Probably the following proposition can be proved using class field theory. But I don't know how.
My question: Is the following proposition true? If yes, how would you prove this?
Proposition Let $K$ be an algebraic number field. Let $L$ be a finite abelian extension of $K$. Let $\mathcal{I}$ be the group of fractional ideals of $K$. Let $\mathcal{P}$ be the group of principal ideals of $K$. Let $\mathcal{H}$ be a subgroup of $\mathcal{I}$ such that $\mathcal{I} \supset \mathcal{H} \supset \mathcal{P}$. Suppose that every prime ideal $P$ of $K$ of absolute degree 1 in $\mathcal{H}$ splits completely in $L$. Then [$L : K$] = [$\mathcal{I} : \mathcal{H}$] and $L/K$ is unramified at every prime ideal of $K$.
EDIT The above proposition is false as David Loeffler pointed out. So I change the assertion: Then [$L : K$] | [$\mathcal{I} : \mathcal{P}$] and $L/K$ is unramified at every prime ideal of $K$.
Motivation
Let me explain that we can get a useful information about the class number of a cyclotomic number field by the above proposition.
Let $k$ be an algebraic number field. Let $K$ be the Hilbert class field over $k$. Let $L$ be a finite extension of $k$. Let $E = KL$. Let $\mathcal{I}$ be the group of fractional ideals of $L$. Let $\mathcal{P}$ be the group of principal ideals of $L$. Let $\mathcal{H}$ = {$I \in \mathcal{I}$; $N_{L/k}(I)$ is principal}. Note that $\mathcal{H} \supset \mathcal{P}$. Let $\mathfrak{P}$ be a prime ideal of absolute degree 1 in $\mathcal{H}$. Then by this result, $\mathfrak{P}$ splits completely in $E$. Hence, by the above proposition, [$E : L$] | [$\mathcal{I} : \mathcal{P}$] and $E/L$ is unramified at every prime ideal of $L$. Suppose $L$ is a cyclotomic number field of an odd prime order and $k$ is the unique quadratic subfield of $L$. Let $h$ be the class number of $k$. Let $h'$ be the class number of $L$. Since $K/k$ is unramified, $K \cap L = k$. Hence [$E : L$] = [$K : k$] = $h$. Hence $h | h'$. Note that $h$ can be computed relatively easily when the disciminant of $k$ is small.
Effort
Let $\mathcal{I}_L$ be the group of fractional ideals of $L$. Perhaps, by the assumption, $\mathcal{H} \subset N_{L/K}(\mathcal{I}_L)\mathcal{P}$. By class field theory, $[L : K] = [\mathcal{I} : N_{L/K}(\mathcal{I}_L)\mathcal{P}]$. Hence, $[L : K] | [\mathcal{I} : \mathcal{H}]$.