TITLE: How to prove that the Galois group of a normal extension transitively permutes the factors of an irreducible polynomial?
QUESTION [5 upvotes]: How to do the following problem?

Let $K$ be a normal extension of $F$, and let $f(x)\in F[x]$ be an irreducible polynomial over $F$. Let $g(x)$
  and $p(x)$ be monic irreducible factors of $f(x)$ in $K[x]$. Prove that there is a $z\in\operatorname{Gal}(K/ F)$,
  with $z(g)=p$.

REPLY [7 votes]: Let $N/F$ be a normal extension, let $X$ be an indeterminate, let $f\in F[X]$ be an irreducible polynomial, and let $g_1,g_2\in N[X]$ be irreducible monic factors of $f$. 
We want to find an $F$-automorphism of $N$ which maps $g_1$ to $g_2$. 
Let $N^a$ be an algebraic closure of $N$, and $\alpha_i$ a root of $g_i$ in $N^a$. 
The minimal polynomial of $\alpha_i$ over $F$ being $f$, there is an $F$-automorphism $\sigma$ of $N^a$ mapping $\alpha_1$ to $\alpha_2$. 
As $N/F$ is normal, we have $\sigma N=N$. 
The minimal polynomial of $\alpha_i$ over $N$ being $g_i$, our automorphism $\sigma$ maps $g_1$ to $g_2$.