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Let $K$ be a Galois extension of $F$ and $p$ be a prime factor of the degree $[K:F]$.

I want to show that there is an intermediate extension $F\subseteq L\subseteq K$ with $[K:L]=p$.

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Do we us for that the fact that the intermediate fields are in one to one correspondence with the subgroups of the Galois group?

Then since $p$ is a prime factor of the degree $[K:F]$ there must be an intermediate field with that degree according to Lagrange theorem.

Is that correct?

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    $\begingroup$ So, it is a matter of finding a subgroup of the Galois group, of order $p$. This is done by Cauchy's theorem. $\endgroup$ – Sungjin Kim Nov 19 '18 at 22:24
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    $\begingroup$ Here is Cauchy'theorem I am referring to: en.wikipedia.org/wiki/Cauchy%27s_theorem_(group_theory) $\endgroup$ – Sungjin Kim Nov 19 '18 at 22:26
  • $\begingroup$ Ah ok! So since $p\mid [K:F]=|\text{Gal}(K/F)|$ the Galois group $\text{Gal}(K/F)$ has an element of order $p$, so there is a subgroup of the Galois group of order $p$, and so there is an intermediate field $L$ with $[K:L]=p$. Is everything correct? $$$$ Do we use the same argument also for the following: Let $K$ be a finite and normal extension of $F$ and suppose that the extension $K/F$ has no proper intermediate extension. Show that the degree $[K:F]$ is a prime number. $$$$ Do we use here again the one correspondence with the subgroups? But we don't have here a Galois group. @i707107 $\endgroup$ – Mary Star Nov 19 '18 at 22:54
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    $\begingroup$ @MaryStar With $G = Aut(K/F)$ then $K/F$ is Galois iff $F = K^G$ (the subfield fixed by $G$). If $K/F$ is normal : if $G$ is non-trivial then look at $K^G$ and $K^H,H \le G$. If $G$ is trivial then show the minimal polynomial of any $a\in K$ is of the form $f(t) = (t-a)^n$ with $f$ irreducible thus $\gcd(f(t),f'(t)) = 1$ thus $n = {p^m}, p= char(F)$. $\endgroup$ – reuns Nov 20 '18 at 1:24
  • $\begingroup$ Could you explain that further to me? I got stuck right now. @reuns $\endgroup$ – Mary Star Nov 24 '18 at 16:44

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