# Abstract algebra

1. Apr 24, 2007

### catcherintherye

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 $$|X|\equiv |X^G|modp$$

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

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

$${x_1,x_2,....x_a}, |X^G|=a, x_a+1,....x_m$$ 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 $$x_1, x_2,.....,x_a fixed G_x_i = G$$ 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

$$x_i not fixed, G_x_i not equal G$$

but $$|G| = p^n so |G_x_i| = p^e_i$$

where e_i < n 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. Apr 24, 2007

### matt grime

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.

3. Apr 24, 2007

### catcherintherye

ta, the last bit was what I was looking for