13.02 (Determine whether or not φ is a homomorphism.) Let φ : R → Z under addition be given by φ(x) = the greatest integer ≤ x. Solution: It is not a homomorphism, since φ(1.5) + φ(1.5) = 1 + 1 = 2 6= 3 = φ(1.5 + 1.5). 13.04 (Determine whether or not φ is a homomorphism.) Let φ : Z6 → Z2 be given by φ(x) = the remainder of x when divided by 2, as in the division algorithm. Solution: It is a homomorphism: φ(a + b) = φ(a) + φ(b) for all a, b ∈ Z6. 13.06 (Determine whether or not φ is a homomorphism.) Let φ : R → R∗, where R is additive and R∗ is multiplcative, be given by φ(x) = 2x. Solution: It is a homomorphism: φ(a + b) = 2a+b = 2a2b = φ(a)φ(b) for all a, b ∈ R. 13.08 (Determine whether or not φ is a homomorphism.) Let G be any group and let φ : G → G be given by φ(g) = g−1 for g ∈ G. Solution: If G is not abelian then φ is not a homomorphism. Because when G is not abelian, there exist a, b ∈ G such that ab 6= ba. Then φ(ab) = (ab)−1 6= (ba)−1 = a−1b−1 = φ(a)φ(b). 13.10 (Determine whether or not φ is a homomorphism.) Let F be the additive group of all continuous functions mapping R into R. Let R be the additive group of real numbers, and let φ : F → R be given by Z 4 φ(f) = f(x) dx. 0 Solution: It is a group homomorphism, since for any f, g ∈ F , Z 4 Z 4 Z 4 φ(f + g) = (f + g)(x) dx = f(x) dx + g(x) dx = φ(f) + φ(g). 0 0 0 13.18 Compute ker(φ) and φ(18) for φ : Z → Z10 such that φ(1) = 6. Solution: For any integer k, φ(k) = φ(1 + 1 + ··· + 1) = φ(1) + φ(1) + ··· + φ(1) = 6k mod 10. | {z } | {z } k copies k copies 23 So ker φ = {k ∈ Z | 6k = 0 mod 10} = 5Z, φ(18) = 6 ∗ 18 = 108 = 8 ∈ Z10. 13.21 Compute ker(φ) and φ(14) for the homomorphism φ : Z24 → S8 where φ(1) = (2, 5)(1, 4, 6, 7). Solution: By homomorphism property, φ(k) = φ(1 + 1 + ··· + 1) = φ(1)φ(1) ··· φ(1) = φ(1)k = (2, 5)k(1, 4, 6, 7)k. | {z } | {z } k copies k copies Therefore, φ(14) = (2, 5)14(1, 4, 6, 7)14 = (1, 4, 6, 7)2 = (1, 6)(4, 7), and ker(φ) = 4Z24. 13.29 Let G be a group, and let g ∈ G. Let φg : G → G be defined by −1 φg(x) = gxg for x ∈ G. For which g ∈ G is φg a homomorphism? Solution: For every x, y ∈ G we have −1 −1 −1 φg(xy) = gxyg = (gxg )(gyg ) = φg(x)φg(y). So for every g ∈ G the map φg is a homomorphism. 13.32 a. T. b. T. Let φ : G → G0 such that φ(g) := e0 for g ∈ G. c. F. d. T. e. F. When |G| is finite, |φ[G]| is a divisor of |G|. f. F. g. T. For example the trivial homomorphism. h. T. For example the trivial homomorphism. i. F. Kernel contains at least the identity. j. F. Let φ : Z2 → Z2 × Z by φ(a) := (a, 0). 24 13.35 (If possible, give an exmaple of a nontrivial homomorphism φ. If not, explain why.) φ : Z2 × Z4 → Z2 × Z5. Solution: Define φ : Z2 × Z4 → Z2 × Z5 by φ((a, b)) := (a, 0) ∈ Z2 × Z5, for (a, b) ∈ Z2 × Z4. It is easy to check that φ is a nontrivial homomorphism. 13.36 (If possible, give an exmaple of a nontrivial homomorphism φ. If not, explain why.) φ : Z3 → Z. Solution: No such nontrivial homomorphism exists. Since if φ : Z3 → Z is a homomorphism, then 0 = φ(0) = φ(3) = 3φ(1) in Z, and so φ(1) = 0, and so φ is trivial. 13.44 Let φ : G → G0 be a group homomorphism. Show that if |G| is finite, then |φ[G]| is finite and is a divisor of |G|. Solution: φ[G] = {φ(a) | a ∈ G}. So |φ[G]| ≤ |G| is finite. Let H := ker φ. Partition G into left cosets of H: G = a1H t a2H t · · · t akH. Then φ[G] = {φ(a1), ··· , φ(ak)} and |φ[G]| = k. Clearly |G| = k|H|. So |φ[G]| is a divisor of |G|. 13.45 Let φ : G → G0 be a group homomorphism. Show that if |G0| is finite, then |φ[G]| is finite and is a divisor of |G0|. Solution: The image φ[G] is a subgroup of G0. By Theorem of La- grange in Section 10, when |G0| is finite, |φ[G]| is a divisor of |G0| and is finite. 13.49 Show that if G, G0 and G00 are groups and if φ : G → G0 and γ : G0 → G00 are homomorphisms, then the composite map γφ : G → G00 is a homomorphism. Solution: Let x, y ∈ G, then γφ(xy) = γ(φ(xy)) = γ(φ(x)φ(y)) = γφ(x)γφ(y) So γφ satisfies the homomorphism property. Thus it is a homomor- phism. 25 13.51 Let G be any group and let a be any element of G. Let φ : Z → G be defined by φ(n) = an. Show that φ is a homomorphism. Describe the image and the possibilities for the kernel of φ. Solution: First we show that φ is a homomorphism. Given m, n ∈ Z, φ(m + n) = am+n = aman = φ(m)φ(n). So φ is a homomorphism. Second we describe the image of φ: φ[Z] = {φ(n) | n ∈ Z} = {an | n ∈ Z} = hai (≤ G). Finally we describe the possibilities for the kernel of φ. Let e be the identity of G. If a has a finite order m (so that |hai| = m), then am = e and so ker φ = φ−1({e}) = {n ∈ Z | an = e} = mZ. If a has infinte order, then there exists no positive integer m such that am = e. So ker φ = φ−1({e}) = {n ∈ Z | an = e} = {0}. 13.52 Let φ : G → G0 be a homomorphism with kernel H and let a ∈ G. Prove the set equality {x ∈ G | φ(x) = φ(a)} = Ha. Solution: Let e0 be the identity of G0. Then {x ∈ G | φ(x) = φ(a)} = {x ∈ G | φ(x)φ(a)−1 = e0} = {x ∈ G | φ(xa−1) = e0} = {x ∈ G | xa−1 ∈ H} = {x ∈ G | x ∈ Ha} = Ha. 26.