Let $K$ be an imaginary quadratic number field (so there are no real embeddings) with ring of integers $\mathcal{O}\_K$ . Let $w$ be the number of units in $K$ and $h$ be the class number of $K$. Let $\mathfrak{f}$ be a non-zero integral ideal in $K$, and let $\mathcal{I}\_{\mathfrak{f}}$ be the ray class group mod $\mathfrak{f}$, i.e.
$$
\mathcal{I}\_{\mathfrak{f}} = \frac{J\_{\mathfrak{f}}}{P\_{\mathfrak{f}}},
$$
where $J\_{\mathfrak{f}}$ is the group of fractional ideals coprime to $\mathfrak{f}$ and, for imaginary quadratic fields, $P\_{\mathfrak{f}}$ is the group of principal fractional ideals $(\alpha)$ with $\alpha \equiv 1 \mod{\mathfrak{f}}$. Let $h(\mathfrak{f})$ denote the order of $\mathcal{I}\_{\mathfrak{f}}$, $\varphi(\mathfrak{f})$ the order of the multiplicative group $(\mathcal{O}\_K/\mathfrak{f})^\*$, and $g$ the number of units $\varepsilon$ with $\varepsilon \equiv 1 \mod{\mathfrak{f}}$. I have two questions pertaining to some literature I have read recently (namely [this paper of Coleman](https://doi.org/10.1112/plms/s3-61.3.433) and [this paper by Khale et. al.](https://doi.org/10.1016/j.jnt.2021.04.004))


1. In both papers linked above, there appears the term "ideal class group mod $\mathfrak{f}$." Is this meant to be the *ray* class group mod $\mathfrak{f}$, or is there some other object this term is referring to? From both papers, it seems the answer is yes, it's the ray class group, but I want to make sure I'm not missing something.
2. In the first paper linked above, there appears implicit in the arguments the identity
$$
h(\mathfrak{f}) = \frac{h\varphi(\mathfrak{f})g}{w},
$$
This appears specifically between displays (4.4) and (4.6). I would like a reference and/or an explanation of why this identity holds.


My algebraic number theory is not the strongest, and my familiarity with the literature and common jargon is low, so forgive me if these are "dumb" questions.



Questions about ray class groups
**1.** Surely, by "class group modulo $\mathfrak{f}$", the authors mean "ray class group modulo $\mathfrak{f}$". I guess the authors omit "ray", because $K$ has no real embedding, so there are no "rays" here.


**2.** Let $K^\times\_\mathfrak{f}$ be the multiplicative group of elements $\alpha\in K^\times$ congruent to $1$ modulo $\mathfrak{f}$, so that $P\_\mathfrak{f}$ is the multiplicative group of fractional ideals generated by such elements. Analogously, let $K^\times\_{\ast,\mathfrak{f}}$ be the multiplicative group of elements $\alpha\in K^\times$ coprime to $\mathfrak{f}$, and let $P\_{\ast,\mathfrak{f}}$ be the multiplicative group of fractional ideals generated by such elements. Then we have the following exact sequence of finite abelian groups, with the obvious homomorphisms:
$$1\to U/U\_\mathfrak{f}\to K^\times\_{\ast,\mathfrak{f}}/K^\times\_\mathfrak{f}\to J\_\mathfrak{f}/P\_\mathfrak{f}\to J\_\mathfrak{f}/P\_{\ast,\mathfrak{f}}\to 1.$$
Indeed, the image of the second map and the kernel of the third map are equal to $UK^\times\_\mathfrak{f}/K^\times\_\mathfrak{f}$, while the image of the third map and the kernel of the fourth map are equal to $P\_{\ast,\mathfrak{f}}/P\_\mathfrak{f}$. Comparing the sizes of the groups in the exact sequence, we get that
$$|U/U\_\mathfrak{f}|\cdot|J\_\mathfrak{f}/P\_\mathfrak{f}|=|K^\times\_{\ast,\mathfrak{f}}/K^\times\_\mathfrak{f}|\cdot|J\_\mathfrak{f}/P\_{\ast,\mathfrak{f}}|.$$
On the other hand, it follows from weak approximation that $J\_\mathfrak{f}/P\_{\ast,\mathfrak{f}}$ is isomorphic to the class group of $K$, hence our equation becomes
$$(w/g)\cdot h(\mathfrak{f})=\varphi(\mathfrak{f})\cdot h.$$


