Proof that finite group contains an element of prime order I tried to prove the following claim but it seemed a bit too easy:


  Let $G$ be a finite group. Then $G$ contains an element of prime order.


Please could someone tell me if my proof is correct or if I'm missing something?

Let $g$ be any non-identity element of $G$. Let $p$ be any prime factor of $|g|$. Then $g^{|g|\over p}$ has prime order $p$. $\Box$
Yes. This is right. To ensure it:

If $|g|$ is a prime, then we are done. Let $p$ be a prime factor of $|g|$, so that $|g|&gt;p$. 

Let $h=g^{\frac{|g|}{p}}$. Then $h^p=g^{|g|}=1$, which implies that order of $h$ divides $p$, so it is $1$ or $p$.

If order of $h$ is $1$ then this means, $h=1$ i.e. $g^{|g|/p}=1$, i.e. $|g|\leq |g|/p$, this is contradiction.
No the proof is not correct. It is possible that $g^{\frac{|g|}{p}}=e$. The theorem you are stating is commonly called Cauchy's theorem.  There is an elementary proof here using the class equation.