TITLE: Finding the 'cube roots' of a permutation
QUESTION [5 upvotes]: How do I find three elements $\sigma \in S_9$ such that $\sigma^3=(157)(283)(469)$?  Since the three $3$-cycles in $\sigma^3$ are disjoint, $|\sigma^3|=\operatorname{lcm}(3,3,3)=3$.  Then since $(\sigma^3)^3=\sigma^9=e$ we have that $|\sigma|=9$, so $\sigma$ is a $9$-cycle.  That is all I'm really able to say.
How do we intelligently/methodically find the desired elements?  A generalized approach is welcome as well.
Thanks.

REPLY [3 votes]: We may assume $\sigma=(1ab5cd7ef)$. Now $2$ can be any of $a,b,c,d,e,f$. Let's say $a=2$. Then we have $\sigma=(12b58d73f)$. Now $4$ can be any of $b,d,f$.