1188: For Group, Inverse of Subset Is Image of Subset Under Inverse Map, and Double Inverse of Subset Is Subset

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description/proof of that for group, inverse of subset is image of subset under inverse map, and double inverse of subset is subset

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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 inverse of subset 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 inverse of any subset is the image of the subset under the inverse map, and the double inverse of the subset is the subset.

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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:
$G$: $\in \{\text{ the groups }\}$, with the inverse operation, $f_2:G\to G$
$S$: $\subseteq G$
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Statements:
$S^{-1}=f_2(S)$
$\wedge$
${S^{-1}}^{-1}=S$
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2: Proof

Whole Strategy: Step 1: see that for each $g\in S^{-1}$, $g\in f_2(S)$, and for each $g\in f_2(S)$, $g\in S^{-1}$; Step 2: see that ${S^{-1}}^{-1}=f_2\circ f_2(S)=id(S)=S$.

Step 1:

Let $g\in S^{-1}$ be any.

$g^{-1}\in S$, so, $g={g^{-1}}^{-1}=f_2(g^{-1})\in f_2(S)$.

Let $g\in f_2(S)$ be any.

There is a $g^{\prime }\in S$ such that $g=f_2(g^{\prime })=g^{\prime -1}$, which means that $g^{\prime }={g^{\prime -1}}^{-1}=g^{-1}$, which means that $g^{-1}\in S$, which means that $g\in S^{-1}$.

So, $S^{-1}=f_2(S)$.

Step 2:

${S^{-1}}^{-1}=f_2\circ f_2(S)$, by Step 1.

But as $f_2\circ f_2=id$, $=id(S)=S$.

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References

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