# Prime index subgroups of $\langle Q^{i}(\mathbb Z^{2}) \mid i \in \mathbb Z \rangle$ that is invariant under matrix $Q$

Let $$Q$$ be a matrix in $$\operatorname{GL}(2, \mathbb{Q})$$ and consider the group $$G = \langle Q^{i}(\mathbb Z^{2}) \mid i \in \mathbb Z \rangle := \langle Q^{i}(v) \mid i \in \mathbb Z, v \in \mathbb Z^{2} \rangle$$ under vector addition.

Question: What are the prime index subgroups $$H$$ of $$\langle Q^{i}(\mathbb Z^{2}) \mid i \in \mathbb Z \rangle$$ such that for all $$v \in H$$, we have $$Q(v) \in H$$?

I am interested in the case when both $$Q$$ and $$Q^{-1}$$ have non-integral coefficients in their characteristic polynomial (Here are some equivalent conditions).

"Easy" case: When $$Q$$ (or $$Q^{-1}$$) is an integer matrix we can start with a subgroup $$H'$$ of $$\mathbb Z^2$$ that has finite index $$p$$ and is invariant under $$Q$$ (such Q-invaraiant subgroup exists iff the char poly of $$Q$$ splits over $$\mathbb F_p$$), then $$H=\langle Q^{i}( H') \mid i \in \mathbb Z \rangle$$ would be a subgroup of $$G$$ which has finite index $$p$$ and is invariant under $$Q$$.

One of the reasons we can do this is because when $$Q$$ is an integer matrix, we have $$\langle Q^{i}(v)\mid i\in\mathbb Z,v\in\mathbb Z^{2}\rangle = \langle Q^{-i}(v)\mid i\in\mathbb N,v\in\mathbb Z^{2}\rangle$$. However, it is quite difficult to show that this holds for the general rational matrices.

Where I get this question from: I am trying to find all the finite index subgroups of $$G$$ that are invariant under $$Q$$. Given a prime $$p$$ that is coprime to all the denominators of the entries in $$Q$$ and $$Q^{-1}$$, we know that $$\langle Q^{i}(p\mathbb Z^{2}) \mid i \in \mathbb Z \rangle$$ would have finite index $$p^2$$ and is invariant under $$Q$$. From this, we can deduce that G has a subgroup of index $$p$$ (and hence a subgroup of index $$p$$ that is preserved under $$Q$$). I was wondering if there are any other finite index invariant subgroups.

Thank you for reading. Any ideas for constructing such a subgroup would be really appreciated.

(This question was copied from StackExchange)

• Your conclusion for $Q\in\mathrm{GL}_2(\mathbf{Z})$ (and more generally in the "fully coprime" case, which is true, for given $Q$, for $p$ large enough) is not correct: you get a $Q$-invariant index-$p$ subgroup iff the char. pol. of your matrix is split modulo $p$. For instance if $p$ is the companion matrix of $X^2+1$, this holds only for $p=2$ and $p=1$ mod $4$.
– YCor
Commented Jun 17 at 16:47
• Thanks @YCor , just to make sure I understand it correctly. Given $Q \in \operatorname{GL}(2, \mathbb{Q})$. Suppose $p$ is a big enough prime (in particular, $p$ is coprime to all the denominators of the entries in $Q$ and $Q^ {-1}$ ). Then $\langle Q^{i}(v) \mid i \in \mathbb Z, v \in \mathbb Z^{2} \rangle$ has a subgroup of index $p$ if and only if that char poly of $Q$ splits over $\mathbb F_p$? (Does this holds for any rational matrix in $\operatorname{GL}(2, \mathbb{Q})$) Commented Jun 17 at 18:53
• Yes, this is what I say (for every matrix $Q$ such that $p$ doesn't appear in denominators of $Q$ and $Q^{-1}$).
– YCor
Commented Jun 17 at 21:50

Let $$p$$ be a prime. Let $$a,b$$ be the eigenvalues of $$Q$$ in $$\mathbf{Q}_p$$ ($$a,b$$ thus belong to some quadratic extension of $$\mathbf{Q}_p$$).

Proposition.

1. If $$|a|$$, $$|b|$$ are both $$\neq 1$$, then $$G$$ has no subgroup of index $$p$$.
1. If exactly one of $$|a|$$, $$|b|$$ is $$1$$, then $$G$$ has a unique subgroup of index $$p$$, thus $$Q$$-invariant.
1. If $$|a|=|b|=1$$, then $$G$$ has $$p+1$$ subgroups of index $$p$$. In this case, $$Q$$ is conjugate to a matrix in $$\mathrm{GL}_2(\mathbf{Z}_p)$$, thus inducing a matrix $$Q'$$ in $$\mathrm{GL}_2(\mathbf{Z}/p\mathbf{Z})$$. [PS see edit]
• 3a) If $$Q'$$ has two distinct eigenvalues in $$\mathbf{Z}/p\mathbf{Z}$$, then $$G$$ has exactly two $$Q$$-invariant subgroups of index $$p$$.

• 3b) If, in $$\mathbf{Z}/p\mathbf{Z}$$, the matrix $$Q'$$ is not a scalar multiple of identity and has a double eigenvalue, then $$G$$ has a unique $$Q$$-invariant subgroup of index $$p$$.

