Let \(L/k\) be a finite Galois extension and let \(F\) be an intermediate field, \(k\subseteq F\subseteq L\). Then:
\(L/F\) is Galois.
\(F/k\) is Galois if and only if \(\Gal(L/F) \normal \Gal(L/k)\).
In that case restriction \(\sigma\mapsto\sigma|_F\) induces an isomorphism \[ \Gal(F/k) \cong \Gal(L/k)/\Gal(L/F). \]
\(L\) is the splitting field over \(k\) of a separable polynomial \(f\in k[x]\), and \(L\) is also generated over \(F\) by the roots of \(f\in F[x]\), so \(L\) is the splitting field of \(f\) over \(F\) and \(L/F\) is Galois.
and (c). Fix an algebraic closure \(\bar k\) containing \(L\). Every \(k\)-embedding \(F\to\bar k\) extends to a \(k\)-embedding \(L\to\bar k\), which has image \(L\) by PR-OZYUC because \(L/k\) is normal; so the \(k\)-embeddings \(F\to\bar k\) are exactly the restrictions \(\sigma|_F\) with \(\sigma\in\Gal(L/k)\). For \(\sigma\in\Gal(L/k)\) we have \(\Gal(L/\sigma(F))=\sigma\Gal(L/F)\sigma^{-1}\), and by the Galois correspondence \(\sigma(F)=F\) if and only if \(\sigma\Gal(L/F)\sigma^{-1}=\Gal(L/F)\). Hence \(\Gal(L/F)\normal\Gal(L/k)\) if and only if \(\sigma(F)=F\) for every \(\sigma\in\Gal(L/k)\), if and only if every \(k\)-embedding \(F\to\bar k\) has image \(F\), which by PR-OZYUC means \(F/k\) is normal. Since \(F/k\) is separable as a subextension of the separable extension \(L/k\), this is equivalent to \(F/k\) being Galois. In that case restriction is a homomorphism \(\Gal(L/k)\to\Gal(F/k)\); it is surjective because every \(k\)-automorphism of \(F\) extends to \(L\), and its kernel is \(\Gal(L/F)\), which gives (c).