Looking for a simple proof that groups of order $2p$ are up to isomorphism $\Bbb{Z}_{2p}$ and $D_p$ for prime $p>2$. I'm looking for a simple proof that up to isomorphism every group of order $2p$ ($p$ prime greater than two) is either $\mathbb{Z}_{2p}$ or $D_{p}$ (the Dihedral group of order $2p$).
I should note that by simple I mean short and elegant and not necessarily elementary. So feel free to use tools like Sylow Theorems, Cauchy Theorem and similar stuff.
Thanks a lot!

Use Cauchy Theorem

Cauchy's theorem — Let $G$ be a finite group and $p$ be a prime. If $p$ divides the order of $G$, then $G$ has an element of order $p$.

then you have an element $x\in G$ of order $2$ and another element $y\in G$ of order $p$. Now you have to show that $xy$ is of order $2p$
using commutativity we get:
$(xy)^2 = y^2$, and hence $ord(xy) \not| 2$
$(xy)^p = x$ , therefore $ord(xy) \not| p$
and  $(xy)^{2p} = y^{2p} = e$, then $ord(xy) | 2p$
hence $1 <2<p<ord(xy) | 2p$
an then $ord(xy) = 2p$ because it doesn't equal to any divisor  different of $2p$.
