Existence of a finite group $G$, that contains two elements of order $7$, whose product is an element of order $14$. $|G|=n \space \text{for some} \space n \in \Bbb N$, $\exists x,y \in G \space \text{such that:}$
$o(x)=o(y)=7$ and $o(xy)=14$.  
Is there such a group?
 A: Here is the link to the Theorem mentioned by Derek Holt, from the lecture notes of J. S. Milne:
THEOREM 1.64 For any integers $m,n, r > 1$, there exists a finite group $G$ with elements $a$ and $b$ such that $a$ has order $m$, $b$ has order $n$, and $ab$ has order $r$.
In the proof, such a group is constructed, e.g., $SL_2(\mathbb{F}_q)$ for appropriate prime power $q$.
A: In $S_{14}$ both the elements (written according to their cycle decomposition)
$$ \rho_1 = (1, 2, 3, 4, 5, 6, 7)(8, 9, 10, 11, 12, 13, 14)$$
$$ \rho_2 = (1, 8, 6, 5, 4, 3, 2)(7, 14, 13, 12, 11, 10, 9)$$
have order $7$ and 
$$ \rho_1 \rho_2 = (1,9)(7,8) $$
has order $2$. Thus by considering the following elements of $A_{21}$ 
$$ \rho_1 = (1, 2, 3, 4, 5, 6, 7)(8, 9, 10, 11, 12, 13, 14)(15,16,17,18,19,20,21)$$
$$ \rho_2 = (1, 8, 6, 5, 4, 3, 2)(7, 14, 13, 12, 11, 10, 9)$$
we have that both $\rho_1$ and $\rho_2$ have order $7$ but $\rho_1\rho_2$ has order $14$.

What is the smallest group $G$ such that $\exists g,h\in G: o(g)=o(h)=7$ and $o(gh)=2$ (or $14$) is a not so trivial question. Clearly $14\mid |G|$ by Lagrange Theorem, and dihedral groups of the $D_{7k}$ kind looks like good candidates.
