Direction: You are required to complete this test within 50 minutes. Please make sure that you have all the 10 pages. GOOD LUCK!

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1 Test 3 November 11, 2005 Name Math 521 Student Number Direction: You are required to complete this test within 50 minutes. (If needed, an extra 40 minutes will be allowed.) In order to receive full credit, answer each problem completely and must show all work. Please make sure that you have all the 10 pages. GOOD LUCK! 1. (12 points) Show that the group G H is abelian if and only if the groups G and H are abelian. Answer: Suppose the groups G and H are abelian. We want to show that G H is abelian. Since (g 1, h 1 ) (g 2, h 2 ) = (g 1 g 2, h 1 h 2 ) = (g 2 g 1, h 2 h 1 ) (because G and H abelian) = (g 2, h 2 ) (g 1, h 1 ), therefore G H is abelian. Next, we prove the converse. Since (g 1 g 2, h 1 h 2 ) = (g 1, h 1 ) (g 2, h 2 ) = (g 2, h 2 ) (g 1, h 1 ) (because G H is abelian) = (g 2 g 1, h 2 h 1 ), therefore g 1 g 2 = g 2 g 1 and h 1 h 2 = h 2 h 1, and G and H are abelian groups.

2 2. (13 points) List the elements of the factor group (Z 4 Z 6 )/ < (0, 1) >. Is this factor group cyclic? Answer: Let H be the subgroup < (0, 1) >. Then the distinct cosets of H in Z 4 Z 6 are (0, 0) + H = { (0, 1), (0, 2), (0, 3), (0, 4), (0, 5), (0, 0) } (1, 0) + H = { (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 0) } (2, 0) + H = { (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 0) } (3, 0) + H = { (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 0) }. Thus the factor group (Z 4 Z 6 )/ < (0, 1) > consists of (Z 4 Z 6 )/ < (0, 1) > = { (0, 0) + H, (1, 0) + H, (2, 0) + H, (3, 0) + H }. Since (1, 0) + H = { (1, 0) + H, (2, 0) + H, (3, 0) + H, (0, 0) + H } ttherefore (Z 4 Z 6 )/ < (0, 1) > is a cyclic group.

3 3. (12 points) Find the factor group (Z Z)/ < (1, 1) >. The group Z Z can be visualized as the points in the plane with both co-ordinates integers. Give a similar geometrical illustration of the factor group (Z Z)/ < (1, 1) >. Is it isomorphic to Z? If so why? Answer: Let H = < (1, 1) >. Then H = { n (1, 1) n Z } = { (n, n) n Z }. For each x Z, (x, 0) + H = { (x + n, n) n Z } is a coset of H in Z Z. The factor group (Z Z)/ < (1, 1) > is given by (Z Z)/ < (1, 1) > = { (x, 0) + H x Z } and it is illustrated below. The mapping φ : (Z Z)/ < (1, 1) > Z defined by φ ( (x, 0) + H ) = x is clearly well-defined, one-to-one and onto. Since φ ( ((x, 0) + H) + ((y, 0) + H) ) = φ ( (x + y, 0) + H ) = x + y = φ ( (x, 0) + H ) + φ ( (y, 0) + H ), the map φ is an isomorphism and (Z Z)/ < (1, 1) > Z.

4 4. (13 points) How many elements of the group Aut(Z 27 ) have order 9? How many elements of the group Aut(Z 27 ) have order 6? Answer: Using Gauss s result, we have Aut(Z 27 ) U(27) Z 18. The number of elements of order 9 in Aut(Z 27 ) is same as the number of elements of order 9 in Z 18. In Z 18, there are φ(9) = 6 elements of order 9. Hence Aut(Z 27 ) has 6 elements of order 9. Simllarly Aut(Z 27 ) has φ(6) = 2 elements of order 6.

5 5. (13 points) Calculate the number of elements of order 4 in each of the group (a) Z 4 Z 4, and (b) Z 4 Z 2 Z 2. Answer: (a) We are given that 4 = lcm ( a, b ) where a Z 4 and b Z 4. Therefore we have the following table a b Number of elements of order φ(1)φ(4) = φ(2)φ(4) = φ(4)φ(1) = φ(4)φ(2) = φ(4)φ(4) = 4 Total 12 Hence the group Z 4 Z 4 has 12 elements of order 4. (b) We are given that 4 = lcm ( a, b, c ) where a Z 4, b Z 2 and b Z 2. Therefore we have the following table a b c Number of elements of order φ(4)φ(1)φ(1) = φ(4)φ(1)φ(2) = φ(4)φ(2)φ(1) = φ(4)φ(2)φ(2) = 2 Total 8 Hence the group Z 4 Z 2 Z 2 has 8 elements of order 4.

