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How to find all elements of S4 that satisfy the equation x^4=e?

  1. Nov 23, 2011 #1
    well, I'm thinking to answer the question in a generalized way. I want to find all the elements of Sn that satisfy the equation xn=e.
    well, if m|n and xm=e then xn=e, hence, if we raise x (which is of order m and m divides n) to the exponent n we'll have xn=e. so the answer will be this:
    S={all cycles of the length m s.t. m|n}

    Is that a true conclusion?

    As an example, in S4, since divisors of 4 are 1,2 and 4 the solutions will be all cycles with lengths 1,2 and 4. namely, {e,(1 2),(1 3), (1 4), (2 3), (2 4), (3 4), (1 2 3 4), (4 3 2 1), (1 3)(2 4), (1 2)(3 4), (1 4)(2 3)}
    Is there anything I'm missing?
     
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  3. Nov 23, 2011 #2

    micromass

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    First of all something pedantic: I wouldn't call (1 2)(3 4) a cycle. It is rather the product of cycles. So I would say that your set would rather be the cycles generated by the cycles of 1, 2 and 4.

    I think the conclusion holds true. To prove it, you should use that if [itex]x_i[/itex] are disjoint cycles of length [itex]k_i[/itex], then the order of [itex]x_1...x_l[/itex] is [itex]lcm(k_1,...,x_l)[/itex].
     
  4. Nov 23, 2011 #3

    Deveno

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    even this statement is not true. for example, (1 3)(1 2) = (1 2 3), so we don't want the set generated by cycles of length 1,2 and 4, but rather the set of products of disjoint cycles generated by cycles of length 1,2 and 4.

    we can disregard cycles of length 1, as these are all the identity map.
     
  5. Nov 23, 2011 #4
    Yea. so, the correct statement would be: S={all cycles of Sn generated by disjoint cycles of length m such that m divides n} would be the solutions of the equation xn=e.
    if we stop being pedantic, have I found the solutions of that equation correctly or there are more?
     
  6. Nov 23, 2011 #5

    I like Serena

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    There are more.
    Did you enumerate and check all of them?
     
  7. Nov 23, 2011 #6
    well, maybe I should include (3 4 1 2) as well?
    I didn't enumerate them, but tried to find them mentally. I considered all transpositions, because they are of order 2. They can be found easily, then I considered all cycles of the length 4 with their inverses (there was only one such cycle), and then I added all cycles that were products of disjoint cycles of order 2. I think I missed (3 4 1 2), it's of order 2 and it's its own inverse too.
    S={e,(1 2),(1 3),(1 4), (2 3), (2 4), (3 4), (1 2 3 4), (4 3 2 1), (3 4 1 2), (1 2)(3 4),(1 3)(2 4), (1 4)(2 3)}
    well, Is there any other solutions to the equation or I'm done?
     
  8. Nov 23, 2011 #7

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    Did you know that (1 2 3 4)=(3 4 1 2)?
    How many cycles of length 4 did you find?

    Do you know how many elements S4 has?
    How many could you eliminate?
    Do you have all the elements that you could not eliminate?
     
  9. Nov 23, 2011 #8
    Oops, Yea, I'm sorry. I thought (3 4 1 2) meant (1 3)(2 4). my bad.

    2.

    4! = 24.

    all elements of order 3 can be eliminated. right?
    well, I have 12 elements. but I guess there must be more. I'm not sure. Could you tell me one element that is not included in my list?
     
  10. Nov 23, 2011 #9

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    You have 12 elements of order 3?
    How did you count them?

    And 2 elements of order 4?
    There should be more.
     
  11. Nov 23, 2011 #10
    No, I have so far counted 12 solutions. I didn't say I have 12 elements of order 3. If I had 12 elements of order 3 then I was done. lol

    That's exactly where I'm stuck.
    Is there any theorem that tells us how many elements of a specific order we have in a group?
     
  12. Nov 23, 2011 #11

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    Suppose you were to create a 4-cycle from scratch.
    How many choices do you have for the 1st number?
    And the following numbers?
     
  13. Nov 23, 2011 #12
    well, I'm not so sure but I will just tell you my guess based on intuition.
    I'll have 4 choices for the first letter, but just after I chose the first number, the arrange of the 3 next numbers would be known, I can either shift all the remaining numbers to right or I can shift them to left depending on the first number I've chosen. Is that right? so if that's right, we might have 8 cycles of order 4 at maximum.
     
  14. Nov 23, 2011 #13

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    Let's not make assumptions about the other 3 numbers yet.
    Let's start with counting how many sequences of 4 different numbers you can make.

    On another angle, consider (1 2 3 4).
    Which other 4-cycles are identical to this one?
     
  15. Nov 23, 2011 #14
    What kind of sequences you mean? just to understand the terminology.

    (1 2 3 4),(2 3 4 1), (3 4 1 2), (4 1 2 3) are identical to this one.
     
  16. Nov 23, 2011 #15

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    I mean sequences like:
    1 2 3 4
    1 2 4 3
    ...

    How many are there?


    Good!
    So a given 4-cycle comes in 4 variants.
     
  17. Nov 23, 2011 #16
    If you're putting no additional conditions, there are 4!=24 permutations on 4 letters.

    True. well?
     
  18. Nov 23, 2011 #17

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    Yes. Both true!

    Since there are 24 sequences of 4 letters and since each 4-cycle comes in 4 flavors.
    How many different 4-cycles do you think there are?

    Could you enumerate them now?
     
  19. Nov 23, 2011 #18
    (1 2 3 4), (2 3 4 1), (3 4 1 2), (4 1 2 3)
    (1 2 4 3), (2 4 3 1), (4 3 1 2), (3 1 2 4)
    (1 3 2 4), (3 2 4 1), (2 4 1 3), (4 1 3 2)
    (1 3 4 2), (3 4 2 1), (4 2 1 3), (2 1 3 4)
    (1 4 3 2), (4 3 2 1), (3 2 1 4), (2 1 4 3)
    (1 4 2 3), (4 2 3 1), (2 3 1 4), (3 1 4 2)

    All the permutations in the same row are identical. so I think we'll have 6 different 4-cycles. Am I right?
     
  20. Nov 23, 2011 #19

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  21. Nov 23, 2011 #20

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    So do you think you can deduce the number of 3-cycles now?
     
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