I'm studying the idele group $\mathcal{I}$ for a number field $K$. My definition of the ray class group attached to a modulus $\mathfrak{m}$ is $$\mathcal{C}_{\mathfrak{m}}= \mathcal{I}/K^*U_{\mathfrak{m}}$$ where $U_{\mathfrak{m}}$ is the open subgroup of $\mathcal{I}$ attached to $\mathfrak{m}$.
Consider the idele class group $\mathcal{C}=\mathcal{I}/K^*$ and let $\pi$ be the projection of $\:\mathcal{I}$ onto $\:\mathcal{C}$. I have to show that $$ \mathcal{C}_{\mathfrak{m}} \simeq \mathcal{C}/\pi(U_{\mathfrak{m}}). $$
Seems to me a direct application of the third theorem of isomorphism to the groups $\mathcal{I},K^*U_{\mathfrak{m}},K^*$.
I think that i have a simple view of the problem and then my proof is far away to be correct. For this reason i ask you if my reasoning it's correct or if there is a more sofisticate (and correct!) proof of my problem.
That looks right to me. Personally I'm more comfortable thinking about it in terms of norms. If $L$ is the class field corresponding to $K^{\ast}U_m$, then $K^{\ast} U_m = K^{\ast} N_{L/K}(\mathcal{I}_L)$, where $\mathcal I_L$ is the ideles of $L$. If $C_K = \mathcal I_K /K^{\ast}$ and $C_L = \mathcal I_L/ L^{\ast}$, then the norm map $N_{L/K}: C_L \rightarrow C_K$ is well defined, the projection of $K^{\ast} U_m = K^{\ast} N_{L/K}(\mathcal{I}_L)$ to $C_K$ is $N_{L/K}(C_L)$, and all you wanted to prove is that $$C_K/N_{L/K}(C_L) \cong \mathcal I_K/ K^{\ast} N_{L/K}(\mathcal I_L)$$ both of which are isomorphic to $Gal(L/K)$.