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I am a beginner, so my notations could be wrong. Consider in disjoint cycle notation:

$ X = (1)(2, 3, 16, 8, 10, 5, 14, 4, 12, 13)(6, 9)(7)(11, 15) $,

$Y = (1, 10, 6, 12, 8, 9, 3)(2, 14, 16, 11, 13, 15, 4, 7) (5)$,

$Z = (1, 3, 14, 15, 16, 12, 11, 13, 7, 4, 6, 8)(2, 9, 10, 5)$.

It can be cross-checked that $X(\lambda) = Y(Z(\lambda))$ $\forall \lambda$. My question is, given $X$, how can we find such $Y$ and $Z$? I found this example after a lot of trial-and-error.

hola
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  • Disjoint cycle notation does not have (i) commas between disjoint cycles; nor (ii) parentheses surrounding all the disjoint cycles. You would normally write, e.g., $Z=(1,11,15,4,16)(2,6,9,8,10,5,7,13)(3)(12,14)$ (assuming you included trivial cycles, which are often elided). – Arturo Magidin Aug 24 '23 at 20:29
  • $X$ is the identity permutation; there are lots of possible values for $X$ and $Y$. In general, given any permutation $\sigma$, there are lots (in fact, if you are looking at permutations of $n$ elements, there are $n!$) of possible permutation $Y$ and $Z$ such that $\sigma=YZ$. Is your question how to find the inverse of $Y$? – Arturo Magidin Aug 24 '23 at 20:30
  • In your (pre-edit) example, since $X$ is the identity permutation, $Y$ and $Z$ must be inverses of each other, and (as you must have noticed) the inverse of a permutation is given by reversing the cycles. Then for arbitrary $X$ and $Y$, the solution to $YZ=X$ is $Z=Y^{-1}X$. Permutations with composition form a group called the symmetric group, which means we can think of permutation composition as a sort of invertible multiplication operation. – Karl Aug 24 '23 at 20:36
  • @ArturoMagidin Thanks. I updated it as per your suggestion – hola Aug 24 '23 at 20:38
  • @Karl I updated to a non-trivial example. I understand there are lots of solutions, just wanted to find an algorithm to generate those. – hola Aug 24 '23 at 20:40
  • @ArturoMagidin About your second comment, I updated to a non-trivial example. Identity permutation as the example was a not a good idea. – hola Aug 24 '23 at 20:41
  • See the second sentence of my comment. – Karl Aug 24 '23 at 20:42
  • You did not read what I wrote in full: given any permutation $X$ on $n$ letters, there are $n!$ pairs of permutations $Y$ and $Z$ such that $X=YZ$. In fact, for any choice of permutation $Y$, you can find a permutation $Z$ such that $X=YZ$, and given any permutation $Z$ you can find a $Y$ such that $X=YZ$. – Arturo Magidin Aug 24 '23 at 20:42
  • @ArturoMagidin So, "generate $Z$ at random and then compute $Y$"? – hola Aug 24 '23 at 20:44
  • Pick $Z$ any way you want, and you can compute a corresponding $Y$. Pick any $Y$ in any way you want, and you can compute a corresponding $Z$. Any two of $X$, $Y$, and $Z$ uniquely determine the third in the equation $X=YZ$. And any one of $X$, $Y$, and $Z$ will not determine the other two. – Arturo Magidin Aug 24 '23 at 20:45

2 Answers2

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In any group $G$, (including the group $S_n$ of permutations of $n$ elements), for any $a,b\in G$ we have that the equations $ax=b$ and $ya=b$ each have a unique solution. Namely, $x=a^{-1}b$ and $y=ba^{-1}$.

That means that given a permutation $X\in S_n$, you can always pick any permutation $Y\in S_n$, and find a permutation $Z\in S_n$ such that $X=YZ$. How? By computing $Z=Y^{-1}X$.

And if you pick any permutation $Z$, then you can always find a permutation $Y\in S_n$ such that $X=YZ$. How? By computing $Y=XZ^{-1}$.

How to find $Y^{-1}$ given an expression for $Y$ in disjoint cycle notation? For a single cycle $(a_1,a_2,\ldots,a_r)$, the inverse if given by the cycle $(a_r,a_{r-1},\ldots,a_2,a_1)$. And for a product of disjoint cycles, the inverse is given by taking the product of the inverses of each cycle. So if $(a_1,\ldots,a_r)$ and $(b_1,\ldots,b_t)$ are disjoint, then the inverse of $(a_1,\ldots,a_r)(b_1,\ldots,b_t)$ is $(a_r,\ldots,a_1)(b_t,\ldots,b_1)$. Etc.

Arturo Magidin
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Here $X = \operatorname{id}$ so the usual problem would be more:
Find $Z$ such that $Z\circ Y = \operatorname{id}$.

This is just finding the inverse (if you look carefully Z and Y differ only from the change of direction of the cycles).

Finding inverse is not so hard, if $Z(i) = j$, $Y(j) = I$. Or just reverse the direction of cycles.

julio_es_sui_glace
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  • Thanks. I actually wanted for a general permutation (not just identity), please see the updated example. – hola Aug 24 '23 at 20:43