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

It's often said that Cauchy's follows easily from Sylow's, 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 $p^n,\ 2<n<r$?

FireGarden
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    A much stronger result holds, actually: there are subgroups of order equal to all the powers of $p$, from tha maximal $r$ down to $0$ (and hence passing by $1$). Not only, but for every power of $p$ their number is congruent to $1$ modulo $p$. See here: https://math.stackexchange.com/q/4494783/1007416 –  Jul 31 '22 at 08:20

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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 $p$?