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walters Group Title

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

  • one year ago
  • one year ago

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  1. beketso Group Title
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    Z3 ={0,1,2} U(9) = {1,2,4,5,7,8}

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

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

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

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

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

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

    • one year ago
  8. walters Group Title
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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

    • one year ago
  9. experimentX Group Title
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    under what operation is it a group?

    • one year ago
  10. walters Group Title
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    multiplication

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

    • one year ago
  12. walters Group Title
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    check question 3

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

    • one year ago
  14. experimentX Group Title
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    question no??

    • one year ago
  15. walters Group Title
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    question 3 3.1 a

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

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

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

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

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

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

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

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

    • one year ago
  24. walters Group Title
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    ok

    • one year ago
  25. walters Group Title
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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)

    • one year ago
  26. experimentX Group Title
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    http://www.physicsforums.com/showthread.php?t=474271

    • one year ago
  27. experimentX Group Title
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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?

    • one year ago
  28. walters Group Title
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    but the way beketso did this it is not easy to get z3 cross u(9)

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

    • one year ago
  30. experimentX Group Title
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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

    • one year ago
  31. beketso Group Title
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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)

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

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

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

    • one year ago
  35. experimentX Group Title
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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?

    • one year ago
  36. walters Group Title
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    First course in abstract algebra by fraleigh

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

    • one year ago
  38. beketso Group Title
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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)

    • one year ago
  39. walters Group Title
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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)

    • one year ago
  40. beketso Group Title
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    on which operation is U9 a group?

    • one year ago
  41. walters Group Title
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    addition

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

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

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

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

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

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

    • one year ago
  48. walters Group Title
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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)

    • one year ago
  49. walters Group Title
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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

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

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

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

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

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

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

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

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

    • one year ago
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