Difference between revisions of "Conjugation"
(Created page with "==Definition== Two elements {{M|g,h}} of a group {{M|(G,\times)}} are ''conjugate'' if: * {{M|1=\exists x\in G[xgx^{-1}=h]}} ===Conjugation operation=== Let {{M|x}}...") |
m (Done automorphism proof) |
||
| Line 16: | Line 16: | ||
Claim: The map {{M|C_x:G\rightarrow G}} given by {{M|g\mapsto xgx^{-1} }} is an automorphism | Claim: The map {{M|C_x:G\rightarrow G}} given by {{M|g\mapsto xgx^{-1} }} is an automorphism | ||
{{Begin Proof}} | {{Begin Proof}} | ||
| − | {{ | + | To be an automorphism, it must be a bijection, which is to say it is both [[Injection|injective]] and [[Surjection|surjective]] |
| + | |||
| + | |||
| + | Let {{M|x\in G}} be given | ||
| + | |||
| + | : '''Proof of injectivity''' | ||
| + | :: We wish to show that {{M|1=c_x(y)=c_x(y')\implies y=y'}} | ||
| + | ::: Suppose {{M|1=c_x(y)=c_x(y')}} then {{M|1=xyx^{-1}=xy'x^{-1} }} | ||
| + | ::: {{M|1=\implies xy=xy'}} | ||
| + | ::: {{M|1=\implies y=y'}} | ||
| + | :: So {{M|1=c_x(y)=c_x(y')\implies y=y'}} is shown | ||
| + | : Since {{M|x\in G}} was arbitrary, we have shown all {{M|c_x}} are injective | ||
| + | |||
| + | |||
| + | : '''Proof of surjectivity''' | ||
| + | :: We wish to show that {{M|1=\forall g\in G\exists y\in G[c_x(y)=g]}} | ||
| + | ::: Let {{M|g\in G}} be given | ||
| + | :::* '''Note:''' we want {{M|1=c_x(y)=g}} which is {{M|1=xyx^{-1}=g\implies xy=gx\implies y=x^{-1}gx}} | ||
| + | :::** This is okay because: | ||
| + | :::**# By hypothesis {{M|x,g\in G}} | ||
| + | :::**# As {{M|x\in G}} we know {{M|\exists x^{-1}\in G}} | ||
| + | :::**# A group is closed under composition, so {{M|x^{-1}gx\in G}} - which is a unique expression as the group is associative | ||
| + | :::**#: That is to say {{M|1=(x^{-1}g)x=x^{-1}(gx)}} | ||
| + | ::: Choose {{M|1=y=x^{-1}gx\in G}} | ||
| + | ::: Then {{M|1=c_x(y) = xyx^{-1} = xx^{-1}gxx^{-1} = ege = g}} | ||
| + | ::: That is {{M|1=c_x(y)=g}} | ||
| + | :: Since {{M|g}} was arbitrary we have shown for a given {{M|x\in G}} that {{M|c_x}} is surjective | ||
| + | : Since {{M|x}} was arbitrary we have shown that all {{M|c_x}} are sujective | ||
| + | |||
| + | Thus all {{M|c_x\in\text{Aut}(G)}} - as required | ||
{{End Proof}}{{End Theorem}} | {{End Proof}}{{End Theorem}} | ||
{{Begin Theorem}} | {{Begin Theorem}} | ||
Revision as of 08:45, 12 May 2015
Definition
Two elements [ilmath]g,h[/ilmath] of a group [ilmath](G,\times)[/ilmath] are conjugate if:
- [ilmath]\exists x\in G[xgx^{-1}=h][/ilmath]
Conjugation operation
Let [ilmath]x[/ilmath] in [ilmath]G[/ilmath] be given, define:
- [ilmath]C_x:G\rightarrow G[/ilmath] as the automorphism (recall that means an isomorphism of a group onto itself) which:
- [ilmath]g\mapsto xgx^{-1} [/ilmath]
This association of [ilmath]x\mapsto c_x[/ilmath] is a homomorphism of the form [ilmath]G\rightarrow\text{Aut}(G)[/ilmath] (or indeed [ilmath]G\rightarrow(G\rightarrow G)[/ilmath] instead)
This operation on [ilmath]G[/ilmath] is called conjugation[1]
TODO: Link with language - "the conjugation of x is the image of [ilmath]c_x[/ilmath]" and so forth
Proof of clams
Claim: The map [ilmath]C_x:G\rightarrow G[/ilmath] given by [ilmath]g\mapsto xgx^{-1} [/ilmath] is an automorphism
To be an automorphism, it must be a bijection, which is to say it is both injective and surjective
Let [ilmath]x\in G[/ilmath] be given
- Proof of injectivity
- We wish to show that [ilmath]c_x(y)=c_x(y')\implies y=y'[/ilmath]
- Suppose [ilmath]c_x(y)=c_x(y')[/ilmath] then [ilmath]xyx^{-1}=xy'x^{-1}[/ilmath]
- [ilmath]\implies xy=xy'[/ilmath]
- [ilmath]\implies y=y'[/ilmath]
- So [ilmath]c_x(y)=c_x(y')\implies y=y'[/ilmath] is shown
- We wish to show that [ilmath]c_x(y)=c_x(y')\implies y=y'[/ilmath]
- Since [ilmath]x\in G[/ilmath] was arbitrary, we have shown all [ilmath]c_x[/ilmath] are injective
- Proof of surjectivity
- We wish to show that [ilmath]\forall g\in G\exists y\in G[c_x(y)=g][/ilmath]
- Let [ilmath]g\in G[/ilmath] be given
- Note: we want [ilmath]c_x(y)=g[/ilmath] which is [ilmath]xyx^{-1}=g\implies xy=gx\implies y=x^{-1}gx[/ilmath]
- This is okay because:
- By hypothesis [ilmath]x,g\in G[/ilmath]
- As [ilmath]x\in G[/ilmath] we know [ilmath]\exists x^{-1}\in G[/ilmath]
- A group is closed under composition, so [ilmath]x^{-1}gx\in G[/ilmath] - which is a unique expression as the group is associative
- That is to say [ilmath](x^{-1}g)x=x^{-1}(gx)[/ilmath]
- This is okay because:
- Note: we want [ilmath]c_x(y)=g[/ilmath] which is [ilmath]xyx^{-1}=g\implies xy=gx\implies y=x^{-1}gx[/ilmath]
- Choose [ilmath]y=x^{-1}gx\in G[/ilmath]
- Then [ilmath]c_x(y) = xyx^{-1} = xx^{-1}gxx^{-1} = ege = g[/ilmath]
- That is [ilmath]c_x(y)=g[/ilmath]
- Let [ilmath]g\in G[/ilmath] be given
- Since [ilmath]g[/ilmath] was arbitrary we have shown for a given [ilmath]x\in G[/ilmath] that [ilmath]c_x[/ilmath] is surjective
- We wish to show that [ilmath]\forall g\in G\exists y\in G[c_x(y)=g][/ilmath]
- Since [ilmath]x[/ilmath] was arbitrary we have shown that all [ilmath]c_x[/ilmath] are sujective
Thus all [ilmath]c_x\in\text{Aut}(G)[/ilmath] - as required
Claim: The family [ilmath]\{C_x\vert x\in G\} [/ilmath] form a group, and [ilmath]x\mapsto c_x[/ilmath] is a homomorphism from [ilmath]G[/ilmath] to this family
TODO: Not done yet
See also
References
- ↑ Algebra - Serge Lang - Revised Third Edition - GTM
