MAT301H1F Groups and Symmetry: Problem Set 2 Solutions October 20, 2017 Questions From the Textbook: for odd-numbered questions, see the back of the book. Chapter 5: #8 Solution: (a) (135) = (15)(13) is even. (b) (1356) = (16)(15)(13) is odd. (c) (13567) = (17)(16)(15)(13) is even. (d) (12)(134)(152) = (12)(14)(13)(12)(15) is odd. (e) (1243)(3521) = (13)(14)(12)(31)(32)(35) is even. Chapter 5: #10 Solution: α = (123)(45678) is a product of disjoint cycles, one of length 3 and one of length 5, the order of α is the least common multiple of 3 and 5, namely 15. Chapter 5: #14 Solution: 5-cycles are even and 6-cycles are odd, so α is odd and β is even. The number of total 2-cycles for α 5 β 4 α 1 β 3 α 5 is of the form 5 odd + 4 even + odd + 3 even + 5 odd, which is the same as odd + even + odd + even + odd, which is odd. Chapter 5: #16 Solution: using the Lemma on p 103. Let α be the product of r 2-cycles, with r even. Let α 1 be the product of s 2-cycles. Then α α 1 = ε, so by the Lemma on p. 103, r + s is even, which means s is even, because r is even. (But if r is odd, then s must be odd as well.) Chapter 5: #26 Solution: from #16 we know that a permutation and its inverse are both even, or both odd. Let α, β S n. The possibilities for the number of 2-cycles in α 1 β 1 αβ are odd + odd + odd + odd, even + odd + even + odd, odd + even + odd + even, or even + even + even + even, which are all even. Chapter 5: #28 Solution: the answer is the number of distinct 5-cycles in S 7. There are 7!/2! = 2520 distinct ways of writing a 5-cycle (abcde) S 7, but for each such choice, (abcde), (bcdea), (cdeab), (deabc) and (eabcd) represent the same 5-cycle in S 7. So the number of distinct 5-cycles in S 7 is 2520/5 = 504. Chapter 5: #38 Solution: let H = {β S 5 β(1) = 1, β(3) = 3}. Claim H S 5. Obviously ε(1) = 1, ε(3) = 3, so ε H.
If β H, then 1 = β 1 (1) and 3 = β 1 (3), so β 1 H. And if α, β H, then α(β(1)) = α(1) = 1 and α(β(3)) = α(3) = 3, so αβ H. Thus H S 5. The order of H is 6, because H consists of all the permutations of the numbers 2, 4 and 5; that is H S 3. In general, for n 3, H S n 2 ; so H = (n 2)! Chapter 5: #50 Solution: 4 5 3 1 2 The cycle (14253) represents a rotation of 3 72 = 216 counter clockwise around the centre of the pentagon. That is, 1 4, 2 5, 3 1, 4 2, 5 3. Or you could describe it as a rotation of 144 clockwise around the centre. The permutation (25)(34) represents a reflection in the line joining vertex 1 with the midpoint of side connecting vertices 3 and 4. Chapter 5: #54 Solution: (1234) is a cyclic subgroup of order 4 in A 8. And {ε, (12)(34), (13)(24), (14)(23)} is a non-cyclic subgroup of order 4 in A 8. (Note that these are the same four permutations that form the non-cyclic subgroup of order 4 in A 4.)
