The idea is to try to deduce the result for $F$ by induction on $n$.
First, to get your feet wet, you want to show that if $[F:\mathbb{Q}]=2$, then all elements of $F$ are constructible. Feel comfortable with that one.
Then, how would the general proof by induction work?
You want to argue that an $F$ that is Galois over $\mathbb{Q}$ with $[F:\mathbb{Q}]=2^{n}$ has a subextension $K$, $\mathbb{Q}\subseteq K\subseteq F$, with $K$ Galois over $\mathbb{Q}$, $[F:K]=2$, and $[K:\mathbb{Q}]=2^{n-1}$. Then you would use induction to show that everything in $K$ is constructible. And then you'd use an argument similar to the one you used for the case $n=1$ in order to show that since everything in $K$ is constructible, and $[F:K]=2$, then everything in $F$ is constructible. So your induction hypothesis would look something like
- If $L$ is Galois over $\mathbb{Q}$, and has $[L:\mathbb{Q}]=2^k$ for some $k\lt n$, then every element of $L$ is constructible.
To do it that way you would want to use a subgroup of $\mathrm{Aut}(F/\mathbb{Q})$ which is of index $2^{n-1}$ (rather than of order $2^{n-1}$), and which is normal (so that the corresponding extension $K$ is normal over $\mathbb{Q}$).
You can approach it going "the other way", with a subgroup of order $2^{n-1}$ (which is necessarily normal, being of index $2$), as youo ask. How would an inductive argument look in that case? The subgroup $H$ of order $2^{n-1}$ gives you an intermediate field $K$, $\mathbb{Q}\subseteq K\subseteq F$, with $[F:K] = 2^{n-1}$, $[K:\mathbb{Q}]=2$. So you would know, from the case $n=1$, that everything in $K$ is constructible, and you would want to argue inductively that everything in $F$ is constructible.
So here, your induction hypothesis should be somewhat different: the induction hypothesis I quote above would not let you conclude that everything in $F$ is constructible from the fact that everything in $K$ is constructible, because $K$ is not $\mathbb{Q}$ so the induction hypothesis would not apply.
You need a different induction hypothesis: one that doesn't "care" what the base field is, as long as everything in it is constructible. So your induction hypothesis should look like:
- If $K$ is an extension of $\mathbb{Q}$ in which all elements are constructible, and $L$ is a field extension of $K$ with $[L:K]=2$, then every element of $L$ is constructible.
Then you could use that induction hypothesis to conclude everything in $F$ is constructible.
But this introduces a new problem: in order to prove the $n=1$ of the induction hypothesis I proposed first, you would need to prove that if $K$ is a field with $[K:\mathbb{Q}]=2$, then everything in $K$ is constructible; this is not too hard (I hope). But in order to use the second induction hypothesis, the case $n=1$ needs to be more general: now you need to show that if $K$ is any extension of $\mathbb{Q}$ in which every element is constructible, and $L$ is a field extension of $K$ with $[L:K]=2$, then every element of $L$ is constructible. Because the second induction hypothesis assumes more than the first, you need to prove more in the base case if you want to use it.
This is exactly what Chris pointed out: his "first" is the proof of the $n=1$ case of the second proposed induction hypothesis above; his "then" is the inductive step.
(Of course, you are just trading when and where you will prove that if everything in $K$ is constructible and $[L:K]=2$, then everything in $L$ is constructible; in the first argument, you need to do that to finish off the inductive step; in the second argument, you need it to get started. The one advantage of the method you propose is that you only need to know that a group of order $2^n$ has a subgroup of order $2^{n-1}$; with the first method, you need to know it has a normal subgroup of order $2$. This is not very hard, though, one just shows the center is nontrivial.)
Added after edit to the question.
