How to prove that the Galois group of a normal extension transitively permutes the factors of an irreducible polynomial? 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 $z(g)=p$. 

 A: 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 $g_2$.
A: Under the hypotheses, $f$ splits completely over $\bf K$, so $g$ and $p$ are linear. Thus the problem reduces to showing that if $\alpha$ and $\beta$ are roots of $f$ in $\bf K$ then there's an element of the Galois group that takes $\alpha$ to $\beta$. The proof of that result is in every exposition of Galois Theory.  
EDIT: as awllower indicates, I misread the problem, and thought it was assumed that $f$ has a root in $\bf K$. 
