Subextension of a field with Galois series of subextensions of prime degree

Let $p$ be a prime number, and $E/F$ be a field extension. Suppose $E/F$ has a finite series of subfields

$$F = E_0 < E_1 < \cdots < E_n = E$$

with $E_i / E_{i-1}$ Galois of degree $p$ for each $i$. In this case, call $E/F$ special.

Claim: If $K$ is an intermediate field of a special extension $E/F$, then $K/F$ is also special.

Things we know:

• $E/F$ has degree $p^n$ from multiplying the degrees of its subextensions, and so the degrees of $E/K$ and $K/F$ must also be $p$-powers.
• $E/F$ is separable, since each extension $E_i / E_{i-1}$ is separable and separability is transitive. Then, $E/K$ and $K/F$ are also separable since transitivity also goes in the other direction.
• If in addition $K/F$ is normal, then the claim is very simple, since $K/F$ will then be Galois, and $\mathrm{Gal}(K/F)$ is then a $p$-group. We can construct a normal series of $p$-index subgroups in the Galois group, and then use the correspondence to bring them back to the desired series of fields between $F$ and $K$.
• But, $K/F$ will not be normal in general (I think). The order of its automorphism group must divide the degree of the extension, so $\mathrm{Aut}(K/F)$ must still be a $p$-group. We can still construct a normal series of $p$-index groups in the automorphism group, but I don't think then we can bring them back to fields between $F$ and $K$ (indeed, if $|\mathrm{Aut}(K/F)| \neq [K : F]$, then there will not be enough intermediate groups to bring back).

Any tips would be appreciated.