Let \(M/K\) be a finite Galois extension and let \(K\subseteq L\subseteq M\) be an intermediate field such that \(L/K\) is Galois. Prove that \[ \Gal(M/L)\trianglelefteq \Gal(M/K). \]
Solution
Set \[ G=\Gal(M/K), \qquad H=\Gal(M/L). \]
1. Every \(\phi\in G\) preserves \(L\) setwise.
Because \(L/K\) is Galois, it is normal. Therefore every \(K\)-embedding of \(L\) into an algebraic closure has image \(L\). The restriction \(\phi|_L\) is such a \(K\)-embedding, so \[ \phi(L)=L. \]
2. For every \(\phi\in G\) and \(\sigma\in H\), the conjugate \(\phi\sigma\phi^{-1}\) fixes \(L\) pointwise.
Let \(\ell\in L\). By 1, \(\phi^{-1}(\ell)\in L\). Since \(\sigma\in H\), it fixes every element of \(L\), hence \[ \sigma(\phi^{-1}(\ell))=\phi^{-1}(\ell). \] Applying \(\phi\) gives \[ (\phi\sigma\phi^{-1})(\ell)=\ell. \] Thus \(\phi\sigma\phi^{-1}\in H\).
3. Therefore \(H\trianglelefteq G\).
By 2, \[ \phi H\phi^{-1}\subseteq H \] for every \(\phi\in G\). Applying the same inclusion to \(\phi^{-1}\) gives the reverse containment, hence equality.