Conjugate to the Permutation

Question

How many elements in $S_{12}$ are conjugate to the permutation $$\sigma=(6,2,4,8)(3,5,1)(10,11,7)(9,12)?$$ How many elements commute with $\sigma$ in $S_{12}$?

I believe I use the equation $n!/|K_{\sigma}|$ for the second question, but I'm not sure. Is anyone aware of how to do these?

Answer 1

Two facts that should help you on your way to solving this:

1. Two elements in $S_n$ are conjugate if and only if they have the same cycle type (in your case, the cycle type of $\sigma$ is $(4,3,2,2)$).
2. If $\sigma=(a_1,\ldots,a_k)$ is a $k$-cycle, and $\tau$ is any other element in $S_n$, then

$$\tau\circ\sigma\circ\tau^{-1}=(\tau(a_1),\ldots,\tau(a_k)).$$

Answer 2

You can use that, given $\tau\in S_{12}$, $$\tau^{-1}\sigma\tau = \left(\tau(6), \tau(2), \tau(4), \tau(8)\right)\left(\tau(3), \tau(5), \tau(1)\right)\left(\tau(10), \tau(11), \tau(7)\right)\left(\tau(9), \tau(12)\right)$$ How many distinct permutations can you make of this form?

For the second question, you want to know how many $\tau\in S_{12}$ commute with $\sigma$, that is, $\sigma^{-1}\tau^{-1}\sigma\tau=\operatorname{id}$. We can also write this $\tau^{-1}\sigma\tau = \sigma$. So, look at the form above, and decide what type of $\tau$ will give you back the same $\sigma$.

Answer 3

Two permutations are conjugate if they have the same cycle structure (same number of cycles with same lengths).

Given a permutation with this cycle structure, you won't change it if you rotate each cycle as much as you want.

So, there are $4$ "rotated cycles" for the first one, $3$ for the next two cycles, and $2$ for the last. So there are $4\times3\times3\times2$ equivalent permutations. But you can also swap the second and the third cycles, so there are actually $(4\times3\times3\times2)\times2=144$ equivalent permutations, for each given permutation.

So there are $12!/144= 3326400$ permutations conjugate to yours.

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A precision about what I mean by "rotated cycle". Since you have

$$\tau^{-1}\sigma\tau = \left(\tau(6), \tau(2), \tau(4), \tau(8)\right)\left(\tau(3), \tau(5), \tau(1)\right)\left(\tau(10), \tau(11), \tau(7)\right)\left(\tau(9), \tau(12)\right)$$

A priori all permutation $\tau \in S_n$ would give you a conjugate of $\sigma$ by this operation, but there are double counts.

If, for the first cycle, the image of $(\tau(6), \tau(2), \tau(4), \tau(8))$ is, say, $(1,2,3,4)$, then you will get the same cycle if instead the image is $(2,3,4,1)$, $(3,4,1,2)$ or $(4,1,2,3)$. So each given conjugate can be associated with $4$ permutations $\tau$ that give the same cycle $(1,2,3,4)$. But you have to consider also the other cycles. And you need also to consider that the two 3-cycles may be switched.