How to show that the norm of a fractional ideal is well-defined?

Thanks to the hints in the comments, I think I now can summarize an answer myself. But I still need help on how to prove the isomorphy in point (2) below. Please still feel free to point out mistakes or to post alternative (easier?) answers.

Let $c,c' \in \mathcal{O}_K \setminus \{ 0 \}$ such that $c\mathfrak{a} \subseteq \mathcal{O}_K$ and $c'\mathfrak{a} \subseteq \mathcal{O}_K$.

1. We have the chain $ c'c\mathfrak{a} \subseteq c\mathfrak{a} \subseteq \mathcal{O}_K $ of $\mathcal{O}_K$-modules. Because of the isomorphy $$ \frac{\mathcal{O}_K / c'c\mathfrak{a} }{ c\mathfrak{a} / c'c\mathfrak{a}} \cong \mathcal{O}_K / c\mathfrak{a} $$ of quotient modules it follows that $[\mathcal{O}_K : c'c\mathfrak{a}] = [\mathcal{O}_K : c\mathfrak{a}] · [c\mathfrak{a} : c'c\mathfrak{a}]$ (note: all indeces involved are indeed finite – something I do not prove here).

2. We have the isomorphy $ c\mathfrak{a} / (c') · c\mathfrak{a} \cong c\mathfrak{a} · (c\mathfrak{a})^{-1} / (c') · c\mathfrak{a} · (c\mathfrak{a})^{-1}$ (why?) of $\mathcal{O}_K$-modules and therefore $$ c\mathfrak{a} / c'c\mathfrak{a} = c\mathfrak{a} / (c') · c\mathfrak{a} \cong c\mathfrak{a} · (c\mathfrak{a})^{-1} / (c') · c\mathfrak{a} · (c\mathfrak{a})^{-1} = \mathcal{O}_K / (c') · \mathcal{O}_K = \mathcal{O}_K / (c'). $$ This implies $[c\mathfrak{a} : c'c\mathfrak{a}] = [\mathcal{O}_K:(c')]$.

3. By (1) and (2) we now get $$ \frac{[\mathcal{O}_K : c'c\mathfrak{a}]}{\left| N_{K/\mathbb{Q}}(c'c) \right|} = \frac{[\mathcal{O}_K : c\mathfrak{a}] · [c\mathfrak{a} : c'c\mathfrak{a}]}{\left| N_{K/\mathbb{Q}}(c') \right| · \left| N_{K/\mathbb{Q}}(c) \right|} = \frac{[\mathcal{O}_K : c\mathfrak{a}] · \overbrace{[\mathcal{O}_K:(c')]}^{=\left| N_{K/\mathbb{Q}}(c') \right|}}{\left| N_{K/\mathbb{Q}}(c') \right| · \left| N_{K/\mathbb{Q}}(c) \right|} = \frac{[\mathcal{O}_K : c\mathfrak{a}]}{\left| N_{K/\mathbb{Q}}(c) \right|}$$

4. Since we analogously (just switch the roles of $c$ and $c'$ in (1-3)) can show $ \frac{[\mathcal{O}_K : c'c\mathfrak{a}]}{\left| N_{K/\mathbb{Q}}(c'c) \right|} = \frac{[\mathcal{O}_K : c'\mathfrak{a}]}{\left| N_{K/\mathbb{Q}}(c') \right|}$ we get $$ \frac{[\mathcal{O}_K : c\mathfrak{a}]}{\left| N_{K/\mathbb{Q}}(c) \right|} = \frac{[\mathcal{O}_K : c'\mathfrak{a}]}{\left| N_{K/\mathbb{Q}}(c') \right|} $$ and therefore the norm $\mathfrak{N}(\mathfrak{a}) $ of a fractional ideal $\mathfrak{a}$ in $K$ is indeed well-defined.