Using Lagrange's theorem, prove that a non-abelian group of order $10$ must have a subgroup of order $5$.
Attempt:
Let $G$ be a group of order $10$. By Lagrange's theorem, if there exist a subgroup $H$ of $G$, the $o(H)=1,2,5~ or~ 10$. Assume that there is no subgroup of order $5$.
Let, $e\neq a\in G$. Then we have $o(a)|o(G)$. If every element of $G$ is of order $2$, then $a^2=e$ i.e $a=a^{-1}$ for all $a\in G$. Then I can show that $G$ is abelian. This is a contradiction.
Therefore, every element of $G$ is not of order $2$. Thus, $o(a)=5 ~or~ 10$. If $o(a)=5$, then $H= \left< a \right>$ is a cyclic subgroup of order $5$.
I don't know the approach is correct or not.