Find an element $\beta \in S_9$ such that $\beta^3 = (35)(1652)(843)(39)$.

Question

Find an element $\beta \in S_9$ such that $\beta^3 = (35)(1652)(843)(39)$.

I know its probably really simple but I can see the relation for the life of me

Edit: ok so $\beta^3 = (16538492)$ in disjoint form, so when we split it into 2 cycles, $\beta^3 = (12)(19)(14)(18)(13)(15)(16) = \beta$ since 3 is congruent to 1 mod 2. Is my logic correct?

Answer 1

Hint: first write $\beta^3$ as the product of disjoint cycles. Then determine the order of $\beta^3$. If you decompose the permutation into disjoint cycles, you should obtain a single cycle of length $8$, so computing $\beta$, given the single cycle $\beta^3$, will easily follow.

Edit, following your edit: The order of $\beta^3 = 8$, there's no need to decompose into transpositions to figure out $\beta \neq \beta^3.\;\;$ And actually, the correct cycle for $\beta^3$ (...it gets tricky; I second guessed myself too, but I'm quite confident now) is $\beta^3 = (1\,6\,3\,9\,8\,4\,5\,2)$.

Answer 2

Given that $\beta^3 = (1,6,3,9,8,4,5,2)$ (assuming you multiply permutations from right to the left, using standard function notation), then you determine $\beta$ as follows:

You know that $\beta(\beta(\beta(1))) = 6$, so in the cycle notation of $\beta$, there will have to be exactly two "gaps" between $1$ and $6$ (why?). That is, $\beta = (1, \_, \_, 6, \_, \_, \_, \_,)$, where the blanks are still to be determined. Similarly, $\beta^3(6) = 3$, so $\beta = (1, \_, \_, 6, \_, \_, 3, \_)$. I'll leave it for you to figure out the rest, starting by finding which blank to put $\beta^3(3) = 9$ in.

Something else to think about - what is it about $3$ = the exponent of $\beta^3$ and $8$ = the order of the lone cycle in $\beta^3$ that guarantees that we'll be able to successfully fill in the blanks to determine $\beta$?