990: 2 Different Prime-Number-Ordered Subgroups Share Only 1

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description/proof of that 2 different prime-number-ordered subgroups share only 1

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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 admits the proposition that any prime-number-ordered group is cyclic and each element except 1 generates the group.

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Target Context

• The reader will have a description and a proof of the proposition that any 2 different prime-number-ordered subgroups share only 1.

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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:
$G$: $\in \{\text{ the groups }\}$
$p$: $\in \{\text{ the prime numbers }\}$
$q$: $\in \{\text{ the prime numbers }\}$
$G_p$: $\in \{\text{ the }p\text{ -ordered subgroups of }G\}$
$G_q$: $\in \{\text{ the }q\text{ -ordered subgroups of }G\}$
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Statements:
$G_p\ne G_q$
$⟹$
$G_p\cap G_q=\{1\}$
//

$p=q$ is not excluded.

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2: Proof

Whole Strategy: Step 1: suppose that there was a $g\in (G_p\cap G_q)\setminus \{1\}$; Step 2: see that $G_p=⟨g⟩=G_q$, a contradiction.

Step 1:

Let us suppose that there was a $g\in (G_p\cap G_q)\setminus \{1\}$.

Step 2:

$⟨g⟩=G_p$, by the proposition that any prime-number-ordered group is cyclic and each element except 1 generates the group.

$⟨g⟩=G_q$, likewise.

So, $G_p=⟨g⟩=G_q$, a contradiction.

So, the supposition that there was a $g\in (G_p\cap G_q)\setminus \{1\}$ is wrong, and $(G_p\cap G_q)\setminus \{1\}=\mathrm{\varnothing }$, which means that $G_p\cap G_q=\{1\}$.

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References

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