Suppose $G$ is solvable, and $\pi(G)= \{2,m,n\}$, $O_{2}(G)=1$. Then can we use the solvability of $G$ to prove that $O_{2^{\prime}}(G) \neq 1$? Let $\bar{G}= G / O_{2^{\prime}}(G)$, what about $O_{2}(\bar{G})$? I think $O_{2}(\bar{G})=\bar{R}$, $R$ is the Sylow 2-subgroup of $G$. Is this right? What is the specific proof?