Solutions to Selected Homework Problems 1.26 Claim: α : S S which is 1-1 but not onto β : S S which is onto but not 1-1. y z = z y y, z S.

Transcript

1 Solutions to Selected Homework Problems 1.26 Claim: α : S S which is 1-1 but not onto β : S S which is onto but not 1-1. Proof. ( ) Since α is 1-1, β : S S such that β α = id S. Since β α = id S is onto, β is onto. β is not invertible since otherwise α = β 1 would be invertible. So β is not 1-1. ( ) Since β is onto, α : S S such that α β = id S. Since β α = id S is 1-1, α is 1-1. α is not invertible since otherwise β = α 1 would be invertible. So α is not onto Proof. Since α : T U is 1-1, δ : U T such that δ α = id T. So 3.31 Proof. We have β = (δ α) β = δ (α β) = δ (α γ) = (δ α) γ = γ. Letting x = e in the above, Hence is commutative. Now x (y z) = (x z) y x, y, z S. y z = z y y, z S. x (z y) = x (y z) = (x z) y x, y, z S. Hence is associative. 4.8 Proof. (a) α a,0, α b,0 B where a, b R, a 0, b 0, α a,0 α b,0 = α ab,0 B where 0 ab R. Hence is an operation on B. It is associative and commutative. α 1,0 is the identity. (b) α 1,b, α 1,c C where b, c R, α 1,b α 1,c = α 1,b+c C. Hence is an operation on C. It is associative and commutative. α 1,0 is the identity. (c) In fact, α a,b = α 1,b α a, Proof. Let G = {2 m 3 n : m, n Z}. 2 m1 3 n1, 2 m2 3 n2 G, 2 m1 3 n1 2 m2 3 n2 = 2 m1+m2 3 n1+n2 G. Hence G is closed under multiplication is the identity of G. a = 2 m 3 n G, a 1 = 2 m 3 n G. Hence G is a group

2 2 Solution. The Cayley table must be x y z x x y z y y z x z z x y 6.4 Solution. (a) ( ) = (2, 6)(3, 4, 5) (b) (1, 4)(1, 3)(1, 2) = (1, 2, 3, 4). (c) (1, 2, 3) 1 (2, 3)(1, 2, 3) = (3, 2, 1)(1, 2)(3, 2, 1) 1 = (2, 1). (d) (2, 4, 5)(1, 3, 5, 4)(1, 2, 5) = (1, 4)(2, 5, 3). 6.9 Proof. Let α = (a 1, a 2,..., a k ), β = (a 1, a k ) (a 1, a 3 )(a 1, a 2 ). x {1,..., n}, we will show that α(x) = β(x). Case 1. x / {a 1,..., a k }. Then α(x) = x and β(x) = x. Case 2. x = a i for some 1 i k 1. We have α(x) = a i+1 and β(x) = ( (a 1, a k ) (a 1, a i ) ) (a i ) = ( (a 1, a k ) (a 1, a i+1 ) ) (a 1 ) = ( (a 1, a k ) (a 1, a i+2 ) ) (a i+1 ) = a i+1. Case 3. x = a k. We have α(x) = a 1 and β(x) = a Proof. (a) It is a subgroup. (b) It is a subgroup. (c) It is not a subgroup. (1, 2, 3, 4, 5) 2 is not in the set. (d) It is a subgroup Proof. Let a, b Z(G). x G, we have (a b) z = a (b z) = a (z b) = (a z) b = (z a) b = z (a b). Thus a b Z(G). Hence Z(G) is closed under. Clearly, e Z(G). Let a Z(G). Then x G, a x = x a. Thus a 1 (a x) a 1 = a 1 (x a) a 1, i.e., x a 1 = a 1 x. Hence a 1 Z(G). 8.5

