Let $K$ be a Galois extension of $F$ with $[K/F] = n$. If $p$ is a prime divisor of $n$, show that there is a subfield $L$ of $K$ with $[K:L] = p$

Question

Let $K$ be a Galois extension of $F$ with $[K/F] = n$. If $p$ is a prime divisor of $n$, show that there is a subfield $L$ of $K$ with $[K:L] = p$.

Is easy show that $\mathrm{Gal}(K/F)$ has a subgroup $H$ of order $p$ and so, the Fundamental Theorem of Galois Theory says that any subfield $L$ such that $H \longleftrightarrow L$, satisfies $[K:L] = p$.

My question is: how can I ensure there is a subfield $L$ which corresponds to $H$?

Answer 1

Let $K/F$ be an extension and put $G = \operatorname{Gal}(K/F)$.

Given $L$ an intermediate field $K \supseteq L \supseteq F$ , we can consider $$H = \operatorname{Gal}(K/L) = \{\sigma\in G: \sigma(l)=l \text{ for all } l\in L\},$$ which is a subgroup of $\operatorname{Gal}(K/F)$.

Conversely, given a subgroup $H$ of $G$, we can consider $$K^H = \{x\in K:\sigma(x)=x \text{ for all }\sigma\in H\}.$$ This is a subfield of $K$ containing $F$.

These correspondences have the following properties:

(i) They are order-reversing:

if $K \supseteq L_1 \supseteq L_2 \supseteq F$ then $\operatorname{Gal}(K/L_1)\subseteq\operatorname{Gal}(K/L_2);$

if $H_1 \le H_2 \le G$ then $K^{H_1} \supseteq K^{H_2}$.

(ii) $$K^{\operatorname{Gal}(K/L)} \supset L\quad\text{and}\quad\operatorname{Gal}(K/K^H ) \supseteq H.$$

Fundamental Theorem of Galois Theory:

(i) \begin{align*} L&\longmapsto \operatorname{Gal}(K/L)\\ K^H&\longleftarrow H \end{align*} are mutually inverse order-reversing bijections between $$\{\text{intermediate fields }F \subseteq L \subseteq K\} \longleftrightarrow \{\text{subgroups of $\operatorname{Gal}(K/F)$}\}$$

(ii) If $F \subseteq L \subseteq K$ is an intermediate extension, then $[K : L] = |\operatorname{Gal}(K/L)|$.

(iii) If $F \subseteq L \subseteq K$ is an intermediate extension, then the extension $L/F$ is Galois if and only if $N = \operatorname{Gal}(K/L)$ is a normal subgroup of $G$, in which case $\operatorname{Gal}(L/F) \cong G/N$.

Now since you can show for $[K:F]=n$ that for a prime $p\mid n$ (by Cauchy's theorem) there exists a subgroup $H$ of order $p$ of $\mathrm{Gal}(K/F)$, then by $(ii)$ of the Fundamental Theorem, and the order reversing properties you also show the existence of $[K:L]=p$, since you don't get one without the other.

Answer 2

The Galois group $G$ contains an element of order $p$ by Cauchy's Theorem, say $x$.

Let $H:=\left\langle x \right\rangle$. Then Galois theory ensures the existence of a field $L$ corresponding to $H$. Namely, the fixed field of $H$ in $K$.