} \\
Z_{(A')^T, (G')^T} &\simeq X_{dd'I}/ ((H')^{\perp_{dB'}}/J_{F_{dd'I}}).
\end{aligned}$$

Follows directly from Corollary \[binicor\].

Recall that we are asking for the conditions in which $Z_{A^T, G^T}$ and $Z_{(A'), (G')^T}$ are birational. This question can be answered if we can show that the groups $H^{\perp_{d'B}}$ and $(H')^{\perp_{dB'}}$ are equal. We now prove that such an equality holds, if we assume that the groups $G$ and $G'$ are equal.

If the groups $G$ and $G'$ are equal, then $H^{\perp_{d'B}}$ and $(H')^{\perp_{dB'}}$ are equal.

Set $\tilde H : = pr^{-1}(H) $ and $\tilde H' : = pr^{-1}(H')$ (Recall these groups from Section 3.2). Note that we have an equality of groups $( \phi_{d'B})_*\circ pr(\tilde H ) = G = G ' = (\phi_{dB'})_*\circ pr(\tilde H')$. This implies that, for any element $\mathbf{h} \in \tilde H$, there exists an element $\mathbf{h'} \in \tilde H'$ so that $d'B\mathbf{h} = dB' \mathbf{h}'$.

Suppose that $\mathbf{s} \in (\tilde H')^{\perp_{dB'}} $. We claim that $\mathbf{s}$ is in $\tilde H ^{\perp_{d'B}}$, i.e., for every $\mathbf{h} \in \tilde H$, that $\langle \mathbf{s}, \mathbf{h} \rangle_{d'B} \in d {\mathbb{Z}}$. Indeed, this is true. Given any $\mathbf{h} \in \tilde H$, there exists some $\mathbf{h'}$ as above where $d'B\mathbf{h} = dB'\mathbf{h'}$, hence $\langle \mathbf{s} , 