Math Exam 1 Solutions October 12, 2010

Transcription

1 Math Exam 1 Solutions October 1, 1 As can easily be expected, the solutions provided below are not the only ways to solve these problems, and other solutions may be completely valid. If you have questions about what went wrong (or right!) with your own solutions, please talk to me about it so that you will improve on the next exam and the final. 1. [14 points] (a) Let H { a m a, m Z, m }. Show that H is a subgroup of Q under addition. (b) Show that Q + is not a subgroup of Q under addition. Solution: (a) To show that H is a subgroup of Q, we need to show that (i) H is closed under addition; (ii) H; and (iii) H is closed under additive inverses. To see (i), we let a b, m H, with a, b Z and m, n Z +. We then see n a m + b n a n + b m m+n. Since a n +b m Z and m+n, we see that a + b m H. For (ii), certainly H. n a 1 For (iii), if H, then its additive inverse is a n a n, which is also in H. n (b) There are essentially two ways to solve this problem. For the first way, we know that if Q + is a subgroup of Q under addition then they must share the same identity element. The identity of Q is, but / Q +. For the second way, we show that Q + is not closed under additive inverses in Q. For example, Q +, but / Q +. {( ) a. [(a) 8 points; (b) 5 points] Let L M c d (R) }. ad (a) Show that L is a group under matrix multiplication. multiplication is associative.) (b) Is L abelian? Why or why not? (You may assume that matrix Solution: (a) Essentially we show that L is a subgroup of GL(, R). So we need to show that (i) L is closed under matrix multiplication; (ii) L contains an identity element; and (iii) L is closed under inverses. For (i), we let ( ) ( a c d, e ) g h L. We multiply and find ( ) ( ) ( ) a e ae. c d g h ce + dg dh Since (ae)(dh) (ad)(eh) and ad, eh, we see that (ae)(dh), and so the product is indeed in L. Thus L is closed under matrix multiplication. For (ii), we see that the identity matrix I ( 1 1 ) is clearly in L. For (iii), if we are given A ( ) a c d L, we want to show that

2 A 1 L. (We know that A is invertible in GL(, R) because it has non-zero determinant, but we need to show that the inverse is actually in L.) To this end, we calculate that ( ) 1 ( A 1 a 1 a c d c ad which is lower triangular and has non-zero determinant. Thus A 1 L. (b) L is not abelian. To see this we provide a counterexample: ( ) ( ) ( ) whereas ( ) ( ) d ), ( ). 1 [It is not sufficient to simply say that matrix multiplication is not commutative, because L does not contain all matrices. For example, if L had happened to be the set of all diagonal matrices (so putting a in the lower left entry as well as the upper right), then those matrices form an abelian subgroup of GL(, R).] 3. [14 points] In this problem let α e πi/5 U 5. (a) Find all solutions of z 5 i for z C. Write your answer(s) in the form e iθ, θ R. (b) Define an isomorphism φ : Z 5 U 5 such that φ(3) α (where Z 5 is a group under + 5 ). Explain how you arrive at your answer, but you need not supply a formal proof. Solution: (a) We know that i e πi/, and so ( e πi/) 1/5 e πi/1 is a 5-th root of i. [In fact, i already a 5-th root of itself, but it was not necessary to observe this.] Therefore, the roots of z 5 i are the five numbers { e πi/1, αe πi/1, α e πi/1, α 3 e πi/1, α 4 e πi/1} { e πi/1, e 5πi/1, e 9πi/1, e 13πi/1, e 17πi/1}. (b) We make use of the homomorphism property of φ, namely that for all x, y Z 5, we have φ(x + 5 y) φ(x)φ(y). Since we require that φ(3) α, it then follows that we must have φ(1) φ( ) φ(3)φ(3) α, φ(4) φ( ) φ(3) 3 α 3, 4. [14 points] Fun with binary operations. φ() φ( ) φ(3) 4 α 4, φ() φ( ) φ(3) 5 α 5 1. (a) Suppose we define a binary operation on R + by a b a + b. associative? Why or why not? (b) Suppose we define a binary operation on R + by a b binary operation? Why or why not? 1 Is this operation bxe ax dx. Is this a well-defined

