I'm going through the proof of Cauchy's Theorem in a text book and I'm stuck on a particular point. I've tried looking at other proofs, but they all use the Orbit-Stabilizer Theorem or induction. The proof presented in the book probably uses the O-S Theorem, but not directly (since it's not been covered yet).(adsbygoogle = window.adsbygoogle || []).push({});

Let G be a finite group and p a prime dividing the order of G. Define the set [itex] A= \{ (g_{1},...,g_{p} ) : g_{i} \in G , g_{1}...g_{p} = 1 \} [/itex] and a permutation [itex] \pi [/itex] on A by [itex] \pi (g_{1},...,g_{p}) = (g_{2},...,g_{p},g_{1}) [/itex]

It shows that if [itex] g^p=1[/itex], then (g,...,g) is in A and is fixed by [itex]\pi[/itex], and that all other elements of A belong to cycles of size p. That's all fine. Then it shows [itex]|A| = |G|^{p-1} [/itex] and so p divides |A|. Since all cycles have size 1 or p, the number of fixed points is also divisible by p. I don't understand this. Why does the fact that |A| = pk imply that the number of fixed points is also a multiple of p?

All I've been able to deduce is that [itex] kp = |A| = an_{1} + bn_{2} [/itex] where [itex] n_{1}, n_{2}[/itex] is the number of 1 cycles and p cycles, respectively.

Could someone clear this up for me, please?

(also in LaTeX, why are \pi and n_{1},n_{2} not aligned with the rest of the line, and the curly brackets not showing up in the code for the definition of A? EDIT: fixed now, thank you Gregg!)

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# Cauchy's Theorem proof

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