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I tried to do this example on my own. My solution was as follows:

  1. First of all we have $6$ disjoint cycles so we can order them in $6!$ ways without changing $x$

  2. Next the cycle $(1\space 2\space 3)$ can be there can be $3$ representatives ie $(1\space 2\space 3),(2\space 3\space 1),(3\space 1\space 2)$ such that it will be the same cycle. If we apply this further we get for $(7\space 8\space 9\space 10)$ , $4$ representatives and for $(4\space 5\space 6)$ $3$.

  3. The total amount would be then : $6! \times 3 \times 3 \times 4$

My answer is off by a lot. I also noticed that the writer only considers swapping the cycles with cycle type $3$ and the cycle type $1$, but why can we not swap all the $6$ cycles. Why do we not use the fact that these $6$ cycles are disjoint which means they are stable if we permute them?

SoG
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muhammed gunes
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1 Answers1

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Hints:

  1. Orbit-stabilizer theorem

  2. $\sigma, \tau\in S_n$ are conjugate iff $\sigma, \tau$ have same cycle structure.

  3. Two disjoint cycles commute.

  4. The formula for counting number of elements.

$\begin{align}\sigma&=(1\space 2\space 3)(4\space 5\space 6)(7\space 8\space 9\space 10)(11)(12)(13)\end{align}$

$\begin{align}|C_{\sigma}|&=\frac{|S_{13}|}{|\mathcal{cl}_{\sigma}|}\\&=\frac{13!}{\frac{13!}{3^22!4^11!1^33!}}\\&=432\end{align}$

SoG
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  • I think I understand it now, it was a misunderstanding. The only $g\in S_{13}$ which can stabilise (78910) are 4 and they are just the 4 representatives of this 4 cycle if and if we had another cycle we could use the same argument, but in this case we just have one 4-cycle and using the fact that disjoint cycles commute is for another question which is in how many ways can we write this permutation such that it will have the same effect on our set then it would have been 6!x3x4, but now we look for elements in $S_{13}$ which stabilise this element and we use conjugation. – muhammed gunes May 05 '23 at 13:57