I have no idea where to even start.

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- Feb 23rd 2011, 10:26 PMCropDusterFixed point theorem p-groups
I have no idea where to even start.

http://img580.imageshack.us/img580/6...10224at125.png - Feb 24th 2011, 12:41 AMSwlabr
Hint: By Burnside's Lemma (aka, the Lemma Which Is Not Burnside's), the order of G must divide the sum of the orders of the orbits.

- Feb 24th 2011, 01:31 AMTheArtofSymmetry
If G acts on a finite set S, then S can be written as a disjoint union (See Hungerford's algebra p 93)

$\displaystyle S=S_0 \cup \bar{x_1} \cup \bar{x_2} \cup \cdots \cup \bar{x_n}$, where $\displaystyle S_0=\{x \in S: g\cdot x=x \text{ for all } g \in G\}$, $\displaystyle |\bar{x_i}|=|G:G_{x_i}|$, and $\displaystyle |\bar{x_i}|>1$ for all i. We see that $\displaystyle |\bar{x_i}|$ is divisible by p by Lagrange's theorem, since a p-group G acts on S. By assumption, S is not divisible by p, so $\displaystyle S_0$ is not empty and $\displaystyle p \nmid|S_0|$. An orbit of $\displaystyle x \in S$, denoted $\displaystyle \bar{x}$, has a single element iff $\displaystyle x \in S_0$. Can you conclude it from here?