Using Lagrange Thm to prove order of subgroup I'd like to answer the following question.

Suppose that I have a group, namely $G$, which has an order of $14$. I
  want to prove that $G$ must have  a subgroup that has order $7$.

So by assumption, I think that there exists a subgroup of order $1,2,7, 14$, but obviously that is no proof, rather an intuition. 
By Lagrange Theorem, if $K$ is a subgroup of a finite group $G$, then the order of $K$ divides the order of $G$. In particular, $|G|=|K|[G:K]$.
I'm not so sure that Lagrange Theorem really helps us solve this problem. 
How do I proceed with a proof? Any reading material or examples would be great. 
 A: *

*If $G$ has an element $g$ of order $14$, then $\langle g^2 \rangle$ is a subgroup of order $7$.

*If $G$ has an element $g$ of order $7$, then $\langle g \rangle$ is a subgroup of order $7$.

*Otherwise, all nontrivial elements of $G$ have order $2$.
Let $a,b \in G \setminus \{e\}$, with $a\ne b$. Then $ab$ has order $2$ and so $ab=ba$.
Therefore, the subgroup generated by $a$ and $b$ has order $4$, which contradicts Lagrange's Theorem.
A: Since $7$ is a prime number, Cauchy's theorem says that $G$ has some element whose order is $7$.
A: Say you want to stay away from big theorems and actually use only Lagrange.
Take a non-trivial element $a$. Its order is either $2$, $7$ or $14$. If it's $7$, you're done. If it's $14$, you're also done by considering $a^2$. 
The only problem is when every element has order $2$, in which case it is classic that $G$ is abelian, since $$abab=e\implies ba=(aa)ba(bb)=a(abab)b=ab$$
Now when $G$ is abelian, there actually can't be more than one element of order $2$, because if there were two of them they would generate a subgroup of order $4$, which doesn't divide $14$.
