Show that an element of $GL_2(\mathbb{Z})$ has order $1,2,3,4,6$ or $\infty$. Show that an element of $GL_2(\mathbb{Z})$ has order $1,2,3,4,6$ or $\infty$.
My idea: $GL_2(\mathbb{Z})=\{A|A_{2 \times 2}$ & $det(A)\neq0\}$.
Let $A\in GL_2(\mathbb{Z})$, and order of $A$ is $n\geq 2$.
$A^n=I$, the characteristic polynomial is $x^n-1=0$.
$x^n-1=(x-1)(x^{n-1}+x^{n-2}+......+x+1)=0$.
I know that the cyclotomic polynomial is an irreducible monic polynomial in $\mathbb{Z}[x]$ of degree $\phi(n)$.
I confused how do I conclude that $n=2,3,4,6$ or $\infty$? Can anyone suggest me some hint?
 A: Let $A \in GL_2(\mathbb Z)$. In particular, $A$ satisfies its characteristic polynomial, which is a quadratic polynomial with integer coefficients, because $A$ is of dimension $2 \times 2$ and all its entries are integers.
Now, suppose that $A$ satisfies $A^n = I$ for some $I$, and for no smaller $n$ is that true. It follows that $x^n-1$ and the characteristic polynomial of $A$ share a common factor, but this is not true for any $m < n$.
Now, when can this happen? We have the celebrated formula $x^n-1 = \prod_{d | n} \Phi_d(x)$, where $\Phi_d$ is the $d$th cyclotomic polynomial. Furthermore, since each cyclotomic polynomial is irreducible, it follows that this is the unique breakdown into irreducibles of $x^n-1$.
Therefore, if the characteristic polynomial of $A$ shares a common factor with $x^n-1$, it is one of these polynomials. However, we know something more : the degree of $\Phi_d(x)$ is $\phi(d)$, the Euler totient function of $d$. In particular, only for $n=1,2,3,4,6$ is $\phi(n) \leq 2$, so only for these values can $\Phi_d(x)$ divide the characteristic polynomial of $A$.
Can you finish from here?
A: Let $A\in GL_2(\mathbb Z)$.
Suppose $\ \exists n \in \mathbb N^* \ :\ A^n =I_2$
Then, $A$ is diagonalizable.
Let $\lambda_1,\lambda_2$ its eigenvalues.
We have:
$\ \lambda_1^n=1\ $ , $\ \lambda_2=\overline{\lambda_1}\ $.
So, $\ \exists \theta \in \mathbb R \ : \ \lambda_1=e^{i\theta} \ \text{ and } \ \lambda_2=e^{-i\theta}$
Then $\ 2\cos\theta =\operatorname{tr}(A) \ $ is in $\mathbb Z$.
Therefore, we have 5 possibilities:
$\cos\theta = -1\ $ :  $\ A^2=I_2$
$\cos\theta = -\dfrac{1}{2}\ $ :  $\ A^3=I_2$
$\cos\theta = 0\ $ :  $\ A^4=I_2$
$\cos\theta = \dfrac{1}{2}\ $ :  $\ A^6=I_2$
$\cos\theta = 1\ $ :  $\ A^1=I_2$
