601: Bijective Group Homomorphism Is 'Groups - Homomorphisms' Isomorphism

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description/proof of that bijective group homomorphism is 'groups - homomorphisms' isomorphism

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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
• 4: Note

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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.

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Target Context

• The reader will have a description and a proof of the proposition that any bijective group homomorphism is a 'groups - homomorphisms' isomorphism.

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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:
$G_1$: $\in \{\text{ the groups }\}$
$G_2$: $\in \{\text{ the groups }\}$
$f$: $G_1\to G_2$, $\in \{\text{ the bijections }\}\cap \{\text{ the group homomorphisms }\}$
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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.

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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 $f^{-1}({g_2}^{-1})={f^{-1}(g_2)}^{-1}$.

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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.

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References

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