Let \(E/F\) be a Galois extension and let \(K\), \(L\) be intermediate fields. Show that \(K\) and \(L\) are \(F\)-isomorphic (i.e. there exists an isomorphism from \(K\) to \(L\) which is the identity on \(F\)) if and only if the subgroups of \(G = \operatorname{Gal}(E/F)\) corresponding to \(K\) and \(L\) are conjugate in \(G\).
Solution
Goal. Show \(K \cong_F L\) iff the corresponding subgroups \(H = \Gal(E/K)\) and \(H' = \Gal(E/L)\) are conjugate in \(G\).
1. (\(\Rightarrow\)) Suppose \(\sigma: K \to L\) is an \(F\)-isomorphism.
1.1. Extend \(\sigma\) to an automorphism \(\tilde\sigma \in G = \Gal(E/F)\).
\(E/F\) is Galois, so any \(F\)-embedding \(K \to E\) extends to an automorphism of \(E\) (extend \(\sigma\) to an \(F\)-embedding \(K \to E\) and use normality).
1.2. \(\tilde\sigma(K) = L\).
\(\tilde\sigma\) extends \(\sigma\), and \(\sigma(K) = L\).
1.3. Hence \(\Gal(E/L) = \Gal(E/\tilde\sigma(K)) = \tilde\sigma \Gal(E/K) \tilde\sigma^{-1}\).
the Galois group of \(\tilde\sigma(K)\) is the conjugate of the Galois group of \(K\) by \(\tilde\sigma\).
1.4. Hence \(H' = \tilde\sigma H \tilde\sigma^{-1}\), so \(H\) and \(H'\) are conjugate.
2.3.
2. (\(\Leftarrow\)) Suppose \(H' = g H g^{-1}\) for some \(g \in G\).
2.1. \(L = E^{H'} = E^{gHg^{-1}} = g(E^H) = g(K)\).
the fixed field of \(gHg^{-1}\) is \(g\) applied to the fixed field of \(H\).
2.2. Hence \(g|_K: K \to L\) is an \(F\)-isomorphism.
\(g\) fixes \(F\) (it is in \(\Gal(E/F)\)) and maps \(K\) onto \(L\).
3. Q.E.D.
1 and 2 give both directions.