# A certain problem concerning a Hilbert class field

Is the following proposition true? If yes, how would you prove this?

Proposition
Let $k$ be an algebraic number field.
Let $K$ be a finite abelian extension of $k$.
Suppose every principal prime ideal of $k$ splits completely in $K$.
Let $L$ be a finite extension of $k$.
Let $E = KL$.
Let $h’$ be the class number of $L$.

Then [$E : L$] | $h’$ and $E/L$ is unramified at every prime ideal of $L$.

Motivation
I thought I could use this proposition to prove the following result.

On the class number of a cyclotomic number field of an odd prime order

Effort

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 $\mathcal{H} \supset \mathcal{P}$.
Then use the following two links.

Related questions

On a certain criterion for unramification of an abelian extension of an algebraic number field

Complete splitting of a prime ideal in a certain abelian extension of an algebraic number field

#### Solutions Collecting From Web of "A certain problem concerning a Hilbert class field"

This is a direct application of class field theory. The assumption on $K$ implies that $K$ is contained in the Hilbert class field of $k$. Thus $E$ is contained in the Hilbert class field of $L$, and so $E$ is unramified over $L$, and $[E:L]$ divides the class number of $L$.