About the order of element in $GL_2(\mathbb{Z})$

Question

It shows that the possible finite order for elements in $GL_2(\mathbb{Z})$ is $1,2,3,4,6$. I only know it's true for $SL_2(\mathbb{Z})$, the modular group, by proving it's a free group generated by two elements $x,y$ where $x^2=y^3=-1$. The method to prove it "free" is to consider rational number like this which can't be appled directly to the case $GL_2(\mathbb{Z})$. Any hint will be appreciated.

Answer 1

If $M\in M_n(\Bbb{Z})$ is of order $k$ then its minimal polynomial divides $X^k-1$ thus it is a product of distinct cyclotomic polynomials $\Phi_d, d| k$. For $n=2$ the minimal polynomial (which divides the characteristic polynomial) is of degree $1$ or $2$ and $\Phi_d$ is of degree $1$ only for $d=1,2$ and it is of degree $2$ only for $d=3,4,6$.

If $\Phi_d$ is of degree $2$ then the minimal polynomial must be $\Phi_d$ so it divides $X^d-1$ thus $M^d=I$ and $k=d \in 3,4,6$.

Otherwise the minimal polynomial is $X-1,X+1$ or $X^2-1$, in every case $M^2=I$ and $k = 1$ or $2$.