Chapter 10: #14 Solution: let φ : Z 12 Z 10 by φ(x) = 3x. Then φ(2) = 6 and φ(10) = 30 0 mod 10, so that φ(2) + φ(10) = 6 + 0 = 6. But in Z 12, 2 + 10 = 12 0 mod 12, so φ(2 + 10) = φ(0) = 0. Thus and φ cannot be a homomorphism. Chapter 10: #24 φ(2 + 10) φ(2) + φ(10), Solution: it is given that φ : Z 50 Z 15 is a homomorphism with φ(7) = 6. Then φ(7) = 7φ(1) 6 mod 15. The only solution m Z 15 to 7m 6 mod 15 is m = 3; thus φ(1) = 3. (a) φ(x) = φ(x 1) = xφ(1) = x 3 = 3x. (b) im (φ) = {φ(x) Z 15 x Z 50 } = {3x Z 15 x Z 50 } = {0, 3, 6, 9, 12} Z 5 (c) ker(φ) = {x Z 50 φ(x) 0 mod 15} = {x Z 50 3x 0 mod 15} = {0, 5, 10, 15, 20, 25, 30, 35, 40, 45} Z 10 (d) φ 1 (3) = {x Z 50 φ(x) = 3} = {x Z 50 3x 3 mod 15} = {1, 6, 11, 16, 21, 26, 31, 36, 41, 46} Chapter 10: #52 Solution: if G = a is a cyclic group generated by a, then every x G, x e, can be written as x = a n, for some positive integer n. If φ is a homomorphism defined on G, then φ(x) = φ(a n ) = φ(a } a.{{.. a a} ) = φ(a) φ(a)... φ(a) φ(a) ) = (φ(a)) n, }{{} n times n times which shows that φ is completely determined by what it does to a generator of a cyclic group G. Chapter 10: #64 Solution: let φ : Z m Z n be defined by φ(x) = x. Note that in Z m, 1 } + 1 + {{ + 1 } = m 0 mod m. If φ is a homomorphism, then m times 0 = φ(0) = φ(1 } + 1 + {{ + 1 } ) = m φ(1) = m 1 = m; m times that is, m 0 mod n; so n divides m. Conversely, if n divides m, then m = kn, for some positive integer k. Then x + y q mod m kn divides x + y q n divides x + y q. Thus x + y q mod m x + y q mod n, and so φ(x + y) = φ(x) + φ(y), which means that φ is a homomorphism.
Chapter 6: #4 Solution: U(8) = {1, 3, 5, 7}, which is not cyclic, since each element has order 2: 1 2 1 (mod 8), 3 2 1 (mod 8), 5 2 1 (mod 8), 7 2 1 (mod 8). But U(10) = {1, 3, 7, 9} is cyclic, since 3 = 4 and U(10) = 3. So U(8) and U(10) cannot be isomorphic groups. Chapter 6: #12 Solution: let α : G G be defined by α(g) = g 1. Then, for all g, h G, α(gh) = α(g)α(h) (gh) 1 = g 1 h 1 (gh) 1 = (hg) 1 gh = hg. So α is a homomorphism if and only if G is Abelian. Chapter 6: #18 Solution: let φ : G H be an isomorphism; let α Aut (G). Define f : Aut (G) Aut (H) by f(α) = φ α φ 1. Then check the following: 1. f(α) is in Aut (H) : i.e. f(α) : H H and f(α) is an isomorphism. Let x, y H. f(α) is a homomorphism: (f(α))(xy) = (φ α φ 1 )(xy) = φ(α(φ 1 (xy))) = φ(α(φ 1 (x) φ 1 (y))) = φ(α(φ 1 (x)) α(φ 1 (y))) = φ(α(φ 1 (x)) φ(α(φ 1 (y))) = (f(α))(x)(f(α))(y). f(α) is one-to-one: f(α)(x) = e H φ(α(φ 1 (x))) = e H α(φ 1 (x)) = φ 1 (e H ) = e G φ 1 (x) = α 1 (e G ) = e G x = φ(e G ) = e H f(α) is onto: for any y H, (f(α))(φ(α 1 (φ 1 (y)))) = (φ α φ 1 )(φ(α 1 (φ 1 (y)))) = y 2. f : Aut (G) Aut (H) is a homomorphism: Let α, β Aut (G). Then f(α β) = φ α β φ 1 = φ α φ 1 φ β φ 1 = f(α) f(β) 3. f is one-to-one: let i G : G G and i H : H H be the identity automorphisms on G and H, respectively. Then f(α) = i H φ α φ 1 = i H φ α = φ i H = φ α = φ 1 φ = i G 4. f is onto: let β Aut (H). Then Chapter 6: #26 f(φ 1 β φ) = φ φ 1 β φ φ 1 = i G β i H = β. Solution: suppose φ : Z 20 Z 20 is an automorphism such that φ(5) = 5. Since φ(5) = 5 φ(1) = 5 we must have 5 φ(1) 5 mod 20 φ(1) = 1, 5, 9, 13 or 17.
