For a given permutation t, does there exist n-cycle s such that $t=s^{k}$

Question

If $$t=(1\; 2)(3\; 4)(5\; 6)(7\; 8)(9\; 10),$$ question is asking determine whether there is a $n$-cycle $s$ ($n \geq10$) with $t=s^{k}$ for some integer $k$? Can someone provide me start to tackle this one.

Answer 1

$\DeclareMathOperator\lcm{lcm}$If you raise an $n$-cycle to the $k$-th power, you will get $\gcd(n,k)$ cycles of length $\frac{n}{\gcd(n,k)}$. The permutation $\tau=(1\ 2)(3\ 4)(5\ 6)(7\ 8)(9\ 10)$ consists of $5$ $2$-cycles, which can be achieved by letting $n=10$ and $k=5$. Indeed $$ \tau = (1\ 3\ 5\ 7\ 9\ 2\ 4\ 6\ 8\ 10)^5. $$

Here are two other examples to get a better understanding of the general phenomenon: \begin{align} (1\ 2\ 3\ 4\ 5\ 6\ 7\ 8\ 9\ 10\ 11\ 12\ 13\ 14\ 15)^6 &= (1\ 7\ 13\ 4\ 10)(2\ 8\ 14\ 5\ 11)(3\ 9\ 15\ 6\ 12),\\ (1\ 2\ 3\ 4\ 5\ 6)^{15} &= (1\ 2\ 3\ 4\ 5\ 6)^3 = (1\ 4)(2\ 5)(3\ 6). \end{align}