There is a theorem that states that for any algebraic number field $K$ and any integral ideal in its ring of integers $\mathfrak{m} \subset \mathcal{O}_K$, there exists an extension $L$ of $K$ (the ray class field) such that $\textrm{Gal}(L/K)$ is isomorphic to the ray class group $J^{\mathfrak{m}}/P^{\mathfrak{m}}$.
This group is necessarily Abelian.
But conversely, if $E$ is any Abelian extension of $K$, does there always exist some integral ideal $\mathfrak{m} \subset \mathcal{O}_K$ such that
$$J^{\mathfrak{m}}/P^{\mathfrak{m}} \cong \textrm{Gal}(L/K)?$$
No, because the LHS is always at least as big as the class group but the RHS could be smaller (for example trivial). The correct claim is that any Abelian extension is a subfield of a ray class field