Degree of factor of resolvent

As always with my questions this is not at research level, but the assertion is made in a research paper, plus no one's been able (or willing) to answer it over at MSE. Here is the original question, but I'll state it again.

Let $f \in k[t]$ be a given separable polynomial of degree $n$ with Galois group $G$. Let $H \subset S_n$ be a subgroup, $\Theta \in k[x_1,...,x_n]^H$, $H_m = \gamma_m H \gamma_m^{-1}$ with $\{\gamma_1,...,\gamma_e\}$ a left transversal of $S_n/H$, $\Theta_m = \gamma_m \Theta$ and $\theta_m$ its evaluation to the roots of $f$. Let $P$ be a simple, separable, irreducible factor of $L=L_{\Theta,f}:= \prod_{i=1}^e (t-\theta_m)$. Suppose the degree of $P$ is $j$. Then there exists $J \subset [e]$ of cardinality $j$ such that $P=\prod_{m \in J} (t-\theta_m)$. The assertion is that $j=[G: G\cap H_m]$ since $\theta_m$ is a simple root of $L$ (in a lemma, if $\theta$ is a simple root of $L_{\Theta,f}$ then $Gal(E/k(\theta))=G \cap H$).

We're interested in $deg(P) = [k(\theta_m):k]$. If the splitting field of $L$ equals that of $f$ then we have $k \subset k(\theta_m) \subset E$, whence the assertion follows from the tower law and the lemma. However, this is not assumed here. Suppose rather that the splitting fields are not the same. Then we have the field inclusions $k \subset E \cap k(\theta_m) \subset E,k(\theta_m) \subset Ek(\theta_m)$. In this situation we have $[k(\theta_m):k]=[k(\theta_m): E \cap k(\theta_m)][E \cap k(\theta_m) : k]$.

I'm not able to draw any conclusions from this tree of fields. I've gone over the basic results of GT several times but I don't find anything I can use here. Some of these results say something about the order of the Galois groups of some of the extensions above, but in these cases the extensions are not necessarily Galois, whence the degree of an extension doesn't equal the order. ($Ek(\theta_m)/k(\theta_m)$ and $E/E\cap k(\theta_m)$ are Galois, the others not necessarily so, at least not to my knowledge.)

If anyone could point me in the right direction that'd be much appreciated, thanks.