794: For Group and Element, if There Is Positive Natural Number to Power of Which Element Is 1 and There Is No Smaller Such, Integers of Which Powers to Which Element Are 1 Are Only Multiples of 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, integers of which powers to which element are 1 are only multiples of 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
• 4: Note

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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 integers of which powers to which the element are 1 are only the multiples of 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:
$\mathrm{\exists }n\in \mathbb{N}\setminus \{0\}(g^n=1\wedge \mathrm{\neg }\mathrm{\exists }{n}^{\prime }\in \mathbb{N}\setminus \{0\}(n^{\prime }<n\wedge g^{n^{\prime }}=1))$
$⟹$
$nz$ s where $z\in \mathbb{Z}$ are the only integers such that $g^{nz}=1$
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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 $n^{\prime }<n\wedge g^{n^{\prime }}=1$, $nz$ s where $z\in \mathbb{Z}$ are the only integers such that $g^{nz}=1$.

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3: Proof

Whole Strategy: Step 1: suppose that $g^{nz+k}=1$ where $0\le k<n$ and see that $k=0$.

Step 1:

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

Let us suppose that $g^{nz+k}=1$ where $0\le k<n$.

$g^{nz+k}=g^{nz}{g}^k=(g^n{)}^z{g}^k=1^z{g}^k=g^k=1$.

By the supposition, $k=0$.

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4: Note

As an immediate corollary, when $g\ne 1$ and $g^p=1$ where $p\in \{\text{ the prime numbers }\}$, there is no $n\in \mathbb{N}\setminus \{0\}$ such that $n<p\wedge g^n=1$, because otherwise, there would be the smallest $n^{\prime }\in \mathbb{N}\setminus \{0\}$ such that $n^{\prime }\le n\wedge g^{n^{\prime }}=1$ ($n^{\prime }$ might be $n$), and $n^{\prime }z$ would be the only integers such that $g^{n^{\prime }z}=1$, which contradicts $g^p=1$, because $p$ could not be $n^{\prime }z$: $n^{\prime }\ne 1$.

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References

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