I have been told that a Galois extension is always a normal extension. The proof goes like below: $F \subset K$ is a Galois extension. Let $g(x)$ be an irreducible polynomial in $F[x]$ and $\beta$ be a root of $g(x)$. $G(K/F)$ acts on the root of $g(x)$ transitively, then the conclusion follows.
I do not know why the conclusion follows.
Edit: $F \subset K$ is Galois if $\ $ $|G(K/F)| = [K:F]$.
I'm assuming the following definition of Galois: we say that $F \subseteq K$ is Galois if for every $x \in K$ which is not in $F$, there exist an $F$-automorphism $\sigma$ of $K$ such that $\sigma(x) \neq x$.
To show that $F \subseteq K$ is normal, we need to show that if $f \in F[X]$ is irreducible of degree $n > 1$, and there is a root $x$ of $f$ in $K$, then all the roots of $f$ are in $K$.
Just follow your nose. There is a $\sigma \in G := \textrm{Gal}(K/F)$ with $\sigma(x) \in K$ not equal to $x$. So at least two distinct roots of $f$ lie in $K$. Let $x = x_1, ... , x_t$ be all the distinct roots of $f$ which are acquired by applying various $\sigma \in G$ to $x$. The coefficients of the polynomial $$g(X) := (X-x_1) \cdots (X- x_t) \in K[X]$$ are symmetric functions of $x_1, ... , x_t$, and they are unchanged by applying any $\sigma \in G$ to them (do you see why?). Hence these coefficients must lie $F$, because $K/F$ is Galois. Thus $g(X)$ actually lies in $F[X]$. Since $g(x) = 0$, $g$ divides $f$ in $F[X]$, hence $g$ is equal to $f$, because $f$ is irreducible.
This shows that all the roots of $f$ lie in $K$, so we are done.