I think you are complicating your life again by trying to attack the entire problem in one fell swoop somehow, rather than just taking it one chunk at a time. Your original idea of trying to use induction somehow has a lot of merit, and means that you don't really need to invoke theorems on solvability and normal series, just the fact that a group of order $2^{n+1}$ must have a normal/central subgroup of order $2$ (or if you want to go by your original attempt, just using that a group of order $2^{n+1}$ must have a subgroup of order $2^n$, see the final comments after the next horizontal rule).
The key is indeed showing that:
Result we hope to prove. If $[F:K]=2$ (and $F/K$ is a Galois extension; but this is immediate, because any extension of degree $2$ in a field of characteristic different from $2$ is always a Galois extension), and all elements of $K$ are constructible, then all elements of $F$ are constructible.
Assume for a moment that you have already managed to prove this. How will the induction argument go? If $[F:\mathbb{Q}]=2^1$ is a Galois extension (that is, $n=1$) then the result will follow because every element of $\mathbb{Q}$ is certainly constructible, so by the result we are assuming you have proven, every element in $F$ is constructible. Done.
Now, assume inductively that for any finite Galois extension $K$ of the rationals, if $[K:\mathbb{Q}]=2^k$, then every element of $K$ is constructible. We want to show that if $F$ is an extension of $\mathbb{Q}$ with $[F:\mathbb{Q}]=2^{k+1}$, then every element of $F$ is constructible. If we can prove this, then this will prove the result you want by induction on $n$.
So, say $F$ is a Galois extension of $\mathbb{Q}$ with $[F:\mathbb{Q}]=2^{n+1}$. Then we know that $\mathrm{Gal}(F/\mathbb{Q})$ is a group with $2^{n+1}$ elements, and thus has nontrivial center; the center is abelian of order $2^r$ for some $r\geq 1$, so there is a central subgroup of order $2$, call it $N$. Then $N\triangleleft G$, $[G:N]=2^k$, so the fixed field of $N$, call it $K$, satisfies $\mathbb{Q}\subseteq K\subseteq F$, $[K:\mathbb{Q}]=[G:N]=2^k$, $[F:K]=2$, and $K$ is Galois over $\mathbb{Q}$ because $N$ is normal in $G$. By the induction hypothesis, every element of $K$ is constructible. Now, we have $[F:K]=2$, and every element of $K$ is constructible, so by the Result-we-hope-to-prove, we conclude that every element of $F$ is constructible, and we are done. QED, RIP, $\Box$.
So, how about that "Result-we-hope-to-prove"? Well, suppose that, as we've suggested, you manage to prove that:
Lemma-we-hope-to-prove. If $K\subseteq F\subseteq\mathbb{C}$ are field extensions, and $[F:K]=2$, then there exists $\xi\in F$ with $\xi^2\in K$ such that $F=K(\xi)=K[\xi]$.
Then: assuming every element of $K$ is constructible, then notice that every element of $F$ can be written (uniquely) as $a+b\xi$ with $a,b\in K$. Now, $a$ and $b$ are both constructible, so if $\xi$ is constructible, then $a+b\xi$ is constructible (product of constructible numbers is constructible, sums of constructible numbers are constructible), so every element of $F$ is constructible. So, assuming the Lemma, it will all fall to showing $\xi$ is constructible. Since $\xi^2 = k\in K$, and $k$ is constructible, then <insert valid argument here>. This proves the Result-we-hope-to-prove, modulo proving the Lemma-we-hope-to-prove.
So now we are down to proving the Lemma-we-hope-to-prove.
Since $[F:K]=2$, if you take $\alpha\in F$, $\alpha\notin K$ (it must exist), then $\{1,\alpha\}$ is a basis for $F$ over $K$ (linearly independent because $\alpha\notin K$, and of the correct cardinality to be a basis). Since $\{1,\alpha,\alpha^2\}$ is linearly dependent (too many vectors), but $\{1,\alpha\}$ is linearly independent, then $\alpha^2$ is a linear combination of $1$ and $\alpha$. So we can find $b,c\in K$ such that $\alpha^2+b\alpha+c = 0$. That is, the minimal polynomial of $\alpha$ over $K$ is $f(x) = x^2+bx+c$.
