Portability of Thompson theorem about solvability to Moufang loops

Question

Say we have a finite Moufang Loop $Q$, $|Q|<\infty$.
There is a theorem proved by Thompson that states:
Group $G$, $|G|<\infty$ is solvable $\iff$ $\forall a, b \in G \langle a, b\rangle$ is solvable.

My question is: can we translate the theorem to the finite Moufang loops (may be with some extra constraints)?

The statement is reasonable because it is well-known that every pair of Moufang loop's elements forms a group structure (Moufang's theorem).
What is interesting, Feit–Thompson theorem about solvability of finite groups with odd order translates to the Moufang loop case. The proof can be found here.

There is an idea to translate the task to the case of $U(R)$, where $R$ is alternative algebra, $1 \in R$.

Answer 1

Well, I've constructed a chain of proposition which conclude Thompson's theorem being translated to the case of finite Moufang loops in general case. Here we go:

"$\Rightarrow$". Consider $Q$ be a Moufang loop. The solvability property inherits to subloops (thanks we have Lagrange property for Moufang loops). Then each two-generated subloop of $Q$ is solvable.

"$\Leftarrow$". Suppose each two-generated subloop is solvable. By induction we can reduce the task of $Q$-solvability to the case of simple (finite) Moufang loop. But such a loop is either a group (if associative) or a Paige loop (if unassociative). For groups we know the result of Thompson theorem. Now we have a Paige loop. Any Paige loop is a non-commutative three-generated loop. We must get the Paige loop to be solvable if any two-generated subloop is solvable. But anyway if Paige loop is solvable, then it is abelian (by virtue of a fact that it is simple and according to the group relative propositions), a contradiction. Thus Paige loops are never solvable, then whether Moufang loop has two-generated subloops solvable it does not contain a Paige loop, which is equal to say that every simple subloop is a group.