Find $\tau$ so $\tau^5 = \sigma$ when $\sigma = (1,4,7)(2,5,3,9)\in S_9$

Question

I looking for the right way of solving the following question:

let $\sigma = (1,4,7)(2,5,3,9)\in S_9$. Find $\tau$ so $\tau^5 = \sigma$.

In the solution they always just show the answer (which is $(1,7,4)(2,5,3,9)$). I think they expect us to "guess" the answer. Is there a way of finding/solving for $\tau$ without guessing? maybe some trick that can help us? Is it possible to do something like $\sigma^{-5}$? how to calculate it?

Answer 1

Since $\sigma$ is written as a pair of disjoint cycles $\alpha = (1\,4\,7)$ and $\beta=(2\,5\,3\,9)$, you can deal with each of them separately: $\tau$ will have two cycles, each corresponding to one of $\alpha$ or $\beta$.

Look at the cyclic subgroup generated by $\alpha$, so $\{\alpha, \alpha^2, \alpha^3=e\}$. Then notice that $$(\alpha^2)^5 = \alpha^{10} = \alpha(\alpha^3)^3 = \alpha e = \alpha\,.$$

So $\alpha^2$ will be one of the cycles of $\tau$. Then you can play the same game with $\beta$ to finish this off. The broad idea here is that $\sigma$ has finite order, the $\tau$ that you're looking for is probably one of $\{\sigma, \sigma^2, \dotsc\}$, and we know that $\tau$ actually exists because $5$ is relatively prime to the order of $\sigma$.

Answer 2

$\sigma=(147)(2539)=\alpha\beta$, where $\alpha=(147)$ and $\beta=(2539)$ are disjoint cycles,

with $\alpha^3, \beta^4$ and $\sigma^{12}$ being the identity.

Since $5$ is relatively prime to $12=3\times4$, there are $n,m$ such that $5n-12m=1.$

Then $(\sigma^n)^5=\sigma^{5n}=\sigma^{12m+1}={\left(\sigma^{12}\right)}^m\sigma^1=\sigma$.

We can take $n=5, m=2,$ and $\tau=\sigma^n$.

$\tau=\sigma^5$ is easy to compute because $\alpha^5=\alpha^2$ and $\beta^5=\beta$.