Suppose *G* is solvable,and π(*G*)= {2,m,n}, O_{2}(G)=1.Then can we use the solvability of *G* proof 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?