1183: Inverse of Subset of Group

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definition of inverse of subset 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: Note

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Starting Context

• The reader knows a definition of group.

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Target Context

• The reader will have a definition of inverse of subset of 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:
$G$: $\in \{\text{ the groups }\}$
$S$: $\subseteq G$
$\ast S^{-1}$: $=\{g\in G|\mathrm{\exists }s\in S(g=s^{-1})\}$
//

Conditions:
//

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2: Note

$S^{-1}:=\{g\in G|g^{-1}\in S\}$ is an equivalent definition.

Let us confirm that fact.

Let $g\in \{g\in G|\mathrm{\exists }s\in S(g=s^{-1})\}$ be any.

As $g=s^{-1}$, $g^{-1}={s^{-1}}^{-1}=s\in S$, so, $g\in \{g\in G|g^{-1}\in S\}$.

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

Let $s:=g^{-1}\in S$.

Then, $s^{-1}={g^{-1}}^{-1}=g$.

So, $g\in \{g\in G|\mathrm{\exists }s\in S(g=s^{-1})\}$.

So, $\{g\in G|\mathrm{\exists }s\in S(g=s^{-1})\}=\{g\in G|g^{-1}\in S\}$.

When $S$ is a subgroup of $G$, $S^{-1}=S$.

Let us see that fact.

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

$g^{-1}\in S$, but as $S$ is a group, $g={g^{-1}}^{-1}\in S$.

Let $s\in S$ be any.

As $S$ is a group, $s^{-1}\in S$, so, $s\in S^{-1}$.

So, $S^{-1}=S$.

But $S^{-1}=S$ does not necessarily imply that $S$ is a subgroup.

For example, let $G=\mathbb{Z}$ be the additive group and $S=\{-1,0,1\}$, then, $S^{-1}=S$, but $S$ is not any subgroup of $G$: $1+1\notin S$.

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References

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