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.

We supply $2$ different proofs, the first of which is essentially the one Pedro had already posted.
Proof #1.
Let $A$ be a $2 \times 2$ integer matrix with $A^n = I$. If $\lambda$ is an eigenvalue of $A$, then $\lambda^n = 1$. A matrix $A$ of finite order is diagonalizable, so if its eigenvalues satisfy $\lambda^n = 1$, then $A$ has order $n$. We show that if $\lambda$ is an eigenvalue of an integer matrix, and if for some $N$, $\lambda^N = 1$, then $\lambda^n = 1$ with $n = 1$, $2$, $3$, $4$, or $6$.
An eigenvalue $\lambda$ is also a root of the characteristic polynomial of $A$, a quadratic polynomial with integer coefficients. So the degree of $\lambda$ over $\mathbb{Q}$ is at most $2$.
Let us assume that $\lambda \neq \pm 1$. We let $\alpha = \lambda + \lambda^{-1}$, and we look at the chain of fields$$\mathbb{Q} \subset \mathbb{Q}(\alpha) \subset \mathbb{Q}(\lambda).$$Then $\alpha$ is real, but $\lambda$ is not real. Therefore $\lambda \notin \mathbb{Q}(\alpha)$. Since $[\mathbb{Q}(\lambda):\mathbb{Q}] = 2$, it follows that $[\mathbb{Q}(\lambda):\mathbb{Q}(\alpha)] = 2$ and $[\mathbb{Q}(\alpha):\mathbb{Q}] = 1$. So $\alpha = \lambda + \lambda^{-1}$ is a rational number.
We notice that $\lambda$ is a root of the polynomial $q(x) = x^2 + \alpha x + 1$, which has rational coefficients, and since $\lambda$ is not real, $q$ is irreducible. But $\lambda$ is also a root of the integer polynomial $x^n - 1$.
The irreducible factors of $\mathbb{Q}[x]$ of a monic integer polynomial have integer coefficients. Therefore the coefficients of $q$ are integers. Then $\alpha = \lambda + \lambda^{-1}$ is an integer. Since $\lambda$ is on the unit circle and is not $\pm1$, $\alpha$ can be $-1$, $0$, or $1$. Then $\lambda$ will be $\zeta_3^{\pm1}$, $\zeta_4^{\pm1}$, or $\zeta_6^{\pm1}$.
Proof #2.
Let $G$ be the cyclic group generated by a matrix $A$ of order $n$ in $GL_2(\mathbb{Z})$. Because $A$ has integer entries, it carries the lattice $L = \mathbb{Z}^2$ to itself. We construct a $G$-invariant, positive definite, symmetric form $\langle\text{ }\,,\,\rangle$ on $\mathbb{R}^2$ by averaging the dot product: say $$\langle X, Y\rangle = X^\text{T}Y + (AX)^\text{T}(AY) + \dots + (A^{n-1}X)^\text{T}(A^{n-1}Y) = X^\text{T}MY,$$where$$M = I + A^\text{T}A + \left(A^2\right)^\text{T}A^2 + \dots + \left(A^{n-1}\right)^\text{T}A^{n-1}.$$We change basis to an orthonormal basis for this norm, using a real matrix $P$. In the new coordinates, the lattice $L$ becomes $L' = P^{-1}L'$, the form becomes dot product $X^\text{T}Y$, and the matrix of the operator $A$ becomes $A' = P^{-1}AP$. Since $\langle\text{ }\,,\,\rangle$ is $A$-invariant, the dot product is $A'$-invariant. So $A'$ is an orthogonal matrix. And of course, if $\left(A'\right)^n = I$ because $A^n = I$. Now we have an orthogonal matrix $A'$ of order $n$ that carries a lattice $L'$ to itself. The Crystallographic Restriction tells us that $n = 1$, $2$, $3$, $4$, or $6$.
