# Artin Algebra 2.8.3 “Does every group whose order is a power of a prime $p$ contains an element of order $p$?”

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 $p$.

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

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.