Is this group solvable?

Question

Let $G=H\times K$ (a direct product of $H , K$) where $H$ is an abelian 2-group and $K$ is a non-abelian simple group. Is $G$ solvable? why?

These three answers are true, and I thought so. But in a book I found a remark that made me confused: T.M. Gagen, Topics in Finite Groups, London Math. Soc. Lecture Note Ser., vol. 16, Cambridge Univ. Press, Cambridge, 1976, Remark, Theorem A on p. $40 $ and Definition $11.3$ on p. $39$. Please see it and say me your ideas.

Answer 1

Hint: Use the definition of a solvable group, and note that if $H$ is a normal subgroup of $G$, then $G$ is solvable if and only if both $H$ and $G/H$ are solvable.

Put differently, a solvable group cannot have a non-abelian simple group as a composition factor.

Answer 2

HINT: Show that $N=\{e\}\times K$ is a normal subgroup of $G$. Argue why $N$ is not solvable, and use the fact that $G$ is solvable if and only if $N$ and $G/N$ are solvable for a normal subgroup $N$.

Answer 3

Subgroups of soluble groups are soluble, which clearly proves your result. To see this, note that if $G$ is soluble of length $n$ then every element of $G^n=[G, G, \ldots, G]$ (so, commutators of length $n$) is trivial. Then clearly if $H\leq G$ then $H^n\leq G^n$, so all commutators of length $n$ in $H$ are trivial. Thus, $H$ is soluble.