Norm of ideal is an element of the ideal (proof)

Question

There is a lemma in my lecture notes that states

Lemma. Let $K$ be a number field and let $\mathfrak{a}$ be a non-zero ideal of $\mathcal{O}_{K}$. Let $A = \mathrm{Norm}(\mathfrak{a})$. Then $A \in \mathfrak{a}$.

Proof. By definition, $A = \mathrm{Norm}(\mathfrak{a}) = \#\mathcal{O}_{K}/\mathfrak{a}$. By Lagrange's Theorem, $A \cdot (1 + \mathfrak{a}) = 0 + \mathfrak{a}$ in $\mathcal{O}_{K}/\mathfrak{a}$. Thus $A \in \mathfrak{a}$.$\hspace{295pt}\square$

I'm not sure how Lagrange's Theorem is being used here.

I know that $\mathcal{O}_{K}/\mathfrak{a}$ is the set of additive cosets of $\mathfrak{a}$, so that $$ \mathcal{O}_{K}/\mathfrak{a} = \left\{ \mathfrak{a}, 1 + \mathfrak{a}, \ldots, r + \mathfrak{a} \right\} $$ which has $A$ elements in it, each of which have the same size. I thought that $\mathfrak{a} = 0 + \mathfrak{a}$ since $0$ is the additive identity, but I'm not sure why we multiply $A$ by the ideal coset $(1 + \mathfrak{a})$, nor why that equals $0 + \mathfrak{a}$, nor why this implies that $A \in \mathfrak{a}$.

Answer 1

Lagrange's theorem says that, if $g\in G$, a group, then $g^{|G|}=e$, right? In additive notation, this is written: $|G|\cdot g = 0$.

Thus, take for $g$ the coset $(1+\mathfrak{a})$, which is an element of the quotient ring, which is an additive group. Then for $|G|$, we have $A$, and the result is what you're looking for.

Answer 2

Lagrange’s theorem tells you that the order of any subgroup divides the order of the group. In particular, the order of any element of the group divides the order of the group. Hence, $A=n\cdot d$ for some $d$, where $n$ is the order of $1 + \mathfrak{a}$. Thus,

$$A(1+ \mathfrak a) = d(n(1 + \mathfrak a)) = d(0 + \mathfrak a) = 0 + \mathfrak a$$