790: For Group and Finite-Order Element, Order Power of Element Is 1 and Subgroup Generated by Element Consists of Element to Non-Negative Powers Smaller Than Element Order

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description/proof of that for group and finite-order element, order power of element is $1$ and subgroup generated by element consists of element to non-negative powers smaller than element order

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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 order of element of group.
• The reader admits the proposition that for any group, the powers sequence of any element that returns back returns to the element.

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Target Context

• The reader will have a description and a proof of the proposition that for any group and its any finite-order element, the order power of the element is $1$ and the subgroup generated by the element consists of the element to the non-negative powers smaller than the element order.

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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:
$G$: $\in \{\text{ the groups }\}$
$g$: $\in G$
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Statements:
$|(g)|=n\in \mathbb{N}\setminus \{0\}$
$⟹$
$g^n=1\wedge (g)=\{1,...,g^{n-1}\}$
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2: Natural Language Description

For any group, $G$, and any element, $g\in G$, if the order of $g$, $|(g)|$, is a finite $n\in \mathbb{N}\setminus \{0\}$, $g^n=1\wedge (g)=\{1,...,g^{n-1}\}$.

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3: Proof

Whole Strategy: Step 1: see that $(g)=\{1,g,g^{-1},g^2,g^{-2},...\}$; Step 2: think of $(g,g^2,...)$ and see that the sequence returns to $g$; Step 3: see that $(g)=\{1,g,...,g^{n-1}\}$.

Step 1:

Let us suppose that $|(g)|=n\in \mathbb{N}\setminus \{0\}$.

As $(g)$ consists of all the finite multiplications of $g$ and $g^{-1}$ by the definition of subgroup generated by subset of group, $(g)=\{1,g,g^{-1},g^2,g^{-2},...\}$: whether $k_j\in \mathbb{Z}$ is positive, 0, or negative, $g^{k_1}...g^{k_l}=g^{k_1+...+k_l}$.

$|(g)|=n$ means that $\{1,g,g^{-1},g^2,g^{-2},...\}$ contains some duplications and has $n$ distinct elements.

Step 2:

Let us think of $(g,g^2,...)$.

There is an $m\in \mathbb{N}\setminus \{0\}$ such that $\{g,...,g^m\}$ is distinct and $g^{m+1}\in \{g,...,g^m\}$, because otherwise, $\{g,g^2...\}$ would be distinct, which would mean that $|(g)|$ was not finite.

In fact, $g^{m+1}=g$, by the proposition that for any group, the powers sequence of any element that returns back returns to the element.

That means that $g^m=1$, because $g^m=g^{m+1}{g}^{-1}=g{g}^{-1}=1$.

Let us think of $\{1=g^0=g^m,g=g^1,...,g^{m-1}\}$, which is distinct.

Step 3:

For each $k\in \mathbb{N}\setminus \{0\}$ such that $m<k$, $k=ml+j$, where $l\in \mathbb{N}$ such that $0\le l$ and $j\in \mathbb{N}$ such that $0\le j<m$. So, $g^k=g^{ml+j}=g^{ml}{g}^j=(g^m{)}^l{g}^j=1^l{g}^j=1g^j=g^j$. So, $g^k\in \{1,g,...,g^{m-1}\}$.

$g^{-1}=g^{m-1}$, because $g{g}^{m-1}=g^{m-1}g=g^m=1$. $g^{-m}=1$, because $g^{-m}=(g^m{)}^{-1}=1^{-1}=1$.

For each $k\in \mathbb{N}\setminus \{0\}$ such that $0<k$, $k=ml+j$, where $l\in \mathbb{N}$ such that $0\le l$ and $j\in \mathbb{N}$ such that $0\le j<m$. So, $g^{-k}=g^{-ml-j}=(g^{-m}{)}^l{g}^{-j}=(g^{-m}{)}^{l}1g^{-j}=(g^{-m}{)}^l{g}^m{g}^{-j}=1^l{g}^{m-j}=1g^{m-j}=g^{m-j}$. But $0<m-j\le m$. When $m-j=m$, $g^{-k}=g^m=1$. So, $g^{-k}\in \{1,g,...,g^{m-1}\}$.

So, $(g)=\{1,g,g^{-1},g^2,g^{-2},...\}=\{1,g,...,g^{m-1}\}$.

As $|(g)|=n$, in fact, $m=n$.

So, $(g)=\{1,g,...,g^{n-1}\}$.

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References

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