How to prove $A^{n\times n}=I_n\Rightarrow A^n=A^{f(n\times n)}$? Let $A\in M_2(\mathbb{Z})$ s.t. there is a positive integer $n$ satisfying $A^n=I_2$.
Show that $A^{12}=I_2$.
I have no idea where to start. Suggestions?
 A: I refer you to the nice notes http://www.math.uconn.edu/~kconrad/blurbs/grouptheory/SL(2,Z).pdf
and in particular Corollary 2.3, which says the modular group $SL(2, \mathbb{Z})$ is generated by two elements $S=\begin{pmatrix}0&-1\\1&0\end{pmatrix}$ and $ST=\begin{pmatrix}0& -1\\1&1\end{pmatrix}$ of order 4 and 6 respectively. One can show that the finite subgroup generated by $A$ must be conjugate to one of the cyclic subgroups $\langle S\rangle$ and $\langle ST\rangle$.
A: The characteristic polynomial of $A$ can only have cyclotomic factors.  But the only cyclotomic polynomials of degree $\leq 2$ are $\Phi_1$, $\Phi_2$, $\Phi_3$, $\Phi_4$, and $\Phi_6$.
A: Let $\alpha,\beta$ be the eigenvalues of $A$. Then $\alpha ^n=\beta ^n=1$; moreover $A$ is diagonalizable over $\mathbb{C}$.
Case 1. $\alpha,\beta\in\{\pm 1\}$. Then $A^2=I$.
Case 2. $\alpha,\beta\notin \mathbb{R}$. Then $\alpha=e^{2ik\pi/n},\beta=e^{-2ik\pi/n}$ where $k$ is a positive integer $<n$. Note that $\alpha+\beta=2\cos(2k\pi/n)\in\mathbb{Z}$ and $\cos(2k\pi/n)\in\{0,\pm 1/2\}$. Finally $2k\pi/n\in\{\pi/2,3\pi/2,\pi/3,2\pi/3,4\pi/3,5\pi/3\}$, that (resp.) gives: $A^4=I,A^4=I,A^6=I,A^3=I,A^3=I,A^6=I$. In each case, $A^{12}=I$.
