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Very quickly, Sylow's first theorem says a sylow p-subgroup of order $p^rm$ exists and Cauchy's theorem says if $p \vert |G|$ then there is an element of order $p$.

It's often said that Cauchys follows easily from Sylows, but I just don't see it! I don't see why a sylow p-subgroup must have an element of order $p$; why couldn't they all be of order $p^n,\ 2<n<r$?

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If you have an element $g$ of order $p^n$, consider $$g^{p^{n - 1}}$$ Can you show that this is an element of order $p$?

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  • $\begingroup$ Oh. That's clever! I like that. Thank you! $\endgroup$ – FireGarden Jan 11 '14 at 20:18

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