A. Let G be the set of the fifth roots of unity. 1. Use de Moivre's formula to verify that the fifth roots of unity form a group under complex multiplication. Under de Moivre's formula, has five roots: ) + cos( cos( cos( G is a subset of complex numbers. G= {1, , } Let * be a binary operation on a nonempty set G. G is said to be a group if the following properties hold. 1. The operation * is associative, that is, for all a, b, and c in G, (a*b)*c=a*(b*c). 2. There is an element e G such that e*a=a*e=a for all aG. 3. For each aG, there is an element bG such that a*b=b*a=e. Based on the table below, the following information can be identified: Closure holds true {G,*}. o To show the ring is closed under multiplication, a*b is in G, it needs to be shown that: A*b is in G; and 1 is the multiplicative identity for G. o The table shows when 1 multiplies any number, the number is left unchanged. All elements have an inverse. o To find the inverse, start by picking a number in the left column. Look through that row until the identity (1) is found, and then go to the top of that row and column. This will show the two complex numbers multiplied by each other to get the identity element. They are also inverses of each other. Some examples are: 1 * 1 * * Cayley Multiplication Table for G: 1 1 1 1 1 1 1 In the following work, let a = cos(a); b = sin(a); c = cos(b); d = sin(b); e = cos(c); f = sin(c) be used to prove associativity. [(a + ib)(c + id)](e + if) = [ac bd + (bc + ad)i](e + if) = e(ac bd) f(bc + ad) +i[e(bc + ad) + f(ac bd)] = ace bde fbc adf + i[bce + ade + acf bdf] = a(ce df) b(de + fc) + i[a(de + cf) + b(ce df)] = (a + ib)[ce df + (de + fc)i] = (a + ib)[(c + id)(e + if)] OR [(cos(a) + isin(a)(cos(b) + isin(b))](cos(c) + isin(c)) = [cos(a)(cos(b)sin(a)(sin(b)+(sin(a)cos(b)+cos(a)sin(b))i](cos(c)+isin(c)) = cos(c)((cos(a)(cos(b)(sin(a)(sin(b)) - sin(c)(sin(a)cos(b)+cos(a)sin(b)) + i[cos(c)(sin(a)cos(b) + cos(a)sin(b)) + sin(c)(cos(a)(cos(b)(sin(a)(sin(b))] = cos(a)cos(b)cos(c) - sin(a)sin(b)sin(c) - sin(c)sin(a)cos(b) - cos(a)sin(b)sin(c) + i[sin(a)cos(b)cos(c) + cos(a)sin(b)sin(c) + cos(a)cos(b)sin(c) - sin(a)sin(b)sin(c)] = cos(a)(cos(b)cos(c) - sin(b)sin(c)) - sin(a)(sin(b)cos(c) + sin(c)cos(b)) + i[cos(a)(sin(b)cos(c) + cos(b)sin(c)) + sin(a)(cos(b)cos(c) - sin(b)sin(c))] =(cos(a) +isin(b))[(cos(b) + isin(b))(cos(c) + isin(c))] This shows {G,*} is associative. A2. Prove that G is isomorphic to Z5 under addition. Groups {G1, 1} and {G2, 2} are said to be isomorphic if there exists a function f : G1 G2 such that: i. f is a bijection, o injective (onetoone) o surjective (onto) ii. f is operation preserving. The function f is called an isomorphism. = Using the function if f(a) = f(b) then: = = ln(a) = ln(b) = substitute function, multiply both sides, divide, and use logarithm rule = a = b (onetoone proven) f(a*b) = logarithm rules = = =f(a) + f(b) (order preserving proven) If (b, +) b= then logarithm rules, combine powers = = = = x and x 5th roots of unity. (onto proven) B. Let F be a field. Let S and T be subfields of F. Prove that ST is a field using the definition of a field and subfield. G is a field under addition and multiplication. The field {G,+} is a commutative group by definition. Definition of a subfield: Under addition and multiplication, all subfields are fields. Definition of field: Under addition, S and T are commutative groups. Definition of field: Under multiplication, S and T are commutative groups. 0 and 1 are elements of S and T by the conditions of subfield. 0 and 1 are members of ST so ST is not an empty set. 0 is the additive identity. Definition of Closure Property of Addition: If x and y are part of S and T, and x + y is in S and x + y is in T, then, x + y is in S and T. This proves addition is closed over ST. If x + y = y + x, it is in S and x + y = y+ x then it is in T by the commutative property of addition. If x + y = y + x, then ST. If x + (y + z) = (x + y) + z is in S and (x + y) + z and x + (y + z) is in T by the associative property of addition. If x + (y + z) = (x + y) + z, then ST. If x is in S and x is in T, where x + (x) = (x) + x = 0 in both S and T. x is in ST where x + (x) = (x) + x = 0, then all elements in ST have additive inverses in ST. ST is a commutative group under addition. 1 is an element of S and T by the conditions of subfield. The multiplicative identity 1 is in ST. ST does not equal an empty set. x * y is in S and x * y is in T by the closure of x over S and T. Therefore, x * y is in ST. If x1 is in S and x1 is in T, where x*x1 = x1*x = 1 in both S and T x1 is in S, where x*x1 = x1*x = 1. All elements in S have multiplicative inverses in S. Multiplication is closed over ST. By the commutative property of multiplication x * y = y * x is in S and x * y = y * x is in T. x*y = y * x is in ST by the commutative property of multiplication. x*(y * z) = (x * y)z is in S and x*(y * z) = (x * y)*z is in T by the associative property of multiplication. x*(y * z) = (x * y)*z is in ST by the associative property of multiplication. ST is a commutative group under multiplication. The result of x(y + z) = xy + xz in S must be the same as x(y + z) = xy +xz in T. Therefore, x(y + z) = xy + xz is in ST. (x + y)z = xz + yz in S must be the same as (x + y)z = xz + yz in T. Therefore, (x + y)z = xz + yz is in ST. Multiplication distributes in ST. The above conditions determine ST is a field. References Gilbert, J., & Gilbert, L. (2000). Elements of modern algebra (6th ed.). Pacific Grove, CA: Brooks/Cole