Fixed point theorem p-groups

**CropDuster** · February 23rd 2011, 10:26 PM

I have no idea where to even start.

**Swlabr** · February 24th 2011, 12:41 AM

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.

**TheArtofSymmetry** · February 24th 2011, 01:31 AM

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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