862: For Permutations Group, Its Element, and Element of Permutations Domain, Sequence of Power Operations of Element on Domain Element Returns Back from Domain Element

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description/proof of that for permutations group, its element, and element of permutations domain, sequence of power operations of element on domain element returns back from domain element

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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: Note
• 3: Proof

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Starting Context

• The reader knows a definition of group.

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Target Context

• The reader will have a description and a proof of the proposition that for any permutations group, its any element, and any element of the permutations domain, the sequence of the power operations of the element on the domain element returns back from the domain element.

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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:
$S$: $\in \{\text{ the finite sets }\}$, with any linear order
$G$: $=\text{ the permutations group of }S$
$\sigma$: $\in G$
$p$: $\in S$
$(\sigma (p),{\sigma }^2(p),...)$:
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Statements:
$\mathrm{\exists }n\in \mathbb{N}\text{ such that }1\le n(\{\sigma (p),...,{\sigma }^n(p)\}\text{ is distinct }\wedge {\sigma }^{n+1}(p)\in \{\sigma (p),...,{\sigma }^n(p)\})$
$\wedge$
${\sigma }^n(p)=p$
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Of course, we can insert $p$ at the beginning of the sequence as $(p,\sigma (p),...,{\sigma }^{n-1}(p),{\sigma }^n(p)=p,...)$

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2: Note

This proposition resembles the proposition that for any group, the powers sequence of any element that returns back returns to the element, but is not any immediate implication of it, because ${\sigma }^{n+1}(p)={\sigma }^k(p)$ does not necessarily mean ${\sigma }^{n+1}={\sigma }^k$, because for a $p^{\prime }\in S$ such that $p^{\prime }\ne p$, it may be that ${\sigma }^{n+1}(p^{\prime })\ne {\sigma }^k(p^{\prime })$. In fact, the sequence may return back earlier than $(\sigma ,{\sigma }^2,...)$ does. The logic of Proof is almost the same though.

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3: Proof

Whole Strategy: Step 1: see that eventually, ${\sigma }^{n+1}(p)\in \{\sigma (p),...,{\sigma }^n(p)\}$, and let ${\sigma }^{n+1}(p)={\sigma }^k(p)$; Step 2: see that $k=1$; Step 3: see that ${\sigma }^n(p)=p$.

Step 1:

As each term in the sequence is in $S$, which is finite, eventually, ${\sigma }^{n+1}(p)\in \{\sigma (p),...,{\sigma }^n(p)\}$, because there cannot be more than the cardinality of $S$ distinct terms.

There is a $k\in \mathbb{N}$ such that $1\le k\le n$ such that ${\sigma }^{n+1}(p)={\sigma }^k(p)$.

Step 2:

$0\le n-k$, so, $2\le n-k+2\le n-1+2=n+1$.

As ${\sigma }^{n+1}(p)={\sigma }^k(p)$, ${\sigma }^{n-k+2}(p)={\sigma }^{n+1-(k-1)}(p)={\sigma }^{-(k-1)}{\sigma }^{n+1}(p)={\sigma }^{-(k-1)}{\sigma }^k(p)={\sigma }^{k-(k-1)}(p)=\sigma (p)$. As $2\le n-k+2\le n+1$, $n-k+2=n+1$, because otherwise, $\{\sigma (p),...,{\sigma }^n(p)\}$ would not be distinct. That implies that $k=1$.

So, ${\sigma }^{n+1}(p)=\sigma (p)$.

Step 3:

But that implies that $\sigma ({\sigma }^n(p))=\sigma (p)$, and as $\sigma$ is bijective, ${\sigma }^n(p)=p$.

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References

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