863: For Permutations Group, Its Element, Element of Permutations Domain, and Sequence of Power Operations of Element on Domain Element, Another Sequence with Another Domain Element Not Contained in 1st Sequence Is Disjoint from 1st Sequence

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description/proof of that for permutations group, its element, element of permutations domain, and sequence of power operations of element on domain element, another sequence with another domain element not contained in 1st sequence is disjoint from 1st sequence

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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.
• The reader admits 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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Target Context

• The reader will have a description and a proof of the proposition that for any permutations group, its any element, any element of the permutations domain, and the sequence of the power operations of the element on the domain element, another sequence with any another domain element not contained in the 1st sequence is disjoint from the 1st sequence.

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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
$G$: $=\text{ the permutations group of }S$
$\sigma$: $\in G$
$p$: $\in S$
$(\sigma (p),{\sigma }^2(p),...)$:
$p^{\prime }$: $\in S\setminus \{\sigma (p),{\sigma }^2(p),...\}$
$(\sigma (p^{\prime }),{\sigma }^2(p^{\prime }),...)$:
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Statements:
$\{\sigma (p),{\sigma }^2(p),...\}\cap \{\sigma (p^{\prime }),{\sigma }^2(p^{\prime }),...\}=\mathrm{\varnothing }$
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2: Note

There may not be such any $p^{\prime }$; in that case, this proposition is vacuously true.

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3: Proof

Whole Strategy: Step 1: see that $(\sigma (p^{\prime }),{\sigma }^2(p^{\prime }),...)$ inevitably comes to $p^{\prime }$ as ${\sigma }^n(p^{\prime })$; Step 2: suppose that ${\sigma }^k(p)={\sigma }^l(p^{\prime })$ and find a contradiction.

Step 1:

$(\sigma (p^{\prime }),{\sigma }^2(p^{\prime }),...)$ inevitably comes to $p^{\prime }$ as ${\sigma }^n(p^{\prime })$, by 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.

Step 2:

Let us suppose that there were a $k\in \mathbb{N}$ such that $1\le k$ and a $l\in \mathbb{N}$ such that $1\le l$ such that ${\sigma }^k(p)={\sigma }^l(p^{\prime })$. We can suppose that $l\le n$, because the sequence is cyclic with the order, $n$.

${\sigma }^{k+n-l}(p)={\sigma }^{n-l}({\sigma }^k(p))={\sigma }^{n-l}({\sigma }^l(p^{\prime }))={\sigma }^n(p^{\prime })=p^{\prime }$, a contradiction against $p^{\prime }$'s not in the 1st sequence.

So, the intersection is empty.

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References

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