Theorem 1   Let $F/K$ be an extension of number fields such that for any intermediate Galois extension $L/K$ , with $K\subsetneq L \subsetneq F$ , there is at least one finite place or infinite place which ramifies in the extension $L/K$ . Then, $h_K$ , the class number of $K$ , divides the class number of $F$ , $h_F$ .

Corollary 1   Let $F/K$ be an extension of number fields which is totally ramified at some prime (or at an archimedean place). Then $h_K$ divides $h_F$ .

Proof. The proof is clear since there cannot be unramified subextensions. The theorem applies.

Corollary 2   Let $F/K$ be a Galois extension of number fields such that $\Gal(F/K)$ is a non-abelian simple group. Then $h_K$ divides $h_F$ .

Proof. In this case, there cannot be subextensions with abelian Galois group and the theorem applies.

Proof. [Proof of the Theorem] Let $H$ be the Hilbert class field of $K$ . By definition, $H$ is the maximal unramified abelian extension of $K$ , $\Gal(H/K)$ is isomorphic to $\Cl(K)$ , the ideal class group of $K$ and $[H:K]=h_K$ . Since there are no nontrivial unramified abelian subextensions of $F/K$ , we have $F\cap H=K$ and so $[FH:F]=[H:K]=h_K$ . One can show that the extension $FH/F$ is unramified and abelian (in fact $\Gal(FH/F)\cong \Gal(H/K)$ ). Therefore $FH$ is contained in $L$ , the Hilbert class field of $F$ . Hence: $$h_F=[L:F]=[L:FH]\cdot[FH:F]=[L:FH]\cdot [H:K]=[L:FH]\cdot h_K$$ and so, $h_K$ divides $h_F$ .