789: For Group, Powers Sequence of Element That Returns Back Returns to Element

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description/proof of that for group, powers sequence of element that returns back returns to 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: Natural Language Description
• 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 group, the powers sequence of any element that returns back returns to the 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:
$G$: $\in \{\text{ the groups }\}$
$p$: $\in G$
$(p,p^2,...)$:
//

Statements:
$\mathrm{\exists }n\in \mathbb{N}\setminus \{0\}(\{p,...,p^n\}\text{ is distinct }\wedge p^{n+1}\in \{p,...,p^n\})$
$⟹$
$p^{n+1}=p$
//

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2: Natural Language Description

For any group, $G$, any element, $p\in G$, and the powers sequence, $(p,p^2,...)$, if there is an $n\in \mathbb{N}\setminus \{0\}$ such that $\{p,...,p^n\}$ is distinct and $p^{n+1}\in \{p,...,p^n\}$, $p^{n+1}=p$.

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3: Proof

Whole Strategy: Step 1: let $p^{n+1}=p^k$ where $1\le k\le n$ and see that $k=1$.

Step 1:

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

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

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

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References

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