• 3c) If, in $$\mathbf{Z}/p\mathbf{Z}$$, the matrix $$Q'$$ is a scalar multiple of identity, then all $$p+1$$ index $$p$$ subgroups of $$G$$ are $$Q$$-invariant.

• 3d) If the matrix $$Q'$$ has no eigenvalue in $$\mathbf{Z}/p\mathbf{Z}$$, then none of the $$p+1$$ index $$p$$ subgroups of $$G$$ is $$Q$$-invariant.

The proof is quite standard but I can post details if needed.

Note that all but finitely many primes fall in case 3. If the characteristic polynomial of $$Q$$ is irreducible over $$\mathbf{Q}$$, both Cases 3a and 3d appear for infinitely many primes.

Edit: in 3 the matrix in $$\mathrm{GL}_2(\mathbf{Z}_p)$$ is not unique and different choices might have reductions mod $$p$$ that fall in different cases. However, one chooses a conjugating matrix mapping $$\mathbf{Z}_p^2$$ onto the closure of $$G$$ (which, in case 3, is a compact open subgroup of $$\mathbf{Q}_p^2$$), and one conjugates by this matrix to get a matrix in $$\mathrm{GL}_2(\mathbf{Z}_p)$$, which one reduces modulo $$p$$. In other words $$Q'$$ is given by the action on $$\bar{G}/p\bar{G}$$, which is a vector space of dimension 2 over $$\mathbf{Z}/p\mathbf{Z}$$, so $$Q'$$ is really well-defined up to conjugation.

Edit 2. In detail. The question is to classify subgroups of index $$p$$ in $$G$$, and then classify the $$Q$$-invariant ones. For the first question, it is enough to determine $$G/pG$$, and for the second one, one has to understand the $$Q$$-action on $$G/pG$$. Since $$G$$ is a subgroup of $$\mathbf{Q}^2$$, $$G/pG$$ is a vector space over $$\mathbf{Z}/p\mathbf{Z}$$ of dimension in $$\{0,1,2\}$$. Call this cases 1, 2, 3.

In Case 1, there is no subgroup of index $$p$$, and in Case 2 there is a single one (hence $$Q$$-invariant). In Case 3, there are $$p+1$$ index $$p$$ subgroups in $$G$$. The action of $$Q$$ on $$G/pG$$ defines a matrix $$Q'$$ in $$\mathrm{GL}_2(\mathbf{Z}/p\mathbf{Z})$$, well-defined up to conjugation. If $$Q'$$ is a scalar multiple of identity (3c), all $$p+1$$ index-$$p$$ subgroups are $$Q$$-invariant. If $$Q'$$ is a scalar multiple of a non-identity unipotent, then there is a single $$Q$$-invariant index-$$p$$ subgroup in $$G$$ (3b). If $$G$$ is diagonalizable over $$\mathbf{Z}/p\mathbf{Z}$$ with distinct eigenvalues, then there are exactly 2 $$Q$$-invariant index-$$p$$ subgroups in $$G$$ (3a). The remaining case is when the characteristic polynomial of $$Q'$$ is irreducible over $$\mathbf{Z}/p\mathbf{Z}$$ and then none of the $$p+1$$ index-$$p$$ subgroups of $$G$$ is $$Q$$-invariant.

What remains is to read from $$Q$$ in which of the cases 1, 2, 3 it falls, and this depends of $$Q$$ viewed as matrix over $$\mathbf{Q}_p$$. The closure of $$G$$ in $$\mathbf{Q}_p^2$$ is an open $$Q$$-invariant subgroup of $$\mathbf{Q}_p^2$$. If both eigenvalues (over an extension of $$\mathbf{Q}_p$$) have modulus $$\neq 1$$, the only invariant open subgroup is $$\mathbf{Q}_p^2$$ itself, so $$G$$ is dense and it is not hard to deduce that $$G=pG$$ (Case 1). If both eigenvalues have modulus 1, one sees that the closure of $$G$$ is compact, and hence we are in Case 3. If exactly one of the eigenvalues has modulus 1, one sees that we are in Case 2.

• Thank you very much Yves, this is really helpful. For case 3, could you explain how you deduced that $Q$ is conjugate to a matrix in $\mathrm{GL}_2(\mathbf{Z}_p)$? I have seen a similar result: Given a rational matrix $Q$, if the characteristic equation of $Q$ has integral coefficients, then $Q$ is conjugated ( over $Q$ ) to a matrix with integral entries. Does the statement in case 3 follow from this result? Commented Jun 19 at 11:17
• @ghc1997 this is because in this case the closure of $G$ is a compact open $Q$-invariant subgroup. So there is $A$ in $\mathrm{GL}_2(\mathbf{Q}_p)$ that sends $\bar{G}$ onto $\mathbf{Z}_p^2$. Since $\bar{G}$ is $Q$-invariant, $A\bar{G}=\mathbf{Z}_p^2$ is $AQA^{-1}$-invariant. This means that $AQA^{-1}\in\mathrm{GL}_2(\mathbf{Z}_p)$. [Here by "$V$ is $Q$-invariant" I mean $QV=V$, not only $QV\subset V$.]
– YCor
Commented Jun 19 at 11:50