Use Galois theory to prove that $H \triangleleft G$

Question

Suppose that $F \leq E \leq \mathbb{C}$, $[E : F] = 100$, $E$ is Galois over $F$ and $G = Gal(E/F)$
contains a subgroup $H$ such that $|H| = 25$. Use Galois theory to prove that $H \triangleleft G$.

$\textbf{My attempt:}$
by Fundamental theorem of Galois theory, there exist maps $K \rightarrow Gal(E/F)$ and $H \rightarrow E^H$ such that they are inverse of each other. Which ​$K = E^H$ if and only if $H = Gal(E/F)$.
(1) $|H| = [E:K]$ and $[K:F]=[G:H]=\frac{|G|}{|H|}$.
(2) The extension $K/F$ is Galois if and only if $H \triangleleft G$.

Since, $|H| = 25$. Then, $[E:K] = 25$.
by Tower law, $[E:F] = [E:K] \times [K:F] = 100$.
Then, by (1), $[K:F] = 100/25 = 4$ and $[G:H] = \frac{|G|}{|H|} = 4$.
So, only need to prove that the extension $K/F$ is Galois.
Then, by (2), I can conclude that $H \triangleleft G$.

$\textbf{I know that I can use Sylow's Theorem to prove that $H \triangleleft G$ directly}$.
$\textbf{but how do I prove the extension $K/F$ is Galois without using Sylow's Theorem ? }$

Answer 1

Using Sylow theory is a natural thing to do. No major worries!

The following is an alternative.

1. Let $K$ be the fixed field of $H$. You already showed that $[K:F]=4$.
2. By the primitive element theorem $K=F(\rho)$ for some element $\rho\in K$. Let $m(x)$ be the minimal polynomial of $\rho$ over $F$. We know that the degree of $m(x)$ is four.
3. Because $E/F$ is normal, $m(x)$ is irreducible over $F$, and $m(x)$ has a zero, namely $\rho$, in $E$, we can deduce that $m(x)$ splits over $E$.
4. Let $M\subseteq \Bbb{C}$ be the splitting field of $m(x)$ over $F$. We know three things: A) $M\subseteq E$. B) Because $M$ is the splitting field of a quartic, $[M:F]$ is a factor of $4!=24$. C) $K\subseteq M$.

I guess you can take it from here.