = {1, a, b, ab} is closed and apply La-grange's theorem.

1. (a) {1}, {a, a³}, {a²}, {b, ba²}, {ba, ba³}. The normal subgroups are the unions {1}, {1, a²} and {1, a, a², a³}, {1, b, a², ba²} and {1, ba, a², ba³}.

7. Let K = g⁻¹Hg, so H = gKg⁻¹. We claim N(K) = g⁻¹N(H)g. Let a N(K) so a⁻¹Ka = K. To show a g⁻¹N(H)g it suffices to show gag⁻¹ N(H). But (gag⁻¹)⁻¹H(gag⁻¹) = ga⁻¹g⁻¹Hgag⁻¹ = ga⁻¹Kag⁻¹ = gKg⁻¹ = H. HenceN(K) ⊆ g⁻¹N(H)g. A similar argument shows that N(H) ⊆ gN(K)g⁻¹ so g⁻¹N(H)g ⊆ N(K).

11. Because H ⊆ N(H) ⊆ G, Exercise 31 §2.6 shows that |G: N(H)| is finite. Hence, Theorem 2 applies.

14. We have a⁻¹Ha = {1, ba²} so ba² ∉ N(H). Continue in this way.

16.

25. H is a union of conjugacy classes. Show that there exists a ≠ 1 such that {a} ⊆ H.

29. Since C G, let Z[G/C] = K/C. Since |G/C| > 1, Theorem 6 shows C ⊂ K. But K G so K H by Exercise 23 §2.8. If k K then kC is in the center of G/C, so h⁻¹k⁻¹hk C H. Hence k⁻¹Hk ⊆ H, and similarly kHk⁻¹ ⊆ H. Thus k N(H) and we have shown K ⊆ N(H).

1. (a) By Cauchy's Theorem let a G, |a| = 5. If then |G · H| = 4, so there is a homomorphism θ: G → S₄ with ker θ ⊆ H. Then ker θ ≠ {1} because |G| = 20 does not divide |S₄| = 24. Because H is simple, ker θ = H, so H G.

10. (a) H₀ ⊆ H because H = 1_G(H). If τ autG then τ⁻¹σ autG for all σ autG, so H₀ ⊆ τ⁻¹σ(H). Thus τ(H₀) ⊆ σ(H) for all σ, so τ(H₀) ⊆ H₀. Simi-larly τ⁻¹(H₀) ⊆ H₀, whence τ(H₀) = H₀. Thus H₀ is characteristic in G.

15. If σ = (k₁k₂ )(m₁m₂ )(n₁n₂ ) , the orbits of the group G in X_n are G · k₁ = {k₁, k₂, . . . }, G · m₁ = {m₁, m₂, . . . }, G · n₁ = {n₁, n₂, . . . }, . Clearly, G · k = {k} if and only if k is fixed by σ.

17. (a) x ≡ x because x = x· 1; if x ≡ y, say y = x · a, a G, then x = y · a⁻¹, so y≡ x; if x ≡ y and y ≡ z, say y = x · a, z = y · b, then z = (x · a) · b = x · (ab), so z ≡ x.

23. (a) If a, b S(x) then (ab) · x = a · (b · x) = a · x = x, so ab S(G). Similarly, a⁻¹ · x = a⁻¹ · (a · x) = (a⁻¹ · a) · x = 1 · x = x, so a⁻¹ S(G). Finally 1 · x = x shows 1 S(x), and we are done.

28. If X = {H ⊆ G|H is a subgroup and |H| = p^k}, let G act on X by conjugation. Use Theorem 4.

32. (a) (1, 1) · x = 1x1⁻¹ = x, and . The orbit is (H × K) · x = {(h, k) · x h H, k K} = {hxk⁻¹ h H, k K} = HxK.

