Conjecture 1.40

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-/

import FormalConjecturesUtil

/-!
# 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](https://arxiv.org/abs/1401.0300v40)
-/

open scoped commutatorElement

namespace 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 := by
  sorry

end Kourovka.«1.40»

• Formalization source ↗

On-site verification needs an executable Lean target. None is attached yet.

A source attachment or link is not a verified proof. The statement’s correspondence with the original question needs human review.