997: For Finite Cyclic Group and Its Prime Factor of Order, There Is at Most 1 Subgroup of Factor Order

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description/proof of that for finite cyclic group and its prime factor of order, there is at most 1 subgroup of factor 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: Note
• 3: Proof

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Starting Context

• The reader knows a definition of cyclic group by element.
• The reader admits the proposition that any prime-number-ordered group is cyclic and each element except 1 generates the group.
• The reader admits the proposition that any 2 different prime-number-ordered subgroups share only 1.

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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 its any prime factor of the order of the group, there is at most 1 subgroup of the factor 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:
$p$: $\in \{\text{ the prime numbers }\}$
$G$: $\in \{\text{ the cyclic groups }\}$ with order $pn$, $=\{g,...,g^{pn}=1\}$
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Statements:
$|\{\text{ the }p\text{ -ordered subgroups of }G\}|\le 1$
//

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2: Note

By 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, there is the cyclic subgroup of the $p$ order mentioned in the proposition, and in fact, that is the only $p$-ordered subgroup of $G$, by this proposition.

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3: Proof

Whole Strategy: Step 1: see that any $p$-ordered subgroup is a cyclic group generated by a $g^r$ where $1\le r<pn$; Step 2: see the condition that $r$ needs to satisfy and that there are only $p-1$ possibilities; Step 3: conclude the proposition.

Step 1:

Any $p$-ordered subgroup is a cyclic group, by the proposition that any prime-number-ordered group is cyclic and each element except 1 generates the group.

So, it is generated by an element of $G$, and let the element be $g^r$ where $1\le r<pn$.

Step 2:

$r$ needs to satisfy $(g^r{)}^p=1$.

$(g^r{)}^p=g^{rp}$. $rp=pnm$ for an $m\in \mathbb{N}\setminus \{0\}$. So, $r=nm$.

As $1\le r<pn$, the only possibilities are $m=1,...,p-1$.

So, only those $p-1$ elements can possibly generate any $p$-ordered subgroup.

Step 3:

If there were 2 different $p$-ordered subgroups, the 2 subgroups would share only $1$, by the proposition that any 2 different prime-number-ordered subgroups share only 1. So, the other different $2(p-1)$ elements would generate a $p$-ordered subgroup, by the proposition that any prime-number-ordered group is cyclic and each element except 1 generates the group, which is a contradiction against the fact that only those $p-1$ elements can possibly generate any $p$-ordered subgroup.

So, there is at most 1 $p$-ordered subgroup.

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References

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