Does the permutation group $S_8$ contain elements of order 14? Does the permutation group $S_8$ contain elements of order $14$?
My answer: If $\sigma =\alpha \beta $where $\alpha$ and $ \beta$ are disjoint cycles, then 
$|\sigma| =lcm(|\alpha|, |\beta|)$. Therefore the only possible disjoint cycle decompositions for a permutation $\sigma \in S_8$ with $|\sigma| =14$ is $(7,2)$. Since $7+2\neq 8$ so there is no element of order 14 in $S_8$.
Is my answer right? If no, what's the right answer?

Hints:
1) Every element in $\,S_n\,$ can be expressed as a product of disjoint cycles
2) The order of a product of disjoint cycles is the least common multiple of their lengths
So...can you see now why there is no element of order $\,14\,$ in $\,S_8\,$? There though are elements of order $\,15\,,\,6\,,\,10\ldots$
