Let $F/K$ be a finite extension and suppose $F$ is the splitting field of a separable polynomial over $K$. Show that if Gal(F/K) is cyclic then for each divisor $d$ of $[F : K]$ there exists exactly one subfield $E$ of $K \subseteq E \subseteq F$ such that $[E : K]=d$.
Is this actually true? what I get is that one can find a subfield $E$ of $F$ such that $[F : E] = d$.
My work:
First, since $F/K$ is Galois then $[F : K] = |Gal(F/K)|$ so that if $d$ divides $[F : K]$ then d divides $|Gal(F/K|)$. By assumption the latter group is cyclic so for each divisor we can find a unique subgroup $H$ of $Gal(F/K)$ such that $|H| = d$.
But now the fundamental theorem implies $[F : F^{H}] = |H| = d$ so that $E=F^{H}$. I get then that $[F : E] = d$ and not $[E : K]=d$. What am I doing wrong?