1. If P is a Sylow 3-subgroup, then where γ is a 3-cycle, say γ = (ijk). If where {1, 2, 3, 4} = {i, j, k, x}, then σ(123)σ⁻¹ = γ. Hence so P is conjugate to

3. P is a Sylow p-subgroup of N(P), being a p-subgroup of maximal order. It is unique because it is normal in N(P).

7. (a) |G| = 40 = 2³ · 5. Thus, n₅ = 1, 2, 4, 8, and n₅ ≡ 1 (mod5). Hence, n₅ = 1, so the Sylow 5-subgroup is normal. (c) |G| = 48 = 2⁴ · 3. If P is a Sylow 2-subgroup then |G: P| = 3, so a homomorphism θ: G → S₃ exists. Clearly, ker θ ≠ {1}.

9. (a) |G| = 70 = 2 · 5 · 7. Then n₅ = 1, 2, 7, 14, and n₅ ≡ 1 (mod5), so n₅ = 1. Similarly n₇ = 1, so let P G and Q G, where |P| = 5 and |Q| = 7. Because P ∩ Q = {1}, PQ ≅ P × Q ≅ C₅ × C₇ ≅ C₃₅. Hence, |G: PQ| = 2, so PQ G.

11. (a) |G| = 105 = 3 · 5 · 7. Then n₇ = 1, 3, 5, 15, and n₇ ≡ 1 (mod7), so n₇ = 1, 15. Similarly, n₅ = 1, 21. Let P and Q by Sylow 7-and 5-subgroups. If neither is normal in G, then G has 21 · 4 = 84 elements of order 5 and 15 · 6 = 90 elements of order 7, a contradiction. So P G or Q G; hence PQ is a subgroup, and |PQ| = |P||Q| = 35 because P ∩ Q = {1}. As |G: PQ| = 3, let θ: G → S₃ be a homomorphism with ker θ ⊆ PQ. Then | ker θ| ≠ 1, 5, 7, so PQ = ker θ G. Finally, P PQ and Q PQ by the Sylow Theorems, so PQ ≅ P × Q ≅ C₇ × C₅ ≅ C₃₅.

13. Let P be a Sylow p -subgroup of G. Since p > m we have |P| = pⁿ, so |G: P| = m. Apply Theorem 1 §8.3.

19. If Q is also a Sylow p -subgroup of G, then Q = a⁻¹Pa by Sylow's second theorem. If g N(Q) then Q = g⁻¹Qg; that is a⁻¹Pa = g⁻¹a⁻¹Pag. This implies that aga⁻¹ N(P) = P, whence g a⁻¹Pa = Q.

1. (a). Write σ = (1 2) S_n and Then A_n ⊆ A_nH ⊆ S_n As S_n/A_n ≅ C₂, either A_nH = S_n or A_nH = A_n. Since σ ∉ A_n we have S_n = A_nH. Similarly, A_n ∩ H ≠ {ε} means A_n ∩ H = H (because H is simple), contradicting h ∉ A_nonce more. Hence A_n ∩ H = {ε} and the result follows from Theorem 2.

3. This is an instance of Theorem 3 (3), where p = 3 and q = 13. We have q ≡ 1 (mod p), so we look for m such that 1 ≤ m ≤ 12 and m⁵ ≡ 1 (mod 13). If m = 1 then G ≅ C₁₃ × C₃ ≅ C₅₅. The first solution with m > 1 is m = 3, whence where o(a) = 11, o(b) = 3 and ab = ba³.

6. (a)

8. (a)

1. (a) 3; C₂, C₂, C₂   (c) 3; C₂, C₂, C₂   (e) 3; C₂, C₂, C₂

3. (a) If H_k is the unique subgroup of order k in C₂₄, the series are

8. (a) Let n = p₁p₂ p_m, where the p_i are distinct primes. Then C_n has length 1 + 1 + + 1 = m by Example 8.

11. Induct on n. If n = 1 then G = G₀ ⊃ G₁ = {1} so G ≅ G₀/G₁ is finite. In gen-eral, G₁ is finite by induction, and G/G₁ = G₀/G₁ is finite by hypothesis. Thus G consists of |G/G₁| cosets, each with |G₁| elements. Hence G is finite. Now |G| = |G₀/G₁| · |G₁|, and the formula follows by induction.

15. (a) If M ⊆ C_n is maximal normal, then C_n/M has order a prime q (being simple and abelian). Hence, q = p_i for some i because q divides |C_n| = n. Thus, for some i = 1, 2, . . ., r. Since C_n is cyclic, it has exactly one subgroup of order by Theorem 9 §2.4.

