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i was looking a proof of the following:
Let $k\subset F, F\subset K$ be Galois field extensions, then $Gal (K/F)$ is a normal subgroup of $Gal(K/k)$.
I understand the proof but it start using the fact that $k\subset K$ is Galois and i can't see it.
Is it trivial?
Thanks in advance.
The relation "$L$ is Galois over $M$" is not transitive. For example if $n$ is a squarefree integer then the extensions $\Bbb Q(\sqrt[4]{n})/{\Bbb Q}(\sqrt{n})$ and ${\Bbb Q}(\sqrt{n})/\Bbb Q$ are both quadratic hence automatically Galois and yet the extension $\Bbb Q(\sqrt[4]{n})/\Bbb Q$ is not Galois (there are nonreal conjugates of $\sqrt[4]{n}$).
If $K/k$ is Galois and $K/F/k$ an intermediate field, then automatically $K/F$ is Galois (if $K$ is the splitting field of some family of $k[x]$ polynomials, then it is also the splitting field of some $F[x]$ polynomials, since $k\subset F$). However $F/k$ is not automatically Galois. Hence the proof would make sense if it began assuming $K/k$ and $F/k$ were Galois with $K/F/k$ intermediate, as you note.
On these hypotheses, all $G(K/k)$-actions fix $F$ setwise (since $F$ is Galois over $k$), so there is a map given by restriction $G(K/k)\to G(F/k):\sigma\mapsto\sigma|_F$ with kernel $G(K/F)$, those automorphisms which fix $F$ pointwise. Hence $G(K/F)\triangleleft G(K/k)$. (I assume this is the proof you're reading.)
