Obivously, if the Hilbert class field $H$ of $K$ is abelian over $\mathbb Q$, then $K$ (which is a subfield of $H$) must be abelian over $\mathbb Q$. So suppose that this is the case. In general, there is then a maximal subfield of $H$ that is abelian over $\mathbb Q$; call it $F$. It certainly contains $K$, and is called the genus field of $K$. By class field theory $Gal(F/K)$ is a quotient of the class group $Cl(K)$ of $K$. Which quotient? At least when $Gal(K/Q)$ is cyclic, it is the maximal quotient on which $Gal(K/Q)$ acts trivially.

If $K$ is quadratic then $Gal(K/Q)$ acts on $Cl(K)$ by inversion, so $F/K$ corresponds to the maximal 2-elementary abelian quotient of $Cl(K)$. In particular, in this case $F = H$ if and only if $Cl(K)$ is an elementary abelian 2-group. (E.g. $\mathbb Q(\sqrt{-5})$, whose class group is of order 2.)

Obivously, if the Hilbert class field $H$ of $K$ is abelian over $\mathbb Q$, then $K$ (which is a subfield of $H$) must be abelian over $\mathbb Q$. So suppose that this is the case. In general, there is then a maximal subfield of $H$ that is abelian over $\mathbb Q$; call it $F$. It certainly contains $K$, and is called the genus field of $K$. By class field theory $Gal(F/K)$ is a quotient of the class group $Cl(K)$ of $K$. Which quotient? At least when $Gal(K/\mathbb Q)$ is cyclic, it is the maximal quotient on which $Gal(K/\mathbb Q)$ acts trivially.

If $K$ is quadratic then $Gal(K/\mathbb Q)$ acts on $Cl(K)$ by inversion, so $F/K$ corresponds to the maximal 2-elementary abelian quotient of $Cl(K)$. In particular, in this case $F = H$ if and only if $Cl(K)$ is an elementary abelian 2-group. (E.g. $\mathbb Q(\sqrt{-5})$, whose class group is of order 2.)

Obivously, if the Hilbet class field $H$ of $K$ is abelian over $\mathbb Q$, then $K$ (which is a subfield of $K$) must be abelian over $\mathbb Q$. So suppose that this is the case. In general, there is then a maximal subfield of $H$ that is abelian over $\mathbb Q$; call it $F$. It certainly contains $K$, and is called the genus field of $K$. By class field theory $Gal(F/K)$ is a quotient of the class group $Cl(K)$ of $K$. Which quotient? At least when $Gal(K/Q)$ is cyclic, it is the maximal quotient on which $Gal(K/Q)$ acts trivially.

If $K$ is quadratic then $Gal(K/Q)$ acts on $Cl(K)$ by inversion, so $F/K$ corresponds to the maximal 2-elementary abelian quotient of $Cl(K)$. In particular, in this case $F = K$ if and only if $Cl(K)$ is an elementary abelian 2-group. (E.g. $\mathbb Q(\sqrt{-5})$, whose class group is of order 2.)

Obivously, if the Hilbert class field $H$ of $K$ is abelian over $\mathbb Q$, then $K$ (which is a subfield of $H$) must be abelian over $\mathbb Q$. So suppose that this is the case. In general, there is then a maximal subfield of $H$ that is abelian over $\mathbb Q$; call it $F$. It certainly contains $K$, and is called the genus field of $K$. By class field theory $Gal(F/K)$ is a quotient of the class group $Cl(K)$ of $K$. Which quotient? At least when $Gal(K/Q)$ is cyclic, it is the maximal quotient on which $Gal(K/Q)$ acts trivially.

If $K$ is quadratic then $Gal(K/Q)$ acts on $Cl(K)$ by inversion, so $F/K$ corresponds to the maximal 2-elementary abelian quotient of $Cl(K)$. In particular, in this case $F = H$ if and only if $Cl(K)$ is an elementary abelian 2-group. (E.g. $\mathbb Q(\sqrt{-5})$, whose class group is of order 2.)