844: For Group, Subgroup, and Element of Group, if k Is 1st Positive Power to Which Element Belongs to Subgroup, Multiples of k Are Only Powers to Which Element Belongs to Subgroup

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description/proof of that for group, subgroup, and element of group, if $k$ is 1st positive power to which element belongs to subgroup, multiples of $k$ are only powers to which element belongs to subgroup

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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: 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, any subgroup, and any element of the group, if $k$ is the 1st positive power to which the element belongs to the subgroup, the multiples of $k$ are the only powers to which the element belongs to the subgroup.

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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:
$G^{\prime }$: $\in \{\text{ the groups }\}$
$G$: $\in \{\text{ the subgroups of }{G}^{\prime }\}$
$g^{\prime }$: $\in G^{\prime }$
$k$: $\in \mathbb{N}\setminus \{0\}$ such that $g^{\prime k}\in G\wedge \mathrm{\forall }j\in \mathbb{N}\setminus \{0\}\text{ such that }j<k(g^{\prime j}\notin G)$
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Statements:
$\mathrm{\forall }l\in \mathbb{Z},\mathrm{\forall }j\in \mathbb{N}\text{ such that }0\le j<k(g^{\prime lk+j}\in G⟺j=0)$
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2: Proof

Whole Strategy: Step 1: let $m=lk+j$ and suppose that $g^{\prime m}\in G$, and see that $j=0$; Step 2: let $m=lk+j$ and suppose that $j=0$, and see that $g^{\prime m}\in G$.

Step 1:

Let $m=lk+j$.

Note that $m$ s cover all the integers.

Let us suppose that $g^{\prime m}\in G$.

$g^{\prime m}=g^{\prime lk+j}=g^{\prime lk}{g}^{\prime j}=(g^{\prime k}{)}^l{g}^{\prime j}$. So, $g^{\prime j}=(g^{\prime k}{)}^{-l}(g^{\prime k}{)}^l{g}^{\prime j}=(g^{\prime k}{)}^{-l}{g}^{\prime m}$. But as $g^{\prime k}\in G$, $(g^{\prime k}{)}^{-l}\in G$, and also as $g^{\prime m}\in G$, $g^{\prime j}=(g^{\prime k}{)}^{-l}{g}^{\prime m}\in G$.

As $0\le j<k$, by the supposition of this proposition, $j=0$.

Step 2:

Let $m=lk+j$.

Let us suppose that $j=0$.

$g^{\prime m}=g^{\prime lk+j}=g^{\prime lk}=(g^{\prime k}{)}^l\in G$ as $g^{\prime k}\in G$.

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References

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