Exercise 14.3.8

Suppose that γ , γ ′ : T → { 1 , … , l } are one-to-one and onto. As explained in the text, these give isomorphisms γ ^ , γ ^ ′ : S ( T ) ≃ S l .

(a)
Explain why σ = γ ∘ ( γ ′ ) − 1 is an element of S l .
(b)
Let σ ∈ S l be as in part (a), and let σ ^ : S l → S l be conjugation by σ . Thus σ ^ ( τ ) = στ σ − 1 for τ ∈ S l . Prove that γ ^ = σ ^ ∘ γ ^ ′ .

This proves that γ ^ and γ ^ ′ differ by conjugation by an element of S l .

Answers

Proof. (a) Since γ , γ ′ : T → { 1 , … , l } are bijective, then σ = γ ∘ ( γ ′ ) − 1 : { 1 , … , l } → { 1 , … , l } is bijective, so σ ∈ S l . (b) For all φ ∈ S ( T ) ,

( σ ^ ∘ γ ^ ′ ) ( φ ) = σ ^ ( γ ′ ∘ φ ∘ ( γ ′ ) − 1 ) = σ ∘ γ ′ ∘ φ ∘ ( γ ′ ) − 1 ∘ σ − 1 = γ ∘ ( γ ′ ) − 1 ∘ γ ′ ∘ φ ∘ ( γ ′ ) − 1 ∘ γ ′ ∘ γ − 1 = γ ∘ φ ∘ γ − 1 = γ ^ ( φ ) ,

thus σ ^ ∘ γ ^ ′ = γ ^ .

Note: Let G be a subgroup of S ( T ) , and G 1 = γ ^ ( G ) , G 2 = γ ^ ′ ( G ) the corresponding subgroups in S l , then G 1 = σ ^ ( G 2 ) = σ G 2 σ − 1 are conjugate subgroups. □

User profile picture
2022-07-19 00:00
Comments