Then since φ(x) = x φ(1), there are five possible homomorphisms: φ(x) = x, φ(x) = 5x, φ(x) = 9x, φ(x) = 13x, φ(x) = 17x. To be an automorpphism, ker(φ) = {0}. This rules out φ(x) = 5x, so the only four possibilities are φ(x) = x, φ(x) = 9x, φ(x) = 13x or φ(x) = 17x. Alternate Solution: since Aut (Z 20 ) U(20) = {1, 3, 7, 9, 11, 13, 17, 19}, there are only 8 possible automorphisms φ of Z 20, defined by if and only if φ(1) = 1, 3, 7, 9, 11, 13, 17 or 19 φ(x) = x, 3x, 7x, 9x, 11x, 13x, 17x or 19x. Only four of these satisfy φ(5) = 5. (The other four satisfy φ(5) 5 mod 20.) Chapter 6: #32 Solution: this is similar to #26. If φ : Z 50 Z 50 is an automorphism with φ(7) = 13, then So the only possibility is φ(x) = 9x. Chapter 6: #36 7 φ(1) 13 mod 50 φ(1) = 9. Solution: since U(20) = U(24) = 8, U(20) and U(24) are candidates for isomorphic groups. But, in fact, they are not isomorphic. You can see this by considering the orders of elements in each group. Every element x U(24) = {1, 5, 7, 11, 13, 17, 19, 23} satisfies x 2 1 mod 24, so every element x U(24), x 1, has order 2. But in U(20), there are four elements of order 4, namely x = 3, 7, 13, 17. For example, 3 4 1 mod 20, but 3 2 9 mod 20. Chapter 6: #40 Solution: since α = (123)(45678) has order lcm(3, 5) = 15, α Z 15. 5 6 4 3 2 1 The cycle α = (12345678) represents a rotation of order 8 of the octagon. Let β = (34)(25)(16)(78), which represents a reflection of the octagon in the vertical axis of symmetry of the octagon shown to the left. Check that 7 8 Thus α, β D 8. βαβ = (18765432) = α 1. As for U(16) : mimic the construction in the proof of Cayley s Theorem. That is, for every g U(16) = {1, 3, 5, 7, 9, 11, 13, 15}, define T g : U(16) U(16) by T g (x) = gx. Each T g is a permutation of the first eight odd numbers, so it can equally well be considered as a permutation of the ordinal numbers 1, 2, 3, 4, 5, 6, 7 and 8. Thus U(16) {T g g U(16)} S 8. Chapter 6: #54 Solution: if φ Aut (D 8 ) then φ(r 45 ) = R 45 = 8. So φ(r 45 ) must be a rotation of order 8 as well. That is, φ(r 45 ) = R 45, R 135, R 225 or R 315.
