Difference between revisions of "Conjugation"
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(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}}...") |
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This operation on {{M|G}} is called '''conjugation'''<ref name="Lang">Algebra - Serge Lang - Revised Third Edition - GTM</ref> | This operation on {{M|G}} is called '''conjugation'''<ref name="Lang">Algebra - Serge Lang - Revised Third Edition - GTM</ref> | ||
{{Todo|Link with language - "the conjugation of x is the image of {{M|c_x}}" and so forth}} | {{Todo|Link with language - "the conjugation of x is the image of {{M|c_x}}" and so forth}} | ||
| + | |||
==Proof of clams== | ==Proof of clams== | ||
{{Begin Theorem}} | {{Begin Theorem}} | ||
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}} | ||
Latest revision as of 14:51, 18 May 2015
Definition
Two elements g,h of a group (G,×) are conjugate if:
- ∃x∈G[xgx−1=h]
Conjugation operation
Let x in G be given, define:
- Cx:G→G as the automorphism (recall that means an isomorphism of a group onto itself) which:
- g↦xgx−1
This association of x↦cx is a homomorphism of the form G→Aut(G) (or indeed G→(G→G) instead)
This operation on G is called conjugation[1]
TODO: Link with language - "the conjugation of x is the image of cx" and so forth
Proof of clams
[Expand]
Claim: The map Cx:G→G given by g↦xgx−1 is an automorphism
[Expand]
Claim: The family {Cx|x∈G} form a group, and x↦cx is a homomorphism from G to this family
See also
References
- Jump up ↑ Algebra - Serge Lang - Revised Third Edition - GTM
