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Math Help - Sylow 3-subgroup of S_4

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    [Solved] Sylow 3-subgroup of S_4

    The number of m-cycle in S_{n} (n>=m) is given by
    \frac{n(n-1)(n-2)..(n-m+1)}{m}.

    For S_{4}, the groups generated by all 3-cylces (order 3) that I found are

    <(1 2 3)>, <(1 2 4)>, <(1 3 4)>, <(2 3 4)>,
    <(1 3 2)>, <(1 4 2)>, <(1 4 3)>, <(2 4 3) >

    According to the third Sylow theorem, the number of Sylow 3-subgroups is congruent 1 modulo p and divides |G|. Since |G|=24 and p=3, the number of Sylow 3-subgroup should be only 4.

    I think the above eight groups generated by 3-cycles are all Sylow 3-subgroups and the number is 8. What went wrong in my procedure?

    ----------------------------------------
    My mistake. Four subgroups are right. <(1 2 3)>, <(1 2 4)>, <(1 3 4)>, <(2 3 4)>
    Other groups are equivalent to those four subgroups.
    Last edited by aliceinwonderland; December 17th 2008 at 03:41 PM.
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