Here is what I assume:
$P$ is a $p$-group and there is a series of subgroups $P_0\unlhd P_1\unlhd \cdots\unlhd P_m = P$ such that $P_i\unlhd P$ for each $0\le i\le m$ and $P_0\le \Phi(P)$. We also have $G\le\text{Aut}(P)$ such that $G$ normalizes each $P_i$. Let $H\le G$ be the subgroup of those $\alpha\in G$ that acts trivially on $P_i/P_{i-1}$ for each $1\le i \le m$.
We note that there is a normal series of $P/\Phi(P)$:
$$1 = P_0\Phi(P)/\Phi(P) \unlhd P_1\Phi(P)/\Phi(P) \unlhd \cdots \unlhd P_m\Phi(P)/\Phi(P) = P/\Phi(P).$$
And here is what I want to show:
- that $H/C_H(P/\Phi(P))$ normalizes $P_i\Phi(P)/\Phi(P)$ for each $i$ and
- that $H/C_H(P/\Phi(P))$ acts trivially on $(P_{i+1}\Phi(P)/\Phi(P))/(P_i\Phi(P)/\Phi(P)) \cong P_{i+1}\Phi(P)/P_i\Phi(P)$ for each $i$.
I think it should be obvious, and I get why $H/C_H(P/\Phi(P))$ normalizes $P_i\Phi(P)/\Phi(P)$ since $H$ acts on $P/\Phi(P)$ and so $H/C_H(P/\Phi(P))$ acts on $P_i\Phi(P)/\Phi(P)$ via
$$xy\Phi(P) \stackrel{\alpha C_H(P/\Phi(P))}{\longmapsto} (x\alpha)(y\alpha)(\Phi(P)\alpha) \in P_i \Phi(P)/\Phi(P),$$ because $x\in P_i$ which is normalized by $\alpha\in H$ and $y\in \Phi(P)$ is normalized by $H$.
But I can't see why $H/C_H(P/\Phi(P))$ acts trivially on $P_{i+1}\Phi(P)/P_i\Phi(P)$... Can someone help?
I'll write $x^\alpha$ instead of $x\alpha$ so that my holomorphs are multiplicative (you may be writing $P/\Phi(P)$ as row vectors, thinking more additively).
Acting trivially ($\alpha$ on $P_i/P_{i-1}$) means $[x,\alpha] := x^{-1} x^\alpha \in P_{i-1}$ for every $x \in P_i$. Normalizing ($\alpha$ on $P_i$) means $[x,\alpha] \in P_i$ for every $x \in P_i$.
Now $\alpha$ acts trivially on $P_i\Phi(P)/P_{i-1}\Phi(P)$ since $[x, \alpha ] \in P_{i-1}$ for every $x \in P_{i-1}$ and $[y,\alpha] \in \Phi(P)$ for every $y \in \Phi(P)$.
Then just use that $[xy,\alpha] = [x,\alpha][[x,\alpha],y][y,\alpha] \in P_{i-1} P_{i-1} \Phi(P)$ since $y$ (and all of $P$) normalizes $P_{i-1}$.