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walters

  • one year ago

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

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

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

  3. experimentX
    • one year ago
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    looks like both of you are on same school

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

  5. walters
    • one year ago
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    yes i must multiply z3 by u(9)

  6. experimentX
    • one year ago
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    what is Z3 and U9 ??

  7. beketso
    • one year ago
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    is it multiply or cross(X)

  8. walters
    • one year 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
    • one year ago
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    under what operation is it a group?

  10. walters
    • one year ago
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    multiplication

  11. experimentX
    • one year ago
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    how?? Z3 is not closed under multiplication

  12. walters
    • one year ago
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    check question 3

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

  14. experimentX
    • one year ago
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    question no??

  15. walters
    • one year ago
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    question 3 3.1 a

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

  17. walters
    • one year ago
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    "by combine wat do u mean "

  18. experimentX
    • one year ago
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    LOL Q 5.4 came in my exam

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

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

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

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

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

  24. walters
    • one year ago
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    ok

  25. walters
    • one year 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
    • one year ago
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    http://www.physicsforums.com/showthread.php?t=474271

  27. experimentX
    • one year 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
    • one year ago
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    but the way beketso did this it is not easy to get z3 cross u(9)

  29. beketso
    • one year 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
    • one year 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
    • one year 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
    • one year ago
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    @experimentX Z is the set of integers and Zn is Z mod n

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

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

  35. experimentX
    • one year 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
    • one year ago
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    First course in abstract algebra by fraleigh

  37. walters
    • one year ago
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    i think it will be closed under *

  38. beketso
    • one year 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
    • one year 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
    • one year ago
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    on which operation is U9 a group?

  41. walters
    • one year ago
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    addition

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

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

  44. walters
    • one year ago
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    so it means the operation is either + or *

  45. walters
    • one year ago
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    7+8 is also 8 wen working with addition

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

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

  48. walters
    • one year 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
    • one year 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
    • one year 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
    • one year ago
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    because u r saying the identity is (0,1) so meaning the operation will be *

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

  53. walters
    • one year ago
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    it is also morning to me see u

  54. experimentX
    • one year ago
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    it's 7 am in the morning!!

  55. walters
    • one year ago
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    to me is 3:12 am

  56. beketso
    • one year ago
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    @walters will do this more later

  57. experimentX
    • one year ago
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    see ya later .. will see this tomorrow.

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