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Proving Cauchy's Theorem in Group Theory
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[QUOTE="fresh_42, post: 6069909, member: 572553"] That's not how I read it. First of all, we do not have generators, only simple group elements. So I read ##S## as a subset of ##G^p## such that its "digit product" is the neutral element. For a group action, we only need a set. My concern is less the action itself, but how do we know that all permutations multiply to ##e## again, i.e. is ##S## closed under this operation? This part is missing in your argument. [/QUOTE]
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Proving Cauchy's Theorem in Group Theory
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