Integer $2 \times 2$ matrices such that $A^n = I$ An earlier question today motivates this slight variant:

For what natural numbers  $n$ does there exist a non-identity integer $2\times 2$ matrix $A$, such that $A^n = I$? (And let's say $A^k \ne I$ for $|k| &lt; |n|$, too.)

Clearly there are solutions for $n=2,3$, because $(-I)^2 = I$, and for $n=3$ we have
$\left( \begin{smallmatrix} -2 &amp; 1\\ -3 &amp; 1 \\\end{smallmatrix} \right)$ 

The solution for $n = 3$ suggests to me that something involving the euclidean algorithm might come into play...and the fact the the determinant is $\pm 1$ suggests that this is really an $SL(2, \mathbb Z)$ (or $PSL(2, \mathbb Z)$)problem...but that's a group I'm woefully ignorant about.

For $n = 4$, there's $\left( \begin{smallmatrix} 0 &amp; -1\\ 1 &amp; 0 \\\end{smallmatrix} \right)$. And right about there I run out of ideas.
For $n=6$ there's $\left( \begin{smallmatrix} 2 &amp; -1\\ 3 &amp; -1 \\\end{smallmatrix} \right)$, not surprisingly. And these are the only such $n$, even if you allow rational entries instead of just integers.

Suppose that $A^n=I$ but $A^k\ne I$ for $k&lt;n$. Then the minimal polynomial of $A$ divides $x^n-1$ but doesn't divide $x^k-1$ for $k&lt;n$. It follows that the minimal polynomial of $A$ divides the $n$th cyclotomic polynomial $\Phi_n(x)$. But that polynomial is irreducible over the rationals, and so the minimal polynomial of $A$ equals $\Phi_n(x)$. Finally, the minimal polynomial of $A$ has degree at most $2$, since it divides the characteristic polynomial of $A$ by the Cayley-Hamilton theorem. Therefore the degre of $\Phi_n$ is at most $2$, which means that $\phi(n)$ is at most $2$, and so $n\in\{1,2,3,4,6\}$.
$$R_{\theta}=\begin{bmatrix}
cos \theta  &amp; -sin \theta \\ 
 sin \theta&amp; cos \theta 
\end{bmatrix}\\$$if $$R_{\theta+2k\pi}=R_{\theta}\\(R_{\theta})^n=R_{n\theta}\\\theta=\frac{2k\pi}{n} \rightarrow (R_{\theta})^n=R_{n\theta}=R_{n\frac{2k\pi}{n}}=R_{2k\pi}=I $$