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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.

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

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    $\begingroup$ What do you mean by it suffices to show this? $\endgroup$ – quid May 31 '15 at 16:20
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    $\begingroup$ @quid If you can prove this, then you've proved the result in the question, because of the correspondence between intermediate fields and intermediate subgroups in the fundamental theorem of Galois theory. $\endgroup$ – James May 31 '15 at 16:27
  • $\begingroup$ @James but how do you show that $F/K$ is Galois? $\endgroup$ – quid May 31 '15 at 16:31
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    $\begingroup$ @quid $F/K$ is Galois iff the subgroup is normal. This is a well-known, fundamental fact in Galois theory $\endgroup$ – Ewan Delanoy May 31 '15 at 16:33
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    $\begingroup$ @EwanDelanoy well, yes. But when I comment there was no mention of "normal." $\endgroup$ – quid May 31 '15 at 16:35
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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 $\mathsf{Aut}(\mathbb{Z}_5^2)$.

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  • $\begingroup$ Every group of nonprime order has a non-trivial subgroup. $\endgroup$ – quid May 31 '15 at 16:38
  • $\begingroup$ Yes, indeed. But now we can even say, which (normal) subgroups can occur, i.e., which intermediate subfields. $\endgroup$ – Dietrich Burde May 31 '15 at 16:40
  • $\begingroup$ Yes, and this useful as one needs a normal one. (Sorry I was not quite clear reagrding my point in the first comment.) $\endgroup$ – quid May 31 '15 at 16:41

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