I'm reading a paper and it says the following theorem implies that the Frobenius complement of any finite Frobenius group is not a non-abelian simple group.
(Zassenhaus 1936) Let $G$ be a finite Frobenius group and $A$ be its Frobenius complement. Then the Sylow $p$-subgroups of $A$ are cyclic for odd $p$ and are either cyclic or generalized quaternion for $p=2$.
I know that if $A$ has cyclic Sylow $2$-subgroups or no Sylow $2$-subgroups, then $G$ is solvable and hence not a non-abelian simple group. That follows from the fact that if all Sylow subgroups of a group are cyclic then this group is solvable. But I don't have any idea about how to deal with the case where the Sylow $2$-subgroups are generalized quaternion. Maybe this question is easy and the reason why I'm stuck is that I'm not familiar with generalized quaternion groups. Any help is sincerely appreciated. Thank you!
The Brauer-Suzuki theorem states that if $G$ has generalized quaternion Sylow $2$-subgroups and $O_{2'}(G)=1$ (i.e., $G$ has no normal subgroups of odd order) then $|Z(G)|=2$. Thus there is no simple group with generalized quaternion Sylow $2$-subgroups.