Let $V$ be a finite-dimensional complex vector space and let $$ \Lambda_1\subseteq\Lambda_2\subseteq V $$ be full-rank lattices (discrete subgroups of $V$ such that $\mathrm{span}(\Lambda_i)=V$). Let $$ \Lambda_i^*:=\{\xi\in V^*:\xi(\Lambda_i)\subseteq\mathbb{Z}\} $$ be the dual lattices. Then, we can consider the abelian groups $\Lambda_2/\Lambda_1$ and $\Lambda_1^*/\Lambda_2^*$.
Question. Is it true that $\Lambda_2/\Lambda_1\cong\Lambda_1^*/\Lambda_2^*$ as abelian groups?
Of course, they are non-canonically isomorphic to each other.
The easiest way of seeing this is to use the structure theorem of finitely generated abelian groups.
Since $\Lambda_1\subseteq \Lambda_2$ are both free and the index is finite, by the structure theorem, there exist a $\mathbb Z$-basis $(e_i)_i$ of $\Lambda_2$ and non-zero integers $(d_i)_i$ such that $(d_ie_i)_i$ is a $\mathbb Z$-basis of $\Lambda_1$.
The space $V$ being $\Lambda_2 \otimes_{\mathbb Z}\mathbb C$, we may use $(e_i)_i$ again as a $\mathbb C$-basis of $V$. Let $(f_i)_i$ be the dual basis of $V^*$.
It is then clear that $(f_i)_i$ is a $\mathbb Z$-basis of $\Lambda_2^*$, and that $(\frac 1 {d_i} f_i)_i$ is a $\mathbb Z$-basis of $\Lambda_1^*$.
Therefore both $\Lambda_2/\Lambda_1$ and $\Lambda_1^*/\Lambda_2^*$ are isomorphic to $\oplus_i\mathbb Z/d_i \mathbb Z$.