# Reask: If $\tau=(1\ 2)(3\ 4)(5\ 6)(7\ 8)(9\ 10)$ determine whether there is $\sigma$ s.t $\sigma^k = \tau$

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?

• 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. – awllower Jun 20 at 2:33

## 1 Answer

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 $$\tau$$.