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Order of this Group

  1. Mar 12, 2016 #1

    RJLiberator

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    1. The problem statement, all variables and given/known data
    Let f,h ∈S_4 be described by:
    f(1) = 1
    f(2) = 4
    f(3) = 3
    f(4) = 2

    h(1) = 4
    h(2) = 3
    h(3) = 2
    h(4) = 1

    Express (f°h) in terms of its behavior on {1,2,3,4} and then find the order of (f°h).

    2. Relevant equations


    3. The attempt at a solution

    So, first I express (f°h) in terms of its behavior.
    (f°h)(1) = 2
    (f°h)(2) = 3
    (f°h)(3) = 4
    (f°h)(4) = 1

    Done!
    Now, find the order of (f°h):

    The order of (f°h) is min{K:g^k-e} or infinity if no such k exists.

    I'm having a bit of difficulty in this. To me, when I search wikipedia: https://en.wikipedia.org/wiki/Order_(group_theory)
    It says that the order is the cardinality or the number of elements in the set.

    (f°h)'s order thus is 4?
    But I have a feeling it might be 3 based on in class examples.
    My answer is 4.

    How did I do?
     
  2. jcsd
  3. Mar 12, 2016 #2

    Orodruin

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    You need to separate the order of a group, i.e., the number of elements it contains, from the order of a group element, i.e., the power you need to raise it to to get the identity element.

    I would also suggest using cycle notation for group elements of ##S_n##, it reduces what you have to write significantly.
     
  4. Mar 12, 2016 #3

    RJLiberator

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    So the number of elements it contains is 4.
    The order of a group element (the power needed to raise it to get the identity element) is 1?
    so 4-1 = 3?
     
  5. Mar 12, 2016 #4

    Orodruin

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    No, the point is not to mix the concepts. The group is ##S_4## and its order is ##4!##, but that was not what the question was about. The question is about the order of the group element fh.
     
  6. Mar 12, 2016 #5

    RJLiberator

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    Ah. So that is the distinction.
    I'm sorry, I made a typo in the question f,h was supposed to be (f°h)

    The order of the group element (f°h) is thus 1 as any element in (f°h) has a power of 1.
     
  7. Mar 12, 2016 #6

    Orodruin

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    No. The order of a group element ##f## is the lowest number ##k## such that ##f^k = e##, where ##e## is the identity element (which is the only element of order one!).
     
  8. Mar 12, 2016 #7

    RJLiberator

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    Hm.
    OH. The order is 4? Because (f°h)(4) = 1!

    That's the e, the identity.

    I think the light went off in my head. Correct?
     
  9. Mar 12, 2016 #8

    Orodruin

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    The order is 4, but the reason is not the one you stated. Try to figure out the order of f and h to start with. Note that also h(4)=1.
    Edit: Are you familiar with cycle notation? It will simplify your life significantly.
     
  10. Mar 12, 2016 #9

    RJLiberator

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    Damn.

    The order of h is 4.
    The order of f is 1.

    f composed of h is thus 4*1 = 4 ?
     
  11. Mar 12, 2016 #10

    Orodruin

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    No. What is h^2?
     
  12. Mar 12, 2016 #11

    Orodruin

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    And again note that the only element with order one is the identity, for no other element ##f## is ##f^1=f=e##.
     
  13. Mar 12, 2016 #12

    RJLiberator

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    So it deals with powers ?

    h^2 = I am not even really sure how to express this. Perhaps this is my problem.

    h^2 = h(2)^2 = 9 ?
     
  14. Mar 12, 2016 #13

    Orodruin

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    No, h^2 is the usual notation for h×h, where × is the group operation.
     
  15. Mar 12, 2016 #14

    RJLiberator

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    Ok.
    h^2 then is

    (h°h)(1) = 1
    (h°h)(2) = 2
    (h°h)(3) = 3
    (h°h)(4) = 4

    (f°f) is similar.
     
  16. Mar 12, 2016 #15

    Orodruin

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    Right, so what is the order of f and h? If you apply the same reasoning to f×h, what is its order?
     
  17. Mar 12, 2016 #16

    RJLiberator

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  18. Mar 12, 2016 #17

    Orodruin

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    No, taking a power of the group operation has nothing to do with taking a power of the elements you rearrange. Still, I suggest you first figure out the orders of f and h.
     
  19. Mar 12, 2016 #18

    RJLiberator

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    Oh, it deals with permutations (from reading the link?)

    {1, 2, 3, 4} = the permutations are thus for the power of say, 2 we get {(1,3)(2,4)}
    And so the permutations of 4 is the identity back.
     
  20. Mar 12, 2016 #19

    Orodruin

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    So what are the orders of the elements f, h, and f×h?
     
  21. Mar 12, 2016 #20

    RJLiberator

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    f = {1, 4, 3, 2}
    so the order is 3
    due to permutations

    h = {4, 3, 2, 1}
    so the order is 4

    fxh = { 2, 3, 4, 1}
    so the order is 4
     
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