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Let $p$ be a prime number and $G$ be a group with $|G|=p^n$. Show that G contains an element of order $p$.

I would immediately say: "use Cauchy's theorem!", but this question is from a course that hasn't introduced this yet. Is there another way to prove this?

It's clear that every element has an order $p^r$ for some $r\leq n$ because of Lagrange's theorem. How can we proceed?

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    $\begingroup$ Consider the cyclic subgroup generated by an element of order $p^{r}$. $\endgroup$ – Alex Wertheim Mar 17 '16 at 19:55
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Take $g \in G$, $g\ne 1$. Then the order of $g$ is $p^k$ and so the order of $g^{p^{k-1}}$ is $p$.

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  • $\begingroup$ Don't you need Lagrange for the fact the "the order of $g$ is $p^k$"? And isn't Lagrange usually shown by making use of Cauchy? $\endgroup$ – jpvee Mar 19 '16 at 20:14
  • $\begingroup$ @jpvee, yes, see the last sentence of the question. $\endgroup$ – lhf Mar 19 '16 at 20:23

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