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Studying class field theory, I have come across the following Proposition:

Proposition. Let $K/E$ be an extension of number fields so that there is no nontrivial unramified subextension $F/E$ with $Gal(F/E)$ abelian. Then $h_E$ divides $h_K$.

In the proof, $H$ denotes the Hilbert class field of $E$ and it is derived that $Gal(HK/K)\cong Gal(H/E)$, which I was able to understand why it holds. Then the author says that this isomorphism also gives that $Gal(HK/K)$ is an unramified abelian extension of $K$.

Edit: Following KConrad's suggestion in the comments, I started looking in the more general context, when $K/E$ a number fields extension and $F/E$ a finite Galois extension, how is the ramification in $F\cdot K/K$ related to the one in $F/E$. Unfortunately, I do not know how to start off.

I am familiar with the notions of Decomposition group, Inertia group and Frobenius element, in case any of these are relevant to the answer.

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    $\begingroup$ If $L$ and $K$ are two quadratic extensions of $\mathbf Q$ then they have isomorphic Galois groups over $\mathbf Q$, but you certainly can't read off the ramification in $L$ just from knowing it in $K$. $\endgroup$
    – KConrad
    Commented Aug 28, 2015 at 12:23
  • $\begingroup$ Thank you very much. But then to what extent do we have any such restrictions and how does this apply to the particular case of the Proposition mentioned? $\endgroup$ Commented Aug 28, 2015 at 12:26
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    $\begingroup$ You should think not about how ramification in two Galois extensions $K/E$ and $L/E$ are related when their Galois groups are isomorphic, but how ramification in $F/E$ and $FK/K$ are related where $F/E$ is a finite Galois extension (not necessarily the Hilbert class field of $E$). $\endgroup$
    – KConrad
    Commented Aug 28, 2015 at 14:39
  • $\begingroup$ I have been trying for a little while now to see how I can use your suggestion, but could not figure it out. Can you please give me a starting point? $\endgroup$ Commented Sep 3, 2015 at 10:58

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