description/proof of that group is 'groups - homomorphisms' isomorphic to reversed operator group of group
Topics
About: group
The table of contents of this article
- Starting Context
- Target Context
- Orientation
- Main Body
- 1: Structured Description
- 2: Natural Language Description
- 3: Proof
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.
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.
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:
\(G\): \(= \{\text{ the groups }\}\), with the operator, \(\cdot\)
\(\widetilde{G}\): \(= \text{ the reversed operator group of } G\), with the operator, \(*\)
\(f\): \(:G \to \widetilde{G}, p \mapsto p^{-1}\)
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Statements:
\(G \cong_f \widetilde{G}\), 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 \(G\), \(\widetilde{G}\), with the operator, \(*\), \(G \cong_f \widetilde{G}\), where \(\cong\) denotes being 'groups - homomorphisms' isomorphic by \(f\), where \(f: G \to \widetilde{G}, p \mapsto p^{-1}\).
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 \(\widetilde{G}\), 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 \(p_1 \neq p_2\). 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, \(f (p_1) \neq f (p_2)\). So, \(f\) is injective.
Let \(p \in \widetilde{G}\) 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. \(f (p_1 \cdot p_2) = (p_2 * p_1)^{-1} = {p_1}^{-1} * {p_2}^{-1} = f (p_1) * f (p_2)\).
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.
