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I really got stuck with the following contradiction.
Say we have a Moufang loop $Q$, $|Q| < \infty$.
To put it briefly, Moufang loops are groups that not necessary be associative, with extra identities. They are also quasigroups with neutral element.

  1. The classical loop solvability is equal to existence for loop $Q$ series $1 = A_0 \triangleleft A_1 \triangleleft\space ... \triangleleft\space A_n = Q$, $A_{i+1} / A_i$ - commutative group;
  2. There is a Feit-Thompson theorem which say that every finite group with odd order is solvable. In this work the theorem is proved for finite Moufang loops;
  3. Simple Moufang loop is a loop without nontrivial proper normal subloops. Finite simple nonassociative Moufang loops are called Paige loops;
  4. There is a nice representation of Paige loops as a 3-generated loops. For Paige loops with odd order $q \neq 9$ the generators can be the following:

    $\begin{pmatrix}1 & (1,0,0)\\\ (0,0,0) & 1\end{pmatrix}$, $\begin{pmatrix}1 & (0,1,0)\\\ (0,0,0) & 1\end{pmatrix}$, $\begin{pmatrix}0 & (0,0,u)\\\ (0,0,u^{-1}) & 1\end{pmatrix}$, where

    $u$ - primitive element of $GF(q)$ (with $q$ elements);

  5. Trivial check on commutativity of second and third generators by rule

    $\begin{pmatrix}a & \alpha\\\ \beta & b\end{pmatrix} \begin{pmatrix}c & \gamma\\\ \delta & d\end{pmatrix} = \begin{pmatrix}ac+\alpha\delta & a\gamma+d\alpha-\beta\times\delta\\ c\beta+b\delta+\alpha\times\gamma & bd+\beta\gamma\end{pmatrix}$

    Gives us that Paige loops are always noncommutative (generally regardless to the parity);

  6. Finally, by definition of solvability
    $Q$ - simple finite solvable loop $\Rightarrow$ $Q$ - commutative (quotient of $Q$ by $1$).
    Particularly by Feit-Thompson theorem, Paige loops with odd order are solvable, hence since they are simple, they must be commutative, but we showed that Paige loops are never commutative. A contradiction.

My question is: where the misfire is? What tricky moment I couldn't see here?

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According to https://groupprops.subwiki.org/wiki/Paige_loop, the order of a Paige loop is $q^3(q^4-1)$ when $q$ is even and $\frac{1}{2}q^3(q^4-1)$ when $q$ is odd. Hence a Paige loop can never have odd order.

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  • $\begingroup$ Yes, this night I realized it, too :) $\endgroup$ – Evgeny Nov 6 '17 at 13:55

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