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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.

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    $\begingroup$ Easy negative answer to Question 2. Take $G=A_4=C_3\ltimes C_2^2$ and $H$ subgroup of order 2. Then $G$ is solvable, acts transitively on $G/H$ which has cardinal 6, but $G$ has no subgroup of index 2. (This is the smallest solvable group of even order with no subgroup of index 2.) Is $A_4$ a $p$-adic Galois group? $\endgroup$ Commented Oct 21, 2023 at 6:42
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    $\begingroup$ I don't think that $A_4$ can be a $p$-adic Galois group for $p$ odd. Suppose that $L/K$ is $A_4$, and let $F$ be the fixed field of $C_2^2$, so $L/F$ is a $C_2^2$ extension. We must have $L = F(\sqrt{\pi}, \sqrt{u})$ where $\pi$ is a uniformizer of $F$ and $u \in \mathcal{O}_F$ is a non-QR modulo $\pi$. The three fields intermediate between $L$ and $F$ are then $F(\sqrt{\pi})$, $F(\sqrt{u\pi})$ and $F(\sqrt{u})$. Then $C_3 = \text{Gal}(L/F)$ is supposed to permute these intermediate fields transitively, but $F(\sqrt{\pi})/F$ and $F(\sqrt{u\pi})/F$ are ramified, whereas $F(\sqrt{u})/F$ is not. $\endgroup$ Commented Oct 21, 2023 at 7:22
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    $\begingroup$ Regarding "must have": Because $F^{\ast}/ (F^{\ast})^2$ has order $4$, with $(1, u, \pi, u \pi)$ representing the cosets. When $p=2$, this isn't true. $\endgroup$ Commented Oct 21, 2023 at 7:23
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    $\begingroup$ Ah, indeed I can read at mathoverflow.net/q/172569/14094 that a finite $p$-adic Galois group has to be supersolvable (= has normal series with cyclic subquotients). This indeed excludes $A_4$. $\endgroup$ Commented Oct 21, 2023 at 7:57
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    $\begingroup$ It's indeed true that a finite supersolvable group of even order has a subgroup of index 2. $\endgroup$ Commented Oct 21, 2023 at 8:17

3 Answers 3

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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.

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  • $\begingroup$ Since the answer to Question 4 is Yes, the answer to Question 3 is Yes as well. Here $L/K$ is a finite Galois extension. The subgroup $H$ of index 2 in the Galois group $G={\rm Gal}(L/K)$ that I have in mind is $H={\rm Gal}(L/F)$ where $F$ is the quadratic subextenion of $L/K$ constructed in the answer. $\endgroup$ Commented Oct 22, 2023 at 16:42
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    $\begingroup$ LSpice: I have edited your answer to make the notations compatible with those of the question. I also stated explicitly that the answer to Question 4 is Yes. Please check that I have not made mistakes. $\endgroup$ Commented Oct 22, 2023 at 16:53
  • $\begingroup$ @MikhailBorovoi, re, yes, it looks correct to me. Sorry, I noticed that the extension field was called $L$, and don't seem to have bothered to have checked the name of the ground field! Re, I understand how we get an index-$2$ subgroup $H$, but don't see why $H$ must contain each $\operatorname{Stab}(x)$, again almost certainly by my own failure. $\endgroup$ Commented Oct 22, 2023 at 17:38
  • $\begingroup$ I don't understand the sentence: "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 $L$." Could you please add details? $\endgroup$ Commented Oct 22, 2023 at 20:13
  • $\begingroup$ Re, let $\varpi'$ be any uniformiser of $K$, and let $q$ be the order of the residue field of $K$. If $\varpi^{-1}\varpi'$ projects to a square in the residue field of $E$, then we are done. Otherwise, since $q$ and $f$ are odd, so is $\frac{q^f - 1}{q - 1}$ (because it's $q f$ modulo $2$); so not every $(q - 1)$st root of unity is a square in $\mathbb F_{q^f}$, so we can adjust $\varpi'$ by a non-square root of unity in $K$ so that now $\varpi^{-1}\varpi'$ is a square. … $\endgroup$ Commented Oct 22, 2023 at 20:48
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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 $H$.

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  • $\begingroup$ I'm not sure this answer should be accepted. I wrote it assuming that Galois groups of $p$-adic fields are supersolvable, but this is false, at least for $p=2$. $\endgroup$ Commented Oct 21, 2023 at 8:57
  • $\begingroup$ Nevertheless, I'm using only the following group property (let's call it $2$-supersolvable): $G$ is finite and every quotient of $G$ has the property that its minimal nontrivial subgroups of even order are cyclic. This is strictly stronger than solvable, but strictly weaker than supersolvable. $\endgroup$ Commented Oct 21, 2023 at 9:12
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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.

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