Let $F$ be a nonarchimedean local field of characteristic zero, and $E$ an extension of $F$ with $[E:F]=2^n$ for some $n$. Is it always possible to find a quadratic extension $M$ of $F$ such that $F\subseteq M\subseteq E$ ?
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4$\begingroup$ Equivalently, if $G$ is a finite group that is a Galois group over a local field of char. zero, and $H$ is a proper subgroup of 2-power index, is it contained in a subgroup of index $2$? $\endgroup$– YCorCommented Dec 17, 2022 at 14:21
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5$\begingroup$ There exist $S_4$-extensions of $\mathbb{Q}_2$, which shows that the answer is no if no further restrictions are made. $\endgroup$– Arno FehmCommented Dec 17, 2022 at 14:36
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Try $f(x)=x^4+2x+2\in \Bbb{Q}_2[x]$ and $K=\Bbb{Q}_2[a]/(f(a))$.
Then $f(x) = (x-a)g(x)\in K[x]$ where $g$ is irreducible ($a^{-3}g(ax-a/3)$ is Eisenstein). So $f$ has only one root in $K$, $Aut(K/\Bbb{Q}_2)$ is trivial and hence there is no subfield $L$ such that $[K:L]=2$ ie. no subfield such that $[L:\Bbb{Q}_2]=2$.
If $F$ is a finite extension of $\Bbb{Q}_p$ with $p$ odd and $[E:F]=2^n,n\ge 1$ then yes there is always a quadratic subfield. Assume the opposite, take an uniformizer $\pi_E$, the residue field of $F$ is $\Bbb{F}_q$, that of $E$ is $\Bbb{F}_{q^m}$, it must be that $m | 2^n$, and it must be that $m=1$ as otherwise $F(\zeta_{q^2-1})/F$ is a quadratic subfield.
Therefore, $E = F(\pi_E)$ where $\pi_E^{2^n} = u \pi_F$ for some $u\in O_E^\times$.
So $u=\zeta_{q-1}^r (1+c\pi_E)$ with $c\in O_E$,
$\varpi_E =(1+c\pi_E)^{-1/2^n} \pi_E\in E$ is again an uniformizer and $\varpi_E^{2^n}=\zeta_{q-1}^r \pi_F\in F$.
ie. $F(\varpi_E^{2^{n-1}})/F$ is a quadratic subextension.
