791: For Group and Element, if There Is Positive Natural Number to Power of Which Element Is 1 and There Is No Smaller Such, Subgroup Generated by Element Consists of Element to Non-Negative Powers Smaller Than Number

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description/proof of that for group and element, if there is positive natural number to power of which element is 1 and there is no smaller such, subgroup generated by element consists of element to non-negative powers smaller than number

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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 and any element, if there is a positive natural number to power of which the element is 1 and there is no smaller such, the subgroup generated by the element consists of the element to the non-negative powers smaller than the number.

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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:
$\mathrm{\exists }n\in \mathbb{N}\setminus \{0\}(g^n=1\wedge \mathrm{\neg }\mathrm{\exists }{n}^{\prime }\in \mathbb{N}\setminus \{0\}(n^{\prime }<n\wedge g^{n^{\prime }}=1))$
$⟹$
$(g)=\{1,g,...,g^{n-1}\}$
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2: Natural Language Description

For any group, $G$, and any element, $g\in G$, if there is an $n\in \mathbb{N}\setminus \{0\}$ such that $g^n=1$ and there is no $n^{\prime }\in \mathbb{N}\setminus \{0\}$ such that $n^{\prime }<n\wedge g^{n^{\prime }}=1$, $(g)=\{1,g,...,g^{n-1}\}$.

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3: Proof

Whole Strategy: Step 1: see that $(g)=\{1,g,g^{-1},g^2,g^{-2},...\}$; Step 2: see that $\{1,g,...,g^{n-1}\}$ is distinct; Step 3: see that $(g)=\{1,g,...,g^{n-1}\}$.

Step 1:

Let us suppose that $\mathrm{\exists }n\in \mathbb{N}\setminus \{0\}(g^n=1\wedge \mathrm{\neg }\mathrm{\exists }{n}^{\prime }\in \mathbb{N}\setminus \{0\}(n^{\prime }<n\wedge g^{n^{\prime }}=1))$.

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, $g^{k_1}...g^{k_l}=g^{k_1+...+k_l}$.

Step 2:

Let us see that $\{1,g,...,g^{n-1}\}$ is distinct.

Let us suppose that $\{1,g,...,g^{n-1}\}$ was not distinct, which would mean that there were some $k,l\in \mathbb{N}$ such that $0\le k<l\le n-1$ and $g^k=g^l$.

$1=g^k{g}^{-k}=g^l{g}^{-k}=g^{l-k}$, where $0<l-k\le n-1$, a contradiction.

So, $\{1,g,...,g^{n-1}\}$ is distinct.

Step 3:

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

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

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

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

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References

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