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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Yes, we can conclude that $O_{2'}(G) \ne 1$ (a minimal normal subgroup is contained in $O_{2'}(G)$), and no we cannot conclude that $O_2(\bar{G}) = \bar{R}$ with $R \in {\rm Sy}l_2(G)$.
For a counterexample, let $M$ be a faithful irreducible module for $S_4$ of dimension $3$ over ${\mathbb F}_5$, and let $G = M \rtimes S_4$ be the semidirect product with the action induced from the module.
Then $|G| = 8 \cdot 3 \cdot 125$, $\pi(G) = \{2,3,5\}$, $O_{2'}(G) = M$, and $O_{2'}(\bar{G})$ has order $4$, and is equal to $\bar{N}$, where $N$ is the normal subgroup of $S_4$ of order $4$.
answered Apr 19, 2022 at 13:54
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$\begingroup$ So we can not determine the $O_{2}(\bar{G})$? $\endgroup$– BobCommented Apr 19, 2022 at 14:34
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$\begingroup$ I don't know what you are asking. What do you mean by "determine the $O_2(\overline{G})$"? $\endgroup$ Commented Apr 19, 2022 at 15:21
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$\begingroup$ I mean,under the conditions that $G$ is solvable, $\pi(G)= \{2,m,n\}$, $O_{2}(G)=1$.Can we figure out $O_{2}(\bar{G})$ by comparing the orders? $\endgroup$– BobCommented Apr 20, 2022 at 0:45