A subgroup of index 2 for a solvable finite group transitively acting on a finite set of cardinality $2m$ where $m$ is odd

Question

Let $G$ be a finite group acting transitively on the finite set $X=\{1,2,\dots, 2m\}$ of cardinality $2m\ge 6$ where $m$ is odd.

Question 1. Is it true that $G$ always has a subgroup $H$ of index 2 containing the stabilizers ${\rm Stab}(x)$ for $x\in X$ ?

Answer: No. A counterexample is the alternating group $A_{2m}$ acting on $X$ for $2m\ge 6$. Since the group $A_{2m}$ is simple, it has no nontrivial normal subgroups, hence no subgroups of index 2.

Question 2. Assume additionally that $G$ is solvable. Is it true that $G$ always has a subgroup $H$ of index 2 containing the stabilizers ${\rm Stab}(x)$ for $x\in X$ ?

If the answer to Question 2 is "No", I would like to consider a special class of solvable groups $G$.

Question 3. Assume additionally that $G$ is the Galois group of a finite Galois extension $L/K$ where $K$ is a $p$-adic field (that is, a finite extension of the field of $p$-adic numbers ${\Bbb Q}_p)$, and $p>2$. Is it true that $G$ always has a subgroup $H$ of index 2 containing the stabilizers ${\rm Stab}(x)$ for $x\in X$ ?

Question 3 is related (equivalent?) to the following question:

Question 4. Let $L/K$ be a finite extension of $p$-adic fields for $p>2$ (not necessarily a Galois extension) of degree $2m\ge 6$ where $m$ is odd. Is it true that $L$ always contains a quadratic extension $F$ of $K$ ?

I expect answers "No" to all questions and would be happy to see counter-examples.

Answer 1

At @MikhailBorovoi's request, I copy here two comments 1 2 from p-adic field extension of degree 2n without a subfield of degree 2? as an answer to Question 4. (The comment suggested that they answer Question 3, but I don't immediately see that, almost certainly by my own failure.)

The answer to Question 4 is Yes.

If the residual degree of $L/K$ is even, then we may choose $F/K$ quadratic unramified; so assume it's odd. If the residual characteristic $p$ doesn't divide $[L : K]$, then $L/K$ is tame, so one can write $L = E(\sqrt[e]\varpi)$ for some unramified $E/K$ and uniformiser $\varpi$ of $E$, where $e$ is the ramification degree of $L/K$. Since $E/K$ has odd degree, there is a uniformiser $\varpi'$ of $K$ such that $\varpi^{-1}\varpi'$ projects to a square in the residue field of $E$, and then $F \mathrel{:=} K(\sqrt{\varpi'})$ is contained in $L$.

Since $p \ne 2$, the maximal tame subextension of $L/K$ is also of even degree, hence admits a quadratic subextension by the above argument.

Answer 2

Let's say that a [pro]finite group $G$ is $p$-supersolvable if for all [closed] normal subgroups $N_1\subset N_2$ of $G$ with $N_1$ maximal $G$-normal proper subgroup of $N_2$, the group $N_2/N_1$ is elementary abelian, and if $N_2/N_1$. is $p$-group then it is cyclic.

If $G$ is a $2$-supersolvable finite group and $H$ is a subgroup with $|G/H|$ in $4\mathbf{Z}+2$, then $H$ is contained in a subgroup of index 2.

Proof: we can suppose that $H$ is maximal for the property of having even index. We can also suppose that $G$ acts faithfully on $G/H$ (i.e. $H$ has trivial core). Let $C$ be a minimal nontrivial normal subgroup of $G$. Since $H$ has trivial core in $G$, $C$ is not contained in $H$ and hence $HC$ strictly contains $H$. Maximality implies that $HC$ has odd index; in particular $C$ must have even order, and hence, since $G$ is $2$-supersolvable, $C$ is cyclic of order $2$, hence is central in $G$. Write $C=\{1,z\}$.

Let $D/C$ be a minimal nontrivial normal subgroup of $G/C$, so it is also cyclic of odd order. Since $D$ is central-by-cyclic, it is abelian. If $D/C$ has odd order, we deduce that $D=C\times C'$ for $C'$ of odd prime order, and thus $C'$ is normal in $G$, which we contradicted in the previous paragraph. Hence $D$ has order $4$. The group $D\cap H$ cannot be trivial, since $|G/H|$ would be a multiple of $4$. Hence $D\cap H$ has order $2$, say equals $\{1,z'\}$; write $D=\{1,z,z',z''\}$. It is normal of order two in $H$, hence is central in $H$. Let $P$ be the centralizer of $z'$ in $G$. Since $H$ has trivial core, $z'$ is not central in $G$ and hence $P\neq G$. Since $z$ is central, we deduce that the only $G$-conjugate of $z'$ is $z''$ and hence $P$ has index 2 and contains $H$.

Answer 3

LSpice answered my Question 4 in the positive. Here I deduce the positive answer to Question 3, which I restate as Question 3'.

Question 3'. Let $G$ be a finite group acting transitively on the finite set $X=\{1,2,\dots, 2m\}$ of cardinality $2m\ge 6$ where $m$ is odd. Assume additionally that $G$ is the Galois group of a finite Galois extension $L/M$, where $M$ is a $p$-adic field (that is, a finite extension of the field of $p$-adic numbers ${\Bbb Q}_p)$, and $p>2$. Is it true that $G$ always has a subgroup $H$ of index 2 containing the stabilizers $G_x$ for all $x\in X$ ?

The answer is Yes.

Let $G_1\subset G$ be the stabilizer of $1\in X$, and let $K_1\subset L$ denote the corresponding subfield, that is, $K_1=L^{G_1}$. Then $[G:G_1]=2m$ and $[K_1:M]=2m$. By the answer of @LSpice, there exists a quadratic subextension $F/M$ in $K_1$, that is, $F\subset K_1$ and $[F:M]=2$. Let $H\subset G$ be the subgroup corresponding to the extension $F/M$, that is, $$ H=\{g\in G\ \,|\ \,{}^g a=a \ \ \forall a\in F\}.$$ Then $H$ is a subgroup of $G$ of index 2, and hence it is normal in $G$. Since $K_1\supset F$, we see that $G_1\subset H$. Since $G$ acts on $X$ transitively, for any $x \in X$ the stabilizer $G_x$ is conjugate in $G$ to $G_1$, and hence is contained in the normal subgroup $H$ of $G$. Thus $H$ is a normal subgoup of $G$ of index 2 containing all stabilizers $G_x\,$, as required.