= Solution
The <Frattini subgroup> is the intersection of the <maximal subgroups> of $G$, with $\Phi(1)=1$. A useful equivalent property is
$$
H\Phi(G)=G\quad\Longrightarrow\quad H=G\qquad(H\le G).
$$
Indeed, a proper <subgroup> $H$ lies in a <maximal subgroup>, which also contains $\Phi(G)$. The <Frattini subgroup> is a <characteristic subgroup>, since <automorphisms> permute the <maximal subgroups>.
The <Fitting subgroup> is the largest <nilpotent normal subgroup>. To prove its existence, first form the <p-core> $O_p(G)$: the product of all normal <p-subgroups>. For two such <subgroups>, their product is normal and has order $|AB|=|A||B|/|A\cap B|$, a power of $p$. There are only finitely many <subgroups>, so $O_p(G)$ exists. For distinct primes $p,r$,
$$
[O_p(G),O_r(G)]\le O_p(G)\cap O_r(G)=1.
$$
Thus the product of the <p-cores> is a <nilpotent normal subgroup>, a <direct product of groups> of its <Sylow subgroups>. Conversely, a finite <nilpotent normal subgroup> $K$ has unique <Sylow subgroups>, each a <characteristic subgroup> of $K$. Each is normal in $G$ and lies in the corresponding <p-core>. Consequently \b[the required largest <subgroup> exists], and
$$
\boxed{F(G)=\prod_{p\mid |G|}O_p(G).}
$$
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