Does this lemma still holds when replacing 'Galois' by ‘normal’?

Question

All of the field extensions involved are assumed to be finite.

This lemma is from Isaacs' Algebra, serving as a key step toward the establishment of Galois' criterion of solvability. The proof of the original lemma is as shown in the pictures.

The original lemma: Suppose $F = F_0 \subseteq F_1 \subseteq \dots \subseteq F_r = L$, where for $1 \leq i \leq r$, each of the extensions $F_{i-1} \subset F_i$ is Galois with an abelian Galois group. Suppose $F \subseteq E \subseteq L$ and $E$ is Galois over $F$. Then $\operatorname{Gal}(E/F)$ is solvable.

It seems to me that the lemma holds still if one replace all 'Galois' in the original version by 'normal'. Then it becomes like this:

The modified lemma: Suppose $F = F_0 \subseteq F_1 \subseteq \dots \subseteq F_r = L$, where for $1 \leq i \leq r$, each of the extensions $F_{i-1} \subset F_i$ is normal with an abelian Galois group. Suppose $F \subseteq E \subseteq L$ and $E$ is normal over $F$. Then $\operatorname{Gal}(E/F)$ is solvable.

But I am not sure if the modified lemma am right. I also write down a proof for the version lemma assuming normality. If it will not take a lot of time, please verify the following proof.

Proof: We work by induction on $r$. Let $E_1=\langle F_1, E\rangle$. The fact that $E_1$ is normal over $F$ follows from normality of $F \subseteq E$ and $F \subseteq F_1$. Thus, $F_1 \subseteq E_1$ is also normal. We have a tower of length $r-1$ of normal extensions from $F_1$ to $L$, and by the inductive hypothesis one concludes $\operatorname{Gal}(E_1/F_1)$ is solvable.

Consider the tower $F \subset F_1 \subset E_1$. Since $F \subset F_1$ and $F \subset E_1$ are normal, $\operatorname{Gal}(E_1/F_1) \lhd \operatorname{Gal}(E_1/F)$, and $\operatorname{Gal}(E_1/F)/\operatorname{Gal}(E_1/F_1) \cong \operatorname{Gal}(F_1/F)$. Given both $\operatorname{Gal}(E_1/F_1)$ and $\operatorname{Gal}(F_1/F)$ are solvable, one knows $\operatorname{Gal}(E_1/F)$ is solvable.

Next consider the tower $F \subset E \subset E_1$. Again we have two normal extensions $F \subset E$ and $F \subset E_1$, which follows $\operatorname{Gal}(E_1/E) \lhd \operatorname{Gal}(E_1/F)$, and $\operatorname{Gal}(E/F) \cong \operatorname{Gal}(E_1/F)/\operatorname{Gal}(E_1/E)$. One deduce $\operatorname{Gal}(E/F)$ is solvable as a quotient of a solvable group.

Answer 1

I try to give a proof based on the sketch given by reuns. His sketch of proof is quoted as follows (reuns states that $Aut(K(E^{p^n})/K(E^{p^n})) = Aut(E/F)$, but I cannot prove the equality) :

Prove that if $E/F/K$ is a finite tower in characteristic $p$ and $E/F$ is normal then taking $n$ large enough such that $p^n∤[E:K]$, $K(E^{p^n})/K$ is separable and $K(E^{p^n})/K(F^{p^n})$ is Galois with $Aut(K(E^{p^n})/K(F^{p^n})) \cong Aut(E/F)$.

Proof: Let $S = \{\gamma \in E \mid \gamma$ is separable over $K \}$, i.d. the maximal separable extension of $K$ in $E$.

We first prove that $K(E^{p^n}) = S$. To do this we must show that $K(E^{p^n})$ is separable over $K$ and $E$ is purely inseparable over $K(E^{p^n})$. $[E:S]$ is a power of $p$, say $p^m$, where $m \lt n$. Each $\alpha \in E$ has minimal polynomial over $S$ of the form $f(X)=X^{p^l}-a$ for some element $a\in S$ and some integer $l\leq m\lt n$. That is, $\alpha ^{p^l} = a \in S$. So $\alpha ^{p^n} \in S$, $K(E^{p^n}) \subseteq S$, hence separable over $K$. On the other hand, for each $\beta\in E$, $\beta^{p^n} \in K(E^{p^n})$, therefore $E/K(E^{p^n})$ is purely inseparable. By the uniqueness of intermediate field that is separable over $K$ and over which $E$ is purely inseparable, we conclude that $S=K(E^{p^n})$.

Similarly we can show that $K(F^{p^n})$ is the maximal separable extension of $K$ if $F$.

By the first part of proof, we know that $K(E^{p^n})/K$ is separable. Since $K \subseteq K(F^{p^n}) \subseteq K(E^{p^n})$, $K(E^{p^n})/K(F^{p^n})$ is finite separable. We shall show that $K(E^{p^n})$ is normal over $K(F^{p^n})$. $E$ is a splitting field over $F$ for some polynomial $g(X) = \Pi (X-\beta_i)^{e_i}$, where $\beta_i \in E$ are distinct. Let $h(X) = \Pi (X - \beta_i^{p^n})^{e_i}$, then $h(X^{p^n}) = g^{p^n}(X) \in F^{p^n}[X]$. So $h(X) \in F^{p^n}[X]$, and $E^{p^n} = F^{p^n}(\beta_i^{p^n})$ is a splitting field of $h(X)$ over $K(F^{p^n})$, hence Galois over $K(F^{p^n})$.

Finally we need to show $Aut(K(E^{p^n})/K(F^{p^n})) \cong Aut(E/F)$. Now that $K(E^{p^n})$ is Galois over $F$, we can define a homomorphism $\rho: Aut(E/F) \to Aut(K(E^{p^n})/K(F^{p^n}))$ to be the restriction of $\sigma \in Aut(E/F)$ to $K(E^{p^n})$. Each $\sigma \in \ker \rho$ fixes $K(E^{p^n}) = S$, and by the properties of such intermediate field $S$, $\sigma$ must fix $E$. It follows that $\rho$ is injective. For the same reason, any $\tau$ that fixes $K(F^{p^n})$ must fix $F$, therefore each $\sigma \in Aut(K(E^{p^n})/K(F^{p^n}))$ extends to some $\tau \in Aut(E)$ that fixed $F$, which follows the surjectivity of $\rho$.