Let $d \in \mathbb{Z} \setminus \{0, 1\}$ square free, $K = \mathbb{Q}(\sqrt{d})$, $\mathcal{O}_K$ the corresponding ring of integers, $\mathfrak{a} \subseteq K$ a fractional ideal $n, n' \in \mathbb{N}$ with $n\mathfrak{a}, n'\mathfrak{a} \subseteq \mathcal{O}_K$.
I'm currently trying to prove that the definition of the idealnorm for fractional ideals is well defined. Inside this proof I need the following result: $[n\mathfrak{a} : nn'\mathfrak{a}] = [\mathcal{O}_K : (n')]$ or more general $n\mathfrak{a} / nn'\mathfrak{a} \cong \mathcal{O}_K / (n')$ as abelian groups.
I don't know how I could approach this step. Any help is welcome!
Duplicate claim: My proof looks similar to the one posted here, but the linked post doesn't prove the question I asked here.
I think both $[n\mathfrak a : nn' \mathfrak a]$ and $[\mathcal O_K : (n')]$ are equal to $(n')^d$, where $d$ is the degree of the number field $K$ over $\mathbb Q$.
For example, if $x_1, \dots, x_d$ is an integral basis for $\mathcal O_K$, then $n'x_1, \dots, n'x_d$ is an integral basis for $(n')$. Thus $\mathcal O_K / (n') \cong \mathbb Z_{n'}^{\oplus d}$ as an abelian group.
Similarly, if $y_1, \dots, y_d$ is an integral basis for the ideal $n\mathfrak a \subset \mathcal O_K$, then $n'y_1, \dots, n'y_d$ is an integral basis for $nn'\mathfrak a$. So $\mathcal n\mathfrak a / nn'\mathfrak a \cong \mathbb Z_{n'}^{\oplus d}$ as an abelian group too.