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Friedrich Knop
Friedrich Knop
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Proofreading
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LSpice
LSpice
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$G$ is a finite solvable group. Let$\{P_{1}, P_{2}, \ldots , P_{s}\}$ $\{P_{1}, P_{2}, \dotsc , P_{s}\}$ be a Sylow basis of $G$. We have that $G=P_{1}P_{2}\cdots P_{s}$$G=P_{1}P_{2}\dotsm P_{s}$. Set \begin{equation} \begin{aligned} T=\prod\limits_{t=1}^{s-1}P_t, H=\prod\limits_{k\neq3}^sP_k, K=\prod\limits_{r\neq2}^sP_r.\nonumber \end{aligned} \end{equation} \begin{equation} \begin{aligned} %% The alignment is never used …. T=\prod\limits_{t=1}^{s-1}P_t, H=\prod\limits_{k\neq3}^sP_k, K=\prod\limits_{r\neq2}^sP_r.\nonumber \end{aligned} \end{equation} Suppose that $T$ is nilpotent  (i.e. $T=P_{1}\times P_{2}\times P_{3}\times \cdots \times P_{s-1}$$T=P_{1}\times P_{2}\times P_{3}\times \dotsb \times P_{s-1}$), $N_H(P_s)=P_s$ and $N_K(P_s)=P_s$.

Can we get that $N_G(P_s)=P_s$?

$G$ is a finite solvable group. Let$\{P_{1}, P_{2}, \ldots , P_{s}\}$ be a Sylow basis of $G$. We have that $G=P_{1}P_{2}\cdots P_{s}$. Set \begin{equation} \begin{aligned} T=\prod\limits_{t=1}^{s-1}P_t, H=\prod\limits_{k\neq3}^sP_k, K=\prod\limits_{r\neq2}^sP_r.\nonumber \end{aligned} \end{equation} Suppose that $T$ is nilpotent(i.e. $T=P_{1}\times P_{2}\times P_{3}\times \cdots \times P_{s-1}$), $N_H(P_s)=P_s$ and $N_K(P_s)=P_s$.

Can we get that $N_G(P_s)=P_s$

$G$ is a finite solvable group. Let $\{P_{1}, P_{2}, \dotsc , P_{s}\}$ be a Sylow basis of $G$. We have that $G=P_{1}P_{2}\dotsm P_{s}$. Set \begin{equation} \begin{aligned} %% The alignment is never used …. T=\prod\limits_{t=1}^{s-1}P_t, H=\prod\limits_{k\neq3}^sP_k, K=\prod\limits_{r\neq2}^sP_r.\nonumber \end{aligned} \end{equation} Suppose that $T$ is nilpotent  (i.e. $T=P_{1}\times P_{2}\times P_{3}\times \dotsb \times P_{s-1}$), $N_H(P_s)=P_s$ and $N_K(P_s)=P_s$.

Can we get that $N_G(P_s)=P_s$?

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Moomoo Angel line
Moomoo Angel line
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The property of self-normalizing subgroup

$G$ is a finite solvable group. Let$\{P_{1}, P_{2}, \ldots , P_{s}\}$ be a Sylow basis of $G$. We have that $G=P_{1}P_{2}\cdots P_{s}$. Set \begin{equation} \begin{aligned} T=\prod\limits_{t=1}^{s-1}P_t, H=\prod\limits_{k\neq3}^sP_k, K=\prod\limits_{r\neq2}^sP_r.\nonumber \end{aligned} \end{equation} Suppose that $T$ is nilpotent(i.e. $T=P_{1}\times P_{2}\times P_{3}\times \cdots \times P_{s-1}$), $N_H(P_s)=P_s$ and $N_K(P_s)=P_s$.

Can we get that $N_G(P_s)=P_s$