description/proof of that bijective group homomorphism is 'groups - homomorphisms' isomorphism
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
- 4: Note
Starting Context
- The reader knows a definition of group.
- The reader knows a definition of bijection.
- The reader knows a definition of %structure kind name% homomorphism.
- The reader knows a definition of %category name% isomorphism.
Target Context
- The reader will have a description and a proof of the proposition that any bijective group homomorphism is a 'groups - homomorphisms' isomorphism.
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:
\(G_1\): \(\in \{\text{ the groups }\}\)
\(G_2\): \(\in \{\text{ the groups }\}\)
\(f\): \(G_1 \to G_2\), \(\in \{\text{ the bijections }\} \cap \{\text{ the group homomorphisms }\}\)
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Statements:
\(f \in \{\text{ the 'groups - homomorphisms' isomorphisms }\}\)
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2: Natural Language Description
For any groups, \(G_1, G_2\), any bijective group homomorphism, \(f: G_1 \to G_2\), is a 'groups - homomorphisms' isomorphism.
3: Proof
As \(f\) is bijective, there is the inverse, \(f^{-1}: G_2 \to G_1\).
The issue is whether \(f^{-1}\) is necessarily group homomorphic.
As \(f\) maps the identity element of \(G_1\) to the identity element of \(G_2\), \(f^{-1}\) maps the identity element of \(G_2\) to the identity element of \(G_1\).
For any \(g_{2, 1}, g_{2, 2} \in G_2\), \(f^{-1} (g_{2, 1} g_{2, 2}) = f^{-1} (g_{2, 1}) f^{-1} (g_{2, 2})\)? \(f (f^{-1} (g_{2, 1}) f^{-1} (g_{2, 2})) = f (f^{-1} (g_{2, 1})) f (f^{-1} (g_{2, 2})) = g_{2, 1} g_{2, 2}\), which means that \(f^{-1} (g_{2, 1} g_{2, 2}) = f^{-1} (g_{2, 1}) f^{-1} (g_{2, 2})\).
For any \(g_2 \in G_2\), \(f^{-1} ({g_2}^{-1}) = {f^{-1} (g_2)}^{-1}\)? \(f^{-1} ({g_2}^{-1}) f^{-1} (g_2) = f^{-1} ({g_2}^{-1} g_2) = f^{-1} (1) = 1\), which means that \(f^{-1} ({g_2}^{-1}) = {f^{-1} (g_2)}^{-1}\).
4: Note
In general, a bijective morphism of a category is not necessarily any %category name% isomorphism. For example, a bijective continuous map, which is a morphism of the 'topological spaces - continuous maps' category, is not necessarily a homeomorphism, which is a 'topological spaces - continuous maps' isomorphism.
