can then define the subgroup $\tilde H^{\perp_B} \subseteq {\mathbb{Z}}^{n+1}$ to be $$\tilde H^{\perp_B} := \left\{ \mathbf{s} \in {\mathbb{Z}}^{n+1} \middle| \langle \mathbf{s}, \mathbf{h}\rangle_B \in d{\mathbb{Z}}\text{ for all } \mathbf{h} \in \tilde H\right\}.$$ Define $H^{\perp_B}$ to be the image of $\tilde H^{\perp_B}$ under $pr$, $pr(\tilde H^{\perp_B})$.

We remark that it is clear that the group $J_{F_{dI}}$ is contained by $H$ as $(\phi_B)_*(e^{2\pi i / d}, \ldots, e^{2\pi i/d}) = \prod_j \rho_j$ is a generator of $ J_{F_A}$.

We have assumed that the group $G$ is a subgroup of $SL(F_A)$. This requires that, for all group elements $\mathbf{h} = (h_k)_k \in \tilde H$, the product $\prod_{j=0}^ne^{\frac{2\pi i }{d} \sum_{k} b_{jk} h_k}$ equals $1$. This implies that the sum $\sum_{j,k = 0}^n b_{jk}h_k$ is an integer divisible by $d$; therefore, $(1,\ldots, 1) \in \tilde H^{\perp_B}$. So, its image $pr(1,\ldots, 1)$ must be in $H^{\perp_B}$. The element $pr(1,\ldots, 1) = (e^{2\pi i/d}, \ldots, e^{2\pi i/d})$ is a generator of the group $J_{F_{dI}}$, hence $H^{\perp_B}$ contains $J_{F_{dI}}$.

Moreover, if one unravels all the definitions, one can see that $(\phi_{B^T})_*(H^{\perp_B}) = G^T$. In order for a monomial $\prod_{i=0}^n x_i^{s_i}$ to be $G$-invariant, we will need, for any $\prod_{i=1}^n \rho_i^{h_i} =(e^{{\frac{2\pi i }{d} \sum_{i=0}^n b_{ij}h_j}})_i \in G$, that 