Instructions. Answer each of the questions on your own paper, and be sure to show your work so that partial credit can be adequately assessed. Credit will not be given for answers (even correct ones) without supporting work. Put your name on each page of your paper. 1. [15 Points] Prove that the formula (1 1n ) 2 = n + 1 2n is valid for all integers n 2. Your proof should be written in grammatically correct complete sentences. Of course, sentences can contain mathematical symbols. Solution. For n P, let P (n) be the statement: (1 1n ) 2 = n + 1 2n. For n = 2, P (2) is the statement: (1 12 2 ) = 2 + 1 2 2 = 3 4, which is clearly a true statement. verified. Thus P (2) is true, and the base step for induction is Now assume that P (k) is true for some fixed k 2. This means that (1 1k ) 2 = k + 1 2k Consider the left hand side of the statement P (n) for n = k + 1. That is: ( ) (1 1k ) ( ) 1 2 1 (k + 1) 2 Because of the assumption that P (k) is true, ( ) can be written as (1 1k ) ( ) 1 2 1 (k + 1) 2 = k + 1 2k = k + 1 2k = 1 2k k2 + 2k k + 1 k + 2 = 2(k + 1). ( ) 1 1 (k + 1) 2 ( (k + 1) 2 ) 1 (k + 1) 2 This is the statement that P (k + 1) is true, provided P (k) is true, and by the principle of induction, we conclude that P (n) is a true statement for all n P with n 2. 2. [10 Points] (a) If a and b are positive integers, what does it mean to say that a and b are relatively prime? That is, give the definition of the term relatively prime. Math 4023 Section 1 May 13, 2004 1
Solution. The positive integers a and b are relatively prime if their greatest common divisor is 1. (b) Give an example of integers a, b, and c such that a divides bc, but a divides neither b nor c. Solution. a = 4, b = 2, c = 2 is one example. (c) Give an example of integers a, b, and c such that a divides c and b divides c, but ab does not divide c. Solution. a = b = c = 2 is one example. 3. [10 Points] How many elements are there in each of the following sets. Proofs are not required. (a) Z n Answer: n (b) G n = Z n Answer: ϕ(n) where ϕ denotes Euler s φ function. (c) S(n) Answer: n! (d) Z 7 S(4) Answer: 7 4! = 7 24 = 168 (e) The set of integers between 1 and 1323 = 49 27 that are relatively prime to 1323. 4. [15 Points] Solution. This is ϕ(1323) = ϕ(7 2 3 3 ) = ϕ(7 2 )(ϕ(3 3 ) = (7 2 7)(3 3 3 2 ) = 42 18 = 756. (a) What is the relationship between a and n which guarantees that [a] n has a multiplicative inverse in Z n? (Just state the condition. It is not necessary to verify it.) Solution. [a] n has a multiplicative inverse in Z n if and only if a and n are relatively prime. (b) Find the multiplicative inverse of [19] 2773 in Z 2773. 5. [15 Points] Solution. Use the Euclidean algorithm to compute that 146 19 2773 = 1. From this equation, we see that [19] 1 2773 = [146] 2773. (a) Precisely state Euler s theorem. Solution. See the text, Page 68. (b) Find the remainder when 35 215 is divided by 11. Solution. From Euler s theorem 35 10 1 mod 11. Since 215 5 mod 10 it follows that 35 215 35 5 mod 11 and since 35 2 mod 11 we further conclude that 35 5 2 5 32 10 mod 11. Hence 35 215 10 mod 11. 6. [15 Points] Math 4023 Section 1 May 13, 2004 2
(a) What does it mean to say that the order of a permutation π S(n) is k? Solution. The order of π is k provided k is the smallest positive integer such that π k = id. (b) Let π = (5, 2, 4, 9, 1)(3, 1, 9, 4)(1, 7, 2, 6, 3) S(9). Write π as a product of disjoint cycles, and use this to find the order of π and determine if it is even or odd. Solution. π = ( 1 7 4 3 ) ( 2 6 5 ). Hence the order of π is the least common multiple of 4 and 3, i.e., o(π) = 12. A 3-cycle is even and a 4-cycle is odd, so pi is an odd permutation. (c) Solve the equation xπ = (1, 3, 7) for x S(9). Express x as a product of disjoint cycles. Solution. x = ( 1 3 7 ) π 1 = ( 1 3 7 ) ( 1 3 4 7 ) ( 2 5 6 ) = ( 1 7 3 4 ) ( 2 5 6 ). 