Chapter 7: #8 Solution: in S 3, let H = {ε, (12)}, K = {ε, (13)}. Then HK = {hk h H, k K} = {ε, (12), (13), (132)} which is not a subgroup of S 3 because (132) 2 = (123) / HK. Chapter 7: #12 Solution: given a and b are nonidentity elements with different orders, in a group G of order 155 = 5 31, we must have a = 5 and b = 31 (or b = 5 and a = 31.) Let H be a subgroup of G that contains both a and b. Then both 5 and 31 divide H, and H divides G = 155. This means 155 H 155, so H = 155 and H = G. Chapter 7: #18 Solution: let a U(n). The order of U(n) is φ(n), so by Corollary 4 to Lagrange s Theorem, Chapter 7: #22 a φ(n) = e a φ(n) 1 mod n. Solution: suppose H = 12 and K = 35. Let d = H K. Since H K is a subgroup of both H and K, we know that d divides both 12 and 35, which are relatively prime. Thus d = 1 and H K = {e}. In general, if gcd( H, K ) = 1, then H K = 1. Chapter 7: #38 Solution: proof by contradiction. Suppose [G : Z(G)] = p, a prime. Pick x / Z(G). We have Z(G) C(x) G, where C(x) is the centralizer of x in G. Using Corollary 1 to Lagrange s Theorem we have [G : Z(G) = G, [G : C(x)] = G Z(G) from which we can conclude There are two possibilities: 1. [G : C(x)] = 1 and [C(x) : Z(G)] = p. C(x), [C(x) : Z(G)] = C(x) Z(G), p = [G : Z(G)] = [G : C(x)][C(x) : Z(G)]. If [G : C(x)] = 1, then C(x) = G, and every element of G commutes with x. But then x Z(G), contradicting our choice of x. 2. [G : C(x)] = p and [C(x) : Z(G)] = 1. In this case, [C(x) : Z(G)] = 1, implying Z(G) = C(x). But then x Z(G), again contradicting our choice of x. Conclusion: [G : Z(G)] is not prime. Chapter 7: #40 Solution: if G = 63 = 3 2 7 and a e is an element of G, then a divides 63. The possibilities are a = 3, 7, 9, 21 or 63. Suppose there is no element a in G with a = 3. Then there can t be any elements b G with order 9, 21, or 63, either. That is:
b = 9 b 3 = 3, b = 21 b 7 = 3, b = 63 b 21 = 3. So every non-identity element a G must have order 7, and so G has 62 elements of order 7. By the Corollary to Theorem 4.4, the number of elements of order 7 in a group is a multiple of φ(7) = 6. Thus we require 6k = 62, which is impossible if k is an integer. Conclusion: every group of order 63 must have an element of order 3. Chapter 7: #44 Solution: D n = R, F R n = F 2 = I, F RF = F 1 where R is a rotation of order n and F is a reflection of order 2. Let G = R be the subgroup of D n consisting solely of rotations; G = n. Now suppose H D n and H is odd. By Corollary 2 of Lagrange s Theorem, H contains no reflections, since every reflection has order 2, and 2 does not divide an odd number. Thus H consists entirely of rotations and H G. Finally, every subgroup of a cyclic group is cyclic, so H is cyclic. Chapter 7: #50 Solution: A 5 contains a copy of A 4, which has order 12. That is, H = {ε, (123), (132), (124), (142), (134), (143), (234), (243), (12)(34), (13)(24), (14)(23)} = A 4 is a subgroup of A 5. (Since 5 choose 4 equals 5, A 5 contains at least five copies of A 4.)