3 3 Solution. Let p k = e k 2π 1 5, k = 0, 1, 2, 3, 4 and let T = {p 0,..., p 4 }. The question is determine M (T ). p 2 y p 1 p 0 x p 3 p 4 Let We claim that α = the counterclockwise rotation of 72, β = the reflection with respect to the x-axis. M (T ) = {α i β j : 0 i 4, j = 0, 1}. Proof of the claim. It suffices to show that every γ M (T ) is of the form α i β j for some 0 i 4 and 0 j 1. There is a j (0 j 4) such that α j (p 0 ) = γ(p 0 ). Thus (α j γ)(p 0 ) = p 0. Since (α j γ) is an isometry such that leaves T invariant, it follows that (α j γ)({p 1, p 4 }) = {p 1, p 4 }. Case 1. (α j γ)(p 1 ) = p 1 and (α j γ)(p 4 ) = p 4. Using the fact (α j γ) is an isometry such that leaves T invariant, it follows that (α j γ)(p k ) = p k for all k = 0, Thus (α j γ) = id. Therefore, γ = α j = α j β 0. Case 2. (α j γ)(p 1 ) = p 4 and (α j γ)(p 4 ) = p 1. Then (βα j γ)(p k ) = p k for k = 0, 1, 4. By the arguments in Case 1, we have (βα j γ)(p k ) = p k = id. Hence, 9.8 γ = α j β 1. Proof. (a) Reflexivity. a N, a = a 10 0, hence a a. Symmetry. Assume a, b N such that a b. Then a = b 10 k for some k Z. We have b = a 10 k, hence b a. Transitivity. Assume a, b, c N such that a b and b c. Then a = b 10 k and b = c 10 l for some k, l Z. It follows that a = c 10 k+l, hence a c. (b) {a N : 10 a} is a complete set of equivalence class representatives. 9.17

4 4 Proof. (a) Routine. (b) [(x, y)] consists of all points in R 2 obtained by shifting (x, y) and integral units horizontally and an integral units vertically. (c) [0, 1) [0, 1) is a complete set of equivalence class representatives Proof. Since a b (mod n) and c d (mod n), we have a = b + un and c = d + vn for some u, v Z. Thus, ac = (b + un) + (d + vn) = bd + bvn + und + uvn 2 = bd + (bv + ud + uvn)n where bv + ud + uvn Z. Hence ac bd (mod n) Solution. Not true. Example: 4 1 (mod 3), but (mod 3 2 ) Solution. (f) [17] [76] = [2] [1] = [2] Disproof. [2] [2] / Z Solution. (a) (1, 2, 3, 4) = {id, (1, 2, 3, 4), (1, 3)(2, 4), (4, 3, 2, 1)}. (b) (1, 2, 3, 4, 5) = {id, (1, 2, 3, 4, 5), (1, 3, 5, 2, 4), (4, 2, 5, 3, 1), (5, 4, 3, 2, 1)}. (c) (1, 2,, n) = n Proof. (a) (ab) mn = a m b n = e. (b) It follows from Theorem 14.3 (b). (c) [2] Z 4 with o([2]) = 2. But o([2][2]) = o([0]) = e a a 2 a 3 a 4 e e a a 2 a 3 a 4 a a a 2 a 3 a 4 e a 2 a 2 a 3 a 4 e a a 3 a 3 a 4 e a a 2 a 4 a 4 e a a 2 a Proof. Since (a 1 ba) n = a 1 b n a, we see that (a 1 ba) n = e a 1 b n a = e b n = e. Thus o(a 1 ba) = o(b) Proof. Since 24, 36, , 36, Since 6 = , 36, , 36, Proof. Z 2 Z 2 is not cyclic since it does not contain any element of order 4.

