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?

Assume that $g$ is an element of $S_8$ of order $14$.
Write $g$ as a product of disjoint cycles and let $a_1,\ldots,a_n$ be the cycle lengths (fixed points are counted as cycles of length 1).
Then you have the equation $$a_1 + a_2 + \ldots + a_n = 8$$ and since the order is the least common multiple of the cycle lengths $$\operatorname{lcm}(a_1,a_2,\ldots,a_n) = 14.$$
So you have to decide if there is a solution in positive integers $a_i$ to this system of equations.
