# Showing existence of an element with order $p$

If a group $G$ has order $p^n$, where $p$ is prime and $n \geq 1$, does there exist some element $a\in G$ s.t. the order of $a$ is $p$?

I happen to know that this is true by Cauchy's theorem, but that theorem has not been presented yet in the book. I only have this so far:

Let $a \in G$, then the order of the cyclic subgroup generated by $a$ divides the order of the group $G$ (by Lagrange's theorem).

But I get stuck because I don't know how to show that the $a$ has order $p$.

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Hint: Consider various powers of $a$. –  Tobias Kildetoft May 29 '13 at 0:02

To ellaborate on Tobias' comment. You have that the order of $a$ divides the order of $G$, and the order of $G$ is $p^n$. This means the order of $a$ can only be $1, p^1, p^2, \cdots, p^n$, as these are the only divisors of $p^n$.
First, you need to pick $a$ such that the order is one of $p^1, \cdots, p^n$ (how?). Then, you need to pick $k$ such that $a^k$ has order equal to $p$.
@AlanH it's hard to ellaborate if you don't first ellaborate on what I should ellaborate on. But first you need to show that there exists some $a$ with order $\ne 1$ (which implies order one of $p^1, p^2,\cdots,p^n$). Pick this $a$. Let $p^l$ be the order of $a$, so that $\displaystyle a^{(p^l)}$ is the smallest power of $a$ equal to $1$. Then find a power of $a$, given by $a^k$, which is also an element of the group, and try to pick $k$ such that $a^k$ has order $p$, i.e. $\displaystyle {(a^k)}^p = a^{(p^l)}$. –  Goos Jun 2 '13 at 18:25
Sorry, I was referring to how I need to pick that $a$ such that the order is one of $p^1, \dots, p^n$. –  AlanH Jun 2 '13 at 21:09
OK. Do you understand why the order of $a$ always has to be one of $1,p^1,\cdots,p^n$? From there, all you have to do is make sure the order of $a$ isn't 1. –  Goos Jun 2 '13 at 23:02
It's because the order of an element of a group has to divide the order of the group. The only possible divisors of $p^n$ are $1,p^2,\cdots, p^n$. The only element that has order 1 is the identity $e$. So we just pick an element not equal to $e$. Is this correct? –  AlanH Jun 2 '13 at 23:23