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Cauchy's Theorem: Let $G$ be a finite group and let $p$ divide the order of $G$. Then $G$ has an element of order $p$ and consequently a subgroup of order $p$ (of course the cyclic subgroup generated by the aforementioned element of order $p$).

Corollary: Let $G$ be a finite group, and $p$ be a prime. Then $G$ is a $p$-group if and only if the order of $G$ is a power of $p$.

Relevant Definition: $G$ is a $p$-group (for a prime $p$) if every element of $G$ has order $p^m$ for some $m\geq 1$.

The proof of the theorem was a beautiful application of using group actions to count. Now I'm trying to use the theorem to prove the corollary which is stated in my text leaving the proof as an exercise, but I am stuck because there doesn't seem to be a natural way to apply the theorem. Please accept my 'request for a hint'.

Thanks very much!

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Think of Lagrange's theorem for one direction, and use Cauchy's for the other. –  Xodarap Jul 3 '12 at 0:16
    
thanks! That was exactly it! –  Kyle Schlitt Jul 3 '12 at 0:46
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up vote 4 down vote accepted

If the order $n$ of the group is not a power of $p$, there is a prime $q\ne p$ that divides $n$. But then $\dots$.

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thanks! i felt silly once i saw how simple the solution was. –  Kyle Schlitt Jul 3 '12 at 0:46
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Missing a little something is no reason to feel silly, else I would feel silly pretty often. –  André Nicolas Jul 3 '12 at 0:48
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