Remark

Let $G_1, G_2$ be two lattices in $\mathbf{R}^n$ (no. 1) such that $G_2 \subset G_1$: then (Proposition 5) $G_1^*$ and $G_2^*$ are lattices in $\mathbf{R}^n$ such that $G_1^* \subset G_2^*$. We have seen in no. 1 that there is an integer $m > 0$ such that $m G_1 \subset G_2$;

consequently for $x \in G_1$ and $u \in G_2^*$ we have $m(u|x) \in \mathbf{Z}$ and therefore $(u|x) \in \mathbf{Q}$. If $x \in G_2$ and $u \in G_2^*$, or if $x \in G_1$ and $u \in G_1^*$, we have $(u|x) \in \mathbf{Z}$ by definition. Consequently on passing to the quotients the $\mathbf{Z}$-bilinear mapping $(x, u) \to (u|x)$ of $G_1 \times G_2^*$ into $\mathbf{Q}$ defines a $\mathbf{Z}$-bilinear mapping $B$ of $(G_1/G_2) \times (G_2^*/G_1^*)$ into $\mathbf{Q}/\mathbf{Z}$. Moreover, it is clear that if $\bar{x}_0 \in G_1/G_2$ (resp. $\bar{u}_0 \in G_2^*/G_1^*$) is such that, for each $\bar{u} \in G_2^*/G_1^*$ (resp. for each $\bar{x} \in G_1/G_2$) we have

then necessarily $\bar{x}_0 = 0$ (resp. $\bar{u}_0 = 0$). It follows that there is a $\mathbf{Z}$-linear bijection $h$ of $G_2^*/G_1^*$ onto the dual of $G_1/G_2$, such that $\langle \bar{x}, h(\bar{u}) \rangle = B(\bar{x}, \bar{u})$ for $\bar{x} \in G_1/G_2$ and $\bar{u} \in G_2^*/G_1^*$; in particular, the finite groups $G_1/G_2$ and $G_2^*/G_1^*$ are isomorphic.