697: For Group, Conjugation by Element Is 'Groups - Homomorphisms' Isomorphism

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description/proof of that for group, conjugation by 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: Natural Language Description
• 3: Proof

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Starting Context

• The reader knows a definition of conjugation for group 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, the conjugation by any 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:
$G$: $\in \{\text{ the groups }\}$
$p$: $\in G$
$f_p$: $=\text{ the conjugation for }G\text{ by }p$
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Statements:
$f_p\in \{\text{ the 'groups - homomorphisms' isomorphisms }\}$
//

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2: Natural Language Description

For any group, $G$, any element, $p\in G$, and the conjugation for $G$ by $p$, $f_p$, $f_p$ is a 'groups - homomorphisms' isomorphism.

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3: Proof

Whole Strategy: Step 1: see that $f_p$ is a group homomorphism; Step 2: see that $f_p$ is bijective; Step 3: conclude that $f_p$ is a 'groups - homomorphisms' isomorphism.

Step 1:

Let us see that $f_p$ is a group homomorphism.

For each $p_1,p_2\in G$, $f_p(p_1{p}_2)=f(p_1)f(p_2)$?

$f_p(p_1{p}_2)=p{p}_1{p}_2{p}^{-1}=p{p}_1{p}^{-1}p{p}_2{p}^{-1}=f_p(p_1)f_p(p_2)$.

$f_p$ 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 2:

Let us see that $f_p$ is bijective.

Let $p_1,p_2\in G$ be any such that $p_1\ne p_2$. Let us suppose that $f_p(p_1)=f_p(p_2)$. $p{p}_1{p}^{-1}=p{p}_2{p}^{-1}$, $p_1=p^{-1}p{p}_1{p}^{-1}p=p^{-1}p{p}_2{p}^{-1}p=p_2$, a contradiction. So, $f_p(p_1)\ne f_p(p_2)$. So, $f_p$ is injective.

Let $p_2\in G$ be any. Let $p_1=p^{-1}{p}_{2}p\in G$. $f_p(p_1)=f_p(p^{-1}{p}_{2}p)=p{p}^{-1}{p}_{2}p{p}^{-1}=p_2$. So, $f_p$ is surjective.

So, $f_p$ is bijective.

Step 3:

$f_p$ 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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