There exist elements of finite order $n$ in $GL_2(\Bbb Z)$ for every positive $n$.

Question

I am trying to prove the title statement. So far, I have tried to do this algebraically. For example, for $n = 2$, and a matrix $\begin{pmatrix} a & b \\ c & d \end{pmatrix},$ I tried to find such $a, b, c, d$ where $ab - cd = 1$ or $-1$ and where $\begin{pmatrix} a^2 + bc & ab + bd \\ ac + dc & cd + d^2 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}.$ I found that $\begin{pmatrix} 4 & 3 \\ -5 & -4 \end{pmatrix}$ works. However, is there a way to do this without algebra for larger $n?$ Any hints as to how I would go about this?

Answer 1

Suppose $\;A\in GL_2(\Bbb Z)\;$ has finite order $\;n\in\Bbb N\;$ . If $\;\lambda\in\Bbb C\;$ is an eigenvalue of $\;A\;$ , then $\;A^n=I\implies \lambda^n=1\implies \lambda\;$ is an $\;n\,-$ th root of unity . Observe that if $\;\lambda\in\Bbb R\implies \lambda=\pm1\;$ , and if $\;\lambda\in\Bbb C\setminus\Bbb R\;$ , then both $\;\lambda,\,\overline\lambda\;$ are the different eigenvalues of $\;A\;$ as they're complex roots of a real polynomial.

If we now take the Jordan Canonical Form of $\;A\;$ ($JCF_A$) , over the complex say, then we find that $\;A\;$ must be diagonalizable: if $\;A\;$ has two different eigenvalues this is immediate, otherwise it has one single eigenvalue which then must be $\;\pm1\;$ , but then if $\;A\;$ isn't diagonalizable we get

$$A\sim JCF_A=\begin{pmatrix}\!-1&1\\0&\!-1\end{pmatrix}\;,\;\;\text{or}\;\;\begin{pmatrix}1&1\\0&1\end{pmatrix}$$

and both these two matrices clearly have no finite order.

Thus we have the two big possibilities (in diagonal form):

Real eigenvalues:

(a) Different eigenvalues:

$$\begin{pmatrix}\pm1&0\\0&\mp1\end{pmatrix}\;,\;\;\text{of order two}$$

(b) Same eigenvalue:

$$\begin{pmatrix}1&0\\0&1\end{pmatrix}\;\;\text{of order one}\;,\;\;\begin{pmatrix}-1&0\\0&-1\end{pmatrix}\;\;\text{of order two}$$

Complex non-real eigenvalues: of the form

$$\begin{pmatrix}\lambda&0\\0&\overline\lambda\end{pmatrix}$$

Observe that the characteristic=minimal polynomial of this matrix is

$$\;x^2-(2\,\text{Re}\,\lambda)\,x+|\lambda|^2=x^2-(2\,\text{Re}\,\lambda)\,x+1\;$$

since $\;\lambda\;$ is a root of unity and thus it is an element of $\;S^1:=\{\,z\in\Bbb C\;/\;|z|=1\}\;$ , and since also $\;\lambda+\overline\lambda=\text{tr.}\,A\in\Bbb Z\;$ , the only possibilities are $\;\lambda+\overline\lambda=0,\pm1\;$, so:

$$\begin{align*}&(1)\;\;\lambda+\overline\lambda=0\;\implies \lambda=-\overline\lambda\implies\lambda=\pm i,\,\,\text{since}\;\;(yi)^n=1\implies|y|=1\,,\,\,\\{}\\ &(2)\;\;\lambda+\overline\lambda=1\,,\,\text{with characteristic polynomial}\;\;x^2-x+1\,\mid x^6-1\implies \lambda=e^{\pi i/3}\\{}\\&(3)\;\;\lambda+\overline\lambda=-1\,,\,\text{with characteristic polynomial}\;\;x^2+x+1\,\mid x^3-1\implies \lambda=e^{2\pi i/3}\end{align*}$$

The order of $\;A\;$ is: in the first case $\;4\;$ , in the second $\;6\;$ and in the third one $\;3\;$ .

Answer 2

As can be seen from this question the contrary is true, even if one replaces $GL_2(\Bbb Z)$ by $GL_m(\Bbb Q)$ for any fixed$~m$: only a finite number of finite orders occur in such a matrix group.

Specialising the answer I gave there to $GL_2(\Bbb Z)$, the essential argument is that any element of order$~n$ must have a minimal polynomial that divides $X^n-1$, while (by Cayley-Hamilon) the degree of that minimal polynomial is at most$~2$. Therefore the irreducible factors of that minimal polynomial must be among the set of cyclotomic polynomials $\Phi_d$ of degree at most$~2$, and that set is finite, consisting of $\Phi_1=X-1$, $\Phi_2=X+1$, $\Phi_3=X^2+X+1$, $\Phi_4=X^2+1$, and $\Phi_6=X^2-X+1$. Then of course the number of ways to multiply elements from this set while keeping the degree${}\leq2$ is very limited (indeed apart from $\Phi_1\Phi_2=X^2-1$ there is no way for a finite order element to have a reducible minimal polynomial) and the set of possible orders is $\{1,2,3,4,6\}$.

The fact that the given list of cyclotomic polynomials of degree at most$~2$ is complete, is easily proved using the formula for the degree $\phi(n)$ (Euler's totient function) of$~\Phi(n)$: it gives a factor $p-1$ for every prime number $p$ dividing$~n$ (which leaves only $p=2$ and $p=3$) and an additional factor $p$ for every occurrence of$~p$ as factor after the first (which forbids a second factor$~3$, or a third factor$~2$, and also rules out $n=12=2^2\times3$ for which $\phi(12)=4$).