$G$ 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 $N\_G(P\_s)=P\_s$?



The property of self-normalizing subgroup
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.


