Is the order of a ray class group bounded by the class number? Let $K$ be an algebraic number field with ring of integers $\mathcal{O}_K$, and $\mathfrak{m}$ a module of $K$.
Let $J$ be the group of fractional ideals in $\mathcal{O}_K$.
Let $P$ be the subgroup of fractional ideals in $\mathcal{O}_K$.
Let $J^{\mathfrak{m}}$ be the group of fractional ideals coprime to $\mathfrak{m}$.
Let $P^{\mathfrak{m}}$ be the subgroup of fractional ideals congruent to $1\ (\textrm{mod}\ \mathfrak{p})$ for every prime in the factorisation of $\mathfrak{m}$.
The class group of $K$ is defined as the quotient
$$
J_K/P_K,
$$
and the class number of $K$ is the order of its class group.
The ray class group of $K$ with respect to the module $\mathfrak{m}$ is defined as the quotient
$$
J_{K}^{\mathfrak{m}}/P_K^{\mathfrak{m}}.
$$
My question is: Is the order of the ray class group always bounded by the class number. Or in other words: For any module $\mathfrak{m}$, do we always have
$$
\lvert J_{K}^{\mathfrak{m}}/P_K^{\mathfrak{m}} \rvert \leq \lvert J_K /P_K\rvert\ ?
$$
I know this seems like a very basic question, but I don't think it's as simple as it seems.
 A: It is rather the other direction: $\lvert J_{K}^{\mathfrak{m}}/P_K^{\mathfrak{m}} \rvert \geq \lvert J_K /P_K\rvert$.
To see this, note that there is a group homomorphism from $J_{K}^{\mathfrak{m}}$ to $J_K /P_K$, sending any ideal to its ideal class.
This homomorphism is surjective, by strong approximation theorem, and its kernel contains $P_K^{\mathfrak{m}}$.
Therefore it induces a surjective homomorphism from $J_{K}^{\mathfrak{m}}/P_K^{\mathfrak{m}}$ to $J_K /P_K$.

Note that this classical language is a bit "outdated". The modern language uses adeles (or ideles).
Restating the above, the ray class group $J_{K}^{\mathfrak{m}}/P_K^{\mathfrak{m}}$ is nothing but the quotient $\Bbb A_K^\times / K^\times U_\mathfrak{m}$, where $U_\mathfrak{m}$ is the open subgroup of $\widehat{\mathcal O_K}^\times \times K_\infty^\times$ of conductor $\mathfrak m$.
Among all the subgroups $U_\mathfrak m$, the largest one is the case $\mathfrak m = 1$, which is simply $U_1 = \widehat{\mathcal O_K}^\times \times K_\infty^\times$. The corresponding ray class group $J_K^1/P_K^1$ is nothing but the usual class group $J_K/P_K$.
From this point of view, it is apparent that $J_K/P_K$ is a quotient of $J_{K}^{\mathfrak{m}}/P_K^{\mathfrak{m}}$ for any $\mathfrak m$.
For details of the idele version, see e.g. the corresponding wiki page.
A: By definition, for a given algebraic number field $K$, the ray class group relative to the modulus $\frak M$ surjects onto the class group. But the appeal to CFT - which translates these groups in terms of Galois groups - gives much more precise information. For an excellent account of the main results and terminologies (in "modulii" as well in "idelic" terms), I recommend D. Garbannati, "CFT summarized", Rocky Mountain J. of Math. 11, 2 (1981). 
The class group of $K$ is isomorphic to the Galois group over $K$ of the maximal abelian unramified extension of $K$, whereas the ray class group relative to $\frak M$ is isomorphic to the Galois group over $K$ of the maximal abelian extension of $K$ which is unramified outside $\frak M$ (NB: ramification at archimedean primes must be precisely defined, but this is just a question of conventions). The class group is always finite, whereas the ray class group can be infinite. A much studied case is when the places in $\frak M$ are just the places defined by the prime ideals of $K$ dividing a given rational prime $p$ (assume $p$ odd to get rid of the archimedean primes). Then CFT asserts that the maximal pro-$p$-quotient of the ray class group is isomorphic to $T \times \mathbf Z_p ^{1+c+d}$, where $T$ is finite, $\mathbf Z_p$ denotes the $p$-adic integers, $c$ is the number of complex places of $K$ and $d$ is conjecturally null (Leopoldt's conjecture). In a perfect world, the $p$-class group would be a quotient of $T$, but no. The relations between them are governed by a complicated combination of isomorphism and duality usually baptized "reflection" ("Spiegelung" in German).
