I have calculated some real quadratic field 's Hilbert class field with class number $2$,and I found they were satisfied $Gal(H\_{K}/Q)\cong Z/2Z\oplus Z/2Z$,here $H\_{K}$ is the Hilbert class field of a real quadratic field $K$ whose class number is $2$.Is it true for all quadratic field with class number $2$? How to prove it?(ps:I'm a beginner of algebraic number theory ,and I'm very interested in Hlibert class field .)



A problem about real quadratic field
This is a consequence of "genus theory". If $K / \mathbf{Q}$ is an abelian extension, then the "genus field" of $K$ is the maximal extension $L / K$ such that $L/K$ is unramified and $L/\mathbf{Q}$ is abelian. You can read more about genus theory here: <https://en.wikipedia.org/wiki/Genus_field>.


In your case, the fact that $H\_K$ is Galois over $\mathbf{Q}$ and $[H\_K : \mathbf{Q}] = 4$ implies that $H\_K$ must be abelian over $\mathbf{Q}$ (since there are no nonabelian groups of order 4); so $H\_K$ coincides with the genus field $L\_K$. The genus field of of a quadratic field is always a composite of quadratic fields, so $H\_K$ must have Galois group $C\_2 \times C\_2$.


(More generally, for any quadratic field (real or imaginary), genus theory completely describes the 2-torsion part of the class group.)


