1. FormalConjectures
  2. Kourovka
  3. «1_40»
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Conjecture 1.40

by Sh. S. Kemkhadze

Is a group a nilgroup if it is the product of two normal nilsubgroups?

Here a nilgroup (Engel group) is a group in which every element is an Engel element. This is the Engel-group analogue of Fitting's theorem, which guarantees that the product of two normal nilpotent subgroups is nilpotent.

Reference: The Kourovka Notebook

open scoped commutatorElementnamespace Kourovka.«1.40»variable {G : Type*} [Group G]

The iterated commutator $[x,, {}_n y]$, defined by $[x,, {}0 y] = x$ and $[x,, {}{n+1} y] = [[x,, {}_n y], y]$.

def engelCommutator (x y : G) : ℕ → G | 0 => x | n + 1 => ⁅engelCommutator x y n, y⁆

An element $y$ of a group $G$ is a (left) Engel element if for every $x \in G$ there is some $n$ with $[x,, {}_n y] = 1$.

def IsEngelElement (y : G) : Prop := ∀ x : G, ∃ n : ℕ, engelCommutator x y n = 1

A nilgroup (Engel group) is a group in which every element is an Engel element.

def IsEngelGroup (G : Type*) [Group G] : Prop := ∀ y : G, IsEngelElement y

Is a group a nilgroup if it is the product of two normal nilsubgroups?

Since $H$ and $K$ are normal, the product $HK$ coincides with the join $H \sqcup K$, so "$G$ is the product of $H$ and $K$" is stated as $H \sqcup K = G$.

@[category research open, AMS 20] theorem kourovka_1_40 : answer(sorry) ↔ ∀ (G : Type) [Group G] (H K : Subgroup G), H.Normal → K.Normal → IsEngelGroup H → IsEngelGroup K → H ⊔ K = ⊤ → IsEngelGroup G := ⊢ True ↔ ∀ (G : Type) [inst : Group G] (H K : Subgroup G), H.Normal → K.Normal → IsEngelGroup ↥H → IsEngelGroup ↥K → H ⊔ K = ⊤ → IsEngelGroup G All goals completed! 🐙end Kourovka.«1.40»