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Sylow 2-subgroups of S4

  1. Nov 6, 2009 #1
    1. The problem statement, all variables and given/known data

    I want to find the Sylow 2-subgroups of the permutation group S4

    2. Relevant equations

    I dont understand why is my application of Sylow's third theorem wrong.

    3. The attempt at a solution

    The order of S4 is 24=233. Thus, there are Sylow 2-subgroups and Sylow 3-subgroups by Sylow's first theorem. By Sylow's third theorem, the number of Sylow 2-subgroups is k=1 mod 2= 1,3,5,.... and must divide the order of the group. Thus, k=1 or 3. But, each permutation (12), (13), (23), (14), (34) together with the identity permutation forms a subgroup of order 2 of S4. Thus, K should not be 3.
     
  2. jcsd
  3. Nov 6, 2009 #2

    Office_Shredder

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    Sylow's subgroups are maximal p-subgroups, not subgroups generated by an element of order p
     
  4. Nov 6, 2009 #3
    Oh, ok. That's true. Sylow p-groups are maximal subgroups. Hence, in my problem my Sylow 2-subgroups are maximal, but they are subgroups of order 8. Thus, I should be looking for three Sylow 2-subgroups of order 8. Is this correct? Thank you for your guidance.
     
  5. Nov 7, 2009 #4
    That's the statement of Sylows Theorem.

    If the order of G equals n = p^k*m, p prime, then there exists a subgroup of order p^k, where k is the largest integer such that p^k divides the order of G.

    You then know as well that there is an element of order 2, though.
     
  6. Sep 4, 2010 #5
    I believe that you have misunderstood the definition of a p-Sylow subgroup, which in this case is not of order 2 but of order 8. The first theorem states that if p^m is the maximum power of p that divides |G| then there is a subgroup of order p^m and this is a p-Sylow subgroup. The 3rd theorem applies to this group, not the groups of order 2.

    In fact there is a subgroup isomorphic to the Dsub4 (order 8), the group of symmetries of the square. In cyclic notation it is ----numbering the vertices in order clockwise (or the reverse) around the square --- {e, (1234), (13(24), (1432), (12)(34), (14)(32), (13), (24)}.

    A few sample conjugations, say of (1234) by (123) shows that this is bot a normal subgroup. But all its conjugates (all 3 of them!) should contain the four-group as a subgroup, since all the elements with 2 2-cycles are present and the cycle structure is preserved.

    BTW, Cauchy's Theorem predicts that a prime-power group like this has an element of order 2, ehich it certainly does. There is no statement of how many such groups there are, however.
     
  7. Sep 4, 2010 #6

    Office_Shredder

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    Dude. This thread is 10 months old
     
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