Thread:

1. Fixed point theorem p-groups

I have no idea where to even start.

2. 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.

3. If G acts on a finite set S, then S can be written as a disjoint union (See Hungerford's algebra p 93)
$S=S_0 \cup \bar{x_1} \cup \bar{x_2} \cup \cdots \cup \bar{x_n}$, where $S_0=\{x \in S: g\cdot x=x \text{ for all } g \in G\}$, $|\bar{x_i}|=|G:G_{x_i}|$, and $|\bar{x_i}|>1$ for all i. We see that $|\bar{x_i}|$ 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 $S_0$ is not empty and $p \nmid|S_0|$. An orbit of $x \in S$, denoted $\bar{x}$, has a single element iff $x \in S_0$. Can you conclude it from here?
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