Answer and Solutions to Section 4.5 Homework Problems S. F. Ellermeyer November 19, 2006

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1 Answer and Solutions to Section 4.5 Homework Problems S. F. Ellermeyer November 19, It has been shown in class that H = fr 0 ; R 1 ; R 2 g is a subgroup of D 3. Since H is a group of order 3 and Z 3 (under addition) is also a group of order 3, and since all groups of order 3 are isomorphic to each other, we conclude that H and Z 3 are isomorphic. 2. U 7 = f1; 2; 3; 4; 5; 6g has Cayley table * and Z 6 = f0; 1; 2; 3; 4; 5g has Cayley table Note that, in U 7, j1j = 1, j2j = 3, j3j = 6, j4j = 3, j5j = 6, and j6j = 1; whereas in Z 6, j0j = 1; j1j = 6, j2j = 3, j3j = 1, j4j = 3, and j5j = 6. Since isomorphisms preserve the orders of elements, these observations give us a clue as how to rewrite the Cayley table of U 7 in order to see 1

2 that U 7 is isomorphic to Z 6. We use the correspondence U 7 Z and rewrite the Cayley table of U 7 as * making it easy to observe that U 7 = Z6. 3. The Klein 4 Group, K = Z 2 Z 2, has Cayley table (0; 0) (0; 1) (1; 0) (1; 1) (0; 0) (0; 0) (0; 1) (1; 0) (1; 1) (0; 1) (0; 1) (0; 0) (1; 1) (1; 0) (1; 0) (1; 0) (1; 1) (0; 0) (0; 1) (1; 1) (1; 1) (1; 0) (0; 1) (0; 0) and the subgroup, G = fr 0 ; R 2 ; H; V g, of D 4 (note the typographical error in the textbook) has Cayley table R 0 R 2 H V R 0 R 0 R 2 H V R 2 R 2 R 0 V H H H V R 0 R 2 V V H R 2 R 0. 2

3 It is clear upon examining these two tables that an isomorphism of K and and G is given by the correspondence (0; 0) $ R 0 (0; 1) $ R 2 (1; 0) $ H (1; 1) $ V. 7. This has been done in class and is also in the notes on the Web page. 8. This has been done in class and is also in the notes on the Web page. 9. Skip this problem. 10. De ne f : Z 3! Z 6 by f ([x] 3 [2x] 6. Then f ([0] 3 [0] 6, f ([1] 3 [2] 6, and f ([2] 3 [4] 6. To see that f is well de ned, note that if [x] 3 = [y] 3, then x = y + 3t for some integer t which means that 2x = 2y + 6t and hence that [2x] 6 = [2y] 6 and hence that f ([x] 3 f ([y] 3 To see that this is a homomorphism, note that for any x and y we have f ([x] 3 + [y] 3 f ([x + y] 3 [2 (x + y)] 6 = [2x] 6 +[2y] 6 = f ([x] 3 )+f ([y] 3 ). 11. De ne f : Z 9! Z 3 by f ([x] 9 [x] 3. To see that f is well de ned, note that if [x] 9 = [y] 9, then x = y + 9t for some integer t which means that x = y + 3 (3t) and hence that [x] 3 = [y] 3 and hence that f ([x] 9 f ([y] 9 To see that f is a homomorphism, note that for any x and y we have f ([x] 9 + [y] 9 f ([x + y] 9 [x + y] 3 = [x] 3 +[y] 3 = f ([x] 9 )+f ([y] 9 ). Note the importance of checking that f is well de ned. If, for example, we were to attempt to de ne f : Z 3! Z 9 according to f ([x] 3 [x] 9, then this would not be legitimate because this function is not well de ned. For example, [1] 3 = [4] 3 but f ([1] 3 ) 6= f ([4] 3 ) (because [1] 9 6= [4] A non trivial homomorphism from Z 6 into Z 4 is given by f ([x] 6 [2x] 4. There is no non trivial homomorphism from Z 6 into Z 5. The 3

4 reasons for the answers to these questions are given in problems 13 and 14 below in which the general question of homomorphisms from Z m into Z n is considered. 13 and 14. Suppose that m and n are positive integers and suppose that q is an integer. We will prove that the following statements are equivalent. (a) qm is divisible by n. (b) The mapping f : Z m! Z n given by f ([x] m [qx] n is well de ned. (c) The mapping f : Z m! Z n given by f ([x] m [qx] n is a homomorphism. Proof: First we prove that the truth of statement a implies the truth of statement b: Suppose that qm is divisible by n. Then qm = tn for some positive integer t. Now let us show that f : Z m! Z n given by f ([x] m [qx] n is well de ned. If [x] m = [y] m, then x = y + ms for some integer s. Therefore qx = qy + qms and hence qx = qy + n (ts) which implies that [qx] n = [qy] n and hence implies that f is well de ned. Now we show that the truth of statement b implies the truth of statement c: Suppose that the mapping f : Z m! Z n given by f ([x] m [qx] n is well de ned. Then for any integers x and y we have f ([x] m + [y] m f ([x + y] m [q (x + y)] n = [qx + qy] n = [qx] n +[qy] n = f ([x] m )+f ([y] m ) which shows that f is a homomorphism. Finally we show that the truth of statement c implies the truth of statement a: Suppose that the mapping f : Z m! Z n given by f ([x] m [qx] n is a homomorphism. Then, since f ([m] m [qm] n and also f ([m] m f ([0] m [0] n and since f is well de ned (because assuming that f is a homomorphism automatically implies that f must be well de ned), we see that [qm] n = [0] n and this tells us that qm = 0+nt for some integer t. Therefore qm must be divisible by n. 15. Let f : G 1! G 2 be a homomorphism and let x 2 G 1. Suppose that jxj = n. Then x n = e 1 and x m 6= e 1 for any positive integer m < n. Now note that, because f is a homomorphism, (f (x)) n = f (x n f (e 1 e 2. 4

5 By Corollary 4 on page 211, we conclude that n is divisible by jf (x)j. Note that if m is any positive integer such that m < n, then (f (x)) m = = f (x m e2 if x ) m 2 ker (f) 6= e 2 if x m =2 ker (f). Thus it may be the case that jf (x)j < jxj (of x m 2 ker (f) for some m < n) or it may be that case that jf (x)j = jxj (if x m =2 ker (f) for any m < n 16 and 17. done in class and in the notes. 5

MA441: Algebraic Structures I. Lecture 26

MA441: Algebraic Structures I. Lecture 26 MA441: Algebraic Structures I Lecture 26 10 December 2003 1 (page 179) Example 13: A 4 has no subgroup of order 6. BWOC, suppose H < A 4 has order 6. Then H A 4, since it has index 2. Thus A 4 /H has order