Let $H$ be a maximal subgroup of a finite group $G$ such that $|G:H|=4$. Then there exists $K\leq H$ such that $|H:K|=3$

Question

Let $H$ be a maximal subgroup of a finite group $G$ such that $|G:H|=4$. Then there exists $K\leq H$ such that $|H:K|=3$.

My attempt: Since maximal subgroups of nilpotent groups have prime index, so $G$ is not nilpotent. (In particular, $G$ is neither Abelian nor a $p$-group.) Also, as $H\leq N(H)\leq G$ and $H$ is maximal, so $N(H)=G$ or $N(H)=H$.

I also know that $G/\bigcap_{g\in G} H^g$ is isomorphic to a subgroup of $S_4$.

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I haven no idea how to proceed and am really lost. Any hints are appreciated.

Answer 1

Since $H$ is maximal, the normalizer must equal $H$ (otherwise, $G/H$ would be of order $4$ and hence contain a proper subgroup of order $2$, yielding a proper subgroup that properly contains $H$).

Thus $H$ has exactly four conjugates. As you note, the action of $G$ on the cosets of $H$ gives a morphism from $G$ to $S_4$, whose kernel is the core of $H$. This means that the image of $G$ in $S_4$ is transitive and has four maximal subgroups of index $4$. The subgroup $H$ maps to one of these.

Of the transitive subgroups of $S_4$ with enough proper subgroups of index $4$ we may exclude $D_8$ (the subgroups of index $4$ are not maximal). This leaves only $A_4$ and $S_4$ for the image of $G$.

The subgroups of index $4$ in $S_4$ are the stabilizers of a single element, isomorphic to $S_3$; these have subgroups of index $3$ which can be lifted to $G$. The subgroups of index $4$ in $A_4$ are isomorphic to $A_3$, which likewise have subgroups of index $3$ which can be lifted to $G$.

Answer 2

$G$ acts transitively on $G/H$. Moreover, since the subgroup $H$ is maximal, the action is primitive. Therefore, we get an morphism $$G \to \mathcal{Sym}(G/H)$$ with kernel the normal core $C(H)$ of $H$, and so $$G/C(H) \hookrightarrow \mathcal{Sym}(G/H)$$ Now $\mathcal{Sym}(G/H) = S_4$, and $G/C(H)$ is a primitive subgroup of $S_4$ ( so also transitive).

Here are classified the transitive groups of degree $4$ ( that is, acting transitively on $4$ elements). Out of these, only $S_4$, and $A_4$ act primitively.

We conclude: $G/C(H)\simeq A_4$, or $S_4$, and so $H/C(H)$ is of order $3$ or $6$. Now we only need a $2$ Sylow $K/C(H)$ of $H/C(H)$... ( all could be done with more details)