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This is problem 3B.12. from the afore mentioned book. It asks to show that, if $G$ is a solvable Frattini-free group (that is, $Φ(G)=1$) and $H$ is a subgroup of some maximal subgroup $M$ of $G$, then $G$ contains a subgroup of index $|M:H|$.
It is easy to see that, if $N \unlhd G$ and $M\cap N=1$, then $G=MN$, $M\cong G/N $ and $H\cong K/N $ for some $ K \leqslant G $, where $K$ contains $N$. Then $|G:K|=|G/N:K/N|=|M:H| $ and we are done. However, the hypothesis will not necessarily hold. But, if $N$ is minimal among normal subgroups of $G$ (in the sense of containment), then $N$ is elementary abelian, hence nilpotent, which implies that $M \cap N \lneq N_G( M \cap N ) \cap N $. Thus, $M \lneq N_G( M \cap N )$ and by the maximality of $M$ we deduce that $M \cap N \unlhd G$. So, for this particular $M$, if it is not true that there exists some normal subgroup of $G$ with which it intersects trivially, then $M \cap N = N$ by the minimality of $N$. Hence $M$ contains every minimal subgroup of $G$.

How should I proceed from here?

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This exercise is false. Here is a counterexample where M does in fact contain every minimal normal subgroup.

Let G be the alternating group of degree 4 (and order 12), let M be the normal Sylow 2-subgroup of order 4 and let H ≤ M be a subgroup of order 2. Then [M:H] = 2, but G contains no subgroup of index 2.

One has better luck if one assumes M is core-free rather than Φ(G)=1, that is assume ∩{ Mg : g in G } = 1 versus ∩{ K : K maximal, even non-conjugate to M } = 1.

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