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$\DeclareMathOperator\Syl{Syl}$$G$ is a finite solvable group and $G=Q\rtimes(P\times R)$, where $P\in \Syl_{p}(G)$ with $P$ is cyclic, $Q\in \Syl_{q}(G)$ with $Q$ is normal elementary abelian, $R\in \Syl_{2}(G)$ with $|R|=2$. Furthermore, $C_{P}(R)=P$ and $C_{Q}(R)=1$.

Let $M_1$ be a maximal subgroup of $PQ$. By the solvability of $G$, we have that $p\mid|PQ:M_1|$ or $q\mid|PQ:M_1|$. If $p\mid|PQ:M_1|$, then $Q\leq M_1$. Can we get that $M_1\unlhd G$ when $p\mid|PQ:M_1|$?

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  • $\begingroup$ In that case $M_1=QP_1$ with $P_1$ maximal in $P$, so yes. (There is no need to assume that $P$ is cyclic.) $\endgroup$
    – Derek Holt
    Commented Mar 3, 2023 at 8:00
  • $\begingroup$ How do we determine that $(P_1Q)^x=P_1Q$ for any $x\in Q$? $\endgroup$
    – Moomoo Angel line
    Commented Mar 3, 2023 at 8:33
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    $\begingroup$ $[x,P_1Q] \le Q$. Or just use the correspondence theorem on subgroups of $PQ/Q$. $\endgroup$
    – Derek Holt
    Commented Mar 3, 2023 at 8:39
  • $\begingroup$ Thank you very much! $\endgroup$
    – Moomoo Angel line
    Commented Mar 3, 2023 at 9:02

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