3 Solution: (a) No this operation is not associative. We provide a counterexample, taking a 1, b 3, c 6. Then a (b c) , whereas (a b) c (b) Yes, it is well-defined. First, given a, b R +, the function f(x) bxe ax is a continuous function and f(x) for all x in [, 1]. Therefore, a b 1 bxe ax dx is always a non-negative real number. Thus we need only show that a b >. There are various ways to do this (integrate by parts, estimation, etc.), but here is one way. Since f(x) only for x in [, 1], the integral above represents the area of between the graph of y f(x) in the first quadrant above the interval [, 1]. This is necessarily positive. 5. [1 points] Consider the group Z 4 under + 4. (a) Determine the order of the subgroup 35. (b) Which of 8, 19, and 1 generate Z 4. Why? (c) Determine all elements of Z 4 of order 7. Solution: (a) By a theorem from class (or Thm in Fraleigh), we know that for a cyclic group a of order n, we have a r n gcd(r, n). Therefore in Z 4 1, which is written additively, we have 35 4 gcd(35, 4) (b) One of the corollaries to the theorem above is that the generators of the cyclic group a are all elements of the form a r with gcd(r, n) 1. Therefore, the elements of Z 4 that are relatively prime to 4 will generate Z 4. Among the numbers we are given, 19 is relatively prime to 4 (and so is a generator), but 8 and 1 are not (and so are not generators). (c) Using the same theorem from part (a), we see that a Z 4 has order 7 if and only if 4 gcd(a,4) 7, which is true if and only if gcd(a, 4) 6. The numbers in Z 4 {, 1,..., 41} that have gcd 6 with 4 are 6, 1, 18, 4, 3, and 36.

4 6. [(a) 4 points; (b) 1 points; (c) 4 points] Suppose G is a group of order 8 containing two elements, A and B, such that A has order 4, and B has order, BAB A 3, G {e, A, A, A 3, B, AB, A B, A 3 B}. (a) Show that BA B A and BA 3 B A. (b) Write down the multiplication table for G. You do not need to justify all of your work, but do write elements of G in the first row and column in the order e, A, A, A 3, B, AB, A B, A 3 B. (c) (Extra Credit) Is G isomorphic to the quaternion group H 8? Why or why not? Solution: (a) We know that A 4 e and B e, and in particular A 1 A 3 and B 1 B. Now using BAB A 3, we see that (BAB)(BAB) (A 3 )(A 3 ), BAB AB A 6, BA B A. Likewise, if we invert both sides of BAB A 3, we see that (BAB) 1 (A 3 ) 1, B 1 A 1 B 1 A 3, BA 3 B A. (b) To write down the multiplication table for G, we use the following relations obtained from part (a) and the given information: A 4 e, B e, BA A 3 B, BA A B, and BA 3 AB. The multiplication table is then the following. e A A A 3 B AB A B A 3 B e e A A A 3 B AB A B A 3 B A A A A 3 e AB A B A 3 B B A A A 3 e A A B A 3 B B AB A 3 A 3 e A A A 3 B B AB A B B B A 3 B A B AB e A 3 A A AB AB B A 3 B A B A e A 3 A A B A B AB B A 3 B A A e A 3 A 3 B A 3 B A B AB B A 3 A A e

5 7. [1 points] Suppose G is an abelian group, and let H {a G a a 1 }. Show that H G. Solution: We need to show that (i) H is closed under multiplication in G; (ii) e H; and (iii) H is closed under inverses. To show (i), we let a, b H. Then (ab) 1 b 1 a 1, ba (because a, b H), ab (because G abelian). Therefore ab H, and so H is closed under multiplication in G. To see (ii), we simply observer that e 1 e, and so yes, e H. For (iii), we let a H. Since a a 1 by the definition of H and a H, indeed a 1 H.