Exercise 14.3.2

Let γ I n , v ∈ AGL ( n , 𝔽 q ) be translation by v ∈ 𝔽 q n , and let γ A , w ∈ AGL ( n , 𝔽 q ) be arbitrary.

Answers

(a)
Prove that γ A , w − 1 = γ A − 1 , − A − 1 w .
(b)
Prove that γ A , w ∘ γ I n , v ∘ γ A , w − 1 = γ I n , Av .
(c)
Part (b) shows that the translation subgroup 𝔽 q n ⊂ AGL ( n , 𝔽 q ) is normal. Prove that the quotient group AGL ( n , 𝔽 q ) ∕ 𝔽 q n is isomorphic to GL ( n , 𝔽 q ) .
(d)
Prove that AGL ( n , 𝔽 q ) is isomorphic to the semidirect product 𝔽 q n ⋊ GL ( n , 𝔽 q ) , where GL ( n , 𝔽 q ) acts on 𝔽 q n by matrix multiplication.

Proof. We give first the product of two elements in AGL ( n , 𝔽 q ) , to generalize the results of Section 6.4A. Using the definition γ A , v ( u ) = Au + v , we obtain, for all u ∈ 𝔽 q n ,

( γ A , v ∘ γ B , w ) ( u ) = γ A , v ( Bu + w ) = A ( Bu + w ) + v = ABu + Aw + v = γ AB , Aw + v ( u ) ,

thus

γ A , v ∘ γ B , w = γ AB , Aw + v (1) (a) For all u , v ∈ 𝔽 q n , v = γ A , w ( u ) ⟺ v = Au + w ⟺ u = A − 1 ( v − w ) = A − 1 v − A − 1 w ,

thus

γ A , w − 1 = γ A − 1 , − A − 1 w . (2) (b) Using the formulas (1) and (2), we obtain γ A , w ∘ γ I n , v ∘ γ A , w − 1 = γ A , w ∘ γ I n , v ∘ γ A − 1 , − A − 1 w = γ A , w ∘ γ A − 1 , − A − 1 w + v = γ I n , A ( − A − 1 w + v ) + w = γ I n , Av .

We have proved

γ A , w ∘ γ I n , v ∘ γ A , w − 1 = γ I n , Av . (3) (c) Consider the map φ { AGL ( n , 𝔽 q ) → GL ( n , 𝔽 q ) γ A , v ↦ A .

The map φ is well defined, since for all A , B ∈ GL ( n , 𝔽 q ) and for all v , w ∈ 𝔽 q n ,

γ A , v = γ B , w ⟺ ( A , v ) = ( B , w ) . (4)

Then

φ ( γ A , v ∘ γ B , w ) = φ ( γ AB , Aw + v ) = AB = φ ( γ A , v ) φ ( γ B , w ) ,

thus φ is a group homomorphism.

Since every A ∈ GL ( n , 𝔽 q ) is the image of γ A , 0 , φ is surjective, and

ker ⁡ φ = { γ I , w | w ∈ 𝔽 q ∗ } ≃ 𝔽 q n .

Therefore, the first Isomorphism Theorem gives

AGL ( n , 𝔽 q ) ∕ 𝔽 q n ≃ GL ( n , 𝔽 q ) ,

with the identification of vectors w with the translations γ I , w . (d) Consider

ψ { AGL ( n , 𝔽 q ) → 𝔽 q n ⋊ GL ( n , 𝔽 q ) γ A , v ↦ ( v , A ) .

By formula (4), the map ψ is well defined. It is a bijection, with inverse ( v , A ) ↦ γ A , v .

Following the formula (6.9), ( h , g ) ( h ′ , g ′ ) = ( h ( g ⋅ h ′ ) , g g ′ ) , which defines the product in H ⋊ G , the product in 𝔽 q n ⋊ GL ( n , 𝔽 q ) is given by

( v , A ) ( w , B ) = ( v + Aw , AB ) , v , w ∈ 𝔽 q n , A , B ∈ GL ( n , 𝔽 q ) . (5)

Therefore, using (5) and (1),

ψ ( γ A , v ) ψ ( γ B , w ) = ( v , A ) ( w , B ) = ( v + Aw , AB ) = ψ ( γ AB , Aw + v ) = ψ ( γ A , v ∘ γ B , w ) ,

thus ψ is a group isomorphism, and

AGL ( n , 𝔽 q ) ≃ 𝔽 q n ⋊ GL ( n , 𝔽 q ) .

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2022-07-19 00:00
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