Determining the number and degree of irreducible factors over Galois extension The following is 14.2.28 from Dummit and Foote:
Let $f(x)\in F[x]$ be an irreducible polynomial of degree $n$ over the  field $F$ and let $L$ be the splitting  field of $f(x)$ over $F$ with $\alpha$ a root of $f(x)$ in $L$.  If $K$ is any Galois extension of $F$ contained in $L$, show that the polynomial $f(x)$ splits into a product of $m$ irreducible polynomials each of degree $d$ over $K$, where $m=[F(\alpha)\cap K:F]$ and  $d= [K(\alpha) :K].$
The text provides the following hint:
If $H$ is the subgroup of the Galois group of $L$ over $F$ corresponding to $K$, then the factors of $f(x)$ over $K$ correspond to the orbits of $H$ on the roots of $f(x)$. Then use Exercise 9 of section 4.1. This exercise states that if $G$ acts transitively on the finite set $A$ and $H\leq G$ is normal, then for $a\in \mathcal{O}_1$ we have $|\mathcal{O}_1|=|H:H\cap G_a|$ and $r=|G:HG_a|$, where $G_a$ denotes the stabilizer of $a$ and $r$ is the number of distinct orbits $\mathcal{O}_1, \mathcal{O}_2, \ldots, \mathcal{O}_r$ of $H$ on $A$. 
Now, I understand  that "if $H$ is the subgroup of the Galois group of $L$ over $F$ corresponding to $K$, then the factors of $f(x)$ over $K$ correspond to the orbits of $H$ on the roots of $f(x)$." But I don't see how to apply the exercise.
Overall, I'm pretty lost and would appreciate an explanation.
 A: Let's assume that $f(x)$ is separable over $F$ (otherwise we cannot talk about the Galois group of $L/F$). 
Let $G = Gal(L/F)$, $H = Gal(L/K)$ the subgroup corresponding to $K/F$; since $K/F$ is Galois, $H$ is normal in $G$. We know that $G$ acts transitively on the roots of $f(x)$. Now $H$ also acts on the roots of $f(x)$ and the $H$-orbits are a partitioning of the roots of $f(x)$. Viewing $f(x) \in K[x]$, $f(x)$ factors into irreducible polynomials $p_i(x)$ over $K$ and these factors $p_i(x)$ correspond to $H$-orbits. 
Since $G$ is transitive on the set of roots, $G$ acts transitively the set of $H$-orbits and the exercise says that there are exactly $[G:HG_\alpha]$ many $H$-orbits and furthermore all $H$-orbits have the same cardinality equaling  $[H:H\cap G_\alpha]$ where $G_\alpha$ is the stablizer of $\alpha$ in $G$. So there are $[G:HG_\alpha]$ many irreducible factors $p_i(x)$ of $f(x)$ over $K$, each with degree $[H:H\cap G_\alpha]$.
Now use the Fundamental Theorem of Galois theory to relate the group indexes to degrees of intermediate extensions and we should have it. Note that $G_\alpha = Gal(L/F(\alpha))$ so $G_\alpha$ corresponds to $F(\alpha)$; also, $HG_\alpha$ is a subgroup of $G$ (because $H$ is normal) and corresponds to the subfield $F(\alpha) \cap K$; and $G_\alpha \cap H$ corresponds to the subfield $K(\alpha)$. 
