2025-07-06

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

Topics


About: group

The table of contents of this article


Starting Context



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.

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.


Main Body


1: Structured Description


Here is the rules of Structured Description.

Entities:
\(G\): \(\in \{\text{ the groups }\}\), with the inverse operation, \(f_2: G \to G\)
\(S\): \(\subseteq G\)
//

Statements:
\(S^{-1} = f_2 (S)\)
\(\land\)
\({S^{-1}}^{-1} = S\)
//


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 \(g' \in S\) such that \(g = f_2 (g') = g'^{-1}\), which means that \(g' = {g'^{-1}}^{-1} = g^{-1}\), 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 \(f_2 \circ f_2 = id\), \(= id (S) = S\).


References


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