TITLE: For a non-normal extension, is the Galois action still transitive on roots that *are* in the extension?
QUESTION [9 upvotes]: Let $L/F$ be a field extension where $L$ is the splitting field of the irreducible $f \in F[X]$. Then $\text{Gal}(L/F)$ acts on the roots of $f$ transitively - this follows from how every isomorphism of fields in $L/F$ can be extended to an automorphism of $L$.
Now let $K$ be an intermediate field of $L/F$, where $K$ contains only some of the roots of $f$. Is it true that the automorphisms of $K$ fixing $F$ act transitively on the roots of $f$ in $K$? I know it holds if each of the the roots of $f$ in $K$ generate $K/F$, but I'm not seeing a counterexample where $K$ is not simply generated by some of the roots, or some inductive proof applying this fact to simple subextensions.

REPLY [3 votes]: Consider a finite Galois extension $K\subset L$ with Galois group $G$.  Let $\alpha \in L$ with $\operatorname{Fix}(\alpha) = H$. For $g\in G$ we have  $\operatorname{Fix}(g \alpha) = gH g^{-1}$. Consider the subfield $F = K(\alpha, \beta)$ with fixator subgroup $H \cap gHg^{-1}$.
Question: Does there exists an element in $G$ that takes $\alpha$ to $g\alpha$ and invariates $K(\alpha, g \alpha)$?
Translation: Does there exist an element in $gH$ that invariates(normalizes)  $H \cap gHg^{-1}$?
We'll produce a counterexample to the above group theoretic fact.
Consider the $G = (\mathbb{Z}/2)^3 \rtimes \mathbb{Z}/3$ inside the wreath product $(\mathbb{Z}/2)^3 \rtimes S_3$, $H = \mathbb{Z}/2^2 \times 0$, and $g = ((0,0,0), (1,2,3) )$.  We have $H \cap gHg^{-1} = 0 \times \mathbb{Z}/2 \times 0$, but every element in $gH$ conjugates $0 \times \mathbb{Z}/2 \times 0$ to $0 \times 0 \times \mathbb{Z}/2$.