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...