Find permutations whose third power is known

I have to find permutations $$a$$ such that $$a^3=(1 \ 2)(3 \ 4)(5 \ 6)(7 \ 8 \ 9 \ 10)$$ and I have to find at least 3 solutions.

So first I must find disjoint cycles.
Those are: 1 [2,2,2][4] and 2 [4,2][4] and 3[6][4] are there any more cycles? and what is the order of numbers (1,2,3,...10) in those new cycles? so is solution (1 2)(3 4)(5 6)(7 10 9 8) sufficient for 1 and (1 3 2 4)(5 6)(7 10 9 8) for 2?

• Why this closing proposal ? This asker has done personal work on a question that is (almost) clearly settled. – Jean Marie Feb 11 at 0:07
• @JeanMarie I have no idea and I have voted to leave it open. – José Carlos Santos Feb 11 at 8:53
• I have taken the liberty to modify your first sentence and your title : this is for attracting ... and retaining readers later on. – Jean Marie Feb 11 at 12:03
• @jose how do you "vote to leave it open" before it's actually closed? Sometimes I want to prevent a question from being closed but feel powerless unless the question gets closed/on hold and I have to vote to reopen – Oscar Lanzi Feb 11 at 15:09
• @OscarLanzi Because the question appeared to me in the close votes review queue. – José Carlos Santos Feb 11 at 15:40

Yes for $$p=(1 \ 2)(3 \ 4)(5 \ 6)(10 \ 9 \ 8 \ 7)$$.

No for $$p=(1 \ 3 \ 2 \ 4)(5 \ 6)(7 \ 10 \ 9 \ 8)$$ ; it isn't a solution because $$p^3$$ would send $$1$$ onto $$4$$ instead of $$2$$.

Here is a way to find many solutions :

First, as you have well seen it, we must take for the last cycle $$\color{red}{the \ reversed \ cycle \ (7 \ 10 \ 9 \ 8)}$$ (because for a cycle $$c$$ on 4 elements, $$c^4=id \ \iff \ (c^{-1})^3=c$$, and there are no other solutions).

A first global solution is :

$$(1 \ 2)(3 \ 4)(5 \ 6)\color{red}{(7 \ 10 \ 9 \ 8)}$$

(You had recognized it). Please note that we use the fact that a transposition $$t$$ is such that $$t^3=t$$.

A family of 8 other solutions are found by considering an order-$$6$$ cycle on elements $$1 \cdots 6$$ :

$$(\underline{1} \ 3 \ 6 \ \underline{2} \ 4 \ 5)\color{red}{(7 \ 10 \ 9 \ 8)}$$

$$(\underline{1} \ 4 \ 6 \ \underline{2} \ 3 \ 5)\color{red}{(7 \ 10 \ 9 \ 8)}$$

$$(\underline{1} \ 5 \ 3 \ \underline{2} \ 6 \ 4)\color{red}{(7 \ 10 \ 9 \ 8)}$$

$$(\underline{1} \ 6 \ 3 \ \underline{2} \ 5 \ 4)\color{red}{(7 \ 10 \ 9 \ 8)}$$

$$(\underline{1} \ 3 \ 5 \ \underline{2} \ 4 \ 6)\color{red}{(7 \ 10 \ 9 \ 8)}$$

$$(\underline{1} \ 4 \ 5 \ \underline{2} \ 3 \ 6)\color{red}{(7 \ 10 \ 9 \ 8)}$$

$$(\underline{1} \ 5 \ 5 \ \underline{2} \ 6 \ 6)\color{red}{(7 \ 10 \ 9 \ 8)}$$

$$(\underline{1} \ 6 \ 5 \ \underline{2} \ 5 \ 6)\color{red}{(7 \ 10 \ 9 \ 8)}$$

with the following building recipe :

• $$1$$ and $$2$$ must be separated by two elements,
• the same for $$3$$ and $$4$$,
• the same for $$5$$ and $$6$$.

(being understood that "separated" is by reference to a cyclic arrangement).

As the positions of $$1$$ and $$2$$ are "frozen", the choice is reduced to the $$2\times2\times2 = 8$$ solutions given upwards, to which we must add the exceptional first one.

Remarks :

a) about the necessity to gather $$1,2 \cdots 6$$ into a cycle :

• no solution can exist of the form $$(1 a b c)(* * )$$ for the reason seen upwards : we should have $$p=(1 a b 2)(* *)$$ but then $$p^3=(1 2 a b)(* *)$$ which is not what we desire.

• no solution can exist of the form $$(a b c d)(1 2)$$ for a similar reason.

b) As remarked by @Robert Shore, it is not evident that no other solution exists by grouping for example $$7$$ with other elements than $$8,9,10$$, even if our intimate conviction says that there none.

• It's not immediately obvious (at least to me) why there can be no solution that includes a cycle of the form $(7 x \ldots) \text{ with } x \neq 8, 9, 10$. I can see why with a little work but it's probably worth including the reasoning in the answer, particularly since such cycles would be possible if we were working within $S_{12}$ instead of $S_{10}$. – Robert Shore Feb 11 at 16:52
• You are right, it is not obvious. I am going to try to give an explanation. – Jean Marie Feb 11 at 16:57
• – Jean Marie Feb 13 at 23:32