If G is finite p-group then $d(G)=d(\frac{G}{\Omega_1(Z(G))})$

Question

Let $G$ be a finite p-group such that $G$ has no non-inner automorphism of order p leaving Φ(G) elementwise fixed
If $\Omega_1(Z(G))\le G'\le \Phi(G)$ how we can get
$d(G)=d(\frac{G}{\Omega_1(Z(G))})$
$\Phi(G)$=frattini subgroup
$G'$=commutator subgroup
$\Omega_m(G)=\langle x\in G|o(x)\le p^m\rangle$

Answer 1

$\Phi(G)$ is subgroup containing non-generator elements like x to see that for every maximal subgroup $M$ we have $\langle M,x\rangle=M$ then non of member of $\Phi(G)$ is generator.
we can assume $d(G)\ge d(\frac{G}{\Phi(G)})$ if we assume $d(G)\lt d(\frac{G} {\Phi(G)})$ then If
$G=\langle x_1,...,x_k\rangle$
$\frac{G}{\Phi(G)}=\langle x_1\Phi(G),...,x_{k'}\Phi(G)\rangle$
when $k\gt k'$ then $G=\langle x_1,...,x_{k'},\Phi(G)\rangle$ then $\Phi(G)$ contain some generator, contradiction. then $d(G)=d(\frac{G}{\Phi(G)})$
now
$\Omega_1(Z(G))\le G'\le \Phi(G)\to \frac{\Phi(G)}{\Omega_1(Z(G))}=\langle a_1\Omega_1(Z(G)),...,a_n\Omega_1(Z(G))\rangle\to \Phi(G)=\langle a_1,...,a_n,\Omega_1(Z(G))$
$\frac{G}{\Omega_1(Z(G))}=\langle a_1\Omega_1(Z(G)),...,a_m\Omega_1(Z(G))\rangle$
$\frac{G}{\Phi(G)}=\langle a_1\Phi(G),...,a_{m'}\Phi(G)\rangle \to m\ge m'$
$\to d(G)\ge d(\frac{G}{\Omega_1(Z(G))})\ge d(\frac{G}{\Phi(G)})=d(G)$
then $d(\frac{G}{\Omega_1(Z(G))})=d(G)$