= Solution
For a finite group, define the <Frattini subgroup> by
$$
\Phi(G)=\bigcap_{M\text{ maximal in }G}M;
$$
the intersection is $G$ when there are no <maximal subgroups>, which for a finite group happens only for $G=1$. A second definition is the set of <non-generators>: an element $x$ such that $\langle S,x\rangle=G$ always implies $\langle S\rangle=G$.
If $x\in\Phi(G)$ and $H=\langle S\rangle<G$, put $H$ in a <maximal subgroup> $M$. Then $x\in M$, so $\langle S,x\rangle\le M<G$. Conversely, if $x\notin\Phi(G)$, some <maximal subgroup> $M$ omits it; maximality gives $\langle M,x\rangle=G$, although $M\ne G$. This proves equivalence.
Every <group automorphism> permutes <maximal subgroups>, so $\Phi(G)$ is a <characteristic subgroup>, in particular a <normal subgroup>. Also
$$
H\Phi(G)=G\quad\Longrightarrow\quad H=G:
$$
a proper $H$ would lie in a <maximal subgroup> containing both factors.
Set $F=\Phi(G)$ and take a <Sylow subgroup> $P$ of $F$. For $g\in G$, normality of $F$ makes $P^g$ a <Sylow subgroup> of $F$. Conjugating inside $F$ therefore gives the <Frattini argument> factorization $G=F N_G(P)$. The preceding generating property forces $N_G(P)=G$. Thus every <Sylow subgroup> of $F$ is normal, and part (i) shows that $F$ is a <nilpotent group>.
Now suppose $G$ is a <finite p-group>. The <normalizer> condition makes every <maximal subgroup> $M$ normal: $M<N_G(M)$ and maximality force $N_G(M)=G$. The quotient $G/M$ has no proper nontrivial <subgroup>, hence is cyclic of prime order; its p-power order makes that prime $p$. Consequently every such $M$ contains
$$
D=[G,G]G^p,
$$
where the two factors are the <commutator subgroup> and the <subgroup> generated by pth powers. Thus $D\le\Phi(G)$.
The quotient $G/D$ is an <elementary abelian group>, hence a <vector space> over $\mathbb F_p$. If $x\notin D$, extend the nonzero <vector> $xD$ to a <basis> and choose a <linear functional> nonzero on it. Its kernel pulls back to a <subgroup> of index $p$, a <maximal subgroup> omitting $x$. Therefore $x\notin\Phi(G)$, proving
$$
\boxed{\Phi(G)\text{ is characteristic and nilpotent},\qquad
\Phi(G)=[G,G]G^p\text{ for finite p-groups}.}
$$
The two factors in the last formula are multiplied; the superscripted expression in the TeX aid is not the formula printed in the PDF.
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