Prove that a finite solvable $T_{nsp}$-group is supersolvable Definition: 
A subgroup $H$ Of a finite group $G$ is called nearly S-permutable in $G$ if for each prime $p$ with $(p,|H|)=1$ and for every subgroup $K$ of $G$ containing $H$ the normalizer $N_K(H)$ contains some Sylow $p$-subgroup of $K$. 
A finite group is called $T_{nsp}$-group if nearly S-permutiability is transitive in the group. 

Problem: Prove that a finite solvable $T_{nsp}$-group is supersolvable.

The only ref I found on the internet is this.
Please help.
 A: Let $G$ be a finite solvable group that is not supersolvable. We need to prove that $G$ is not $T_{\rm nsp}$.
Since $G$ is not supersolvable, it has a chief factor $M/N$ which is elementary abelian or order $p^k$ for some prime $p$ and $k>1$. Let $P$ be a Sylow $p$-subgroup of $M$. By the Frattini Argument, $G = NN_G(P)$, so we can assume that $N_G(P) = G$ i.e. $P \unlhd G$.
Choose a $p$-subgroup $L$ with $P \cap N < L < P$ such that $L/N$ is central in a Sylow $p$-subgroup of $G$. We will prove that $L$ is not S-permutable in $G$ which, since $L$ is S-permutable in $P$, and $P$ is S-permutable in $G$, will contradict transitivity of S-permutability. 
Since $M/N$ is a chief factor of $G$, $L$ cannot be normal in $G$ so, since $N_G(L)$ contains a Sylow $p$-subgroup of $G$, there must be a prime $q \ne p$ and $q$-element $g \in G$ such that $g \not\in N_G(L)$.
Let $K = \langle P,g \rangle$. So $|K| = |P|q^t$, where $g$ has order $q^t$. Then $N_G(L)$ does not contain a Sylow $q$-subgroup of $K$, so $L$ is not S-permutable in $G$, as claimed.
