# Find the elements of $S_6$ that commute with $(1234)$ [duplicate]

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!

• Does it suffice to find all the transpositions that commute with it? Not sure, actually curious. – Vik78 Mar 28 '16 at 3:51
• Someone asked the same question with the exact same title 2 days ago. Please use the search feature. – Richard D. James Mar 28 '16 at 17:29

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 $c(a_1 \ a_2 \dots \ a_k)c^{-1}=(c(a_1) c(a_2)… c(a_k))$?

• 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. math.stackexchange.com/questions/48134/… – 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

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.