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Math Help - Orders of elements in Symmetric groups

  1. #1
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    Orders of elements in Symmetric groups

    I'm reviewing for a midterm, and one of the questions is:

    "What is the smallest such n so S_n contains an element of order 30? Give reasons."

    I don't know how to solve this. Clearly the order of the group must be at least 30, so n is greater than 4. Also 30 must divide the order of the group, by Lagrange's Theorem, but it does for every n greater than 4.

    Thanks for any help!
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  2. #2
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by kimberu View Post
    I'm reviewing for a midterm, and one of the questions is:

    "What is the smallest such n so S_n contains an element of order 30? Give reasons."

    I don't know how to solve this. Clearly the order of the group must be at least 30, so n is greater than 4. Also 30 must divide the order of the group, by Lagrange's Theorem, but it does for every n greater than 4.

    Thanks for any help!
    Here's a little hint:

    Suppose  \sigma\in S_n has order  30 .
    Then  30\mid |S_n| \implies 30\mid n! .
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  3. #3
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    Doesn't 30\mid n! for every n that is 5 or greater, though? Does that mean that n=5, and if so, why?
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  4. #4
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by kimberu View Post
    Doesn't 30\mid n! for every n that is 5 or greater, though? Does that mean that n=5, and if so, why?
    Right, so start looking in  S_5 .

    I suggest finding  \sigma\in S_5 with order  5!=120 . Then  \sigma^4 will be an element with order  30 .

    Off the top of my head I can't think of such  \sigma , but I imagine it's not to hard to find.
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  5. #5
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by chiph588@ View Post
    Right, so start looking in  S_5 .

    I suggest finding  \sigma\in S_5 with order  5!=120 . Then  \sigma^4 will be an element with order  30 .

    Off the top of my head I can't think of such  \sigma , but I imagine it's not to hard to find.
    Pardon? S_5 is not cyclic! It is two-generated!

    What you are supposed to figure out in this question is what the order of a permutation is. You know that a 2-cycle has order 2, a 3-cycle has order 3, etc. but you need to figure out what happens when you concatenate two or more cycles.

    So, what is the order of (12)(345)? Why?

    Similarly, what is the order of (12)(34), and why?

    What about (1234)(56)?

    Your solution will involve splitting 30 down into prime factors.

    -----

    (I would like to point out at this stage that the element in S_5 with highest order is an element if the form (12)(345)).
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  6. #6
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    Quote Originally Posted by Swlabr View Post
    So, what is the order of (12)(345)? Why?

    Similarly, what is the order of (12)(34), and why?

    What about (1234)(56)?

    Your solution will involve splitting 30 down into prime factors.
    Those orders are 6, 2, and 4, I think, because those are the LCMs of the elements in the disjoint cycles. So since 30 = 2 x 3 x 5, there must be disjoint cycles of those amounts, and the smallest such case would be S_{10}?
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  7. #7
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by kimberu View Post
    Those orders are 6, 2, and 4, I think, because those are the LCMs of the elements in the disjoint cycles. So since 30 = 2 x 3 x 5, there must be disjoint cycles of those amounts, and the smallest such case would be S_{10}?
    Exactly. The thing you should notice most is that it does not matter what numbers are in the disjoint cycle, it is what the disjoint cycle looks like. The same thing, curiously, happens for conjugacy classes,

    Two permutations are conjugate in S_n if and only if they have the same disjoint cycle shape.

    For instance, (12)(345), (13)(245) and (123)(45) are all conjugate in S_n where n \geq 5.
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  8. #8
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by Swlabr View Post
    Pardon? S_5 is not cyclic! It is two-generated!
    Whoops! How embarrassing!
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