Elements of $GL_{2}(\mathbb{Z})$ of finite order

Prove that any element of $GL_{2}(\mathbb{Z})$ of finite order has order $1,2,3,4,6$ using Field Theory. My idea is to reduce such a finite order matrix say $A$ with order $n$ to modulo a prime $p$.
$\det A=\pm1$ so $A$ will land inside $G=\{M\in GL_{2}(\mathbb{F}_{p}):\det M=\pm 1\}$. $|G|=2p(p^{2}-1)$. So $n\mid 2p(p^{2}-1)$ for all prime $p$. I don't know how to proceed after this.

Solution 1:

Let $G$ be a finite subgroup of $GL(2,\mathbb{Z})$. Then the natural mapping $G\cap SL(2,\mathbb{Z})\rightarrow SL(2,\mathbb{F}_3)$ is an injection. This can be used to find all finite subgroups of $GL(2,\mathbb{Z})$. The first step is to list all finite subgroups of $SL(2,\mathbb{F}_3)$, which has $24$ elements. Then one has to "go back" to $GL(2,\mathbb{Z})$. For the details see the article Finite groups of $2\times 2$ matrices. From the list of subgroups we see that the orders are $1,2,3,4,6$.
Of course, this proves much more than required, but it involves some "field theory", i.e., the study of the modular group over finite fields $\mathbb{F}_p$.

Remarks:
1.) The group $GL(2,\mathbb{F}_p)$ has $(p^2-p)(p^2-1)$ elements, and not $2p(p^2-1)$.

2.) It is easy to see that the order $n$ of finite order elements in $GL(2,\mathbb{Z})$ has to satisfy $\phi(n)\le 2$, because the minimal polynomial of an element of finite order must be divisible by the minimal polynomial over the rationals of a primitive $n$-th root of unity. This forces $n=1,2,3,4,6$.