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Sylow's theorem

  1. Nov 8, 2009 #1
    1. The problem statement, all variables and given/known data
    LET G BE A FINITE GROUP WHOSE ORDER IS DIVISIBLE BY THE PRIME P. SUPPOSE P^M IS THE HIGHEST POWER OF P WHICH IS A FACTOR OF |G|AND SET K=(|G|/P^M), THEN THE GROUP G CONTAINS AT LEAST ONE SUBGROUP OF |P^M|.

    I have the proof but can someone explain it in simpler terms? Maybe even a few examples please.



    2. Relevant equations



    3. The attempt at a solution
    Let X denote the collect of all subsets of G which have p^m elements and let G act on X by left translation so that the group element g is in G sends the subset A in X to gA. The size of X is the binomial coefficient kp^m choose p^m which is not divisible by p. Hence, there must be an orbit G(A) whose size is not a multiple of p. We have |G|=|G(A)|*|G(subA)|, consequently |G(subA)| is divisible by p^m. Now G(subA) is the stabilizer of A, so if a is in A and g is in G(subA), the ga is in A. This means that the whole right coset of G(subA)a is contained in A whenever a is in A and |G(subA)| cannot exceed p^m. Therefore, G(subA) is a subgroup of G which has order p^m.
     
    Last edited: Nov 8, 2009
  2. jcsd
  3. Nov 9, 2009 #2
    The statement is pretty simple. It says that if you have a Group of order n, where n has factoriaztion p^k * m, where p is a prime number, then there exists a subgroup of order p^k.

    The proof is a bit tricky, since it's in direct. What about it is confusing you?
     
  4. Nov 9, 2009 #3
    could you maybe give me a few examples? It makes sense but the proof is a bit rough for me.
     
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