I have a certain class of lattices, and given $\Lambda$ in this class I would like to find obstructions to the existence of a finite-index lattice embedding $\Lambda \rightarrow \mathbb{Z}^n$ for some $n$, where $\mathbb{Z}^n$ has the standard positive definite symmetric bilinear form.
I understand, for example from Nikulin's paper Integral symmetric bilinear forms and some of their applications, that if such an embedding exists, then one can find a corresponding isotropic subgroup $H$ inside the discriminant group $G$ of $\Lambda$. Moreover, the restriction of the determinant bilinear form $b \colon G \times G \rightarrow \mathbb{Q}/\mathbb{Z}$ to $H^{\perp}/H$ must be equal to the determinant bilinear form of $\mathbb{Z}^n$, which should be trivial.
The discriminant group of $\Lambda$ is of the form $G=\frac{\mathbb{Z}}{m^2n^2\mathbb{Z}}\oplus\frac{\mathbb{Z}}{m^2\mathbb{Z}}\oplus\frac{\mathbb{Z}}{n^2\mathbb{Z}}$, with $m,n$ coprime, and the determinant bilinear form $b$ has associated matrix $$ \begin{bmatrix} \frac{1}{m^2n^2} & & \\ & \frac{q}{m^2} & \\ & & \frac{q'}{n^2} \end{bmatrix} $$ for some integers $q,q'$ coprime respectively with $m$ and $n$. It seems to me that the subgroup $H=\frac{\mathbb{Z}}{mn\mathbb{Z}}\oplus\frac{\mathbb{Z}}{m\mathbb{Z}}\oplus\frac{\mathbb{Z}}{n\mathbb{Z}}$ is of the required form.
However, I know for sure that at least some of these lattices do not admit a finite-index embedding in $\mathbb{Z}^n$. Is it possible to determine a stronger condition which $H$ must satisfy if it actually corresponds to an embedding in $\mathbb{Z}^n$?