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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 $S_4$.

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

Shaun
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Guest
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    $H$ cannot be normal: if it were, then $G/H$ would be of order $4$, hence have a subgroup of order $2$ which gives you a proper subgroup of $G$ that properly contains $H$, contradicting maximality. So in fact you know $N(H)=H$. – Arturo Magidin Jul 11 '22 at 17:20
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    Since you are asked to prove this in general you may assume that $\bigcap H^g=1$, since if it is true in that case it is true in general (by taking $K$ for the quotient and taking the preimage). So $G\leq S_4$. – David A. Craven Jul 11 '22 at 17:26

2 Answers2

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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 $G$.

Arturo Magidin
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    You also need to mention $C_4$ and $V_4$ as transitive subgroups of $S_4$ that you exclude for the same reason as $D_8$. – David A. Craven Jul 11 '22 at 17:55
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    @DavidA.Craven They don't have enough subgroups of index $4$, which is why I didn't think about them. But I've added "enough" to that part. – Arturo Magidin Jul 11 '22 at 18:04
  • @ArturoMagidin What does "transitive subgroups" of $S_4$ mean and what does it mean by "image of $G$ in $S_4$ is transitive"? I only know the term "transitive" for an action and I understand that the action of $G$ on cosets of $A$ is transitive. How does transitivity help here? – Guest Jul 11 '22 at 18:19
  • @Guest A subgroup of $S_n$ is transitive if it is transitive in its natural action on ${1,2,\ldots,n}$. Because the morphism from $G$ to $S_4$ is determined by the action of $G$ on the cosets, which is a transitive action, the subgroup you get must be a transitive subgroup of $S_4$. These are known: they are cyclic of order $4$, isomorphic to the Klein $4$ subgroup, isomorphic to $D_8$, $A_4$, or $S_4$. So we know the quotient must be one of those. Then we eliminate all but the last two by the properties we know $G$ had. – Arturo Magidin Jul 11 '22 at 18:29
  • @Guest Others may disagree, but I really think that you have your definitions the wrong way round - you are putting the cart before the horse. The principal and original meaning of the property "transitive" in group theory is that a permutation group (i.e. a subgroup of ${\rm Sym}(X)$ for some set $X$) is transitive if, for all $x,y \in X$, there exists $g \in G$ such that $g$ maps $x$ to $y$. The idea of a group action came much later (when I studied group theory in the early 1970s, the concept of group action was never introduced). ctd. ... – Derek Holt Jul 11 '22 at 18:56
  • An action of a group $G$ on a set $X$ is transitive if its image in ${\rm Sym}(X)$ is a transitive permutation group. – Derek Holt Jul 11 '22 at 18:57
  • @DerekHolt This is how I was taught in class, so I wouldn't really say that "I" am putting the cart before the horse. In fact, in class we never covered the property of being "transitive" in terms of permutation groups. – Guest Jul 11 '22 at 19:06
  • @DerekHolt I think your comment was supposed to go to me... I also learned groups without discussing actions (late 80s, out of Herstein). I think some books still follow the order you suggest, while others define group actions first in abstract and then relate them to permutation groups. Six of one, half a dozen of the other as long as both are covered and connected. – Arturo Magidin Jul 12 '22 at 17:02
  • @ArturoMagidin Last line in the second paragraph: would it be at least $4$ maximal subgroups of index $4$? – user371231 Aug 10 '22 at 10:39
  • @user371231 It says "has four subgroups of index 4", not "has exactly four subgroups of index 4". – Arturo Magidin Aug 10 '22 at 13:35
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$G$ 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)

orangeskid
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