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I need to find the centralizer of the permutations $\begin{pmatrix} 1 & 2 & \cdots & n \end{pmatrix}$ and $\begin{pmatrix} 1 & 2 & \cdots & n-1\end{pmatrix}$ in $S_{n}$. Now, perhaps I'm confusing it with the center, but I thought that individual elements didn't have centralizers. I thought that only sets of elements did. The definition I have in my class notes is the following:

Let $G$ be a group and $Y$ any set of elements of $G$. The centralizer $C_{G}(Y)$ is the set of elements of $G$ which commute with all $y \in Y$.

So, is this problem asking me to list which elements of $S_{n}$ commute with each other? I remember learning that any permutation can be written as a product of disjoint cycles and disjoint cycles commute, so is it true that two permutations commute iff they share no cycles in common?

If so, how would I prove this (conjugacy classes, perhaps?)? And if not, or if this statement is incorrect, how would I go about finding the centralizers of these two permutations in $S_{n}$?

Thank you.

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  • $\begingroup$ I think the question wants you to compute $C_{S_n}(\{\sigma_1\})$ and $C_{S_n}(\{\sigma_2\})$ separately. Unfortunately, disjointness is not necessary: elements in $\langle \sigma \rangle$ commute with $\sigma$ but are rarely disjoint with one another. Do you have the orbit-stabilizer theorem? $\endgroup$
    – pjs36
    Jan 12 '17 at 20:38
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    $\begingroup$ @pjs36 that's in the next chapter, so that's off limits for me right now. $\endgroup$
    – user100463
    Jan 12 '17 at 20:54
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I will assume permutations act on the right. Throughout $i,j$ will refer to any integers with $1\le i,j\le n$.

Let $\sigma,\omega\in S_n$. Denote $\sigma^\omega=\omega^{-1}\sigma\omega$ and notice $(i^\omega)^{(\sigma^\omega)}=(i^\sigma)^\omega$ so if, for example, $\sigma=(x_1,\ldots,x_r)$ is a cycle then $\sigma^\omega=(x_1^\omega,\ldots,x_r^\omega)$.

Write $\sigma_1=(1,2,\ldots,n)$ and $\sigma_2=(1,2,\ldots, n-1)$. I will answer both interpretations of the question, so I will calculate $C_{S_n}(\sigma_1)$, $C_{S_n}(\sigma_2)$ and $C_{S_n}(\{\sigma_1,\sigma_2\})=C_{S_n}(\sigma_1)\cap C_{S_n}(\sigma_2)$.

Suppose $\omega\in C_{S_n}(\sigma_1)$, so $\sigma_1^\omega=\sigma_1$. From the above, $\sigma_1=\sigma_1^\omega=(1^\omega,\ldots,n^\omega)$. Say $1\le i\le n-1$, then $i^{\sigma_1}=i+1$ and $n^{\sigma_1}=1$. In particular if $i^\omega=j$ then (modulo $n$) $(i+1)^\omega=j+1=i^\omega+1$. Hence $\omega\in\langle \sigma_1\rangle$, but $C_{S_n}(\sigma_1)\supseteq\langle \sigma_1\rangle$ so $C_{S_n}(\sigma_1)=\langle \sigma_1\rangle$.

Suppose $\omega\in C_{S_n}(\sigma_2)$, then by the above $(n^\omega)^{\sigma_2}=(n^\omega)^{(\sigma_2^\omega)}=(n^{\sigma_2})^\omega=n^\omega$. The only fixed point of $\sigma_2$ is $n$ so $n^\omega=n$. Therefore $\omega\in S_{n-1}$ so we are reduced to the $\sigma_1$ case (but for $n-1$) so $C_{S_n}(\sigma_2)=\langle \sigma_2\rangle$.

Finally we have $C_{S_n}(\{\sigma_1,\sigma_2\})=C_{S_n}(\sigma_1)\cap C_{S_n}(\sigma_2)=\{1_{S_n}\}$

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  • $\begingroup$ would you mind explaining some of your notation? In particular, what is $i$ here (identity permutation?) ? What then does $(i^{\omega})^{(\sigma^{\omega})}$ mean? And raising things to the $\omega$ power? I'm wondering if the parentheses $( )$ mean "the order of", but I'm not sure. So, if you would mind clarifying what these things mean, I would very much appreciate it. $\endgroup$
    – user100463
    Jan 12 '17 at 21:21
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    $\begingroup$ @RobertChamberlain ouch! I just accidentally stabbed my arm with a pencil reaching over to accept this answer! It was worth it though...thank you for this answer. It's extremely helpful! $\endgroup$
    – user100463
    Jan 13 '17 at 16:58
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    $\begingroup$ Best response I have ever had. $\endgroup$ Jan 13 '17 at 16:59
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    $\begingroup$ @RobertChamberlain one thing though. All of the $\sigma$'s in the part about $\sigma_{2}$ should be $\sigma_{2}$'s, right? $\endgroup$
    – user100463
    Jan 13 '17 at 19:14
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    $\begingroup$ @RobertChamberlain also, by $\{1\}$ did you mean the identity permutation? $\endgroup$
    – user100463
    Jan 13 '17 at 19:23
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Look at your definition of centralizer, and note that you were given a set of elements containing just two permutations. So you have to find the set of all elements of $S_n$ that commute both with the cycle $A\equiv (1\to2\to \cdots \to n\to 1)$ and the permutation $B\equiv(1\to2\to \cdots \to [n-1] \to 1),(n\to n)$.

This group might not be the trivial group containing only the identity. But it might be.

You could interpret the question as two separate questions, finding the centralizer of the set contaning just one element $A$, and the centralizer of the set contaning just one element $B$ -- but that would be two questions.

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  • $\begingroup$ well the question specifically said "centralizers" plural. $\endgroup$
    – user100463
    Jan 12 '17 at 20:55
  • $\begingroup$ @JessyunBourne Plural, because two permutations are given. For each its centralizer. $\endgroup$
    – ahulpke
    Jan 12 '17 at 21:17

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