Let \(F \subseteq E \subseteq K\) be a tower of fields such that \(K/F\) is a finite cyclic Galois extension. Suppose \(E/F\) is normal. Prove that both \(E/F\) and \(K/E\) are cyclic Galois extensions.
Solution
Let \[ G=\operatorname{Gal}(K/F),\qquad H=\operatorname{Gal}(K/E). \] Since \(K/F\) is cyclic Galois, \(G\) is cyclic. By the Galois correspondence, \[ \operatorname{Gal}(K/E)=H\le G, \] so \(H\) is cyclic because every subgroup of a cyclic group is cyclic. Hence \(K/E\) is cyclic Galois.
Also every subgroup of a cyclic group is normal, so \(H\trianglelefteq G\). Therefore \(E/F\) is Galois and restriction gives \[ \operatorname{Gal}(E/F)\cong G/H. \] A quotient of a cyclic group is cyclic, so \(E/F\) is cyclic Galois.
Thus both \(K/E\) and \(E/F\) are cyclic. In fact, the stated assumption that \(E/F\) is normal is redundant: every intermediate field of a finite cyclic Galois extension is Galois over the base field.