I have this problem, and its solution:
If $\tau = (1,2)(3,4,5)$ determine whether there is a $n$-cycle $\sigma$ ($n\geq 5$) with $\tau = \sigma^k$ for some integer $k$.
And its solution:
But in the second para I can't understand two things:
What does $\{1,2,3,4,5\}$ can't be fixed by $\sigma$ mean and
Why without loss of generality, we have to consider only two cases $\sigma(1)=6$ and $\sigma(4)=6$?
$\{1,2,3,4,5\}$ is not fixed by $\sigma$ means that $\sigma(\{1,2,3,4,5\}) \not \subset \{1,2,3,4,5\}$ or equivalently $\exists a \in \{1,2,3,4,5\}: \sigma(a) \not \in \{1,2,3,4,5\}$. If you want a high-level explanation, any action of group $G$ on a set $X$ also induces an action on the power set $\mathcal{P}(X)$ given by $gA=\{ga \mid a \in A\}$ for $A \subset X$. If a subset is fixed by the action of a group or an element, it means that it is a fixed point with respect to this induced action on the set of all subsets.
In the elements of $\{1,2,3,4,5\}$ there are two types: those that are part of a 2-point orbit under the action of $\sigma^k=(1\textrm{ }2)(3\textrm{ }4\textrm{ }5)$ and those that are part of a 3-point orbit. Because of the way conjugation works in $S_n$, if we have any element of the first type or of the second type, we can just conjugate everything (which amounts to renaming all elements) so that we have to deal with $1$ or $4$.
I think that this statement is better seen in general context, rather than relying on ad-hoc methods: if $\sigma$ is a $N$-cycle, then $\sigma^k$ is a product of $d$ disjoint $\frac{N}{d}$-cycles, where $d = \gcd(N,k)$.
Intuitively it would be strange if we had a permutation $\sigma$ that acts on some elements in a similar way, e.g. $N$ elements all lie in one orbit if $\sigma$ is a $N$-cycle, and some power of $\sigma$ acts on those elements in fundamentally different ways, e.g. some of those $N$ elements are fixed, some are part of a 2-point orbit and some are part of a 3-point orbit.