Let $p$ is an odd prime and $G$ is a group which has $2p$ elements. Show that there exists at least one element with order $p$.
I tried showing in 2 parts as $G$ is cyclic and not cyclic, but I couldn’t show it for not cyclic groups. I think I must do something different. Could you please help me in the easiest way becasue we have learnt only elemantry theorems and definitions?
Let, $p$ be an odd prime number and $G$ be a group of order $2p$. Then $G$ has a subgroup of order $p$, which is indeed cyclic! Consequently, the subgroup contains an element of order $p$ and so does the group $G$.