1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Dude. This thread is 10 months old
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook