843: For Group and Subgroup, Conjugation for Subgroup by Group Element Is 'Groups Homomorphisms' Isomorphism

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description/proof of that for group and subgroup, conjugation for subgroup by group element 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: Note
• 3: Proof

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Starting Context

• The reader knows a definition of conjugate subgroup of subgroup by element.
• 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 for any group and its any subgroup, the conjugation for the subgroup by any group element 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:
$G^{\prime }$: $\in \{\text{ the groups }\}$
$G$: $\in \{\text{ the subgroups of }{G}^{\prime }\}$
$g^{\prime }$: $\in G^{\prime }$
$f_{g^{\prime }}$: $:G\to g^{\prime }G{g}^{\prime -1}$, $=\text{ the conjugation for }G\text{ by }{g}^{\prime }$
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Statements:
$f_{g^{\prime }}\in \{\text{ the 'groups - homomorphisms' isomorphisms }\}$
//

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2: Note

Compare with the proposition that for any group, the conjugation by any element is a 'groups - homomorphisms' isomorphism, which is about the conjugation for the whole group, while this proposition is about the conjugation for a subgroup. This proposition is naturally expected from that proposition, but let us prove it for sure.

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3: Proof

Whole Strategy: Step 1: see that $g^{\prime }G{g}^{\prime -1}$ is a group; Step 2: see that $f_{g^{\prime }}$ is a groups homomorphism; Step 3: see that $f_{g^{\prime }}$ is bijective; Step 4: conclude the proposition.

Step 1:

$g^{\prime }G{g}^{\prime -1}$ is indeed a group, because it is a conjugate subgroup of subgroup by element: see Note for the definition of conjugate subgroup of subgroup by element.

Step 2:

Let us see that $f_{g^{\prime }}$ is a group homomorphism.

For each $g_1,g_2\in G$, $f_{g^{\prime }}(g_1{g}_2)=g^{\prime }{g}_1{g}_2{g}^{\prime -1}=g^{\prime }{g}_1{g}^{\prime -1}{g}^{\prime }{g}_2{g}^{\prime -1}=f_{g^{\prime }}(g_1)f_{g^{\prime }}(g_2)$.

$f_{g^{\prime }}$ is a group homomorphism, 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:

Let us see that $f_{g^{\prime }}$ is bijective.

Let $g_1,g_2\in G$ be any such that $g_1\ne g_2$. Let us suppose that $f_{g^{\prime }}(g_1)=f_{g^{\prime }}(g_2)$. $g^{\prime }{g}_1{g}^{\prime -1}=g^{\prime }{g}_2{g}^{\prime -1}$, $g_1=g^{\prime -1}{g}^{\prime }{g}_1{g}^{\prime -1}{g}^{\prime }=g^{\prime -1}{g}^{\prime }{g}_2{g}^{\prime -1}{g}^{\prime }=g_2$, a contradiction. So, $f_{g^{\prime }}(g_1)\ne f_{g^{\prime }}(g_2)$. So, $f_{g^{\prime }}$ is injective.

$f_{g^{\prime }}$ is surjective, because the codomain is determined to be the range.

Step 4:

$f_{g^{\prime }}$ 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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