1. No, Z(S₄) = {ε}.

3. No. S₄ is solvable (Example 4) but is not abelian. Indeed because S₄/A₄ is abelian. Thus , {ε} or K = {ε, (1 2)(3 4), (1 3)(2 4), (1 4)(2 3)}. But S₄/{ε} and S₄/K are not abelian (see Exercise 30 §2.9).

8. (a) This is because α[a, b] = [α(a), α(b)] for every commutator [a, b] from G.

9. By Exercise 14 §8.4, let K G, K ≠ {1}, G. Then both |K| and |G/K| are in {p, q, p², pq}. Hence, both are either abelian or of order pq and thus are solvable. Use Theorem 4.

15. If G is solvable and G = G₀ ⊃ G₁ ⊃ ⊃ G_p = {1} is a composition series, each simple factor is abelian and hence finite. Hence, is finite (see Exercise 11 §9.1). The converse holds because every finite group has a composition series.

19. HK is a subgroup as K G, and is solvable by Theorem 3 (H is solvable). Done by Theorem 4.

21. (a) Because G ≠ {1}, G′ ≠ G by Theorem 5. Thus G/G′ is nontrivial and abelian.

23. (a) Write {K G G/K solvable} = {K₁, K₂, . . ., K_m}. This set is nonempty as it contains G. Then is normal and G/R is solvable by Exercise 18. If K G and G/K solvable, then R ⊆ K by definition.

27. (a) Write . Then V G because the intersection of normal subgroups is normal. Note that the intersection is not empty because G G and G/G is in . If V = K₁ ∩ K₂ ∩ ∩ K_n then G/V embeds in (as in Exercise 18) and is in by induction because is closed under taking direct products. Hence G/V is in , being isomorphic to a subgroup of a group in

2. If H G and K G then a⁻¹[h, k]a = [a⁻¹ha, a⁻¹ka] [H, K] for all h H and k K.

6. (a) By induction on n, it suffices to show that Γ_i(G × H) ⊆ Γ_i(G) × Γ_i(H). Do so by induction on i. If i = 0, then Γ₀(G × H) = G × H = Γ₀(G) × Γ₀(H). If the relation holds for i ≥ 0, then Γ_i+1(G × H) = [Γ_i(G × H), G × H] ⊆ [Γ_i(G) × Γ_i(H), G × H], so it suffices to show that, if A ⊆ G, B ⊆ H, then [A × B, G × H] ⊆ [A, G] × [B, H]. This outcome follows because [(a, b), (g, h)] = ([a, g], [b, h]).

9. If n = 2^k then |D_n| = 2^k+1 so D_n is nilpotent by Example 3. Conversely, suppose n = 2^km, m > 1 odd. Show that so D_m is nilpotent by Theorem 1. But in this case {1, b} is a Sylow 2-subgroup that is not normal, contradicting Theorem 4.

13. K ∩ Z(G) ≠ {1} by Exercise 11. Thus, K ∩ Z(G) = K by the condition on K. Now every subgroup of K is normal in G, so |K| is prime.

18. (a) H is itself nilpotent by Theorem 1 so, by Theorem 4, H is a product of p -groups. Now apply Theorem 6 §8.2.

21.

24. (1)⇒(2). We show that G′ ⊆ Φ, that is G′ ⊆ M for every maximal subgroup M of G. Show that this follows by (1) because M G.

26. (a) Write Φ(G) = Φ and , where F G. If M is maximal in G then K ⊆ M (because K ⊆ Φ ⊆ M) and is maximal in . Hence whence F ⊆ M. It follows that F ⊆ Φ. Conversely, let α: G → G/K be the coset map. Then α(Φ) ⊆ Φ(G/K) by the preceding exercise, so x Φ implies ; so x F. Thus Φ ⊆ F.

1. ε, σ⁻¹ and στ all fix F when σ and τ do.

3. because σ(a_i) = a_i for a_i F.