Other Questions: 1. As in Example 6 of the lecture notes for Chapter 5, find how many elements of each possible order there are in S 6. Indicate which elements are even and which are odd. Solution: Cycle structure Order Number of Permutations in S 6 Parity (1) 1 1 even (ab) 2 (6 5)/2 = 15 odd (ab)(cd) 2 (6 5)/2 (4 3)/2 1/2 = 45 even (ab)(cd)(ef) 2 (6 5)/2 (4 3)/2 (2 1)/2 (1/3!) = 15 odd (abc) 3 (6 5 4)/3 = 40 even (abc)(def) 3 (6 5 4)/3 (3 2 1)/3 (1/2) = 40 even (abc)(de) 6 (6 5 4)/3 (3 2)/2 = 120 odd (abcd) 4 (6 5 4 3)/4 = 90 odd (abcd)(ef) 4 (6 5 4 3)/4 (2 1)/2 = 90 even (abcde) 5 (6 5 4 3 2)/5 = 144 even (abcdef) 6 (6 5 4 3 2 1)/6 = 120 odd 2. Let φ : G G be a homomorphism. Let F = {x G φ(x) = x} be the set of fixed points of φ. Show that F is a subgroup of G. (Compare with Chapter 6, #24.) Solution: every homomorphism takes the identity to the identity, so e F. F is closed under multiplication: x, y F φ(xy) = φ(x) φ(y), since φ is a homomorphism φ(xy) = x y, since x, y F xy F F is also closed under inverses: suppose x F. Thus F G. x 1 x = e φ(x 1 x) = φ(e) φ(x 1 ) φ(x) = e φ(x 1 ) x = e, since x F φ(x 1 ) = x 1 x 1 F
3. Consider the cube with vertices (±1, ±1, ±1), with respect to the usual x, y and z axes, and centre at the origin (0, 0, 0). Let S(C) be the symmetry group of this cube. Recall that S(C) consists of the following 48 matrices, with usual matrix multiplication:,,,, with usual matrix multiplication. Let A = (a ij ) be a matrix in S(C) and define Φ : S(C) S(C) ( ) by Φ ((a ij )) = a 2 ij. That is, the effect of Φ on a matrix A is to square each entry of A. (a) Prove that Φ is a homomorphism. (This is tricky!) Solution: let A = (a ij ), B = (b ij ). Then AB = (a i1 b 1j + a i2 b 2j + a i3 b 3j ) and Φ(AB) = Φ(A)Φ(B) ( (a i1 b 1j + a i2 b 2j + a i3 b 3j ) 2) = ( a 2 i1b 2 1j + a 2 i2b 2 2j + a 2 i3b 2 ) 3j (a i1 b 1j + a i2 b 2j + a i3 b 3j ) 2 = a 2 i1b 2 1j + a 2 i2b 2 2j + a 2 i3b 2 3j a i1 a i2 b 1j b 2j + a i1 a i3 b 1j b 3j + a i2 a i3 b 2j b 3j = 0 which last statement is true since for any matrix A or B in S(C) every row of A has two zeros, as does every column of B. (b) Find the kernel and image of Φ. Solution: ker(φ) = {A S(C) Φ(A) = I} =,,. The image of Φ consists of all the matrices in S(C) that have no negative entries, that is the 6 matrices 1 0 0 0 1 0 0 0 1 1 0 0 0 0 1 0 1 0 0 1 0, 0 0 1, 1 0 0, 0 0 1, 0 1 0, 1 0 0, 0 0 1 1 0 0 0 1 0 0 1 0 1 0 0 0 0 1 which as a group is isomorphic to both S 3 and D 3. (c) Prove that (1) Φ Φ = Φ and (2) Φ( A) = Φ(A). Solution: let A = (a ij ) be a matrix in S(C). Then 1. each non-zero entry of Φ(A) is +1, so applying Φ to Φ(A) changes nothing. 2. Φ( A) = Φ (( a ij )) = ( ( a ij ) 2) ( ) = a 2 ij = Φ(A). (d) Find the group of fixed points of Φ. Solution: Φ(A) = A a 2 ij = a ij a ij = 0 or 1. Thus the group of fixed points of Φ is im (Φ).
