In $S_9$: for given $\sigma$ is there $\tau$ with $\tau^2=\sigma, \; \tau^3 = \sigma$? Struggling with these:
Let $\sigma$ in $S_9$ be given by $\sigma=(8\,9)(5\,6\,7\,1\,2\,3\,4)$.
1: Is there a $\tau\in S_9$ with $\tau^2=\sigma\,?\;$ Tip: think of $ \epsilon ( \sigma)$.
2: Is there a $\tau\in \langle\sigma \rangle$ with $\tau^3=\sigma\,?\;$ Tip: think of $ \operatorname{order}( \sigma)$.
Thanks in advance!
 A: *

*Perhaps you are more familiar with the notation $\operatorname{sgn}(\sigma)$? $\epsilon(\sigma)$ is a more general notation which is sometimes used to denote the sign of a permutation. What is the sign of $\sigma$? Is it an even, or odd permutation? What can you conclude about the existence then, (or lack thereof), of a permutation $\tau$ such that $\tau^2 =\sigma$? Edit: I just saw the follow-up comment, and your understanding, conclusions, and argument, about $(1)$ is spot on! (I think you understand more than your original post suggests you do.)

*Recall that $\operatorname{order}(\sigma) = |\langle \sigma\rangle|$, that is, it is equal to the order of the cyclic group generated by $\sigma$. The order of a permutation that is expressed as the product of disjoint cycles is equal to the least common multiple ("lcm") of the lengths of its cycles.  So what is $\operatorname{order}(\sigma) = \operatorname{lcm}(2, 7)\,?\;$ Use this information to argue that exactly one such $\tau \in \langle\sigma\rangle$ exists such that $\tau^3 = \sigma$. Edit: You've correctly found that the order of $\sigma = 14$. So $\sigma^{14} = \text{id} \implies \sigma^{15} = (\sigma^5)^3 = \sigma$. So we simply put $\;\bf \tau = \sigma^5 \implies \tau^3 = (\sigma^5)^3 = \sigma$
A: Use the tips you were given.
What is $\epsilon(\sigma)$? What do you know about $\epsilon$? Can you use what you know about $\epsilon$ to find such a $\tau,$ or prove that no such $\tau$ exists? (Edit: Nice work!)
What is the order of $\sigma$? What then do you know about the structure of the group $\langle\sigma\rangle$? Can you then find such a $\tau$?
