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

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?

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?

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Bob
Bob
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The intersection of product of Sylow subgroups

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?