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In the symmetric group $S_{13}$ how do I find all the permutations $\tau$ that: $$\tau\alpha\tau^{-1}=\gamma$$ Where $\alpha = (1,2,4,8)(3,5,7,9,11,13)(6,12)$ and $\gamma = (1,2,3,4)(5,6,7,8,9,10,11)(12,13)$

I thought firstly that all the permutations are standing that condition, because $o(\alpha)=o(\gamma)=12$ and that's why $\tau^{12}\alpha^{12}\tau^{{12}^{-1}}=\gamma^{12}$ and that's true for any $\tau$. But there is a smaller number which is the answer. Correct me what wrong in my last conclusion and how to find the exact number of such permutations ?

Generally, I just know, how to calculate conjugacy class in the whole symmetric group (the number of k-permutations with the same structure) and the conjugacy class of permutation in sub-group of $S_{13}$. I never knew that there is a method to calculate how much permutations are a conjugacy to specific permutation, if you understand, what I mean (like in this example).

Martin Sleziak
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Ilya.K.
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  • see Joseph Rotman's Introduction to Group Theory – Adelafif Oct 12 '15 at 14:52
  • Maybe answers to this question might help you: Why are two permutations conjugate iff they have the same cycle structure? – Martin Sleziak Oct 12 '15 at 14:56
  • Note that you want, for example $\tau(1,2,4,8)\tau^{-1}=(1,2,3,4)$. Similarly for other cycles. – Martin Sleziak Oct 12 '15 at 15:00
  • Ok, nope, that didn't give me an idea. I stil don't see how can I know how much permutations there is that transform one permutation to another by congacy. – Ilya.K. Oct 12 '15 at 15:07
  • So $(\tau(1),\tau(2),\tau(3),\tau(8))=(1,2,3,4)$. Now it is combinatorics, and my knowledge isn't so good, but as I see, it is 4! of that permutations, isn't it? And all of them will be $6!4!2!$ from that, but the right answer is 48. How do I get this number of such permutations? – Ilya.K. Oct 12 '15 at 15:15
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    Keeping in mind @Martin Sleziak's (true) remark, are you sure the question was rendered correctly? Note that $(3,5,7,9,11,13)$ is a cycle of length 6, while $(5,6,7,8,9,10,11)$ has length 7. – Jacopo Stifani Apr 03 '16 at 07:48
  • Observe that $\alpha = (1,2,4,8)(3,5,7,9,11,13)(6,12)$ is an odd permutation while $\gamma = (1,2,3,4)(5,6,7,8,9,10,11)(12,13)$ is an even permutation. Recall that one cannot arrive at an even permutation by conjugating an odd permutation. Did you mean $\alpha = (1,2,4,8)(3,5,7,9,10,11,13)(6,12)$? – Frenzy Li Apr 07 '17 at 06:16

1 Answers1

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Assumption: $\alpha = (1,2,4,8)(3,5,7,9,11,13)(6,12)$ and $\gamma = (1,2,3,4)(5,6,7,8,9,11)(12,13)$ where we changed the cycle structure of $\gamma$ such that both $\alpha$ and $\gamma$ are odd permutations. Then,
$$\text{Number of $\tau$ that conjugates $\alpha$ to $\gamma$} = \frac{\text{Order of group $S_{13}$}}{\text{Size of conjugacy class for element $\alpha$}}.$$

(I think such a relation holds. I need to learn more group theory to know why this indeed holds.)

The order of group $S_{13}$ is $13! = 6,227,020,800$ and the next critical step is to calculate the size of conjugacy class containing element $\alpha$. This requires some combinatorics.

How to find the number of permutations that contain 3 disjoint cycles of length 4, 6 and 2?

$$\left[\binom{13}{4}(4-1)!\right]\cdot \left[\binom{13-4}{6}(6-1)!\right]\cdot \left[\binom{13-4-6}{2}(2-1)!\right] = 129,729,600.$$

The process is as follows:

  1. Take a combination of 4 distinct vertices out of the 13 available vertices.
  2. There are $(4-1)!$ ways to write a cycle of length 4 using these 4 vertices.
  3. Take a combination of 6 distinct vertices out of the remaining 9 vertices.
  4. There are $(6-1)!$ ways to write a cycle of length 6 using these 6 vertices.
  5. Take a combination of 2 distinct vertices out of the remaining 3 vertices.
  6. There are $(2-1)!$ ways to write a cycle of length 2 using these 2 vertices (which is only 1 way of doing so).

Therefore, given $\alpha$ and a certain $\gamma$ in the conjugation class of $\alpha$, the number of permutations $\tau$ that can specifically conjugate $\alpha$ to $\gamma$ is

$$\frac{13!}{\left[\binom{13}{4}3!\right]\cdot \left[\binom{9}{6}5!\right]\cdot \left[\binom{3}{2}1!\right]} = 48.$$

Frenzy Li
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