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

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?

• note: $(2,5,3,9)^4$ is the identity – J. W. Tanner Jul 10 '19 at 15:49
• Well, $\sigma^{12}$ is the identity. – Lord Shark the Unknown Jul 10 '19 at 15:50
• Note that $\sigma$ is written in disjoint cycle notation, and that $3$ and $4$ are coprime to $5$. – TastyRomeo Jul 10 '19 at 15:54
• @LordSharktheUnknown: so ${\left(\sigma^5\right)}^5=\sigma^{25}={\left(\sigma^{12}\right)}^{2}\sigma=\sigma$ – J. W. Tanner Jul 10 '19 at 16:13

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 $$\sigma$$.

• The key point being that $\gcd(5,3)=\gcd(5,4)=1$. – lhf Jul 10 '19 at 16:15

$$\sigma=(147)(2539)=\alpha\beta$$, 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 $$\beta^5=\beta$$.