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  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 . 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?!