714: For n-Symmetric Group and n-Cycle, Centralizer of Cycle on Symmetric Group Is Cyclic Group by Cycle

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description/proof of that for n-symmetric group and n-cycle, centralizer of cycle on symmetric group is cyclic group by cycle

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Topics

About: group

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Natural Language Description
• 3: Proof

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Starting Context

• The reader knows a definition of n-symmetric group.
• The reader knows a definition of m-cycle on n-symmetric group.
• The reader knows a definition of centralizer of element on group.
• The reader knows a definition of cyclic group by element.

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Target Context

• The reader will have a description and a proof of the proposition that for any n-symmetric group and any n-cycle, the centralizer of the cycle on the symmetric group is the cyclic group by the cycle.

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Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

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Main Body

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1: Structured Description

Here is the rules of Structured Description.

Entities:
$S_n$: $=\text{ the }n\text{ -symmetric group }$
$p$: $=({\sigma }_1,...,{\sigma }_n)$, $\in \{\text{ the }n\text{ -cycles on }{S}_n\}$
$C_G(p)$: $=\text{ the centralizer of }p\text{ on }G$
$⟨p⟩$: $=\text{ the cyclic group by }p$
//

Statements:
$C_G(p)=⟨p⟩$
//

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2: Natural Language Description

For the $n$-symmetric group, $S_n$, any $n$-cycle on $S_n$, $p=({\sigma }_1,...,{\sigma }_n)$, the centralizer of $p$ on $G$, $C_G(p)$, and the cyclic group by $p$, $⟨p⟩$, $C_G(p)=⟨p⟩$.

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3: Proof

Whole Strategy: Step 1: see that for each element of $C_G(p)$, the element is the identity element or does not preserve any element of $\{1,...,n\}$; see that the element that does not preserve any element of $\{1,...,n\}$ shifts $\{{\sigma }_1,...,{\sigma }_n\}$ by a constant; Step 3: see that $⟨p⟩\subseteq C_G(p)$.

Let $[j]$ denote the residue of $j$ divided by $n$ with $0$ replaced by $n$.

So, $1\le [j]\le n$.

For example, when $n=2$, $[1]=1,[2]=2,[3]=1,...$.

The purpose is to avoid having to say like "when $j+k\le n$, $j+k$, but otherwise, $j+k-nm$ such that $1\le j+k-nm\le n$"; instead, we can just say $[j+k]$.

Someone may want to say that that is just a modulo $n$, and in idea, that is so, but strictly speaking, the modulo of $1$ by $2$ is not $1$ but the equivalence class of $1$, and so, we do not say so.

Step 1:

Let us see that for each element of $C_G(p)$, the element is the identity element or does not preserve any element of $\{1,...,n\}$.

Let us suppose that a $p^{\prime }\in C_G(p)$ preserves ${\sigma }_j$.

$p^{\prime }p{p}^{\prime -1}=p$, so, $p^{\prime }p=p{p}^{\prime }$.

$p^{\prime }p({\sigma }_j)=p^{\prime }({\sigma }_{[j+1]})=p{p}^{\prime }({\sigma }_j)=p({\sigma }_j)={\sigma }_{[j+1]}$. That means that $p^{\prime }$ preserves also ${\sigma }_{[j+1]}$. Then, $p^{\prime }$ preserves also ${\sigma }_{[j+2]},...,{\sigma }_{[j+n-1]}$, which means that $p^{\prime }$ preserves all the elements of $\{{\sigma }_1,...,{\sigma }_n\}$, so, $p^{\prime }=id$.

Step 2:

Let us $p^{\prime }\in C_G(p)$ be any that does not preserve any element of $\{1,...,n\}$.

$p^{\prime }p({\sigma }_j)=p^{\prime }({\sigma }_{[j+1]})={\sigma }_{[j+1+k]}$ for a $k\in \mathbb{N}\setminus \{0\}$ such that $1\le k<n$.

$p{p}^{\prime }({\sigma }_j)=p({\sigma }_{[j+l]})={\sigma }_{[j+l+1]}$ for an $l\in \mathbb{N}\setminus \{0\}$ such that $1\le l<n$.

${\sigma }_{[j+1+k]}={\sigma }_{[j+l+1]}$, which implies that $k=l$. So, $p^{\prime }$ shifts ${\sigma }_{[j+1]}$ and ${\sigma }_j$ by $k$.

Likewise, $p^{\prime }$ shifts ${\sigma }_{[j+2]}$ and ${\sigma }_{[j+1]}$ by $k^{\prime }$, but $k^{\prime }=k$, because ${\sigma }_{[j+1]}$ is already known to be shifted by $k$.

So, $p^{\prime }$ shifts $\{{\sigma }_j,...,{\sigma }_{[j+n-1]}\}=\{{\sigma }_1,...,{\sigma }_n\}$ by $k$.

That means that $p^{\prime }=({\sigma }_1,...,{\sigma }_n{)}^k$.

So, $C_G(p)\subseteq ⟨p⟩$.

Step 3:

Let us see that $⟨p⟩\subseteq C_G(p)$.

$p^{k}p(p^k{)}^{-1}=p^{k}p{p}^{-k}=p^{k+1}{p}^{-k}=p$.

So, $C_G(p)=⟨p⟩$.

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References

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