Let $\tau = (1 2)(3 4 5)$. Does there exist an $n$-cycle $σ$ (where $n \ge 5$) such that $\tau=\sigma^k$ for some $k \in \mathbb{Z}$?

Question

Let $\tau = (1 2)(3 4 5)$. Does there exist an $n$-cycle $σ$ (where $n \ge 5$) such that $\tau=\sigma^k$ for some $k \in \mathbb{Z}$?

I was trying to solve it using the orders of permutations, $$ O( \sigma ) = n \\ O(\tau) = 6 = O(\sigma^k) $$ But I am not reaching anywhere.

Answer 1

To get a $2$-cycle times a $3$-cycle from the $k$-th power of an $n$-cycle, we'll need $n=2(k\pmod n)$ and $n=3(k \pmod n)$. So we'd need $k\equiv 0\pmod n$, which gives $\sigma ^k=e$. Thus it's not possible.