On Gluck’s conjecture for wreath product type groups

With the notation above, we have

1. ⟨ a , b ⟩ ≅ Q 8 , a 2 = − I = b 2 and b − 1 ⁢ a ⁢ b = a − 1 ;

2. ⟨ c , d ⟩ ≅ S 3 , c 3 = I = d 2 and d − 1 ⁢ c ⁢ d = c − 1 ;

3. 𝑋 is the semidirect product of ⟨ a , b ⟩ and ⟨ c , d ⟩ with ⟨ a , b ⟩ normal; in particular, c − 1 ⁢ a ⁢ c = b ; c − 1 ⁢ b ⁢ c = b ⁢ a ; d − 1 ⁢ a ⁢ d = a − 1 , d − 1 ⁢ b ⁢ d = ( b ⁢ a ) − 1 .

4. Let v = ( 0 , 1 ) , u = ( 1 , 0 ) ∈ V . Then

   C X ⁡ ( v ) = { x ∈ X ∣ v ⁢ x = v } = ⟨ c , d ⟩ ;

   hence O 2 ⁡ ( X ) = ⟨ a , b ⟩ acts transitively on V ∖ { 0 } . Moreover, we have

   C X ⁡ ( v ) ∩ C X ⁡ ( u ) = 1 ,

   so 𝑋 has a regular orbit on V ⊕ V .

Proof

The results follow from direct calculation. ∎

With the notation above, let u = ( 1 , 0 ) , v = ( 0 , 1 ) ∈ V and set

1. 𝐺 is the unique primitive subgroup (up to conjugacy) in GL ⁡ ( V ⊗ V ) of order 2304, which has three orbits { 0 } , λ G , μ G on V ⊗ V of length 1, 32, 48 respectively. Moreover,

   C G ⁡ ( λ ) = ⟨ τ , c ⊗ I , d ⊗ I , I ⊗ c , I ⊗ d ⟩ ≅ S 3 ≀ S 2 , C G ⁡ ( μ ) = ⟨ a ⊗ a − 1 , b ⊗ ( b ⁢ a ) − 1 , c ⊗ c − 1 , − ( d ⊗ d ) , τ ⟩ ≅ S 4 × S 2 .

2. Let 𝐻 be a subgroup of 𝐺 with S 0 ⊆ H and | G : H | = 2 . Then H = G 0 , G 1 or G 2 . Moreover, for i = 0 , 2, G i has three orbits { 0 } , λ G i , μ G i on V ⊗ V of length 1, 32, 48 respectively; G 1 has four orbits on V ⊗ V of length 1, 32, 24, 24 and

   C G 1 ⁡ ( λ ) = G 1 ∩ C G ⁡ ( λ ) = ⟨ c ⊗ 1 , 1 ⊗ c , d ⊗ d , τ ⟩ = ⟨ c ⊗ c , ( d ⊗ d ) ⁢ τ ⟩ × ⟨ c 2 ⊗ c , τ ⟩ ≅ S 3 × S 3 .

Proof

Part (1) follows from a direct calculation and GAP [14]. For part (2), it is easy to see that G / S 0 ≅ D 8 . Hence there exist three maximal subgroups of 𝐺 containing S 0 , that is, G 0 , G 1 , G 2 . For i = 0 , 2 , we see that C G ⁡ ( λ ) and C G ⁡ ( μ ) is not contained in G i as τ ∉ G i . So

Hence G i has the same orbits as 𝐺 for i = 0 , 2 . But for i = 1 , we see that

Hence λ G 1 = λ G and μ G is the union of two distinct G 1 -orbits of the same length equal to 24, as desired. ∎