Suppose that $M\subseteq \mathbb{Z}^n$ is a module such that $\mathbb{Z}^n/M$ is free and $S\subseteq \mathbb{R}^n$ is a bounded, symmetric (around $0$) convex set. Let $M'$ be the module generated by $S\cap M$. 


Question: Is $\mathbb{Z}^n/M'$ free?


I think it is free if the following is true: for any $x\in M\setminus M'$ and $a\neq 0$, it holds that $ax\not\in M'$. (Basically this is saying that $M/M'$ is free)


Is this true?


It seems true to me, but I haven't found a proof yet... 



sublattice generated by lattice points intersecting a convex set
This is true if $M$ has rank at most $2$ but not beyond that.
For a counterexample in rank $r \geq 3$, choose coordinates
so that $M$ is the body-centered cubic lattice, that is, the
subgroup of ${\bf Z}^r$ consisting of all-even and all-odd vectors;
and let $S$ be the $l^1$ ball of radius $2$, that is,
$$
S = \{ (x\_1,\ldots,x\_r) \in {\bf R}^r : \sum\_{i=1}^r \left|x\_i\right| \leq 2 \}.
$$
Then $S \cap M$ spans $(2{\bf Z})^r$, which is a proper finite-index
subgroup of $M$ (it does not contain $(1,1,\ldots,1)$).


