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Abstract algebra

  1. Apr 24, 2007 #1
    G is a finite group, |G| =p^n, p prime
    *:GxX -> X is group action. X is a finite set,

    I am required to prove the following [tex] |X|\equiv |X^G|modp [/tex]

    Now we start by asserting that [tex] x_1, x_2, .....,x_m [/tex]
    is the set of m orbit representatives. That orbit x [tex] <x_i> = {x_i} \\
    iff x_i [/tex] is a fixed point.

    we arrange the x_i's so that fixed points precede the non-fixed points.

    [tex] {x_1,x_2,....x_a}, |X^G|=a, x_a+1,....x_m [/tex] are the remaining orbit reps.

    numerical form of class eqn says
    |X| = \sum_{i=1}^a \frac{|G|}{|G_x_i|} + \sum_{i=a+1}^m \frac{|G|}{|G_X_i}

    since [tex] x_1, x_2,.....,x_a fixed G_x_i = G [/tex] for 1<=i<=a

    |G|/|G_x_i| =1 for 1<=i<=a
    |X| = a+ \sum_{i=a+1}^m \frac{|G|}{|G_X_i}
    for i=a+1,.....,m

    [tex] x_i not fixed, G_x_i not equal G [/tex]

    but [tex] |G| = p^n so |G_x_i| = p^e_i [/tex]

    where e_i < n :confused: but where does this fact come from?? I don't see how it follows that order of the stability subgroup must be a power of a prime??
  2. jcsd
  3. Apr 24, 2007 #2

    matt grime

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    Science Advisor
    Homework Helper

    Do you know what the class equation is?

    It follows directly from that.

    To be honest, I can't be bothered to work through your post's latex. The pricinple is easy: X is the disjoint union of the orbits. Orbits have size dividing p^n, i.e. 1,p,p^2,..,p^n. X^G is precisely the set of orbits of size 1.

    As to your last question. A stab subgroup is a subgroup of G, and G has order p^n. And the order of a subgroup divides the order of the group.
  4. Apr 24, 2007 #3
    ta, the last bit was what I was looking for
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