Say we have a finite [Moufang Loop](https://en.wikipedia.org/wiki/Moufang_loop) $Q$, $|Q|<\infty$.  

There is a [theorem proved](http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.561.4372&rep=rep1&type=pdf) 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](https://en.wikipedia.org/wiki/Feit%E2%80%93Thompson_theorem) about solvability of finite groups with odd order translates to the Moufang loop case. The proof can be found [here](http://www.sciencedirect.com/science/article/pii/0021869368900501). 


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



Portability of Thompson theorem about solvability to Moufang loops
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](https://www.researchgate.net/publication/231964785_Lagrange%27s_theorem_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](https://groupprops.subwiki.org/wiki/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](http://www.cs.du.edu/~petr/data/papers/finite_simple_Moufang_loops.pdf). 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](http://finitesimple.blogspot.ru/2013/01/solvable-simple-groups-have-prime-order.html) (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.


