My doctor gave me the following theorem:
Theorem: Suppose that $G$ is a finite non-abelian simple group. Then there exists an odd prime $p \in \pi (G)$ such that $G$ has no $\{2,p\}$-Hall subgroup.
My question is the following:
Suppose that $G$ is a finite group with $2$ divides the order of $G$ and $G$ has a $\{2,p\}$-Hall subgroup for every odd prime $p \in \pi (G)$. Does this imply that either $G$ is abelian or $G$ has a normal subgroup $N$ such that $\{1\} \neq N \neq G$ where $1$ is the identity of $G$?
Thanks in advance.