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Let $A\in GL_2\left(\mathbb{Z}\right)$, the group of invertible matrices with integer coefficients, and denote by $\omega(A)$ the order of $A$.

How we prove that $$\left\{\omega(A);A\in GL_2\left(\mathbb{Z}\right)\right\}=\{1,2,3,4,6,\infty\}.$$

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Are you sure that this is true? According to Theorem $3$ of if you have a finite subgroup $G \leq GL_2(\mathbb{Z})$, then $G$ must be isomorphic to a subgroup of $D_4$ or $D_6$. – Thomas Mar 2 '13 at 21:49
@Thomas: Doesn't that show what the OP wants? – anon Mar 2 '13 at 21:51
@anon: Arg.... I read the set as all numbers $1,2,3,4,5,\dots$. You are right! – Thomas Mar 2 '13 at 21:52
+1 nice question Sami – S. Snape Oct 29 '13 at 6:35
up vote 11 down vote accepted

Let's assume that $A\in GL_2\left(\mathbb{Z}\right)$ has finite order, let say $n$. Then $A^n=I$. Now look at this matrix as a matrix in $M_2(\mathbb Q)$: we have $\det A=\pm 1$ and if consider its minimal polynomial $m_A\in\mathbb Q[X]$ this must divide $X^n-1$ and $\deg m_A\le 2$.

If $\deg m_A=1$, then $m_A=X-1$, that is, $A=I$ and $n=1$ or $m_A=X-1$, that is, $A=-I$ and $n=2$. (The only rational roots of unity are $\pm1$.)

If $\deg m_A=2$, then $m_A(X)=(X-\epsilon_1)(X-\epsilon_2)$, $\epsilon_1\neq\epsilon_2$, where $\epsilon_i$ is an $n$th root of unity. Since $m_A\in\mathbb Q[X]$ we have $\epsilon_1+\epsilon_2\in\mathbb Q$ and $\epsilon_1\epsilon_2\in\mathbb Q$. In fact, $m_A(X)=X^2-\operatorname{tr}(A)X\pm1$, and thus we get $\epsilon_1+\epsilon_2\in\mathbb Z$ and $\epsilon_1\epsilon_2=\pm1$. If $\epsilon_1=1$, then $\epsilon_2=-1$, $m_A=X^2-1$ and therefore $A^2=I$, so $n=2$. Now suppose that $\epsilon_i\neq 1$. Some easy remarks show that the only case to consider is $\epsilon_1\epsilon_2=1$, that is, $\epsilon_2=\overline{\epsilon}_1$. Then $m_A(X)=X^2-2\cos\frac{2k\pi}{n}X+1$, with $n\ge 2$ and $1\le k<n$. But $2\cos\frac{2k\pi}{n}\in\mathbb Z$ implies $2\cos\frac{2k\pi}{n}=0,\pm1,\pm2$ so we have five possible minimal polynomials: $X^2+1$, $X^2\pm X+1$ and $X^2\pm2X+1$. (In fact only three, since the last two can't appear.) These give: $A^2+I=0$, so $n=4$; $A^2+A+I=0$, so $n=3$ and $A^2-A+I=0$, so $n=6$.

I leave the OP the pleasure to find for each $n\in\{1,2,3,4,6,\infty\}$ a matrix $A\in GL_2\left(\mathbb{Z}\right)$ of order $n.$

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Nice proof +1, thanks. – user63181 Mar 3 '13 at 17:41

Hint: consider the minimal polynomial for such a matrix. If $A$ has order $n$, the minimal polynomial must divide $x^n-1$ and no smaller $x^m-1$.

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These arguments work for matrices over fields. But if you want to look at the matrix over $\mathbb Q$, then this is another story and the hint can be very useful. – user26857 Mar 3 '13 at 10:07
@YACP No, the matrices over $\mathbb Z$ are a subset of the matrices over $\mathbb Q$, and they have minimal polynomials that are monic with integer coefficients (the roots are algebraic integers.) Basically, the minimal polynomial is either $x^2-tr(A)x\pm 1$ or $x\pm 1$. If $A^n=I$ then the roots of the polynomial have to be $x^n-1$. – Thomas Andrews Mar 3 '13 at 13:24
@YACP It ends up resolving to the question; "When is $\phi(n)=1,2$, since the cyclotomic polynomials are prime and of degree $\phi(n)$ – Thomas Andrews Mar 3 '13 at 13:58
No, my hint works over the integers, not just over a field. @YACP – Thomas Andrews Mar 3 '13 at 16:34
But they do over the integers. @YACP – Thomas Andrews Mar 3 '13 at 16:38

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