Let G be a finite solvable group, and assume that $\Phi(G) = 1$ where $\Phi(G)$ denotes the Frattini subgroup of G. Let M be a maximal subgroup of G, and suppose that $H \subseteq M$. Show that $G$ has a subgroup with index equal to $|M:H|$.
This is question 3B.12 from Finite Group Theory, by M. Isaacs.
Here is my approach so far. I am completely stuck and would welcome any hints or ideas.
Suppose otherwise. Among all of the counter examples choose $G$ of minimum order. Since $G$ is a counterexample it must be the case that $|G| > 1$. Since $G$ is a counter example there is a maximum subgroup $M$ and a subgroup $H \subset M$, such that every subgroup of $G$ does not have the same index as $|M:H|$. So it must be the case that $H$ is properly contained within $M$.
This is where I get stuck. I want to use a minimal normal subgroup $N$ of $G$ which exists. But my argument devolves into a series of cases about whether or not $N$ intersects $H$ and/or $M$ non-trivially.
I do know that $G$ must have a non-normal maximal subgroup, since if they all were normal then it would be nilpotent and since G is finite this implies supersolvable, then $G$ would have a subgroup for any divisor of its order. Since $\Phi(G)=1$ is the intersection of all the maximal subgroups of $G$ I suspect this should help but I'm not sure where to go from here.