The property of self-normalizing subgroup

Question

$G$ 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 $N_G(P_s)=P_s$?

Answer 1

Yes. Nilpotency of $T$ is not needed. Assume that $N_G(P_s)>P_s$. Then for some $i<s$, $N_G(P_s)$ has a Sylow $p_i$-subgroup $Q_i\ne 1$. Assume without loss that $p_i$ divides $|H|$. Then $P_iP_s$ is a Hall $\{p_i,p_s\}$-subgroup of $H$ and $G$. By Hall's extension of Sylow's theorems to solvable groups, $(Q_iP_s)^g\le P_iP_s$ for some $g\in G$. Then $P_s$ and $P_s^g$ are Sylow $p_s$-subgroups of $P_iP_s$, so for some $k\in P_iP_s$, $P_s^{gk}=P_s$. Now $Q_i^{gk}P_s=(Q_iP_s)^{gk}\le P_iP_s\le H$. But $Q_i$ normalizes $P_s$, so $Q_i^{gk}$ normalizes $P_s^{gk}=P_s$. Therefore $Q_i^{gk}$ is a non-identity $p_i$-subgroup of $N_H(P_s)=P_s$, an impossibility.