Math 3560 HW Set 6. Kara. October 17, 2013

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1 Math 3560 HW Set 6 Kara October 17, 013 (91) Let I be the identity matrix 1 Diagonal matrices with nonzero entries on diagonal form a group I is in the set and a b a 1 b a 0 0 b 0 0 a b a n 0 0 b n 0 0 a n b n and a a a n 1 1/a /a /a n (Note that a i 0) Therefore this set is closed under multiplication and taking inverses Associativity follows from associativity of matrix multiplication Zero matrix is symmetric but not invertible Thus symmetric matrices do not form a group 3 I is invertible, with integer entries, but its inverse has rational not integer entries Thus this set is not a group 4 Linear combination of products of rational numbers is rational therefore this set is closed under multiplication Also, as A 1 1 A adj(a) we get that set is closed under taking inverses It contains I and thus forms a group (95) Let cos θ sin θ A θ sin θ cos θ cos ϕ sin ϕ and B ϕ sin ϕ cos ϕ 1

2 To prove Reading the angles in the matrices mod π, A θ A ϕ A θ+ϕ, A θ B ϕ B θ+ϕ, B θ A ϕ B θ ϕ, B θ B ϕ A θ ϕ Proof We can do this via direct computation and trigonometric identities cos θ sin θ cos ϕ sin ϕ A θ A ϕ sin θ cos θ sin ϕ cos ϕ cos θ cos ϕ sin θ sin ϕ (cos θ sin ϕ + sin θ cos ϕ) sin θ cos ϕ + cos θ sin ϕ cos θ cos ϕ sin θ sin ϕ cos(θ + ϕ) sin(θ + ϕ) A sin(θ + ϕ) cos(θ + ϕ) θ+ϕ cos θ sin θ cos ϕ sin ϕ A θ B ϕ sin θ cos θ sin ϕ cos ϕ cos θ cos ϕ sin θ sin ϕ cos θ sin ϕ + sin θ cos ϕ sin θ cos ϕ + cos θ sin ϕ (cos θ cos ϕ sin θ sin ϕ) cos(θ + ϕ) sin(θ + ϕ) B sin(θ + ϕ) cos(θ + ϕ) θ+ϕ cos θ sin θ cos ϕ sin ϕ B θ A ϕ sin θ cos θ sin ϕ cos ϕ cos θ cos ϕ + sin θ sin ϕ sin θ cos ϕ cos θ sin ϕ sin θ cos ϕ cos θ sin ϕ (cos θ cos ϕ + sin θ sin ϕ) cos(θ ϕ) sin(θ ϕ) B sin(θ ϕ) cos(θ ϕ) θ ϕ cos θ sin θ cos ϕ sin ϕ B θ B ϕ sin θ cos θ sin ϕ cos ϕ cos θ cos ϕ + sin θ sin ϕ (sin θ cos ϕ cos θ sin ϕ) sin θ cos ϕ cos θ sin ϕ cos θ cos ϕ + sin θ sin ϕ cos(θ ϕ) sin(θ ϕ) A sin(θ ϕ) cos(θ ϕ) θ ϕ Recall that A θ represents anticlockwise rotation through θ, while B ϕ represents reflection in a line at angle ϕ/ to the positive x-axis Then A θ A ϕ represents anticlockwise rotation first through ϕ, then through θ, which is the same as rotation through ϕ+θ A θ B ϕ B θ+ϕ implies that reflection in a line at angle ϕ/ followed by rotation through θ is the same as reflection in a line at angle (θ + ϕ)/ B θ A ϕ B θ ϕ implies rotation by ϕ followed by

