Hilbert class field application

Question

If $K$ is an imaginary quadratic field and $M$ is an unramified Abelian extension of $K$, the prove that $M$ is Galois over $\mathbb{Q}$

Let see... If $L$ is the Hilbert class field of $K$, then $L$ is the maximal extension unramified of $K$, then $\mathbb{Q} \subset K \subset M \subset L$ and $L$ is Galois over $K$...

Thanks!

Answer 1

Partial answer:

We have $K \subset L$, $M \subset L$ and $\mathbb{Q} \subset K$, Galois extensions. Also, $K \subset M$ Galois, because $Gal(L/M)$ is normal subgroup of $Gal(L/K)$.

What can we say about $Aut(M/\mathbb{Q})$?

$|Aut(M/\mathbb{Q})|= 2|Gal(M/K)|$?

Answer 2

The following simple argument seems to work for any $K$ that is Galois over $\mathbb{Q}$ (could it be that $K$ is specified to be imaginary quadratic in order to ensure one understands that $M/K$ is unramified not only at the finite but also at the infinite primes?): since $M$ is unramified, abelian, it must be contained in $CF(K)$. As $CF(K)/K$ is Galois and abelian, all the intermediate fields must be Galois and abelian. So $M/K$ is Galois. Since $K/\mathbb{Q}$ is also Galois, $M/\mathbb{Q}$ must be Galois as well.