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Let $σ=(1234)∈S_6$. List all elements in $N(σ)={α∈S_6∣σα=ασ}$.

I know that this is related to cycles and orders, but I am having trouble finding the order of $N(σ)$ initially.

I also know that $σ$ has 8 different permutations.

And I also know that $e(1234)e^{-1}=(1234)$

And I do know that one is $α=(13)(24)$

I would just like some help getting to the rest of the elements in $N(σ)={α∈S_6∣σα=ασ}$.

Thank you!

user26857
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rachelh2012
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2 Answers2

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Are you familiar with a formula for conjugation of cycles in $S_n$ which goes like this?

$$\sigma \circ (a_1 \cdots a_p) \circ \sigma^{-1}=(\sigma(a_1) \cdots \sigma(a_p)).$$

This formula practically lays it down for you.

You may wish to do it otherwise, i.e. test that a permutation that commutes with your cycle satisfies $\{\sigma(5),\sigma(6)\}=\{5,6\}$ (this follows from $\sigma \circ c = c\circ \sigma$). Then you can play around with the remaining elements, and see what happens to them.

1) Let me elaborate on the identity above. Suppose that you have a wheel, which lists the numbers $1,2,3,4$ clockwise. Every time you apply the permutation $(1 2 3 4)$ it's like you spin them once clockwise, thus $1$ goes to $2$, $2$ goes to $3$, ... and $4$ goes to $1$. The numbers $5,6$ are left fixed (this is what the permutation given does).

2) Now, you have to make sense of the permutation $\sigma \circ (1 2 3 4)\circ \sigma^{-1}.$ Can you check that it sends $\sigma(1)$ to $\sigma(2)$, for instance? This is really important.

3) Let $1,2,3,4$ be considered as classes (modulo $4$) for ease of notation. We thus see that the numbers $(1,2,3,4)$ and $(\sigma(1), \sigma(2), \sigma(3), \sigma(4))$ are equal as cycles, i.e. their wheels are equal (they are equal modulo cyclic order). Therefore, there exists a number $k$ modulo $4$ such that:

$$\sigma(i)\equiv i+k (\mbox{mod }4).$$ What happens to the values $\sigma(5), \sigma(6)$? Well, neither can take a value from $1$ to $4$, so $\{\sigma(5), \sigma(6)\}=\{5, 6\}.$

Let us put all of this together! $\sigma$ acts on $1,2,3,4$ as a power of our starting cycle $(1 2 3 4)$, and it acts on $5,6$ as a power of the transposition $(5 6).$ These two objects commute with each other, since they permute disjoint subsets of the total set they act on.

4) The total answer is then: $\sigma$ commutes with our cycle if and only if $\sigma=(1 2 3 4)^i (5 6)^j.$ The index $i$ runs between $0$ and $4-1$, and the index $j$ runs between $0$ and $2-1.$ Therefore we have $4\ldotp 2=8$ permutations that commute with our cycle.

There is a link that may be useful, as it deals with the same question. Why $c(a_1 \ a_2 \dots \ a_k)c^{-1}=(c(a_1) c(a_2)… c(a_k))$?

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Theon Alexander
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  • Sorry...I don't know what this is. I'm kind of behind in the class and am really struggling – rachelh2012 Mar 28 '16 at 15:27
  • I just added a bit more. It is key that you learn step by step to operate, so I suggest you really try to complete the steps suggested, and to ask questions when you get stuck. I'll fill in a bit more later. – Theon Alexander Mar 28 '16 at 15:35
  • Here is a link that may help you, but we'll come to this again later. http://math.stackexchange.com/questions/48134/why-are-two-permutations-conjugate-iff-they-have-the-same-cycle-structure – Theon Alexander Mar 28 '16 at 15:37
  • I think I kind of understand now. I also found this $\alpha(1;2;3;4)\alpha^{-1} = (\alpha(1);\alpha(2);\alpha(3);\alpha(4)) = (1;2;3;4)$ but Im not sure how to get Alpha I guess. Could you walk me through an example? – rachelh2012 Mar 28 '16 at 15:51
  • Also, would the order of $N(σ)$ be 8? – rachelh2012 Mar 28 '16 at 15:52
  • Would the elements be: { e,(12),(34),(12)(34),(1324),(13)(24),(1423),(14)(23) }, with e being (56)? – rachelh2012 Mar 28 '16 at 15:58
  • I will add the rest now, once you have spotted this important fact. Now, let us think the following. Where is $1$ in the cycle you have described? For instance, $\alpha(4)=1$, or $\alpha(3)=1$. You will find that each of those cases leads you to one of the permutations! The thing is, the numbers $1,2,3,4$ will appear in the wheel, but with a different order, and that's the whole key to the identity you've just shown! – Theon Alexander Mar 28 '16 at 17:09
  • You're right about the order (i.e. it is $8$); let's see why this is the right one now :) – Theon Alexander Mar 28 '16 at 17:09
  • You will see that the elements $(1 2), (3 4), (1 2) (3 4)$ are not powers of the cycle $(1 2 3 4)$ and thus they cannot be there. Check that you understand the process well so that you can list them all, say in the form given in step 4), and then you can rewrite them. – Theon Alexander Mar 28 '16 at 17:25
  • I corrected a typo. In reality, $\sigma(5), \sigma(6)$ as a set must go to ${5,6}.$ This is simply because $1$ to $4$ are 'busy enough mapping among themselves', as we have seen by visualising the wheels/cycles. – Theon Alexander Mar 28 '16 at 17:34
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In order to improve understanding of the answer, I would suggest you try with the cycle $(1 2 3 4 5 6)$ as a permutation in $S_{11}.$ Which ones are the permutations commuting with this cycle?

What would be the essential requirement, and what role do the numbers $1$ to $6$, and those bigger than $6$ have in this slightly bigger picture? Just go by the commutation condition, and you will see what it imposes on $\sigma$. You just need to look at that condition and add nothing extraneous to it!

We will get later to the different ways of writing (some) powers of a cycle.

Theon Alexander
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