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 .)

$\begingroup$ If there exist a prime $p$,that $Q(\sqrt p)$ has class number $2$,this could be a counterexample of my problem. $\endgroup$ – fool rabbit Mar 29 at 13:02
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 2torsion part of the class group.)