The following problem is from the book "Finite Group Theory" by Martin Isaacs.
(3.B.12) Let $G$ be solvable and assume that $\Phi(G) = 1$. Let $M$ be a maximal subgroup of $G$ and suppose that $H \leq M$. Show that $G$ has a subgroup with index equal to $[M:H]$
I've been trying to solve this problem for quite some time, but all the time I'm coming up empty. As the lecture is about complements, especially the Schur-Zassenhaus Theorem I tried to prove that $M$ has a normal complemented $L$ in $G$, as then $[G:HL] = [M:H]$ and $HL$ is the wanted subgroup. Unfortunatelly I can't seem to prove this one.
I took a look at two cases. In the first assume that there is a minimal normal subgroup $L$ of $G$, s.t. $L \not \leq M$. Then by the maximality of $M$ we get $G = ML$. Also we have $L \cap M \unlhd M$ and $L \cap M \unlhd L$ (as $L$ is elementary abelian, as $G$ is solvable), which gives us that $L \cap M \unlhd G$. As $L \cap M \not = L$ from the minimality of $L$ we get $L \cap M = 1$, so $L$ is a complement of $M$. But I can't solve the case when all minimal normal subgroups of $G$ are in $M$. I tried using induction to solve this question, but the problem arrises when $H$ doesn't contain any normal non-trivial subgroup of $G$, as then I can't reduce the problem to a factor group from smaller order.
Anyway I beleive the problem has to do something with a minimal normal group, as if for some $T \leq G, [G:T] = [M:H]$, then $|T| = p^k \cdot |H|$. But I can't prove the existence of a minimal normal group of order $p^k$, nor I can prove that it intersects trivially with $H$.
I notice that I'm not using the condition $\Phi(G) = 1$, but I can't see it how. The only way I managed to achieve something by using it is to prove that every minimal normal subgroup of $G$ has a complement in $G$, but unfortunatelly this doesn't prove that every maximal subgroup is complemented.