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Cauchy's Theorem proof

  1. Jul 1, 2010 #1
    I'm going through the proof of Cauchy's Theorem in a text book and I'm stuck on a particular point. I've tried looking at other proofs, but they all use the Orbit-Stabilizer Theorem or induction. The proof presented in the book probably uses the O-S Theorem, but not directly (since it's not been covered yet).

    Let G be a finite group and p a prime dividing the order of G. Define the set [itex] A= \{ (g_{1},...,g_{p} ) : g_{i} \in G , g_{1}...g_{p} = 1 \} [/itex] and a permutation [itex] \pi [/itex] on A by [itex] \pi (g_{1},...,g_{p}) = (g_{2},...,g_{p},g_{1}) [/itex]
    It shows that if [itex] g^p=1[/itex], then (g,...,g) is in A and is fixed by [itex]\pi[/itex], and that all other elements of A belong to cycles of size p. That's all fine. Then it shows [itex]|A| = |G|^{p-1} [/itex] and so p divides |A|. Since all cycles have size 1 or p, the number of fixed points is also divisible by p. I don't understand this. Why does the fact that |A| = pk imply that the number of fixed points is also a multiple of p?

    All I've been able to deduce is that [itex] kp = |A| = an_{1} + bn_{2} [/itex] where [itex] n_{1}, n_{2}[/itex] is the number of 1 cycles and p cycles, respectively.
    Could someone clear this up for me, please?

    (also in LaTeX, why are \pi and n_{1},n_{2} not aligned with the rest of the line, and the curly brackets not showing up in the code for the definition of A? EDIT: fixed now, thank you Gregg!)
    Last edited: Jul 2, 2010
  2. jcsd
  3. Jul 1, 2010 #2

    use \{

    [tex] A = \{ x : x^2=1\} [/tex]

    second, use itex tags instead of tex:

    I like to talk about the number[itex]\pi [/itex] during a sentence.
  4. Jul 2, 2010 #3
    The sequences g1,g2,...,gp (if any) where g1g2...gp=e and it isn't the case that g1=g2=...=gp fall into sets of p that are cyclic permutations of each other, because if g1g2...gp=e then g2g3...gpg1=e etc.

    If we remove these from A, since |A| = pk that leaves some multiple m=rp of p sequences ggg...g of elements such that gp=e. One of these is eee...e, so r is not zero. Therefore there must be rp-1>0 elements of order p.

    (Incidentally these fall into subgroups of order p that intersect only in the identity and each contains p-1 elements of order p, so you can also say rp-1=0 (p-1) or r=1 (p-1), so that the number of elements of order p is actually
    (h(p-1)+1)p-1=(hp+1)(p-1) for some integer h.)

    By the way, I believe this is McKay's proof of Cauchy's theorem , first published in American Mathematical Monthly, Vol.66 (1959), page 119.
    Last edited: Jul 2, 2010
  5. Jul 3, 2010 #4
    I'm assuming you mean sets of order p. Ok, it was not obvious to me that the set of non-fixed was also divisible by p (but I see how the result follows from this).
    Thanks for the help (and the reference to McKay)!
  6. Aug 4, 2011 #5
    Found this thread in google (yes, old, i know), and I'm having trouble accepting this as a trivial statement (bold). Is there an easy way to just "see" this? I believe its truth rests in the fact p is prime, but it wasn't mentioned here...
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