Write $( 1\ 2\ 3\ 4)(2\ 3\ 4\ 5)(4\ 5\ 6\ 7) = \pi^3$ for some $\pi \in S_7$. Consider the permutation $\tau = ( 1\ 2\ 3\ 4)(2\ 3\ 4\ 5)(4\ 5\ 6\ 7) \in S_7 $.
Write $\tau = \pi^3$ for some $\pi \in S_7$.
My attempt: First we write $\tau$ as a product of disjoint cycles, $\tau = (1\ 2\ 4\ 5\ 3)(6\ 7)$.
So we want $\pi^3 = (1\ 2\ 4\ 5\ 3)(6\ 7)$.
Then, we write $(\pi^3)^2 = (1\ 4\ 3\ 2\ 5)$, reason being is that I want to use the following lemma \begin{array}{l}\text { We have } \rho\left(a_{1} a_{2} \ldots a_{\ell}\right) \rho^{-1}=\left(\rho\left(a_{1}\right) \rho\left(a_{2}\right) \ldots \rho\left(a_{\ell}\right)\right) \text { for any } \rho \in S_{n} \text { and any } \ell \text { -cycle } \\ \left(a_{1} a_{2} \ldots a_{\ell}\right) \in S_{n}\end{array}
So then $\pi (\pi^3)^2 \pi^{-1} = \pi (1\ 4\ 3\ 2\ 5) \pi^{-1} = (\pi(1)\ \pi(4)\ \pi(3)\ \pi(2)\ \pi(5) )$. But at this point I feel like this is the wrong way to tackle this. Any help would be appreciated.
 A: Ignore fancy lemmas, and instead view this as a problem about cyclic groups.
As @bof has pointed out, the disjoint cycle form of the permutation $\tau$ is $(1\ 2\ 4\ 3)(5\ 6\ 7)$ rather than $\sigma=(1\ 2\ 4\ 5\ 3)(6\ 7)$. This matters, as the question has different answers for these two permutations. I'll answer the question first for $\sigma=(1\ 2\ 4\ 5\ 3)(6\ 7)$, which is easier, and then for $\tau=(1\ 2\ 4\ 3)(5\ 6\ 7)$.

The permutation $\sigma = (1\ 2\ 4\ 5\ 3)(6\ 7)$ has order $5\times2=10$, and so we can view $\sigma$ as a generator for the cyclic group of order $10$. In this view we are simply solving the equation $3x=1\pmod{10}$, as $1$ is the generator of $\mathbb{Z}/10\mathbb{Z}$ and here the notation is additive. We see that $x=7$.
Therefore, $\sigma^7$ is the element you're looking for as: $$(\sigma^7)^3=\sigma^{21}=\sigma.$$

The key fact used above is that $10$ is coprime to $3$. The permutation $\tau=(1\ 2\ 4\ 3)(5\ 6\ 7)$ has order $4\times3=12$, and $12$ is not coprime to $3$. So a different method is needed.
Write $T=\langle\tau\rangle$ for the subgroup of $S_7$ generated by $\tau$, and $P=\langle \pi\rangle$ for the subgroup generated by some element $\pi$ such that $\pi^3=\tau$.
Claim. $\pi$ does not exist.
As $12$ is not coprime to $3$, the element $\pi$ is not a power of $\tau$, and so $\pi\not\in T$. On the other hand, we know that $\tau\in P$. Therefore $T\lneq P$, and so $P$ is a cyclic subgroup of $S_7$ of order greater than $12$ (infact $|P|=36$, but this is not needed).
However, by considering possible forms of disjoint cycles, we see that the largest possible order of an element of $S_7$ is $12$. Therefore no such subgroup $P$ exists, and so there is not element $\pi\in S_7$ such that $\pi^3=\tau$. QED
