Let $f = (1 2 3 4 5 6) ◦ (7 8 9 10) ∈ S_{10}$. Does there exist a positive integer n such that, when $f^n$ is decomposed into disjoint cycles, one of the cycles has length 5? Justify your answer.
Any directions? Because I don't have the experience to play with powers of cycles...
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Let $G$ be a finite group, and $m,n$ positive integers such that gcd$(m,n)=1$. Let $x,y$ be two elements of order $m$ and $n$ respectively.
Then the two cyclic subgroups generated by $x$ and $y$ respectively will have only the identity element in common.
Now coming to symmetric groups, if a cycle of length $p$, a prime number, occurs in the disjoint cycle decomposition of a permutation $\sigma$, then some power of $\sigma$ will be of order $p$. [Verify this].
Now combining these two statements you can see that the answer is negative for your question.
Just an example for the powers of a cycle: \begin{align} f&=(1\,2\,3\,4\,5\,6)& f^2&=(1\,3\,5)(2\,4\,\,6) \\ f^3&=(1\,4)(2\,5)(3\,6)&f^4&=(1\,5\,3)(2\,6\,4) \\ f^5&=(1\,6\,5\,4\,3\,2)=f^{-1}& f^6&=()=\operatorname{id} \end{align} Can see why, if $f$ is a cycle of length $\ell$, $f^k$ is decomposed as the product of $\gcd(\ell,k)$ disjoint cycles of length $\frac{\ell}{\gcd(\ell, k)}$?
Furthermore, powers of disjoint cycles are disjoint.