Is there a way to find a $2 \times 2$ integer-valued matrix of any arbitrary multiplicative order? I guess firstly this comes down to two things. Firstly, given any $n \in \mathbb{N}$, is there always an integer-valued matrix of order $n$? I feel like this is true, so secondly, how could we find such a matrix?
If we let there be complex numbers, we could take a primitive $n$th root of 1, say $\varphi$, and then the matrix $$\begin{bmatrix} \varphi & 0 \\ 0 & 1 \end{bmatrix}$$ would satisfy the requirements. I know that matrices similar to this will also have order $n$, but I can't find any nice way to find an integer-matrix.
If there isn't a way for integer-valued matrices, can we lessen the restriction to real-valued?
 A: We need a few key observations.  First, the characteristic polynomial of $A\in M_{2\times 2}$ is a monic degree 2 polynomial. Second, by Cayley-Hamilton, $A$ satisfies its characteristic polynomial.  Third, if $f(A)=0$ and $g(A)=0$, and if $h=\gcd(f,g)$ then by the division algorithm,  $h(A)=0$.  Fourth, the gcd of two monic integral polynomials is monic and integral (or slightly weaker, the gcd of two rational polynomials is rational).
So suppose that $A$ has order $n$, so that $g(A)=0$, where $g(x)=x^n-1$. Note that $x^n-1=\prod_{d\mid n} \phi_d(x)$ where $\phi_n(x)$ is the $n$th cyclotomic polynomial, which is irreducible.  Let $f(x)$ be the characteristic polynomial of $A$.    Then $\gcd(f(x),g(x))$ is a product of cyclotomic polynomials which is of degree at most 2. But the degree of $\phi_n$ is $\phi(n)$ (here, $\phi$ is Euler's $\phi$ function).  We have $\phi(n)=1$ when $n=1,2$ and $\phi(n)=2$ when $n=3,4,6$.  Putting this together, the order can only be $1, 2, 3, 4,$ or $6$ (although for larger matrices, we can have things combine in more complicated ways).
