Now to get the Sylow theorems let be a group of order with . Then acts on the set of its subsets of order by translation. By the above paragraph, does not divide the size of and so not all stabilizers have index divisible by . Let be a subset of size whose stabilizer has index prime to . Since acts freely on it follows the order of is a -power so we conclude is a -Sylow subgroup.

]]>The last sentence of the proof of Wilson’s theorem? Apply the fixed point theorem.

]]>It doesn’t. The claim is just that they’re equivalent .

]]>Thanks for the reference! I wasn’t aware of it. My understanding is that the Lucas proof is well-known (at least, it’s on the Wikipedia article) but I didn’t know a reference for the Fermat or Wilson proofs.

]]>Yes, just $S_2$ among even-indexed cases is needed. You don’t need $S_n$ for even $n > 2$. I meant to suggest delaying the even case for $n > 2$.

]]>But don’t I need to get the induction to work?

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