Let $P$ be a finite $p$-group and $Q$ a proper subgroup of $P$. Define the normalizer tower of $Q$ in $P$ as follow:
\begin{equation} N^0(Q) = Q \mbox{ and } N^i(Q) = N_P(N^{i-1}(Q)) \end{equation} for $1 \leq i \leq m$, where $m$ is the smallest positive integer satisfying $N^m(Q) = P$.
We say that $Q$ has maximal normalizer tower in $P$ if $[P : Q] = p^m$ (i.e. $[N^i : N^{i-1}] = p$ for all $1 \leq i \leq m$).
Denote the Frattini subgroup of a group $G$ by $\Phi(G)$.
I want to show that if $\Phi(N^i) \leq N^{i-2}$, then $N^{i-2} \vartriangleleft N^i$.
I try to show that $N^{i-2}$ is a maximal subgroup of $N^i$ in this case (since $N^i$ is nilpotent, every maximal subgroup is normal and we are done). But $N^{i-2} \lneq N^{i-1} \lneq N^i$, so it doesn't seem like a good idea...
What can I do? Thanks.
In a $p$-group, the quotient by the Frattini subgroup is elementary abelian. It follows that every subgroup containing the Frattini is normal. What you want immediately follows. As you pointed out, this eventually leads to a contradiction.