Galois extension of degree $ 2^n $ I'm trying to find a way to prove the following statement:
Assume $ \mathbb{Q} \subset E $ is a Galois extension of degree $ 2^n $. Show that there are fields $ \mathbb{Q} = E_0 \subset E_1 \subset \dots \subset E_n = E $ such that $ [E_{i} : E_{i-1}] = 2 $ for $ i = 1, 2 \dots n $.
My attempt: we know that the Galois group $ G$ is of order $ 2^n $, hence there exists $ \sigma \in G $ of order $ 2 $ (by Cauchy theorem). By fundamental theorem of Galois we know that there must be $ \mathbb{Q} \subset E_{n-1} \subset E $ such that $ [E_{n-1}: \mathbb{Q}] = 2^{n-1} $. If I knew that $ \mathbb{Q} \subset E_{n-1} $ was Galois, I'd be done. I'm not sure if that's true though  - and if it isn't, I don't have any good idea on how to prove that.
I'd appreciate some hints
 A: A group $G$ of order $2^n$ is solvable —this should be proved in pretty much any good textbook on finite groups— so in particular it has a chain of subgroups $$1=G_0\subseteq G_1\subsetneq G_2\subsetneq\cdots\subsetneq G_k=G$$ such that each $G_i$ is normal in $G_{i+1}$ and $G_{i+1}/G_i$ is cyclic of prime order, so in your case of order $2$.
A: How about instead of going up, you go down? The Sylow theorems tell us that there's a normal subgroup of the galois group of order $2^i$ for every $i \in \{0, \dots, n\}$. In particular one of order $2^{n-1}$ will be normal. So then this subgroup will correspond to an intermediate subfield which is Galois over the base field. Then just use induction from here.
A: The Galois group G is hence a 2 group. Then we can find a chain of subgroup ${1} < H_{1} \cdots \lhd H_{k}  \lhd G=H_{k+1}$ where $H_{i}$ has index 2 in $H_{i+1}$, and then you consider their fixed fields, and $H_{m} \lhd G$. 
Let G be a p group, then G has a non trivial center, (you can make G act on itself by conjugation,use the orbit stabilizer theorem to prove that there is more than one single orbit). Now, we prove our theorem. Assume G has order $p$. Then the result is ovbious. Let n>1, and consider a group G of order $p^{n+1}$. The center of $G$ is non trivial, and has order divisible by p (lagrange). So it has an element of order $p$, hence a subgroup H of order $p$, and $H$ is a proper normal (included in the center) of $G$. Then $G/H$ is a group of order $p^{n-1}$. It has a non trivial center, so let $y$ be an element of order $p$ in $G/H$. Now,  using the projectiom $\pi:G \to G/H$ as well as the third isomorhism theorem for group, $\pi^{-1}(<y>)$ is a normal subgroup of $G$, and H has index $p$ in it. So we've proven, given a proper normal subgroup $H$ in G, we can find a subgroup $K$ that is normal in $G$ and $[K:H]=p$. Now let $H={1}$, and apply this process inductively.
