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Homework Help: P-Sylow subgroup question

  1. May 16, 2010 #1
    1. The problem statement, all variables and given/known data

    Suppose K is a normal subgroup of a finite group G and S is a p-Sylow subgroup of G. Prove that K intersect S is a p-Sylow subgroup of K. So I know that K is a unique p-sylow group by definition, is that enough to prove that the intersection of K with S is a p-sylow subgroup of K?

    2. Relevant equations



    3. The attempt at a solution

    Since S is a Sylow-p subgroup of G and K intersect S is a subgroup of S and K, we see that K intersect S is a p-subgroup of K. It remains to show that K intersect S is a maximal p-subgroup of K, which implies that K intersect S is a Sylow p-subgroup of K.

    We know that [G:S] is relatively prime to p.
    THe hints I was given to finish this off was that [K: K intersect S] = [KS:S] and [G:S] = [G:KS][KS:S] and from that we are suppose to get that [K:K intersect S] is relatively prime to p.
    I am not getting how we know that [K: K intersect S] = [KS:S] and that [G:S] = [G:KS][KS:S] and how this implies that [K: K intersect S] is relatively prime to p.
    I know that KS is a subgroup of G since K is a normal subgroup of G. If anyone could help me with this problem I will be entirnaly gratefull
     
  2. jcsd
  3. May 16, 2010 #2
    A general fact is that |KS| = |K| |S| / |K intersect S|. Since [G:H] = |G|/|H| (this is Lagrange's theorem), this statement is equivalent to [K: K intersect S] = [KS:S].

    The equation [G:S] = [G:KS][KS:S] is again a consequence of Lagrange. The left side is |G|/|S| and the right side is (|G|/|KS|) (|KS|/|S|). Now, since [G:S] doesn't contain a factor of p and [KS:S] is a factor of [G:S], p cannot be a factor of [KS:S] = [K: K intersect S].
     
  4. May 16, 2010 #3
    A general fact is that |KS| = |K| |S| / |K intersect S|. I forgot about this fact!!! THanks so much, this makes a whole lot of more sense now.
     
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