Invariant basis of some free $\mathbb{Z}$-module

Question

Let $G$ be a group acting on $M:=\mathbb{Z}^{r}$ a $\mathbb{Z}$-module.

My question is that, if a submodule $N$ of $M$ is invariant under $G$, does $N$ have to have a basis which is invariant under $G$?

Are there any result related to my goal?

Answer 1

As pointed out in the comments, the answer to your question is no, even when we take $N=M$ and when $M=\mathbb{Z}^2$. Thinking about it more carefully, if $N$ is invariant under $G$, we can think about it as $G$ acts on $N$ and forget completely about $M$. So the question is really about the case $N=M$.

Now if $(e_i)_{i\in I}$ is a basis of $M$ that is invariant under $G$. Then the matrices of the actions of $G$ only have entries 0 or 1, so they have determinant equal to $\pm 1$. I don't know whether the determinant is independent of basis, but this is not something every automorphism of $M$ has. For example, if the ring is commutative, and $\mathfrak{m}$ is a maximal ideal, then $M/\mathfrak{m}M$ is a vector space with $([e_i])_{i\in I}$ be a basis. Moreover, each automorphism of $M$ gives a canonical automorphism of $M/\mathfrak{m}M$. In this new situation, the determinant is independent of basis, and we obtain a necessary condition for your claim to be true.