On the action of galois groups in towers of fields I would like some confirmation on certain statements I believe to be true:
Let $K\subset L\subset M$ be a tower of fields such that both extensions $L/K$ and $M/K$ are  galois. Let $f(x) \in K[x]$ be an irreducible polynomial over $K$ that splits completely in $M$ and let it split into factors $f_1f_2\dots f_k$ over $L$ such that each $f_i$ is irreducible over $L$.
Let $G = \operatorname{Gal}(M/K)$ and $H = \operatorname{Gal}(M/L)$.
I think it is true and easy to prove all the $f_i$ have equal degree (say $l$). Let $f_i$ have roots $S_i = \{x_{i1},\dots,x_{il}\}$. Then $H$ acts transitively on $S_i$ while $G/H = G = \operatorname{Gal}(L/K)$ acts on the set $\{S_1,\dots,S_k\}$. I am most interested in the following question:
Is the action of $H$ on $S_i$ the "same" as it's action on $S_j$? That is, is there a bijection between $S_i$ and $S_j$ that is invariant under the action of $H$? I believe this is true and further, that the bijection can in fact be given as an element of $G$. 
 A: This is not a definite answer, but it's too long for a comment.
Let $S$ be the set of roots of $f$ in $M$. Then $G$ acts transitively on $S$, as is well-known. You may consider the $H$-orbits of $S$ ; call them $S_i$. Then you can form the polynomial $f_i = \prod_{\lambda\in S_i} (X-\lambda)\in L[X]$. 
It's easy to see that $f=\prod f_i$ is a decomposition in irreducible factors in $L[X]$ (because $H$ acts transitively on the roots of each $f_i$).
Of course this is the same construction as yours, just in the reverse order. The point is that any transitive $G$-set $S$ comes up as the set of roots in $M$ of some irreducible $f$, so there is no loss in considering transitive $G$-sets.
But in general it is clear that if a group $G$ acts transitively on a set $S$, and $H$ is a normal subgroup of $G$, then $G/H$ acts transitively on the $H$-orbits.
Indeed, let $A$ and $B$ be two $H$-orbits, with $a\in A$ and $b\in B$, and let $g\in G$ be such that $ga=b$. Then for any $h\in H$, $(gh)A= gh\cdot Ha = gHa = Hga = Hb = B$. So the action of $G$ on the $H$-orbits induces an well-defined action of $G/H$.
On the other hand, in general I don't think the $H$-orbits are isomorphic as $H$-sets. 
Indeed, suppose there is a $H$-isomorphism $\varphi: A\to B$ for some $H$-orbits $A$ and $B$. Then take $a\in A$, and put $b = \varphi(a)$ ; you get that for any $h\in H$, $\varphi(ha)=hb$. This is possible if and only if $H_a = H_b$ where $H_a$ and $H_b$ are the stabilizer subgroups of $a$ and $b$. But if $b=ga$ for some $g\in G$ then $H_b = H\cap (gG_a g^{-1})$. So we have to find an example where $H\cap (gG_ag^{-1})\neq H\cap G_a$ for all $a$ in an orbit $A$ and all $g$ such that $gA\neq A$.
I'd have to think harder to construct such an example : it amounts to finding a subgroup$K$ of $G$ such that, for any $g\in G$ satisfying $gK\neq hK$ for all $h\in H$, then $H\cap (hKh^{-1})\neq H\cap \left( (gh)K(gh)^{-1}\right)$ for all $h\in H$. But I think this kind of set up exists (not for any $G$, of course, but for some $G$, and it is well-known that any finite $G$ can appear as a Galois group).