5 Proof. ( ) Since [1] Z n = [a] [1] = u[a] for some a Z. Thus ua 1 (mod n), i.e., ua + vn = 1 for some v Z. It follows that gcd(a, n) = 1. ( ) We have 1 = gcd(a, n) = ua + vn for some u, v Z. Thus [1] = u[a] [a]. Hence Z p = [1] [a] Solution. The distinct right coset of Z in R are Z + x, x [0, 1). Z + x, x [0, 1) is obtained by shifting Z x units to the right Solution ((1, 2), [1]) (id, [0]) ={(id, [0]), (1, 2), [1])}, ((1, 2), [1]) (id, [1]) ={(id, [1]), (1, 2), [0])}, ((1, 2), [1]) ((1, 3), [0]) ={((1, 3), [0]), ((1, 3, 2), [1])}, ((1, 2), [1]) ((1, 3), [1]) ={((1, 3), [1]), ((1, 3, 2), [0])}, ((1, 2), [1]) ((2, 3), [0]) ={((2, 3), [0]), ((1, 2, 3), [1])}, ((1, 2), [1]) ((2, 3), [1]) ={((2, 3), [1]), ((1, 2, 3), [0])}, Proof. Assume that HaK HbK. Then h 1, h 2 H and k 1, k 2 K such that h 1 ak 1 = h 2 bk 2. So a = h 1 1 h 2bk 2 k1 1. Now for any h H and k K, we have hak = (hh 1 1 h 2)b(k 2 k1 1 ) HbK since hh 1 1 h 2 H and k 2 k1 1 K. So HaK HbK. By symmetry, HbK HaK Solution. Trivial subgroups: {([0], [0])}, Z p Z p. Subgroups of order p: ([1], y), y Z p and ([0], [1]) Proof. Define f : Z mn Z m Z n by f([x] mn ) = ([x] m, [x] n ), x Z. Note that if [x] mn = [y] mn, then x y (mod mn). Hence x y (mod m) and x y (mod n), i.e., [x] m = [y] m and [x] n = [y] n. Thus f is well defined. For any [x] mn, [y] mn Z mn, we have Hence f is a homomorphism. f([x] mn [y] mn ) = f([xy] mn ) = ([xy] m, [xy] n ) = ([x] m [y] m, [x] n [y] n ) = ([x] m, [x] n ) ([y] m, [y] n ) = f([x] mn )f([y] mn ).

6 6 Suppose f([x] mn ) = f([y] mn ). Then [x] m = [y] m and [x] n = [y] n, i.e., x y (mod m) and x y (mod n). Since gcd(m, n) = 1, we have x y (mod mn), i.e., [x] mn = [y] mn. Therefore, f is 1-1. Since Z mn and Z m Z n are finite of the same cardinality, f is a bijection Reason. S 3 Z 2 has 2 elements of order 3; A 4 has 8 elements of order Proof. Let G = a be an infinite cyclic group. Then o(a) =. Define f : Z G n a n Since f(m + n) = a m+n = a m a n = f(m)f(n), f is a homomorphism. Clearly f is onto. Assume f(m) = f(n) for some m, n Z. Then a m = a n, i.e., a m n = e. Since o(a) =, it follows that m n = 0, i.e., m = n. Thus f is Proof. Clearly, θ is a bijection. a, b G, we have So θ is a homomorphism. θ(ab) = (ab) 1 = b 1 a 1 = a 1 b 1 = θ(a)θ(b) Solution. an element α (1, 2, 3) id (1, 2, 3) (1, 3, 2) 21.5 the permutation θ(α) ( id ) id (1, 2, 3) (1, 3, 2) (1, 2, 3) (1, 3, 2) id ( id (1, 2, 3) ) (1, 3, 2) (1, 3, 2) id (1, 2, 3) Proof. (a) Assume that [a] 6 = [b] 6. Then a b (mod 6). Hence a b (mod 3), i.e., [a] 3 = [b] 3. (b) [a] 6, [b] 6 Z 6, we have θ([a] 6 + [b] 6 ) = θ([a + b] 6 ) = [a + b] 3 = [a] 3 + [b] 3 = θ([a] 6 ) + θ([b] 6 ). (c) ker θ = [3] Proof. x, y G, we have (β α)(xy) = β(α(xy)) = β(α(x)α(y)) = β(α(x))β(α(y)) = ( (β α)(x) )( (β α)(y) ). Hence β α is a homomorphism

7 7 Proof. (a) E is the identity. I 2 = J 2 = K 2 = E. Additional rules of multiplication in Q 8 are illustrated by the diagram below. The product of any two elements in the counterclockwise direction equals the third element; the product of any two elements in the clockwise direction equals the third element with a negative sign. E.g, IJ = K, JK = I, IK = J, etc. I K Thus Q 8 is closed under multiplication and (±E) 1 = ±E, (±I) 1 = I, (±J) 1 = J, (±K) 1 = K. Hence Q 8 is a subgroup of M(2, C). (b) Subgroup of order 1: {E}. Subgroup of order 2: {±E}. Subgroups of order 4: I = {±E ± I}, 22.6 Subgroup of order 8: Q 8. J = {±E ± J}, K = {±E ± K}. Proof. Let G = a and N G. Every element in G/N is of the form Nx for some x G. Since x = a n for some n Z, we have Nx = N(a n ) = (Na) n. Hence G/N = Na Proof. (a) Since G/N = p is a prime, G/N Z p is cyclic. (b) Disproof. Z/6Z Z 6 is cyclic. Yet [Z : 6Z] = 6 is not a prime Proof. Every element of Q/Z is of the form Z + r for some r Q. Write r = m n, where m, n Z and n 0. Then Hence o(z + r) n. n Z with n > 0, o(z + 1 n ) = n n(z + r) = Z + nr = Z + 0. Solution. The homomorphic images of S 3 are S 3, Z 2, and {e} S 3 S 3 /{id}, Z 2 S 3 /A 3, {e} S 3 /S 3. J

