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Can you give me a hint on the first part of the exercise?

Let $p$ be a prime and let $X$ be a finite $G$-set, where $|G| = p^n$ and $|X|$ is not divisible by $p$. Prove that there exists $x \in X$ with $\tau x = x$ for all $\tau \in G$.

I'm trying to prove that there's only one orbit on the $G$-set but the only "special feature" that I see is that the order of $G$ is a power of a prime. However, I cannot figure out how to use that in order to solve prove that there's only one orbit.

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  • $\begingroup$ This is excercise 3.46, not 3.45. Leaving it here for future references. $\endgroup$ – cabo Jan 5 '18 at 3:58
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The question asks you to show that there is an orbit of size $1$, not to show that there is only one orbit. Note that orbits having size not equal to 1 must have size divisible by $p$ by orbit-stabilizer theorem.

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  • $\begingroup$ OK, I can see both things: I have to prove that there is one orbit of size $1$ and every orbit has size equal to a power of $p$. But now I don't know what to do next. Do I have to use Burnside's theorem? $\endgroup$ – Edward May 13 '13 at 2:36
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Hint: Decompose $X$ into a union of orbits and use the orbit-stabilizer theorem (or prove it), and the fact that any subgroup of a $p$-group has as its index a power of $p$.

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