# Showing equality of two groups.

Consider $S_n$, the permutation group. Let $a\in S_n$. I want to show that if $a$ is an $n$ or an $n-1$ cycle, then $\langle a\rangle = \{c \in S_n : ca=ac\}$. Any help will be veru much appreciated.

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I changed $<a>$ to $\langle a\rangle$. That's standard usage. – Michael Hardy Nov 14 '12 at 3:13
Hints: One of the two inclusions you need to prove should be pretty clear. For the other, note that $ca = ac$ iff $cac^{-1} = a$. If you know the cycle decomposition of $a$, there's a simple way to get the cycle decomposition of $cac^{-1}$. – Michael Joyce Nov 14 '12 at 3:16
Comment and answer erased since I read "...if $\,a\,$ is NOT an..." – DonAntonio Nov 14 '12 at 3:19
@Stefan: It's perfectly fine that things don't seem trivial yet ... but the hint is there to get you started while letting you discover what you need to know. For the first statement, ask yourself the following two questions: (1) If I know that a permutation $c$ is a power of $a$, i.e. $c = a^n$, do I know that $ca = ac$? (2) If I know that a permutation $c$ commutes with $a$, i.e. $ca = ac$, then do I know that $c$ is a power of $a$? By thinking through the definitions, you'll hopefully realize that one direction is significantly easier than the other... – Michael Joyce Nov 14 '12 at 3:27
...That's important when you are learning new material. In fact, parsing the definitions to grasp what is really being asked is by far the most important first step is tackling problems where you are stuck. – Michael Joyce Nov 14 '12 at 3:29

Define an action $\,S_n\times S_n\to S_n\,$ by conjugation. Since there are $\,(n-1)!\,$ different $\,n-\,$cycles , we get that if $\,a\,$ is an $\,n-\,$ cycle then

$$(n-1)!=|\mathcal Orb(a)|=[S_n:C_{S_n}(a)]=\frac{|S_n|}{|C_{S_n}(a)|}\Longrightarrow |C_{S_n}{a}|=n$$

and since clearly every power of $\,a\,$ commutes with $\,a\,$ , these powers thus are the only elements in $\,S_n\,$ that do so.

Added on request of the OP: There are $\,(n-2)!\,n\,$ different $\,(n-1)-\,$ cycles in $\,S_n\,$ , so if $\,a\,$ is such a cycle:

$$|\mathcal Orb(a)|=[S_n:C_{S_n}(a)]\Longrightarrow |C_{S_n}(a)|=\frac{n!}{(n-2)!n}=n-1$$

and again we have that the only elements that commute with this cycle are its $\,n-1\,$ powers.

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While this solution is correct, I think it would be preferable to avoid the orbit-stabilizer theorem insofar as one can perform a direct computation. In a first course in group theory, one often learns how to compute the centralizer of an element before one learns about the orbit-stabilizer theorem. – Michael Joyce Nov 14 '12 at 3:31
Perhaps you're right, yet I think groups actions are so basic and important that it is worth knowing this stuff early in the course and, at least where I studied, it was taught before we went into depth in permutation groups. – DonAntonio Nov 14 '12 at 3:48
Would the same argument be applicable for $(n-1)$cycles? – user44069 Nov 14 '12 at 5:52
Yes @Stefan . I added something to my answer. – DonAntonio Nov 14 '12 at 12:23