8 8 Proof. 1 A < G. x, y A, θ(a) B and θ(a) B. Hence θ(xy) = θ(x)θ(y) B, i.e., xy A. Thus A is closed under multiplication. We also have θ(x 1 ) = (θ(x)) 1 B. Hence x 1 A. Thus A is also closed under inversion. It follows that A < G. 2 A G. a A and xg G, we have θ(g 1 ag) = (θ(g)) 1 θ(a)θ(g) B since θ(a) B and B H. Therefore, g 1 ag A, which proves that A G Proof. (a) h 1 k 1, h 2 k 2 HK, where h 1, h 2 H and k 1, k 2 K, we have (h 1 k 1 )(h 2 k 2 ) = (h 1 h 2 )(h 1 2 k 1h 2 )k 2 HK since h 1 h 2 H, (h 1 2 k 1h 2 ) K and k 2 K. Thus HK is closed under multiplication. We also have (h 1 k 1 ) 1 = k1 1 h 1 1 = h 1 1 (h 1k1 1 h 1 1 ) HK. Hence HK is also closed under inversion. It follows that HK < G. Clearly, K HK, hence K < HK. g HK, since K G, we have g 1 Kg = K. Hence K HK. (b) h 1, h 2 H, we have θ(h 1 h 2 ) = Kh 1 h 2 = (Kh 1 )(Kh 2 ) = θ(h 1 )θ(h 2 ). So θ is a homomorphism. h H and k K, we have Khk = hkk = hk = Kh = θ(h). So θ is onto. (c) a H K, we have θ(a) = Ka = K. So a ker θ. Thus H K ker θ. a ker θ, we have K = θ(a) = Ka. So a K. Since the domain of θ is H, we also have a H. So a H K. Thus ker θ H K. By the fundamental homomorphism theorem, 53.6 Proof. (a) (b) HK/H K = HK/ ker θ = HK/K. Orb(1) = Orb(2) = Orb(3) = Orb(4) = {1, 2, 3, 4}. Orb(5) = Orb(6) = {5, 6}. G 1 = G 2 = G 3 = G 4 = {id}. G 5 = G 6 = {id, (1, 3)(2, 4)}. (c) Orb(1) = 4, G / G 1 = 4/1 = 4. Orb(5) = 2, G / G 5 = 4/2 = Proof. a, b G s, we have π a (s) = s and π b (s) = s. Hence π ab (s) = (π a π b )(s) = π a (π b (s)) = π a (s) = s. Thus ab G s. So G s is closed under multiplication. We also have π a 1(s) = πa 1 (s) = s since π a (s) = s. Thus a 1 G s. Therefore G s is closed under inversion. It follows that G s < G.

9 Proof. (a) a, b G, we want to show that π ab = π a π b. gh S, where g G, we have ( (π a π b )(gh) = π a πb (gh) ) ( ) = π a (bg)h = (abg)h = πab (gh). Hence π ab = π a π b. (b) Let a ker π. Then π a = id. Thus for any g G, π a (gh) = gh, i.e., (ag)h = gh. It follows that g 1 (ag) H, i.e., a ghg 1. Since this is true for all g G, we have a {ghg 1 : g G}. So, ker π {ghg 1 : g G}. Let a {ghg 1 : g G}. Then g G, a ghg 1. Thus g 1 (ag) H, hence (ag)h = gh, i.e., π a (gh) = gh. Therefore π a = id. Hence a ker π. So, {ghg 1 : g G} ker π.