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Math 402 Assignment 1. Due Wednesday, October 3, 2012. 1 Math 402 Assignment 1. Due Wednesday, October 3, 2012. 1(problem 2.1) Make a multiplication table for the symmetric group S3. Solution. (i). Iσ = σ for all σ ∈ S3. (ii). (12)I = (12); (12)(12) = I; (12)(13) = (132); (12)(23) = (123); (12)(123) = (23); (12)(321) = (13). (iii). (13)I = (13); (13)(12) = (123); (13)(13) = I; (13)(23) = (321); (13)(123) = (12); (13)(321) = (23). (iv). (23)I = (23); (23)(12) = (321); (23)(13) = (123); (23)(23) = I; (23)(123) = (13); (23)(321) = (12). (v). (123)I = (123); (123)(12) = (13); (123)(13) = (23); (123)(23) = (12); (123)(123) = (321); (123)(321) = I. (vi). (321)I = (321); (321)(12) = (23); (321)(13) = (12); (321)(23) = (13); (321)(123) = I; (321)(321) = (123). 2(3.2) Let a and b be positive integers that sum to a prime number p. Show that the greatest common divisor of a and b is 1. Solution. Begin with the equation a + b = p, where 0 <a<p and 0 <b<p. Since the g.c.d. (a, b) divides both a and b, (a, b) divides their sum a + b = p, i.e., (a, b) divides the prime p; hence, (a, b)=1or(a, b) = p. If(a, b) = p, then p =(a, b) divides a, which is not possible since 0 <a<p. 3(3.3) (a). Define the greatest common divisor of a set {a1, a2, ..., an} of integers, prove that it exists, and that it is an integer combination of a1, ..., an. Assume that the set contains a nonzero element. Solution. The g.c.d. (a1, a2, ..., an) of the set of integers {a1, a2, ..., an} is the largest integer m that divides all the integers in the set {a1, a2, ..., an}. There is a largest integer of that description since each integer that divide each element of the set is no bigger than |a1|. 1 The set of all integer combinations {m1a1 + m2a2 + ··· + mnan|mi ∈ Z} of {a1, a2, ..., an} is a subgroup of the additive group of integers, and so, it equals Zt, where t is the smallest positive integer in the subgroup. We can show that t divides all the elements of {a1, a2, ..., an} as follows: Divide aj by t: aj = st + r, where 0 ≤ r < t. Then r = aj − st is in the subgroup since aj and t are. If r> 0, that would violate the minimality of t, and so, r = 0. Since t is a common divisor and m is the greatest common divisor, t divides m. On the other hand, since m divides all the elements of {a1, a2, ..., an}, m divides all the elements {m1a1 + m2a2 + ··· + mnan|mi ∈ Z} in the subgroup; in particular, m divides the element t of the subgroup. Since m and t divide each other, m = t.Therefore, since t is an integral combination of {a1, a2, ..., an}, the greatest common divisor m = t is an integer combination of {a1, a2, ..., an}. (b). Let d be the greatest common divisor of a set {a1, a2, ..., an} of integers. Show that the greatest common divisor of a set {a1/d,a2/d, ..., an/d} is 1. Solution. By part (a), the g.c.d. d equals s1a1 + s2a2 + ··· + snan for certain integers s1,...,sn. Dividing the equation d = s1a1 + s2a2 + ··· + snan by d, we get the equation 1= s1(a1/d)+ s2(a2/d)+ ··· sn(an/d), where each aj/d is an integer since d divides each aj. Thus, 1 is an integral combination of {a1/d,a2/d, ..., an/d}. 1 is then the smallest positive integral combination of {a1/d,a2/d, ..., an/d}, and so, by the argument i (a), 1 is the greatest common divisor of {a1/d,a2/d, ..., an/d}. 4(4.1) Let a and b be elements of a group G. If a3b = ba3 and if a has order 7, show that ab = ba. Solution. Since a3 and b commute, a6 =(a3)2 and b also commute. Since a7 = 1, a6 = a−1, and so, a−1 and b commute, i.e., a−1b = ba−1. Combining that equation with a on the left and with a on the right, we get the equation ba = ab. 5(4.2) An nth root of unity is a complex number z such that zn = 1. (a). Prove that the set of nth roots of unity form a cyclic subgroup of C× of order n. 2mπi Solution. The distinct nth roots of unity are G = {e n |0 ≤ m < n}. For each integer t, 2tπi 2mπi e n is an nth root of unity. To match it up with a particular element of {e n |0 ≤ m < n}, divide t by n to produce t = sn + m, where 0 ≤ m < n. Then 2tπi 2 2mπ 2mπ e n = e sπe n = e n . Define the composition of elements of G using complex multiplication: ′ ′ ′ 2mπi 2m πi 2mπi 2m πi 2(m+m )πi e n ◦ e n = e n · e n = e n . 2 0 2mπi 2(−m)πi The identity element of G is 1 = e , and the inverse of e n is e n . 2πi 2πi 2mπi Take the particular nth root of unity e n . The other nth roots are (e n )m = e n , with 0 ≤ m < n. Thus, the nth roots of unity form a cyclic group with n elements that is 2πi generated by e n . (b). Determine the product of all the nth roots of unity. Solution. The product of the nth roots of unity equals the product of the on-real roots times the product of the real roots. The non-real roots come in complex-conjugate pairs, where each pair has product 1. Hence, the product the non-real roots equals 1. Thus, the product of the nth roots of unity equals the product of the real roots. If n is odd, the only real root is 1, and if n is even, the real roots are 1 and -1. Hence, when n is odd, the product equals 1, and when n is even, the product equals -1. 6(4.3) Let a and b be elements of a group G. Prove that ab and ba have the same order. Solution. Let e be the identity element of G. ba is related to ab by the equation ba = a−1(ab)a. To show that ba and ab have the same order, we need only show that ab and a−1(ab)a have the same order. For a simpler appearance, let ab = c. Then we want to show that c and a−1ca have the same order, i.e., cn = e if and only if (a−1ca)n = e. But (a−1ca)n = a−1cna, as one can check by multiplication of the left hand side. Thus, we need to show that cn = e if and only if a−1cna = e, which you show by composing the equation a−1cna = e on the left with a and on the right with a−1 to get the equation cn = aea−1 = 1. 3.
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