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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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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0Yes @Stefan . I added something to my answer. – 2012-11-14