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I'm not sure whether or not my answer and proof for this question are valid. Could you point out any flaw?

Let $G$ be an arbitrary group, an arbitrary element of $G$ be $g$ and $|G|=p^n$. Since a cyclic subgroup $H$ of $G$ generated by $g$ has order $p^k$ $(1\leq k\leq n)$, cyclic subgroup $I$ of $H$ such that $I=\{e,g^{p^{k-1}},g^{2p^{k-1}},...,g^{(p-1)p^{k-1}} \}$ has order $p$; hence, $g^{p^{k-1}}$ has order $p$. This shows that there is an element of order $p$.

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1  
I think your proof works. –  Crostul Jul 14 at 9:54
2  
You need to require that $g \neq e$. –  Vincent Pfenninger Jul 14 at 10:04
    
Thanks for your collection. –  Aran Komatsuzaki Jul 14 at 10:07

1 Answer 1

up vote 2 down vote accepted

You are indeed correct. However, I would've avoided some notational acrobatics and simply said that we know $G$ has a cyclic subgroup isomorphic $\mathbb{Z}_{p^k}$, and $\mathbb{Z}_n$ has a subgroup of order $m \iff m|n$.

Well, obviously $p|p^k$.


And finally, Vincent in the comments is correct (+1). You need to make sure to select a non-identity element $g$ to generate the cyclic subgroup we're working with.

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Yes, that looks more elegant. –  Aran Komatsuzaki Jul 14 at 10:49

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