998: For Finite Cyclic Group and Its Prime Factor of Order, Cyclic Subgroup of Factor Order Can Be Extracted in Certain Way

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description/proof of that for finite cyclic group and its prime factor of order, cyclic subgroup of factor order can be extracted in certain way

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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: Note
• 3: Proof

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Starting Context

• The reader knows a definition of cyclic group by element.

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Target Context

• The reader will have a description and a proof of the proposition that for any finite cyclic group and any prime factor of the order of the group, a cyclic subgroup of the factor order can be extracted in a certain way.

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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:
$p$: $\in \{\text{ the prime numbers }\}$
$G$: $\in \{\text{ the cyclic groups }\}$ with order $pn$, $=\{g,...,g^{pn}=1\}$
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Statements:
$G_p:=\{g^n,g^{2n},...,g^{pn}\}\in \{\text{ the }p\text{ -ordered cyclic groups }\}$
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2: Note

By Cauchy's theorem, a $p$-ordered subgroup (inevitably cyclic, by the proposition that any prime-number-ordered group is cyclic and each element except 1 generates the group) can be extracted without $G$ required to be cyclic; the point of this proposition is that a $p$-ordered subgroup can be extracted in the specific way, not "a $p$-ordered subgroup is hidden somewhere unknown".

This proposition does not claim that $G_p$ is the only $p$-ordered subgroup, but it is indeed the only $p$-ordered subgroup, by the proposition that for any cyclic group and its any prime factor of the order of the group, there is at most 1 subgroup of the factor order.

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3: Proof

Whole Strategy: Step 1: see that $G_p$ is indeed a $p$-ordered subgroup.

Step 1:

Let us see that $G_p$ is a subgroup.

For any $g^{jn},g^{kn}\in G_p$, $g^{jn}{g}^{kn}=g^{(j+k)n}$. $j+k=mp+l$ for an $m,l\in \mathbb{N}$ such that $0\le l<p$. $g^{(j+k)n}=g^{(mp+l)n}=g^{mpn+ln}=g^{mpn}{g}^{ln}=(g^{pn}{)}^m{g}^{ln}=1^m{g}^{ln}=1g^{ln}=g^{ln}$. When $0<l<p$, $g^{ln}\in G_p$. When $l=0$, $g^{ln}=g^0=1=g^{pn}\in G_p$. So, $g^{jn}{g}^{kn}\in G_p$ anyway.

$g^{pn}=1\in G_p$.

For each $g^{jn}\in G_p$, $g^{(p-j)n}\in G_p$, and $g^{jn}{g}^{(p-j)n}=g^{jn+(p-j)n}=g^{pn}=1$; $g^{(p-j)n}{g}^{jn}=g^{jn+(p-j)n}=g^{pn}=1$.

$G_p$ is $p$-ordered, because $\{g^n,g^{2n},...,g^{pn}\}$ is distinct, because $\{g,...,g^{pn}=1\}$ is distinct.

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References

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