(e) Show that both p : S(C) {1, 1} and q : S(C) {1, 1} defined by p(a) = det (Φ(A)) and q(a) = det (A Φ(A)) are homomorphisms, where det : S(C) {1, 1} is the usual determinant function. Solution: use properties of determinants and part (a). Watch your brackets! similarly, p(ab) = det(φ(ab)) = det (Φ(A)Φ(B)) = det(φ(a)) det(φ(b)) = p(a) p(b); q(ab) = det (AB Φ(AB)) = det (AB Φ(A)Φ(B)) = det(a) det(φ(a)) det(b) det(φ(b)) = det(a Φ(A)) det(b Φ(B)) = q(a) q(b) (f) Prove that f : S(C) S(C) defined by f(a) = p(a) A is an automorphism of S(C). Solution: first show that f is a homomorphism. Note: p(a) = ±1, which is just a scalar in terms of matrix multiplication. f(ab) = p(ab) (AB) = p(a) p(b) A B = (p(a) A) (p(b) B) = f(a) f(b). To show f is an isomorphism we need to show it is one-to-one and onto. f is one-to-one: show ker(f) = {I}. f(a) = I p(a) A = I A = p(a) I, since p(a) 2 = 1 A = ±I, since p(a) = ±1 But f( I) = p( I) ( I) = det(φ( I))( I) = det(i) ( I) = I, so the only solution is A = I. f is onto: let B be a matrix in S(C). We need to find a matrix A in S(C) such that f(a) = B. Take { B, if p(b) = 1 A = B, if p(b) = 1 Verification: 1. if p(b) = 1, then f(b) = p(b) B = (1)B = B 2. if p(b) = 1, then f( B) = p( B) ( B) = det(φ( B))( B) = det(φ(b))( B) = p(b)( B) = ( 1)( B) = B.
OR first show that f has order 2: f(f(a)) = f(p(a) A) = f(det(φ(a)) A) = p (det(φ(a)) A) det(φ(a)) A = det (Φ (det(φ(a)) A))) det(φ(a)) A { det (Φ (1 A)) (1) A, if det(φ(a)) = 1, = det (Φ ( A)) ( 1) A, if det(φ(a)) = 1 { det (Φ (A)) A, if det(φ(a)) = 1, = det (Φ (A)) ( 1) A, if det(φ(a)) = 1 { 1 A, if det(φ(a)) = 1, = ( 1) ( 1)A, if det(φ(a)) = 1 = A f is one-to-one: f(a) = I f(f(a)) = f(i) A = I f is onto: for any A S(C), f(f(a)) = A. (g) Explain why the group of fixed points of f is ker(p). What are the matrices in ker(p)? Solution: f(a) = A p(a) A = A p(a) = 1. Thus the group of fixed points of f is {A S(C) p(a) = 1} = ker(p) = {A S(C) det(φ(a)) = 1} =,,. (h) Show that f(ker(q)) = ker(det) and conclude that ker(q) ker(det). Solution: A ker(q) q(a) = 1 det(a Φ(A)) = 1 det(a) det(φ(a)) = 1. Consequently, if A ker(q), then det(f(a)) = det (p(a) A) = (p(a)) 3 det(a), since A is a 3 3 matrix = (det(φ(a))) 3 det(a) = (det(φ(a))) 2 (det(φ(a)) det(a)) = (±1) 2 1 = 1, and so f(a) ker(det). This shows that f(ker(q)) ker(det). Then you could count and find that both ker(q) and ker(det) contain 24 matrices, implying that f(ker(q)) = ker(det). OR you could prove the other inclusion: f(ker(det)) ker(q). Suppose A ker(det), so that det(a) = 1. Then q(f(a)) = q (p(a) A) = q (det(φ(a)) A) = det [det(φ(a)) A Φ (det(φ(a)) A)] = det [det(φ(a)) A Φ(A)], since det(φ(a)) = ±1, = (det(φ(a)) 3 det [A Φ(A)], since A Φ(A) is a 3 3 matrix = (det(φ(a)) 3 det(φ(a)) det(a) = (det(φ(a)) 4 det(a) = (±1) 4 1 = 1
and so f(a) ker(q). This shows that f(ker(det)) ker(q). Since f has order 2 it follows that ker(det) f(ker(q)). Thus: f(ker(q)) = ker(det), and since f is an isomorphism, ker(q) ker(det). OR there are other ways. Since f is one-to-one, we know 1. f(ker(q)) ker(det) ker(q) = f(ker(q)) ker(det). 2. f(ker(det)) ker(q) ker(det) = f(ker(det)) ker(q). Combining these two inequalities gives ker(q) = ker(det), and hence f(ker(q)) = ker(det).