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walters

  • 2 years ago

consider the groups Z3 and U(9) list the element of Z3 multiply U(9)

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  1. beketso
    • 2 years ago
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    Z3 ={0,1,2} U(9) = {1,2,4,5,7,8}

  2. beketso
    • 2 years ago
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    now are u supposed to multiply Z3 by U(9)?

  3. experimentX
    • 2 years ago
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    looks like both of you are on same school

  4. walters
    • 2 years ago
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    List the element of\[z _{3}\]multiply U(9)

  5. walters
    • 2 years ago
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    yes i must multiply z3 by u(9)

  6. experimentX
    • 2 years ago
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    what is Z3 and U9 ??

  7. beketso
    • 2 years ago
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    is it multiply or cross(X)

  8. walters
    • 2 years ago
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    z3 set of integers and u(9) means u find the number from 1 to 9 that cannot divide 9 or and those number must never divide each other

  9. experimentX
    • 2 years ago
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    under what operation is it a group?

  10. walters
    • 2 years ago
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    multiplication

  11. experimentX
    • 2 years ago
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    how?? Z3 is not closed under multiplication

  12. walters
    • 2 years ago
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    check question 3

  13. walters
    • 2 years ago
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    i don't know whether tht x mean multiplication or cross

  14. experimentX
    • 2 years ago
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    question no??

  15. walters
    • 2 years ago
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    question 3 3.1 a

  16. beketso
    • 2 years ago
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    you should take an element from Z3 and combine it with an element from U9

  17. walters
    • 2 years ago
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    "by combine wat do u mean "

  18. experimentX
    • 2 years ago
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    LOL Q 5.4 came in my exam

  19. walters
    • 2 years ago
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    z3={0,1,2} u(9)={1,2,4,5,7,8}

  20. beketso
    • 2 years ago
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    Take an the first element from z3 and combine it with all elements in u9

  21. walters
    • 2 years ago
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    then z3 x u(9)={0,1,2,4,5,7,8}

  22. beketso
    • 2 years ago
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    u get (0,1) (0,2) (0,4) (0,5) (0,7) (0,8)

  23. beketso
    • 2 years ago
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    then u have to do the same with the second element of Z3

  24. walters
    • 2 years ago
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    ok

  25. walters
    • 2 years ago
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    oh becuase u came with a new set how will u get the identity because the new set is different to z3 and u(9)

  26. experimentX
    • 2 years ago
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    http://www.physicsforums.com/showthread.php?t=474271

  27. experimentX
    • 2 years ago
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    looks like by group U(n), you mean \( \Bbb Z \mod n \) what is the convention followed for Z in your book?

  28. walters
    • 2 years ago
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    but the way beketso did this it is not easy to get z3 cross u(9)

  29. beketso
    • 2 years ago
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    by U(n) we mean a set of numbers which are relatively prime to n and are smaller than n

  30. experimentX
    • 2 years ago
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    you sure on that?? (how is it closed on operation if it has no mechanism to reduce elements greater than it) and what is Z?? check your book again ... what convention is followed

  31. beketso
    • 2 years ago
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    z3 x U9 ={(0,1) (0,2) (0,4) (0,5) (0,7) (0,8) (1,1) (1,2) (1,4) (1,5) (1,7) (1,8) (2,1) (2,2) (2,4) (2,5) (2,6) (2,8) } The first element is from z3 and the second is from U(9)

  32. beketso
    • 2 years ago
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    @experimentX Z is the set of integers and Zn is Z mod n

  33. experimentX
    • 2 years ago
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    what's the operation? it talks about cyclic group

  34. walters
    • 2 years ago
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    i think since we r in z multiplication and addition are closed .

  35. experimentX
    • 2 years ago
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    since you have taken Cartesian product, \( \Bbb Z_3 \times U(9) \) how do you define this new group? What book are you following for abstract algebra?

  36. walters
    • 2 years ago
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    First course in abstract algebra by fraleigh

  37. walters
    • 2 years ago
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    i think it will be closed under *

  38. beketso
    • 2 years ago
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    am not sure but i think it is the combination of the identity element from Z3 and the identity element from U9 So, the identity element is (0,1)

  39. walters
    • 2 years ago
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    because by addition ther is a high possibility wen u add 2 integar u find the resul will no longer suits u(9)

  40. beketso
    • 2 years ago
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    on which operation is U9 a group?

  41. walters
    • 2 years ago
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    addition

  42. walters
    • 2 years ago
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    because by multiplication it won't hold ie 5*7 is not in u(9)

  43. beketso
    • 2 years ago
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    5*7 = 8 which is in U(9)

  44. walters
    • 2 years ago
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    so it means the operation is either + or *

  45. walters
    • 2 years ago
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    7+8 is also 8 wen working with addition

  46. beketso
    • 2 years ago
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    7+8 = 6 in U(9) which is not in the set

  47. beketso
    • 2 years ago
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    that is why i think the identity element is (0,1)

  48. walters
    • 2 years ago
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    it can't be the values will circulate in {1,2,4,5,7,8} only there is no way we can get any value except those who are in u(9)

  49. walters
    • 2 years ago
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    we can use the fact that says most of the time wen the first elementin a set is 0 in the the the operation that likely to be closed is + and if the fist element is 1 * is likely to be closed

  50. experimentX
    • 2 years ago
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    your text book is very nice. the problem looks like it's related to Group Action on sets

  51. walters
    • 2 years ago
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    because u r saying the identity is (0,1) so meaning the operation will be *

  52. experimentX
    • 2 years ago
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    but it's morning here ... i gotta get some sleep. will see later

  53. walters
    • 2 years ago
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    it is also morning to me see u

  54. experimentX
    • 2 years ago
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    it's 7 am in the morning!!

  55. walters
    • 2 years ago
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    to me is 3:12 am

  56. beketso
    • 2 years ago
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    @walters will do this more later

  57. experimentX
    • 2 years ago
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    see ya later .. will see this tomorrow.

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