Orders of elements of $\mathrm{SL}_2(\ZZ)$

The possible orders are \[ 1,2,3,4,6,\quad\text{and }\infty. \]

Let \(A\in SL_2(\ZZ)\) have finite order. Its eigenvalues are roots of unity and their product is \(1\). Thus \[ \chi_A(x)=x^2-(\operatorname{tr}A)x+1. \] Since the eigenvalues lie on the unit circle, \[ |\operatorname{tr}A|\le2. \] The trace is an integer, so \[ \operatorname{tr}A\in\{-2,-1,0,1,2\}. \] Because \(A\) has finite order in characteristic \(0\), its minimal polynomial divides a separable polynomial \(x^m-1\), so \(A\) is diagonalizable over \(\CC\).

• If \(\operatorname{tr}A=2\), both eigenvalues are \(1\), hence \(A=I\) and \(|A|=1\).
• If \(\operatorname{tr}A=-2\), both eigenvalues are \(-1\), hence \(A=-I\) and \(|A|=2\).
• If \(\operatorname{tr}A=-1\), Cayley–Hamilton gives \(A^2+A+I=0\), so \(A^3=I\) and \(|A|=3\).
• If \(\operatorname{tr}A=0\), then \(A^2+I=0\), so \(|A|=4\).
• If \(\operatorname{tr}A=1\), then \(A^2-A+I=0\), whence \(A^3=-I\) and \(|A|=6\).

All five finite orders occur, for example via \[ I,\quad -I,\quad \begin{pmatrix}0&-1\\1&-1\end{pmatrix},\quad \begin{pmatrix}0&-1\\1&0\end{pmatrix},\quad \begin{pmatrix}0&-1\\1&1\end{pmatrix}. \] Finally, \[ \begin{pmatrix}1&1\\0&1\end{pmatrix}^k =\begin{pmatrix}1&k\\0&1\end{pmatrix} \] for \(k\ge1\), so infinite order also occurs.