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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.

user149792
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2 Answers2

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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 $2,3,4,6$

$$\begin{pmatrix}-1&0\\0&1\end{pmatrix},\begin{pmatrix}-2&-3\\1&1\end{pmatrix}, \begin{pmatrix}2&1\\-5&-2\end{pmatrix},\begin{pmatrix}3&-7\\1&-2\end{pmatrix}$$

  1. Here you can find several proofs.
Pedro
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  • Good answers, but one remark. The way you formulate it, it is not so obvious that you you are using the (well known but not so obvious) theorem that cyclotomic polynomials are always irreducible over $\Bbb Q$. – Marc van Leeuwen Jan 19 '15 at 05:06
  • @MarcvanLeeuwen I'm confused, I say "...factors into the irreducible cyclotomic polynomials." – Pedro Jan 19 '15 at 05:07
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    Yes that is what I meant. It suggests that there are cyclotomic polynomials that are reducible and others that are irreducible, and that somehow you know only the latter are needed here. Or that irreducibility is a property that cyclotomic polynomials have by definition. I just think you could mention more explicitly the fact you are using a theorem here. – Marc van Leeuwen Jan 19 '15 at 05:15
  • @MarcvanLeeuwen Oh. I mean to say "...it factors into cyclotomic polynomials, which are irreducible." – Pedro Jan 19 '15 at 05:18
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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 $6$.

user149792
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  • It is curious the Wikipedia link has no concrete statement of what the theorem says, or at least I couldn't find one. – Pedro Jan 19 '15 at 04:58
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    See page 171-172 of the 2nd edition of Artin's Algebra. $``$Let $L$ be a discrete subgroup of $V^+$ or $\mathbb{R}^{2+}$, and let $H \subset O_2$ be a subgroup of the group of symmetries of $L$. Suppose that $L$ is not the trivial group. Then (a) every rotation in $H$ has order $1$, $2$, $3$, $4$, or $6$, and (b) $H$ is one of the groups $C_n$ or $D_n$, and $n = 1$, $2$, $3$, $4$, or $6$.$"$ – user166854 Jan 19 '15 at 05:04
  • @user141494 I meant I couldn't find one in the page of Wikipedia. Thanks for the reference, though. – Pedro Jan 19 '15 at 05:23