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Math Help - Abelian subgroups of S_{n}

  1. #1
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    Abelian subgroups of S_{n}

    I am finding all abelian subgroups of a symmetric group S_{n} for n>=5.

    What I have found so far is
    1. {e} : trivial group
    2. a cyclic group of order n
    3. a quotient group of order 2: S_{n}/A_{n}.

    I am wondering if below 4 is correct.
    4. "a cyclic group of order k which is a divisor of n"

    Any more subgroup exists for S_{n}?
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  2. #2
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    Quote Originally Posted by aliceinwonderland View Post

    I am finding all abelian subgroups of a symmetric group S_{n} for n>=5.
    i don't think this is basically possible to do because we even don't know all cyclic subgroups of S_n. no formula is known for the maximum order of elemens of S_n. however, there's a result due

    to Landau that if f(n) is the maximum order of elements of S_n, then \lim_{n\to\infty} \frac{\ln f(n)}{\sqrt{n \ln n}} = 1. it's not surprising that there's no closed form for f(n), because f(n) is related to the partition function.
    Last edited by NonCommAlg; January 6th 2009 at 10:08 PM. Reason: a typo!
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  3. #3
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    Quote Originally Posted by NonCommAlg View Post
    i don't think this is basically possible to do because we even don't know all cyclic subgroups of S_n. no formula is known for the maximum order of elemens of S_n.
    Any concrete example of an element of S_n that has an order bigger than n? It is easy to find an element of order n in S_n, but it is hard for me to find an element that has an order bigger than n.
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  4. #4
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    Quote Originally Posted by aliceinwonderland View Post
    Any concrete example of an element of S_n that has an order bigger than n? It is easy to find an element of order n in S_n, but it is hard for me to find an element that has an order bigger than n.
    Sure, take S_7 and consider (1234)(567).
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