Show that $G$ has an element of order $p$.

Question

Let $G$ be a group of order $p^n$, for some prime $p$ and positive integer $n$. Show that $G$ has an element of order $p$.

I know this question has been asked before, but I have not found a complete answer and I am quite confused.

I have tried to solve it and so far this is what I have achieved:

If $|G|=p^n$ and $e\neq a\in G$, we have that order of $a$ divides $p^n$, then $|a|\in \{p^m:1\leq m \leq n\}$. Also, if order of $G$ is $p^n$, for all $a\in G$, $a^{p^{n}}=e$. Then write $p^n=p^{m}p^s$ , $1\leq s\leq n$, we have that $(a^{p^{m-1}p^s})^p=e$. Then we can conclude that $b=a^{p^{m-1}p^s}\in G$ is an element of orden $p$.

Is what I did correct?

Answer 1

According to the_fox hint.

If $G$ is cyclic then it must have an element of order $p$.

So let $G$ be not cyclic and $a\in G$. Let $H=\left \langle a \right \rangle$ then $|H|$ divides $|G|$ and so that it is of the form $p^m$ where $m<n$.Then, as $H$ is cyclic hence it must have an element of order $p$ and this element is then required element of order $p$ in $G$

Answer 2

This follows from Cauchy's theorem.

The proof of Cauchy that I remember is that by Sylow we have, for any prime $p$ dividing the order of the group, a subgroup of order $p^n$, where $p^{n+1}$ does not divide the order of the group. Then, as in the other answer, either that Sylow subgroup is cyclic, in which case it has a subgroup of every order dividing $p^n$, including one of order $p$, or take the cyclic subgroup generated by any element, and apply Lagrange. Then we get a cyclic subgroup of order $p^m$ for some $m\lt n$, and we are done.