I’ll first give the statement and part of the proof I’m unsure about.
Let $K$ be a number field of degree $n$. Let $H$ be a finitely generated subgroup of $\mathcal{O}_K $ (viewed as an additive abelian group) of rank $n$. Suppose $\omega_1 ,...,\omega_n $ and $\eta_1 ,...,\eta_n $ are two bases for $H$. Then $\Delta (\omega_1,...,\omega_n )=\Delta (\eta_1,...,\eta_n ). $
Start of proof is the trouble. It says that for each $i$ we have $\omega_i = \sum_{j=1}^{n} c_{ij}\eta_j $ where the $n\times n $ matrix $(c_{ij})_{n \times n } $ is unimodular.
My question is, why should the matrix $(c_{ij}) $ is unimodular?
The group $H$ is free abelian of rank $n$, which means that $H$ is isomorphic to $\mathbb Z^n$. Choose an isomorphism $f : H \to \mathbb Z^n$. It follows that $f(\omega_1),...,f(\omega_n)$ and $f(\eta_1),...,f(\eta_n)$ are two bases for $\mathbb Z^n$.
As you can easily deduce using properties of isomorphisms, we have an equation $$f(\omega_i) = \sum_{j=1}^n c_{ij} f(\eta_j) $$ where $c_{ij} \in \mathbb Z$ for all $i,j$. Let $C = (c_{ij})$ be the $n \times n$ matrix of coefficients from this equation, so $C$ is a matrix of integers.
We can do the same thing in reverse: $$f(\eta_i) = \sum_{j=1}^n d_{ij} f(\omega_j) $$ where $d_{ij} \in \mathbb Z$ for all $i,j$. Let $D=(d_{ij})$ be the $n \times n$ matrix of coefficients for this second equation, also a matrix of integers.
Now you can further deduce that $C$ and $D$ are inverses: $CD = DC$ is equal to the $n \times n$ identity matrix.
Now apply the product formula for determinants, using that $CD$ is the identity matrix: $$\det(C) \cdot \det(D) = \det(CD) = 1 $$ Since the coefficients of $C$ and $D$ are all integers, it follows that $\det(C)$ and $\det(D)$ are both integers. Therefore, either $\det(C)=\det(D)=1$ or $\det(C)=\det(D)=-1$. In either case, $C$ is unimodular.