TITLE: Finite index subgroup of $\mathbb{Z}^2$ that is invariant under a non-singular matrix
QUESTION [5 upvotes]: Let $M $ be a matrix in $ \operatorname{GL}(2, \mathbb{Z})$ that has at least one eigenvalue of absolute value strictly bigger than $1$. What are the finite index subgroups $H$ of $\mathbb{Z}^2$ such that for all $v \in H$, we have $M(v) \in H$?
Question: One example of such subgroup has the form  $H=\langle (a,0),(0,a)\rangle$, for a non-zero integer $a$, this example works for any matrix in  $ \operatorname{GL}(2, \mathbb{Z})$. I was wondering if there are any other such subgroups.
Thought so far: I think the finite index subgroups of $\mathbb{Z}^2$ have the form $H = \langle (a,b),(c,d)\rangle$, where $a,b,c,d \in \mathbb{Z}$ such that $ad-bc\neq 0$. Hence, in order for $H$ to be a subgroup, we need $M(a,b)= \alpha_1(a,b) + \alpha_2(c,d)$ and $M(c,d)= \beta_1(a,b) + \beta_2(c,d)$ for some integers $\alpha_i$ and $\beta_i$, but I can't see any obvious solutions for these equations.
I was also thinking maybe we can use the eigenvectors, but unfortunately the eigenvectors might not have integers entries.
Any ideas for constructing such a subgroup would be really appreciated.

REPLY [2 votes]: To illustrate how such a question is of arithmetic nature (and what a complete answer should look like), here is a partial answer in a specific case.
Namely: I specify to $M=\begin{pmatrix}2 & 1 \\ 1 & 1\end{pmatrix}$ and I address the question: what are invariant finite index subgroups of prime index.
The characteristic polynomial of this matrix is $X^2-3X+1$, which has discriminant 5. By the quadratic reciprocity formula, for an odd prime $p\neq 5$, 5 is a square mod $p$ iff $p=\pm 1$ mod $5$ (i.e. the decimal expansion of of $p$ terminates with $1$ or $9$). And this polynomial is irreducible mod $2$, and has a double root mod $5$.
Hence, for a prime $p$:

for $p=\pm 2$ mod $5$, there is no $M$-invariant subgroup of index $p$ in $\mathbf{Z}^2$;
for $p=\pm 1$ mod $5$, there are exactly two $M$-invariant subgroups of index $p$ in $\mathbf{Z}^2$;
there is a single $M$-invariant subgroup of index $5$ in $\mathbf{Z}^2$.


Remarks:
For a general matrix $M$, the characteristic polynomial is $X^2+nX\pm 1$ for some $n$ and there should be a similar discussion.
For a general index $q$, one should boil down to when $q$ is a power of a prime $p$. Then in turn one should boil down to when the quotient is cyclic (the quotient being isomorphic to $C_{p^a}\times C_{p^b}$ for some $a\le b$, the subgroup is contained in $p^a\mathbf{Z}^2$ and we can then "replace" $\mathbf{Z}^2$ with $p^a\mathbf{Z}^2$ to assume $a=1$).