Exercise 14.4.16

Let V , W be vector spaces over a field F , and let Aut F ( V ) be the group of vector space isomorphisms V ≃ V . Also let T : V → W be a vector space isomorphism.

(a)
Prove that ϕ ↦ T ∘ ϕ ∘ T − 1 induces a group isomorphism γ T : Aut F ( V ) ≃ Aut F ( W ) .
(b)
Let T ′ : V → W be another isomorphism. Prove that there is Φ ∈ Aut F ( W ) such that T ′ = Φ ∘ T . In the notation of part (a), γ Φ : Aut F ( W ) ≃ Aut F ( W ) is conjugation by Φ .
(c)
In the situation of part (b), prove that γ T ′ = γ Φ ∘ γ T .

Answers

Proof. (a) Let T : V → W be a vector space isomorphism, and consider the map

γ T { Aut F ( V ) → Aut F ( W ) ϕ ↦ T ∘ ϕ ∘ T − 1 .

γ T is a group homomorphism: if ϕ , ψ ∈ Aut F ( V ) ,

γ T ( ϕ ) ∘ γ T ( ψ ) = T ∘ ϕ ∘ T − 1 ∘ T ∘ ψ ∘ T − 1 = T ∘ ϕ ∘ ψ ∘ T − 1 = γ T ( ϕ ∘ ψ ) .

Moreover γ T − 1 : Aut F ( W ) → Aut F ( V ) is such that, for all ϕ ∈ Aut F ( V ) ,

( γ T − 1 ∘ γ T ) ( ϕ ) = T − 1 ∘ ( T ∘ ϕ ∘ T − 1 ) ∘ T = ϕ ,

thus γ T − 1 ∘ γ T = 1 Aut F ( V ) , and similarly γ T ∘ γ T − 1 = 1 Aut F ( W ) . This proves that γ T is bijective, and that γ T − 1 = γ T − 1 . Thus γ T is a group isomorphism. (b) Let T ′ : V → W be another isomorphism, and put Φ = T ′ ∘ T − 1 . Then Φ : W → W is bijective and linear, so Φ ∈ Aut F ( W ) , and T ′ = Φ ∘ T .

By definition,

γ Φ { Aut F ( W ) → Aut F ( W ) ψ ↦ Φ ∘ ψ ∘ Φ − 1 ,

and by part (a), γ Φ is the conjugation by Φ , so is an inner group automorphism:

γ Φ ∈ Int ( Aut F ( W ) ) ⊂ Aut ( Aut F ( W ) ) .

(c) If ϕ ∈ Aut F ( V ) , then

( γ Φ ∘ γ T ) ( ϕ ) = Φ ∘ ( T ∘ ϕ ∘ T − 1 ) ∘ Φ − 1 = ( Φ ∘ T ) ∘ ϕ ∘ ( Φ ∘ T ) − 1 = T ′ ∘ ϕ ∘ ( T ′ ) − 1 = γ T ′ ( ϕ ) ,

so that

γ T ′ = γ Φ ∘ γ T .

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2022-07-19 00:00
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