7. [15 Points] (a) Show that the group G = Z 13 is cyclic by showing that [2] 13 is a generator of G. Solution. Here is a table of the powers of 2 modulo 13: n 1 2 3 4 5 6 7 8 9 10 11 12 2 n mod 13 2 4 8 3 6 12 11 9 5 10 7 1 Since every element of G appears in the second row, it follows that G = [2] 13. That is, G is cyclic with generator [2] 13. (b) Now list all of the generators of Z 13. Hint: The calculations that you did for part (a) should be of use. Solution. If the order of g is n, then the order of g k is n/(n, k). Thus, if G = g, then g k is a generator of G if and only if the order of g k = n, which only occurs if (n, k) = 1. Since the order of [2] 13 in G is 12, it follows that [2 k ] 13 is a generator of G if and only if (k, 12) = 1. Hence we must have k = 1, 5, 7, or 11. Therefore, from the above table of powers of 2 mod 13, we see that 2 = 2 1, 2 5 = 6, 2 7 = 11, and 2 11 = 7 are the generators of G. 8. [15 Points] Find all solutions to the system of simultaneous congruences x 4 mod 24 x 7 mod 11. Math 4023 Section 1 May 13, 2004 3
Solution. By inspection or using the Euclidean algorithm, find the equation ( 5)24 + 11 11 = 1. Then the solutions to the simultaneous congruence are given by x 4 (11 11) + 7(( 5)24) = 356 mod 264. 9. [15 Points] For each of the following groups with four elements, determine whether it is isomorphic to Z 4 or Z 2 Z 2. Reasons for your answer are required. (a) The multiplicative group G 8 = Z 8 of invertible congruence classes modulo 8. Solution. G 8 = Z2 Z 2 since every element of G 8 satisfies the equation x 2 = 1. (b) The subgroup ρ of the group D(4) of symmetries of the square, where ρ is the rotation of the square by π/2 radians. Solution. This is a cyclic group of order 4 with generator ρ, and hence it is isomorphic to Z 4. (c) The subgroup G = {( ), (1, 2)(3, 4), (1, 3)(2, 4), (1, 4)(2, 3)} of S(4). Solution. This group is isomorphic to Z 2 Z 2 since the square of every element is the identity. (d) The groups with multiplication tables as shown below (where the identity element does not necessarily head the first row and column). G 1 G 2 a b c d a d c a b b c d b a c a b c d d b a d c a b c d a b a d c b a b c d c d c b a d c d a b Solution. G 1 has identity c since cx = xc = x for all x G 1. Then a 2 = d, a 3 = ad = b and a 4 = ab = c. Hence G 1 = a is a cyclic group of order 4 and hence isomorphic to Z 4. G 2 has identity b since bx = xb = x for all x G 2. Moreover, x 2 = b for all x G 2 so there is no element of order 4. Hence, G 2 = Z2 Z 2. 10. [25 Points] Let C be the binary linear code with generator matrix 1 0 0 0 1 1 1 G = 0 1 0 0 1 1 0 0 0 1 0 1 0 1. 0 0 0 1 0 1 1 Math 4023 Section 1 May 13, 2004 4
(a) List all of the codewords of C. Solution. The codewords are the elements in B 7 of the form wg where w B 4. Thus there are 2 4 = 16 codewords obtained by taking all of the choices for w B 4. Thus, the list of codewords is: Word w B 4 Codeword wg B 7 0000 0000000 0001 0001011 0010 0010101 0011 0011110 0100 0100110 0101 0101001 0110 0110011 0111 0111000 1000 1000111 1001 1001100 1010 1010010 1011 1011010 1100 1100001 1101 1101010 1110 1110100 1111 1111111 (b) Write a parity check matrix H for C. Solution. 1 1 1 1 1 0 1 0 1 H = 0 1 1. 1 0 0 0 1 0 0 0 1 (c) How many errors can C detect and how many errors can C correct? Solution. The minimum weight of this linear code is 3, so it can detect 2 errors and correct 1 error. (d) Compute a syndrome table for C. Math 4023 Section 1 May 13, 2004 5
Solution. The syndrome table for this code is: Syndrome Coset Leader 000 0000000 111 1000000 110 0100000 101 0010000 011 0001000 100 0000100 010 0000010 001 0000001 (e) Assuming that a message is encrypted using the following number to letter equivalents: 0000 A 0001 B 0010 C 0011 D 0100 E 0101 F 0110 G 0111 H 1000 M 1001 N 1010 P 1011 R 1100 S 1101 T 1110 U 1111 V decode the following received message: Answer: THEEND 1100010 0111001 0110110 1100110 1101100 0011100 Math 4023 Section 1 May 13, 2004 6