724: Group Is 'Groups - Homomorphisms' Isomorphic to Reversed Operator Group of Group

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description/proof of that group is 'groups - homomorphisms' isomorphic to reversed operator group 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: Natural Language Description
• 3: Proof

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Starting Context

• The reader knows a definition of reversed operator group of group.
• The reader knows a definition of %category name% isomorphism.
• The reader admits the proposition that any map between any groups that maps the product of any 2 elements to the product of the images of the elements is a group homomorphism.
• The reader admits the proposition that any bijective group homomorphism is a 'groups - homomorphisms' isomorphism.

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Target Context

• The reader will have a description and a proof of the proposition that any group is 'groups - homomorphisms' isomorphic to the reversed operator group 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:
$G$: $=\{\text{ the groups }\}$, with the operator, $\cdot$
$\tilde{G}$: $=\text{ the reversed operator group of }G$, with the operator, $\ast$
$f$: $:G\to \tilde{G},p\mapsto p^{-1}$
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Statements:
$G{\cong }_f\tilde{G}$, where $\cong$ denotes being 'groups - homomorphisms' isomorphic by $f$.
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2: Natural Language Description

For any group, $G$, with the operator, $\cdot$, and the reversed operator group of $G$, $\tilde{G}$, with the operator, $\ast$, $G{\cong }_f\tilde{G}$, where $\cong$ denotes being 'groups - homomorphisms' isomorphic by $f$, where $f:G\to \tilde{G},p\mapsto p^{-1}$.

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3: Proof

Whole Strategy: Step 1: see that $f$ is bijective; Step 2: see that $f$ is group homomorphic; Step 3: conclude the proposition.

Note that $p^{-1}$ is the inverse of $p$ both in $G$ and in $\tilde{G}$, as is shown in Note for the definition of reversed operator group of group.

Step 1:

Let us see that $f$ is bijective.

Let $p_1,p_2\in G$ be any such that $p_1\ne p_2$. Let us suppose that $f(p_1)=f(p_2)$. ${p_1}^{-1}={p_2}^{-1}$. $p_1\cdot {p_1}^{-1}\cdot p_2=p_1\cdot {p_2}^{-1}\cdot p_2$. $p_2=p_1$, a contradiction. So, $f(p_1)\ne f(p_2)$. So, $f$ is injective.

Let $p\in \tilde{G}$ be any. $p^{-1}\in G$, and $f(p^{-1})=p$. So, $f$ is surjective.

Step 2:

Let $p_1,p_2\in G$ be any. $f(p_1\cdot p_2)=(p_2\ast p_1{)}^{-1}={p_1}^{-1}\ast {p_2}^{-1}=f(p_1)\ast f(p_2)$.

So, $f$ is group homomorphic, by the proposition that any map between any groups that maps the product of any 2 elements to the product of the images of the elements is a group homomorphism.

Step 3:

$f$ is a 'groups - homomorphisms' isomorphism, by the proposition that any bijective group homomorphism is a 'groups - homomorphisms' isomorphism.

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References

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