But we know exactly what the roots of $f(x)$ are: they are
$$\frac{-b+\sqrt{b^2-4c}}{2}\quad\text{and}\quad\frac{-b-\sqrt{b^2-4c}}{2}.$$
So $\alpha$ must be one of them. Letting $r$ be the first root, note that the second root is $-r-b$, so replacing $\alpha$ by $-\alpha-b$ if necessary (remember that $b$ is in $K$), we may assume that $\alpha = \frac{-b+\sqrt{b^2-4c}}{2}$.
Now... what about that $\sqrt{b^2-4c}$ ? Is it in $F$? Is it in $K$? What is $K\left(\sqrt{b^2-4c}\right)$, anyway?
Note. If you change what you are trying to prove to the apparently more general:
Let $F\subseteq\mathbb{C}$ be a Galois extension of $K$ such that $[F:K]=2^n$. If all elements of $K$ are constructible then all elements of $F$ are constructible.
Then you get a result which implies the one you want, by taking $K=\mathbb{Q}$ (since certainly every element of $\mathbb{Q}$ is constructible). In this approach, the base of the induction would be the Result-we-hope-to-prove, and the induction hypothesis would be:
Induction hypothesis. If $M\subseteq\mathbb{C}$ is a Galois extension of $L$, $[M:L]=2^k$, and every element of $L$ is constructible, then every element of $M$ is constructible.
Then instead of using a central subgroup $N$ of $\mathrm{Aut}(F/K)$ of order $2$, you can take the subgroup $H$ of order $2^k$ that you know exists inside $\mathrm{Gal}(F/K)$. Letting $L$ be the fixed field of $H$, you have that $[L:K]=2$, $[F:L]=2^k$, and $F$ is Galois over $L$. By the Result-we-hope-to-prove you know every element of $L$ is constructible (since $H\triangleleft G$, so $L$ is Galois over $K$), and then by the induction hypothesis applied to $F/L$ you know that every element of $F$ is constructible. It all still comes down, though, to the Result-we-hope-to-prove.
Best Answer
Let $n := [K : \Bbb Q]$. Let $p$ be a prime with $p - 1 > n$.
Let $\alpha = 2^{1/p} \in \Bbb R$. Note that $[\Bbb Q(\alpha) : \Bbb Q] = p > n$ and thus, $\alpha \notin K$. Let $E := K(\alpha)$. Note that in particular, we have $$p \mid [E : \Bbb Q]. \tag{1}$$
Claim. $E/K$ is not normal.
Proof. For the sake of contradiction, assume that $E/K$ is normal.
Consider the polynomial $f = X^p - 2 \in K[X]$. Note that $\alpha$ satisfies $f$.
Let $g \in K$ be the minimal (irreducible) polynomial satisfied by $\alpha$ over $K$. Then, we can write $$f = g h$$ for some $h \in K[X]$.
Let $d := \deg(g) \le p$. Then, we have $[E : K] = d$.
Since $\alpha \notin K$, it follows that $g$ is not linear. By assumption of normality, we see that $g$ has another root $\beta \in E$.
Define $\gamma := \beta/\alpha \in E$. Then, we have $\gamma \neq 1$ and $\gamma^p = 1$. Thus, $\gamma$ is a primitive $p$-th root of $1$ and hence, $$(p - 1) \mid [E : \Bbb Q]. \tag{2}$$
(This is because $[\Bbb Q(\gamma) : \Bbb Q] = p - 1$.)
From $(1)$ and $(2)$, we conclude that $$p(p - 1) \mid [E : \Bbb Q]. \tag{3}$$ On the other hand, we have $$[E : \Bbb Q] = [E : K][K : \Bbb Q] = dn < p(p - 1),$$ contradicting $(3)$. $\Box$