Galois extension of degree 75

Question

I'm studying Galois Theory and I stumbled upon this question:

Let $E/K$ be a Galois extension of degree $75$. Show that there exists an intermediate subfield $K \subsetneq F \subsetneq E$ such that $F/K$ is also a Galois extension.

I started like this: let $G = \text{Gal}(E/K)$. Because $E$ is Galois over $K$, $|G| = 75$. Then, by the fundamental theorem of Galois theory, any intermediate field $F$ has the form $F = E_H$ for some subgroup $H$ of $G$ and $[E:F] = |H|$ while $[F:K] = [G:H]$.

Any tips on how to progress further? Any help would be greatly appreciated.

Answer 1

Hint : It suffices to show that the Galois group has nontrivial normal subgroups.

Answer 2

Does every group of order $75$ have a nontrivial subgroup ? We could use Cauchy's theorem to see this.

Remark: Since the order is of the form $pq^2$ for $p<q$ we have even an easy classification of such groups. There are three non-isomorphic groups of order $75$. The cyclic one, $C_{75}$, the direct product $C_5\times C_{15}$ and the nonabelian one $\mathbb{Z}_5^2 \rtimes_\varphi \mathbb{Z}_3$, where $\mathbb{Z}_3 = \langle x \rangle$ and $\varphi(x)$ is any element of order $3$ in $\mathsf{Aut}(\mathbb{Z}_5^2)$.