3 reflection at angle θ/ is the same as reflection at angle (θ ϕ)/ And B θ B ϕ A θ ϕ implies reflection at ϕ/ followed by reflection at θ/ is the same as rotation by θ ϕ (97) Consider the matrix 1/ 0 A 0 1 1/ 0 We want to complete the entries of A so that it is in SO 3 For this, A should satisfy A t A I and det(a) 1 If we do the computations, we see that 1/ 0 1/ B / 0 1/ is in S0 3 We know that any matrix in SO 3 represents a rotation of R 3 about an axis which passes through the origin It is easy to check that B fixes the vector (0, 1, 0) in R 3, ie it represents a rotation about the y-axis In order to find the rotation angle, we can look at the images of the standard basis vectors of R 3 under B We can see that B is an anticlockwise rotation about the y-axis through angle π/4 Now, we want to complete the entries of A so that it is in O 3 For this, A should satisfy A t A I and det(a) 1 If we do the computations, we see that 1/ 0 1/ C / 0 1/ is in O 3 By using the explanation and the notation of the book(page 47), the transformation f C corresponding to C is f C f CU f U f B f U since C BU Hence, C represents a reflection in the (x, y) plane followed by the rotation B (91) Consider the matrices I 0 1 0, J 0 1 0, K 0 1 0, L We see that their determinants are 1 and they are orthogonal matrices, hence they are in SO 3 From their multiplcation table, it is clear that they form a subgroup I J K L I I J K L J J I L K K K L I J L L K J I 3

4 Consider the set {(x, y, x) R 3 x + (y 3) 5, x + (y + 3) 5, 1 z 1} The first equation gives a disk centered at (0, 3) with radius 5, and the second equation gives another disk centered at (0, 3) with radius 5 When we take z between 1 and -1 we get two cylinders and their intersection is a regular two-sided shape This shape has three axes of rotational symmetries The rotational symmetries are the following: rotation about the z-axis through angle π, rotation about the y-axis through angle π and rotation about the x-axis through angle π The order of each of these rotations is since if you apply them twice you get the identity We know that every rotation of R 3 which fixes the origin is represented by a matrix in SO 3 Observe that the rotation about z-axis is represented by L(it fixes the z-axis and rotates a vector in R 3 through π about the z-axis) Similarly, the rotation about the y-axis is representes by K and the rotation about the x-axis is represented by J We showed above that {I, J, K, L} is a subgroup of SO 3 and it is the rotational symmetries of the given shape (106) To prove If A G and B H, then A B G H Proof Let (a, b), (c, d) A B Then (a, b)(c, d) (ac, bd) A B, since ac A and bd B Further, (a 1, b 1 ) A B, because a 1 A and b 1 B Since (a, b)(a 1, b 1 ) (aa 1, bb 1 ) (e, e) (a 1 a, b 1 b) (a 1, b 1 )(a, b), we have A B is closed under multiplication and inverses, which suffices to show that A B is a subgroup To prove There do not exist A, B Z such that A B {(n, n) n Z} Z Z Proof Notice H {(n, n) n Z} is cyclic, with generator (1, 1) Suppose there exist A, B Z such that A B H Since (1, 1) A B, 1 A and 1 B 1 generates Z, so A B Z, but H Z Z, achieving the desired contradiction (107) There are many methods to solve this problem We can see that Z 4 and Z 1 Z 1 are the only abelian groups, but first one is cyclic while second is not Thus they cannot be isomorphic to any other group in the given collection Now probably the easiest way to proceed is to compare the number of elements of order in all groups Notice that, D 4 Z 3 has 5 elements of order, D 1 has 13, A 4 Z has 7, Z D 6 has 15, and S 4 has 9 Therefore none is isomorphic (1011) We want to show that D n is isomorphic to D n Z when n is odd By definition, D n < a, b a n e, b e, aba b > and D n < r, s r n e, s e, rsr s > 4

5 Since n is odd, then n + 1 m(note that this means, m 1 mod n, ie m, 1 m n 1, is the inverse of mod n) Let s define a map from D n to D n Z as follows: φ :D n D n Z r (a m, 1), s (b, 1) We see that φ(r) n (a m, 1) n ((a m ) n, n) ((a m ) n, 0) since n 0 mod ((a 1 ) n, 0) since m 1 mod n (a n, 0) (e, 0) Since m is the inverse of mod n, we have (m, n) 1 This means a m has order n The order of (a m, 1) is the lcm of the orders of a m and 1(its order is ) Hence, order of φ(r) (a m, 1) is n Also, (b, 1) (b, 0) (e, 0), ie the order of φ(s) (b, 1) is Let s also check whether the following relation holds: φ(r)φ(s)φ(r) φ(s) φ(r)φ(s)φ(r) (a m, 1)(b, 1)(a m, 1) (a m b, 0)(a m, 1) (a m ba m, 1) (a m a n m b, 1) (b, 1) φ(s) Now, extend φ to D n and verify that φ is actually an isomorphism(ie a bijection which preserves the group structures) 5