6 6. (12 points) What is the order of the group U(1617)? Express U(1617) as an external direct product of cyclic additive groups of the form Z n. How many subgroup of order one U(1617) have? Answer: Using Fundamental Theorem of Arithmetic, we have 1617 = The order of U(1617) is given by ( U(1617) = φ(1617) = ) ( 1 1 ) ( 1 1 ) = Hence the group U(1617) U(3) U(7 2 ) U(11). Using Gauss s result, we have U(1617) U(3) U(7 2 ) U(11) Z 2 Z 42 Z 10. There is only one subgroup of order 1.

7 7. (12 points) Find all the left and the right cosets of the subgroup H = {(0, 1), (1, 2), (2, 4), (3, 3)} in the group Z 4 U(5). Is H a normal subgroup? Answer: The groups Z 4 and U(5) are abelian groups. Hence their direct product Z 4 U(5) is also abelian. We are given that H = {(0, 1), (1, 2), (2, 4), (3, 3)} is a subgroup of Z 4 U(5). Since every subgroup of an abelian group is normal, therefore H is a normal subgroup of Z 4 U(5). Since H is a normal subgroup of Z 4 U(5), therefore by definition of normal subgroup the left cosets of H are same as the right cosets of H. Hence the cosets of H in Z 4 U(5) are: (0, 1) + H = { (0, 1), (1, 2), (2, 4), (3, 3) } = (1, 2) + H = (2, 4) + H = (3, 3) + H (0, 2) + H = { (0, 2), (1, 4), (2, 3), (3, 1) } = (1, 4) + H = (2, 3) + H = (3, 1) + H (0, 3) + H = { (0, 3), (1, 1), (2, 2), (3, 4) } = (1, 1) + H = (2, 2) + H = (3, 4) + H (0, 4) + H = { (0, 4), (1, 3), (2, 1), (3, 2) } = (1, 3) + H = (2, 1) + H = (3, 2) + H.

8 8. (13 points) Prove that the center Z(G) of a group G is a normal subgroup of G. Answer: Let G ge a group, and Z(G) G. We want to show that Z(G) G. Let ghg 1 gz(g)g 1. We want to show that ghg 1 Z(G). Since ghg 1 = gg 1 h (because h Z(G)) = h Z(G), we have gz(g)g 1 Z(G). Hence by normal subgroup test, we get Z(G) G.

9 9. (12 points) Let S 3 be the symmetric group of permutations. Let H = {(1), (1, 2, 3), (1, 3, 2)} be a subgroup of S 3. Find all the left and right cosets of H. Is H a normal subgroup of S 3? Answer: The group S 3 consists of S 3 = { (1), (12), (13), (23), (123), (132) } and H = { (1), (123), (132) }. The distinct left cosets of H in S 3 are: (1)H = { (1), (123), (132) } and (12)H = { (12), (23), (13) }. The distinct right cosets of H in S 3 are: H(1) = { (1), (123), (132) } and H(12) = { (12), (13), (23) }. Since ah = Ha for all a S 3, the subgroup H normal in S 3.

10 10. (13 points) TRUE or FALSE: (a) Any two groups of order 3 are isomorphic. TRUE (b) If 7 divides the order of the group G, then G has a subgroup of order 7. FALSE (c) Z 2 Z 3 is isomorphic to Z 6. TRUE (d) Z 3 Z 19 is a cyclic group. TRUE (e) Every group of even order has an element of order 2. FALSE (f) Any group of order 8 is isomorphic to a permutation group of order 8. TRUE (g) The group Z 360 is isomorphic to Z 2 Z 2 Z 2 Z 9 Z 5. FALSE (h) The element (4, 2) has order 4 in the group Z 12 Z 8. FALSE (i) Every subgroup generated by a group element is a cyclic group. TRUE (j) The subgroup SL(2, R) is a normal subgroup of the group GL(2, R). TRUE (k) A symmetric group of permutations is an abelian group. FALSE (l) The group U(315) is isomorphic to U(3) U(5) U(7). FALSE (m) The group Aut(Z 6 ) has order 2. TRUE

MATH 28A MIDTERM 2 INSTRUCTOR: HAROLD SULTAN

NAME: MATH 28A MIDTERM 2 INSTRUCTOR: HAROLD SULTAN 1. INSTRUCTIONS (1) Timing: You have 80 minutes for this midterm. (2) Partial Credit will be awarded. Please show your work and provide full solutions,