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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?

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    $\begingroup$ Why this closing proposal ? This asker has done personal work on a question that is (almost) clearly settled. $\endgroup$ – Jean Marie Feb 11 at 0:07
  • $\begingroup$ @JeanMarie I have no idea and I have voted to leave it open. $\endgroup$ – José Carlos Santos Feb 11 at 8:53
  • $\begingroup$ I have taken the liberty to modify your first sentence and your title : this is for attracting ... and retaining readers later on. $\endgroup$ – Jean Marie Feb 11 at 12:03
  • $\begingroup$ @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 $\endgroup$ – Oscar Lanzi Feb 11 at 15:09
  • $\begingroup$ @OscarLanzi Because the question appeared to me in the close votes review queue. $\endgroup$ – José Carlos Santos Feb 11 at 15:40
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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.

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    $\begingroup$ 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 $S_{10}$. $\endgroup$ – Robert Shore Feb 11 at 16:52
  • $\begingroup$ You are right, it is not obvious. I am going to try to give an explanation. $\endgroup$ – Jean Marie Feb 11 at 16:57
  • $\begingroup$ Connected : math.stackexchange.com/q/475105 and math.stackexchange.com/q/978613 $\endgroup$ – Jean Marie Feb 13 at 23:32

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