PROBLEM SET # 2 SOLUTIONS


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1 PROBLEM SET # 2 SOLUTIONS CHAPTER 2: GROUPS AND ARITHMETIC 2. Groups.. Let G be a group and e and e two identity elements. Show that e = e. (Hint: Consider e e and calculate it two ways.) Solution. Since e is an identity element for G, we have e g = g for every g G, and so in particular e e = e. On the other hand, since e is an identity element for G, we also have g e = g for every g G, and in particular e e = e. Thus, e = e e = e. 2. Let G be a group with identity e and such that a 2 = e for every element a in G. Show that G is commutative, i.e., a b = b a for every two elements a and b in G. (Hint: Consider (a b) 2.) Solution. First notice that, since a 2 = e for every a G, we must have a = a for every a G. It then follows that a b = (a b) = b a = b a for every pair of elements a, b G. 9. Fix a positive integer n, and let nz = {..., 2n, n, 0, n, 2n,...} be the set of all integer multiples of n. Show that nz is a subgroup of Z with respect to addition as binary operation. Solution. We need to check that nz is closed under addition and inverse. So, suppose a, b nz, so that a = jn and b = kn for some integers j, k Z. Then a + b = jn + kn = (j + k)n nz and a = (jn) = ( j)n nz, and hence nz is indeed closed under addition and inverse. 2.2 Congruences.. Show that a positive integer m is divisible by if and only if the alternating sum of its digits is divisible by. (Hint: Notice that 0 mod.) Solution. If b 0,..., b k are the decimal digits of m (from the ones digit up to the 0 k  digit), then m = b 0 + b b k 0 k. It follows that m b 0 + b (0) + + b k (0) k mod = b 0 + b ( ) + + b k ( ) k mod,
2 and so m is congruent (modulo ) to the alternating sum of its digits. Thus, m is divisible by if and only if the alternating sum of its digits is. 2.3 Modular Arithmetic. 3. Write down the multiplication table for Z/Z, the set of integers modulo. The subset of invertible integers modulo is denoted by (Z/Z). Extract the multiplication table for (Z/Z). Solution. By a straightforward calculation, we find From this table, we see that all of the nonzero elements of Z/Z are invertible, and the multiplication table for (Z/Z) is Write down the multiplication table for Z/0Z, the set of integers modulo 0. The subset of invertible integers modulo 0 is denoted by (Z/0Z). Extract the multiplication table for (Z/0Z). How many invertible elements do we have here? Solution. By a straightforward calculation, we find 2
3 From this table, we see that the invertible elements of Z/0Z are, 3, 7, and 9, and the the multiplication table for (Z/0Z) is Write down the multiplication table for Z/2Z, the set of integers modulo 2. The subset of invertible integers modulo 2 is denoted by (Z/2Z). Extract the multiplication table for (Z/2Z). How many invertible elements do we have here? Solution. By a straightforward calculation, we find From this table, we see that the invertible elements of Z/2Z are, 5, 7, and, and that the multiplication table for (Z/2Z) is 3
4 Use the Euclidean algorithm to compute the multiplicative inverse of 3 modulo 979. Solution. We first run the Euclidean algorithm: 979 = = = = = 4. The secondtolast convergent in the continued fraction algorithm for 979/3 is therefore = Observe, then, that = 5542 and = 5543, and so 979 ( 28) + 3 (423) =. Thus, the inverse of 3 modulo 979 is Use the Euclidean algorithm to compute the multiplicative inverse of 27 modulo 09. Solution. We first run the Euclidean algorithm: 09 = = = = = = = 5. The secondtolast convergent in the continued fraction algorithm for 09/27 is therefore 8 + = Observe, then, that = and = 24003, and so 09 ( 22) + 27 (89) =. Thus, the inverse of 27 modulo 09 is 89. 4
5 2.4 Theorem of Lagrange.. Let G be a group and g an element in G of order n. Let m be a positive integer such that g m = e. Show that n divides m. (Hint: Write m = qn + r with 0 r < n.) Solution. Following the hint, we write m = qn + r with 0 r < n. Then observe that e = g m = g qn+r = (g n ) q g r = e q g r = g r. Since r < n and n is the order of g, we must therefore have r = 0. Thus m = qn, and hence n divides m. 2. Repeat the argument of Lagrange s theorem with G = (Z/3Z) and g = 5. Solution. Following the proof, we first note that the order of g = 5 is 4. We then compute e, g, g 2, g 3. We obtain e g g 2 g We see that x = 2 is missing from the list. We then compute x, xg, xg 2, xg 3, finding x xg xg 2 xg We see that y = 4 has yet to appear in either of the above lists, so we next compute y, yg, yg 2, yg 4, finding y yg yg 2 yg Looking over the three lists of numbers, we see that we have now accounted for every element of G exactly once, and that 2 = G = 3 4 = 3 ord(g), i.e., org(g) divides G. 2.5 Chinese Remainder Theorem. d Put φ() =. Compute d 000 φ(d). Solution. Since 000 = , the set of divisors of 000 is {, 2, 2 3, 2 3 2, 2 3 3, 2 2, 2 2 3, , , 2 3, 2 3 3, , }. We therefore have (using our known properties of φ) φ(d) = φ() + φ(2) + + φ(2 3 3 ) + φ(2 2 ) + + φ( ) + φ(2 3 ) + + φ( ) = ( φ() + φ(2) + φ(2 2 ) + φ(2 3 ) ) ( φ() + φ(3) + φ(3 2 ) + φ(3 3 ) ) = ( + (2 ) + (2 2 2) + ( ) ) ( + (3 ) + (3 2 3) + ( ) ) = =
6 (Can you see how to prove the equality d n φ(d) = n in general?) 3. Solve the system of congruences x 5 mod x 7 mod 3. Solution. We follow the notation used in lecture. By a simple calculation, one can easily check that y = 6 is the inverse of M = 3 modulo m =, and that y 2 = 6 is the inverse of M 2 = modulo m 2 = 3. The solution to the system of equations is therefore x mod ( 3), which simplifies to x 37 mod Solve the system of congruences x mod 6 x 6 mod 27. Solution. We follow the notation used in lecture. By a simple calculation, one can easily check that y = 3 is the inverse of M = 27 modulo m = 6, and that y 2 = 22 is the inverse of M 2 = 6 modulo m 2 = 27. The solution to the system of equations is therefore x mod (6 27), which simplifies to x 43 mod Find the last two digits of Do not use a calculator. Solution. Let x = We wish to compute the remainder of x modulo 00 = We first observe that we obviously have x = mod 4. Next observe that 2 0 mod 25, and so x = = (2 0 ) ( ) 999 (52) mod 25 ( ) (2) mod 25 3 mod 25. We now wish to solve the system of equation x 0 mod 4 x 3 mod 25. By a simple calculation, we see that y = is the inverse of M = 25 modulo m = 4, and y 2 = 9 is the inverse of M 2 = 4 modulo m 2 = 25. Thus, the solution to this system of equations is x mod (4 25), which simplifies to x 88 mod 00. Thus, the last two digits of x = are 88. 6
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