5. If σ: E → E is an automorphism, show that σ(q) = q for all

7. C₂

9. C₂ × C₂

11. Show that E = F(u) if u E  F. If m is the minimal polynomial of u over F, show °m = 2.

13. Construct σ and τ in gal with σ(u) = iu, σ(i) = i, and τ(u) = u, τ(i) = − i.

15. (a) and in Theorem 6.

16. and in Theorem 6.

17. See the Hint.

19. We proceed by induction on n. If n = 1 then E = F(u₁) = {f(u₁) f(x) F[x]}. Hence σ(f(u₁)) = g(σ(u₁)) = g(τ(u₁)) = τ(f(u₁)) for all f, as required. In gen-eral, write K = F(u₁, u₂, . . ., u_n−1) so that E = K(u_n). By induction, σ = τ on K, so σ, τ gal(K: F). Since σ(u_n) = τ(u_n) the result follows from the case n = 1.

21. See the Hint.

22. (a) For (3)⇒(1), if f has a repeated root u in E ⊇ F, let 1 = fg + f′h in F[x] by (3).

23. If d = gcd (f, f′), show d = 1.

25. If E ⊇ F and q is an irreducible factor of f in E[x], write f = p₁p₂ p_r in F[x], p_i irreducible, and show that q|p_i for some i.

27. If not, and u is a root of f in a splitting field E ⊇ F, show f = (x − u)^p in E[x]. If q is an irreducible factor of f in F[x], show q = (x − u)^t.

29. (a) If F is perfect, and a F, let E be the splitting field of f = x^p − a. If u E is a root of f show f = (x − u)^p. If q is an irreducible factor of f in F[x] show that q = x − u. Use Theorem 4.

30. (a) Let q be the minimal polynomial of u over F. If K = F(u^p) let m K[x] be the minimal polynomial of u over K. Then q K[x] and q(u) = 0, so m|q. But q has distinct roots by hypothesis, so m has distinct roots. On the other hand, x^p − u^p K[x] and x^p − u^p = (x − u)^p in E[x]. Hence m|(x − u)^p so m = (x − u)^r. Since m has distinct roots, r = 1 and so u K.

32. (a) Let p and q be the minimal polynomials of u over F and K respectively. Then p K[x] and p(u) = 0, so q|p. Since p has distinct roots in some splitting field L ⊇ K, q is separable over K.

1.

2. (a) Either :

or :

5. (a) Let show f(t)g(− t) = f(− t)g(t). If char F ≠ 2, write h(t) = f(t)g(− t). Show that h(t) = k(t²) for some polynomial k Similarly and g(t)g(− t) = l(t²) for some polynomial l. Continue.

7. Clear.

9. (a) An intermediate field K is closed if

11. Exercise 34 §2.6.

15. (a) Use the Galois connection.

17. If K = σ(K₁) show If show K₁ ⊆ σ⁻¹(K), so σ(K₁) ⊆ K, and similarly K ⊆ σ(K₁).

19. (a) Let E = F(u₁, u₂, . . ., u_m) where u₁, . . ., u_m are the distinct roots of f, use Theorem 3 §10.1.

20. (a) Apply σ to the formulas for N(u) and T(u).

21. If show f^τ = ∏ _σG[x − τσ(u)] = f.

1. (a)

2. (a) f′ = 5x⁴ − 4 has roots ±a and ±ia, where Then f(a) < 0 and f(− a) > 0, so f has three real roots and two (conjugate) nonreal roots. As f is irreducible (Eisenstein), its Galois group is S₅, as in Example 1.

3. Show that p = x⁷ − 14x + 2 has three distinct real roots and two (conjugate) complex roots. If is the splitting field, view as a subgroup of S_X where are the roots. Then conjugation is a transposition and, if u is a real root, then because p is the minimal polynomial of u over Proceed as in Example 1.

5. Let X denote the set of roots of p in a splitting field E ⊇ F where p F[x]. View G = gal(E: F) ⊆ S_X, so G embeds in S₄.

7. Since f′ = 3(x² − 1), conclude that f has three real roots. In the cubic formula, p³ and q³ are roots of x² + x + 1 which satisfy p³ + q³ = − 1 and pq = 1. The roots are and Show that p = e^2πi/9 and . The roots are and Finally