# Lower bound for order of matrix modulo $n$

For a positive integer $n$ and a square matrix $A$ with integral entries, let $\text{ord}(A, n)$ be the smallest positive integer $k$ such that $A^k \equiv \mathbf{1} \bmod n$, if such integer $k$ exists, otherwise put $\text{ord}(A, n) := +\infty$. So $\text{ord}(A, n)$ is the order of $A$ modulo $n$.

We have the following result [1, Theorem 17].

Theorem. If $\delta > 0$ and $A$ is $2 \times 2$ matrix with determinant equal to $1$ and trace not less than $2$, then $$\text{ord}(A,n) \gg n^{1/2} \exp((\log n)^\delta),$$ for all positive integers $n$ but a set of asymptotic density $0$.

My question is: Are there other results of this kind for square matrices with arbitrary sizes and determinants? Are there stronger results?

Thanks.

[1] P. Kurlberg and Z. Rudnick, On Quantum Ergodicity for Linear Maps of the Torus, Commun. Math. Phys. (2001) 222, 201-227.

• The case of the $1 \times 1$ matrix $A = (2)$ already seems difficult... – js21 May 11 '17 at 15:39
• The matrix $A$ has to be hyperbolic and there exists $\delta > 0$ such that ... – jjcale May 11 '17 at 18:09