Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
Are you sure that this is true? According to Theorem $3$ of mathdl.maa.org/images/cms_upload/George_Mackiw20823.pdf 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 –  Babak S. Oct 29 '13 at 6:35

2 Answers 2

up vote 9 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 $n.$

share|improve this answer
Nice proof +1, thanks. –  Sami Ben Romdhane 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 $x^m-1$.

share|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.