Suppose $p$ is an odd prime and $G$ is a non-abelian group of order $2p$. Prove that $G$ contains an element of order $p$.
By Lagrange’s theorem, for all $x \in G$ , $o(x) = 1, 2, p , ~or~ 2p$
Case 1: There will not be any element of order $2p$ otherwise $G$ become cyclic and thus abelian (but given, $G$ is a non-abelian group).
Case 2: If there exist $x\in G$ such that $o(x)=p$ then there is nothing to prove.
Case 3: $e$ is the only element of order 1. $p\neq 1$, as $p$ is odd prime.
Case 4: Let there is an element $g\in G$ such that$o(g)=2$.
If none of the cases 1, 3, 4 hold, then proof will be completed. How to show that this case 4 is not possible?