SMALL 4-QUASINORMAL SUBGROUPS

Microsoft Engineering Excellence A subgroup H of a group G is QUASINORMAL if HK = KH for all subgroups K of G, i.e. ⟨H, K ⟩ = HK. Write H qn G. In finite groups, quasinormal subgroups are always subnormal. If G is finite and H qn G, then H /HG lies in the hypercentre of G/HG . Quasinormal subgroups of finite p – groups can be a rarity. Microsoft Confidential

Microsoft Engineering Excellence Say a subgroup H of a group G is 4-quasinormal (H qn4 G) if ⟨H, K ⟩ = HKHK for all cyclic subgroups K of G. Every subgroup of a nilpotent group of class 2 is 4-qn. Consider H = ⟨h ⟩ qn4 G , a finite p –group with p odd. When H is cyclic and quasinormal in G (p ≥ 3), then (i) [H, G ] is abelian and (ii) H acts on [H, G ] as a group of power automorphisms. If H is quasinormal & of order p , then Hᴳ is elementary abelian. What happens when H is 4-quasinormal & of order p ? Microsoft Confidential

Microsoft Engineering Excellence From now on, suppose that H = ⟨h ⟩ qn4 G and that |H | = p (≥3). If G has exponent p , then for all g ϵ G , ⟨h , g ⟩ is either ELEMENTARY ABELIAN of rank ≤ 2 or EXTRASPECIAL of order p 3(Ep ). So h commutes with [h , g ] and therefore with h g. Thus H G is ELEMENTARY ABELIAN. (*) Let p ≥ 5. For arbitrary G it can be shown that for all g ϵ G , g p commutes with h. Thus G p centralizes H and hence also H G. By (*), H G is elementary abelian modulo G p. Also H G ∩ G p lies in the centre of H G and therefore H G has class ≤ 2. Then H G is regular and hence has exponent p. Microsoft Confidential

Microsoft Engineering Excellence Let ⟨h ⟩, ⟨c ⟩ and ⟨g ⟩ be cyclic groups of order p , p and p ² respectively. Then let G = ⟨h ⟩ ⋉ {⟨c ⟩ × ⟨g ⟩}, where [h , g ] = c and [c , h ] = g p. Then H qn4 G and H G ≅ Ep . But is [H , G ] always abelian? Let G = ⟨ h , k1, k2 ⟩ be a finite p – group (p ≥ 5) with H = ⟨h ⟩ of order p and 4-qn in G. Is [[h , k 1] , [h , k2]] = 1? Let K = ⟨k1, k2 ⟩. When K ≅ C p × Cp , then |G | ≤ p 5 & G has class ≤ 2. Therefore YES! Microsoft Confidential

Microsoft Engineering Excellence Now let K = ⟨k1, k2 ⟩ ≅ Ep . Then |G | ≤ p 7 & G has class ≤ 3. Therefore G is metabelian. Therefore again YES! If K is abelian, then K p lies in the centre of G ; and G /K p has class ≤ 2. So again G has class ≤ 3 and YES! The answer is also YES when (i) |k 1| = p ; or (ii) K/K p is abelian ; or (iii) K /K p is extraspecial. Let K /K p have class 3. Then . . . . .? Microsoft Confidential