Finite cyclic subgroups of $GL_{2} (\mathbb{Z})$ How could we prove that any element of $GL_2(\mathbb{Z})$ of finite order has order 1, 2, 3, 4, or 6?
I am aware of the proof supplied here at this link: https://www.maa.org/sites/default/files/George_Mackiw20823.pdf. But I am curious if there are any other proofs.

Suppose that $A$ has finite order in ${\rm GL}(2,\Bbb Q)$. Then we know there is $k$ such that $X^k-1$ annihilates $A$. If $m_A$ is the minimal polynomial of $A$, $\deg m_A\leqslant 2$. Moreover, $m_A$ divides $X^k-1$ so $m_A$ has as its irreducible factors the irreducible factors of $X^k-1$. Since everything is monic, irreducibility over $\Bbb Q$ is the same as irreducibility over $\Bbb Z$. Of course $X^k-1$ factors into the cyclotomic polynomials, which are irreducible$^{1}$. We aslo know the $n$-th cyclotomic polynomial has degree $\varphi(n)$. And it happens that $\varphi(1)=1,\varphi(2)=1,\varphi(3)=2,\varphi(4)=2,\varphi(6)=2$, but any other number fails to have $\varphi(n)$ at most $2$. 
For the sake of it, here are (invertible) matrices over $\Bbb Z$ of orders $2,3,4,6$
$$\begin{pmatrix}-1&0\\0&1\end{pmatrix},\begin{pmatrix}-2&-3\\1&1\end{pmatrix}, \begin{pmatrix}2&1\\-5&-2\end{pmatrix},\begin{pmatrix}3&-7\\1&-2\end{pmatrix}$$



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*Here you can find several proofs.

