A new question about solvability of a direct product I had asked that if $G$ is a direct product of a $2$-group and a simple group, then is it possible that $G$ be a solvable group. That the answer is no!
But by a Remark on T.M. Gagen, Topics in Finite Groups, London Math. Soc. Lecture Note Ser., vol. 16, Cambridge Univ. Press,
Cambridge, 1976. I am confused! I will write the Remark below:

Definition 11.3. A group $G$ with an abelian Sylow $2$-subgroup is
  said to be an $A^*$-group, if $G$ has a normal series $1\subseteq  N \subseteq  M \subseteq  G$ where $N$ and $G/M$ are of odd order and $M/N$ is a direct product of a $2$-group and simple groups of type $L_2(q)$ or $JR$.
Theorem A. Let $G$ be a finite group with an abelian Sylow
  $2$-subgroup. Then $G$ is an $A^*$-group.
Remark. If a group $G$ has an abelian Sylow $2$-subgroup $T$ of
  rank $1$, then $G$ is solvable and $2$-nilpotent and clearly an $A^*$-group. If
  $T$ has rank $2$, then $G$ is $2$-nilpotent unless $T$ is of type $( 2^\alpha , 2^\alpha  )$ . But then if $\alpha > 1$, $G$ is solvable by [7] and clearly an $A^*$-group since it has $2$-length $1$. Thus we may assume that $|T | = 4$ and then we can apply
  the results of [13]. Again $G$ is an $A^*$-group.

By the Remark, it is possible that $G$ (with abelian sylow $2$-group) be a solvable group. In this case $M$ and then $M/N$ must be solvable, and can not be have a simple group as its direct factor. a contradiction with Definition!
Am I right?! 
 A: This is a matter of convention. When $G$ is solvable, $M/N$ is an (abelian) 2-group. It is still the case that $M/N$ is the direct product of a 2-group and a set of simple groups, each isomorphic to $L_2(q)$ or $J_1$ or $^2G_2(q)$. In case $G$ is solvable, that set of simple groups is empty.
The remark is trying to clarify what happens in the low rank cases, where typically the result is a solvable group. In the $2^a$ case, Cayley showed you can take $G/M=1$, $M/N$ the cyclic Sylow 2-subgroup. In the $2^a \times 2^a$ ($a>1$) case Brauer's result shows $G/M$ can be taken to have order 1 or 3, and $M/N$ to be the homocyclic Sylow 2-subgroup. In the non-homocyclic case, Burnside or Frobenius shows you can take $G/M=1$ and $M/N$ to be the Sylow 2-subgroup. Only in the case $C_2 \times C_2$ (amongst rank 1 or 2) do you get to the case supporting a non-solvable $M/N$.
Rank 3 is interesting because of $J_1$ and $^2G_2(q)$.
I'll mention that Gagen, Bender, and Gorenstein all use the same phrasing, and all intend to allow the groups to be solvable as well as non-solvable.
Another small point: Gagen and Bender use the name “$A^*$-group” rather than “$A$-group” used by Gorenstein. The definitions are actually nearly identical; I believe they are trying to avoid confusion with a similar (weaker) result classifying A-groups in the sense of Hall: groups in which all Sylow subgroups are abelian.
