975: Intersection of Subgroups of Group Is Subgroup of Group

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description/proof of that intersection of subgroups of group is subgroup of group

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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, the intersection of any possibly uncountable number of subgroups of the group is a subgroup of the group.

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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:
$G^{\prime }$: $\in \{\text{ the groups }\}$
$S$: $\subseteq \{\text{ the subgroups of }{G}^{\prime }\}$
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Statements:
$\cap S\in \{\text{ the subgroups of }{G}^{\prime }\}$
//

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2: Proof

Whole Strategy: Step 1: see that $\cap S$ satisfies the requirements to be a group.

Step 1:

For the identity element, $1\in G^{\prime }$, $1\in \cap S$, because for each $G\in S$, $1\in G$.

For any $g_1,g_2\in \cap S$, $g_1{g}_2\in \cap S$, because for each $G\in S$, $g_1,g_2\in G$, and so, for each $G\in S$, $g_1{g}_2\in G$.

For any $g\in \cap S$, $g^{-1}\in \cap S$, because for each $G\in S$, $g\in G$, and so, for each $G\in S$, $g^{-1}\in G$.

The associativity of multiplications holds, because it holds in the ambient $G^{\prime }$.

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References

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