Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $p$.

share|cite|improve this question
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
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.

share|cite|improve this answer
Yes, that looks more elegant. – Math.StackExchange Jul 14 '14 at 10:49

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.