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Question regarding sylow subgroups

  1. Oct 16, 2008 #1
    Hello everybody,

    I I have an exercise here that I'm really stuck with..

    Let P be in Syl_p(G) and assume N is a normal subgroup of G. Use the conjugacy part of Sylow's theorem to prove that P intesect N is a Sylow p-subgroup of N.

    The "conjugacy part" in this book ('Algebra' by Dummit, Foote) is about
    p-subgroups of G being subgroups of a conjugate of a Sylow p-subgroup
    in G.

    I tried some different approaches but can't get nowhere..
    Using the "conjugacy part of the thm" for P intersect N has no
    use since already P intersect N is a subgroup of P itself, so
    we have to use it for another subgroup.

    I also considered Q a sylow p-subgroup of N. Eventually I would
    like to show that Q and (P intersect N) have the same cardinality.
    Well, (P intersect N) is a p-subgroup of N and Q is a sylow p-subgrp of N
    so that from "the conjugacy part of Sylow's thm" we have that
    (P intersect N) is a subgroup of a conjugate of Q in G, that is,
    (P intersect N) <= gQg^-1, some g in G.
    That seems to be something but can't go more far..


    Finally, the fact that N is normal smells like 2nd iso theorem..
    From this, we deduce that (P intersect N) is normal sbgrp of P
    and N is a normal sbgrp of NP.. Also
    P / (P intersect N) is isomorphic to NP / N.
    So we can play with the cardinalities.
    But can't think of anything else :(

    Any ideas or hints highly appreciated!
    Thanks in advance!

    PS. I wasn't sure if I should post this in the 'homework' section. This is an exercise
    given in a first year graduate course in Algebra and I think that it shouldn't be put
    with "calculus and beyond" questions..


    EDIT:
    Actually the exercise has another part: "Deduce that PN/N is a sylow p-sbgrp of G/N".
    Well, there is a chance that we don't need the normality of N to show the first part of the question (the thing that I asked)
     
    Last edited: Oct 16, 2008
  2. jcsd
  3. Oct 16, 2008 #2

    mathwonk

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    im a little sleepy but it seems kind of obvious. take any p subgroup of N and enlarge it to a sylow subgroup of G which is conjugate to your sylow subgroup, so your p group is conjugate to a subgroup of your p group intersected with N, so that bthing is maximal, hence sylow. ???
     
  4. Oct 16, 2008 #3
    Oh well..
    'm sleepy too, maybe that's why I didn't understand... :(

    Could you please use some letters?

    Let's take an p-subgroup H<N.
    When you say 'enlarge it' you mean 'see it as' right (forgive my english)?
    So, H<K where K = g P g^-1.

    So H is conjugate to a subgroup of H intersected with N?
    I don't understand that..

    Sorry for the lack of understanding, maybe it will be more clear
    tomorrow when I wake up :)
     
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