Definition
Two elements g,h of a group (G,×) are conjugate if:
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:
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
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Claim: The map Cx:G→G given by g↦xgx−1 is an automorphism
To be an automorphism, it must be a bijection, which is to say it is both injective and surjective
Let x∈G be given
- Proof of injectivity
- We wish to show that cx(y)=cx(y′)⟹y=y′
- Suppose cx(y)=cx(y′) then xyx−1=xy′x−1
- ⟹xy=xy′
- ⟹y=y′
- So cx(y)=cx(y′)⟹y=y′ is shown
- Since x∈G was arbitrary, we have shown all cx are injective
- Proof of surjectivity
- We wish to show that ∀g∈G∃y∈G[cx(y)=g]
- Let g∈G be given
- Note: we want cx(y)=g which is xyx−1=g⟹xy=gx⟹y=x−1gx
- This is okay because:
- By hypothesis x,g∈G
- As x∈G we know ∃x−1∈G
- A group is closed under composition, so x−1gx∈G - which is a unique expression as the group is associative
- That is to say (x−1g)x=x−1(gx)
- Choose y=x−1gx∈G
- Then cx(y)=xyx−1=xx−1gxx−1=ege=g
- That is cx(y)=g
- Since g was arbitrary we have shown for a given x∈G that cx is surjective
- Since x was arbitrary we have shown that all cx are sujective
Thus all cx∈Aut(G) - as required
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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