Let \(F/K\) be a finite Galois extension and let \(A,B\) be intermediate fields. Prove \[ \operatorname{Aut}_{AB}(F)=\operatorname{Aut}_A(F)\cap\operatorname{Aut}_B(F) \] and \[ \operatorname{Aut}_{A\cap B}(F)=\operatorname{Aut}_A(F)\vee\operatorname{Aut}_B(F), \] where \(\vee\) denotes the subgroup generated by the two subgroups.
Solution
Let \[ G=\operatorname{Gal}(F/K),\qquad H_A=\operatorname{Gal}(F/A),\qquad H_B=\operatorname{Gal}(F/B). \]
An automorphism fixes the compositum \(AB\) exactly when it fixes both \(A\) and \(B\). Hence \[ \operatorname{Gal}(F/AB)=H_A\cap H_B. \]
For the second identity, let \[ H=\langle H_A,H_B\rangle=H_A\vee H_B. \] Its fixed field is \[ F^H=F^{H_A}\cap F^{H_B}=A\cap B, \] because an element is fixed by the generated subgroup exactly when it is fixed by both generating subgroups. By the finite Galois correspondence, \[ H=\operatorname{Gal}(F/A\cap B). \] Thus \[ \operatorname{Aut}_{A\cap B}(F)=\operatorname{Aut}_A(F)\vee\operatorname{Aut}_B(F). \]