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.
Showing equality of two groups.
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group-theory
group-actions
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0I changed $$ to $\langle a\rangle$. That's standard usage. – 2012-11-14
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1Hints: One of the two inclusions you need to prove should be pretty clear. For the other, note that $ca = ac$ iff $cac^{-1} = a$. If you know the cycle decomposition of $a$, there's a simple way to get the cycle decomposition of $cac^{-1}$. – 2012-11-14
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0Comment and answer erased since I read "...if $\,a\,$ is NOT an..." – 2012-11-14
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0@MichaelJoyce Would it be possible to elaborate a bit more. I am a beginner on this material and some things which might be trivial do not seem that aparent to me. – 2012-11-14
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1@Stefan: It's perfectly fine that things don't seem trivial yet ... but the hint is there to get you started while letting you discover what you need to know. For the first statement, ask yourself the following two questions: (1) If I know that a permutation $c$ is a power of $a$, i.e. $c = a^n$, do I know that $ca = ac$? (2) If I know that a permutation $c$ commutes with $a$, i.e. $ca = ac$, then do I know that $c$ is a power of $a$? By thinking through the definitions, you'll hopefully realize that one direction is significantly easier than the other... – 2012-11-14
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1...That's important when you are learning new material. In fact, parsing the definitions to grasp what is really being asked is by far the most important first step is tackling problems where you are stuck. – 2012-11-14
1 Answers
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Define an action $\,S_n\times S_n\to S_n\,$ by conjugation. Since there are $\,(n-1)!\,$ different $\,n-\,$cycles , we get that if $\,a\,$ is an $\,n-\,$ cycle then
$$(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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3While 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. – 2012-11-14
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1Perhaps 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. – 2012-11-14
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0Would the same argument be applicable for $(n-1)$cycles? – 2012-11-14
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0Yes @Stefan . I added something to my answer. – 2012-11-14