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Math Help - How to find all Sylow 2-subgroups of S_5 efficiently?

  1. #1
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    How to find all Sylow 2-subgroups of S_5 efficiently?

    What I did is list all permutations of S5, 120 in tatol. And then find 7 nonidentity elements of order a power of 2. With the identity element, I only need to check these 8 elements form a subgroup, then it's a Sylow 2-subgroup of S5. To find others, I just need to conjugate it with every elements of S5.

    However, it's easy to say, but hard to do. Listing them all and conjugating them together are huge work to do.
    So I keep wondering... except using Sylow's theorems about the orders and checking if they're normal or not (I've done them all), is there any efficient way to find all Sylow 2-subgroups of S5?
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  2. #2
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    Re: How to find all Sylow 2-subgroups of S_5 efficiently?

    ok, the first thing to do, is find all elements x of S5 with x8 = e.

    these are:

    transpositions (of order 2)
    double disjoint transpositions (of order 2)
    4-cycles (of order 4)

    (S5 has no elements of order 8).

    now we need to count how many of each of these we have in S5.

    we have 10 transpositions, 15 double transpositions, and 30 4-cycles. note that every sylow 2-subgroup must have the same kinds of cycle types in it (since sylow subgroups are conjugate).

    note as well, that we can only have 3,5 or 15 sylow 2-subgroups, since these are the odd divisors of 120/8 (i am not counting 1 since we obviously have more than one sylow 2-subgroup, since there are more than 7 elements of order 2 and 4). the 4-cycles also must belong to the sylow 2-subgroups in inverse-pairs (we have 15 such pairs).

    clearly, one such subgroup is {e, (1 3), (2 4), (1 2 3 4), (1 3)(2 4), (1 4 3 2), (1 4)(2 3), (1 2)(3 4)}.

    so every sylow 2-subgroup of S5 contains 2 transpositions, 3 double transpositions, and 2 4-cycles. furthermore, the 2 transpositions must be disjoint, since if they weren't the group would contain a 3-cycle, of order 3.

    since we have 15 disjoint pairs of 2-cycles, 3 or 5 sylow 2-subgroups would not be enough to contain them all, so we must have 15 sylow 2-subgroups.

    so each sylow 2-subgroup contains a unique 4-cycle inverse-pair: (a b c d) and (a d c b), and thus the disjoint double 2-cycle (a c)(b d).

    but if g(1 2 3 4)g-1 = (a b c d) then either g(1 3)g-1 = (a c) or g(1 3)g-1 = (b d), so in any case our sylow 2-subgroup contains:

    {e,(a b c d),(a c)(b d),(a d c b),(a c),(b d)} and thus must contain (a b c d)(a c) = (a d)(b c) and (a b c d)(b d) = (a b)(c d). we can summarize this as:

    every 2-sylow subgroup of S5 is isomorphic to D4, and is uniquely determined by any 4-cycle it contains.

    now knock yourself out.
    Last edited by Deveno; December 16th 2012 at 12:33 AM.
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  3. #3
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    Re: How to find all Sylow 2-subgroups of S_5 efficiently?

    Thank you so much, you always save me when I need help.

    Your answer is clear enough but it makes me wonder... S5 is just one of cases and you solved it by analyzing the structure of cycles in S5. For Sn, especially when n gets larger, is there any general way for finding Sylow subgroups? Even finding them for other groups?

    Or sadly we can only analyze one group at a time to obtain them with Sylow's theorems?
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  4. #4
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    Re: How to find all Sylow 2-subgroups of S_5 efficiently?

    we have to do it on a case-by-case basis, since the prime factorization of n (and thus n!) can vary so widely. furthermore, as k increases, the number of different kinds of p-subgroups of order pk increases (p = 2 is particularly bad).
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