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If $\tau=(1\ 2)(3\ 4)(5\ 6)(7\ 8)(9\ 10)$ determine whether there is a $n$-cycle $\sigma$ $(n\ge 10)$ with $\tau=\sigma^k$ for some integer $k$.

This question is an exercise from Dummit and Foote, and it has been asked before here. The answer suggests that $\sigma^k$ should be a product of $\frac{n}{\text{gcd}(n,k)}$-cycles, where the number of such cycles is $\text{gcd}(n,k)$. Since the length of each component of $\tau$ is $2$, the number of such cycles is $5$, we see that $\text{gcd}(n,k) = 5$ and $n = 10$. It follows that $k = 5$.

Question: I think this should be a necessary condition, i.e., if such $n$-cycle $\sigma$ exists, we must have $n = 10$ and $k = 5$. But we haven't found a particular $\tau$ yet. Could anyone give me a hint on how to continue?

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    $\begingroup$ I think it is easy to find one by arranging the pairs at a distance of $5$, for example, $\sigma=\pmatrix{1 & 3 &5 &7 &9 &2 &4 &6 &8 &10}$ will do. $\endgroup$ – awllower Jun 20 at 2:33
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Expanding on awllower's comment:

You want $\sigma^5$ to map $1$ to $2$, so $\sigma$ looks like

$$ \sigma = \begin{pmatrix} 1 & ? & ? & ? & ? & 2 & ? & ? & ? & ? \end{pmatrix}$$

You want $\sigma^5$ to map $3$ to $4$, so $\sigma$ looks like

$$ \sigma = \begin{pmatrix} 1 & 3 & ? & ? & ? & 2 & 4 & ? & ? & ? \end{pmatrix}$$

and so on.

Once you finish this process, it should be straightforward to check that $\sigma^5$ is actually equal to $\tau$.

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