I recently read the statement "up to conjugacy there are 4 nontrivial finite subgroups of ${\rm SL}_2(\mathbb{Z})$." They are generated by

$$\left(\begin{array}{cc} -1&0 \\\ 0&-1\end{array}\right), \left(\begin{array}{cc} -1&-1 \\\ 1&0\end{array}\right), \left(\begin{array}{cc} 0&-1 \\\ 1&0\end{array}\right), \left(\begin{array}{cc} 0&-1 \\\ 1&1\end{array}\right) $$ and are isomorphic to $\mathbb{Z}_2$, $\mathbb{Z}_3$, $\mathbb{Z}_4$, and $\mathbb{Z}_6$, respectively. Does someone know a reference for this statement? (Or, is it easy to see?) My attempt at a Google search turned up this statement, but I wasn't able to find a reference.

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    $\begingroup$ Here is a sketch of a proof: (i) every compact group of $\operatorname{GL}_n(\mathbb{R})$ is conjugate to a subgroup of the orthogonal group $O_n(\mathbb{R})$ (e.g. Thm. 3 of math.uga.edu/~pete/8410Chapter9.pdf). (ii) So any finite subgroup of $SL_2(\mathbb{R})$ is conjugate to a subgroup of $SO_2(\mathbb{R}) \cong S^1$. (iii) The finite subgroups of $S^1$ are $\langle \zeta_n \rangle$, where $\zeta_n$ is an nth root of 1. (iv) The charpoly of $\zeta_n$ is $T^2 - (\zeta_n + \zeta_n^{-1})T + 1$. The values of $n$ for which the coeff. of $T$ is rational are exactly those you give. $\endgroup$ – Pete L. Clark Jun 6 '10 at 18:19
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    $\begingroup$ Regarding (iii), note more generally that any finite subgroup of the multiplicative group of a field is cyclic. $\endgroup$ – Qiaochu Yuan Jun 6 '10 at 23:59
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    $\begingroup$ You asked for a reference and got real and interesting answers instead. I am sure there are a million places to find this information, but one gentle entree to the question (with pointers to other references) is J. Kuzmanovich and A. Pavlichenkov "Finite groups of matrices whose entries are integers" from American Math. Monthly, vol. 109, #2 (2002) 173-186. jstor.org/stable/2695329 $\endgroup$ – Skip Jun 7 '10 at 14:05
  • $\begingroup$ Thanks for the reference, it looks like a nice read. I mainly asked for a reference because I thought that the result was probably well-known enough to be written down nicely somewhere, and it seems easier to reference an answer than write it down. But reading people's answers here is often nicer. In particular, thanks for all the good answers. $\endgroup$ – Peter Samuelson Jun 8 '10 at 14:06

A finite subgroup will map to a finite subgroup of $PSL_2(\mathbb Z)$, which is a free product $Z_2 * Z_3$. I believe I have been told that finite subgroups of free products of finite groups are conjugate to subgroups of the factors being free-producted together.

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    $\begingroup$ That's right. It follows, in particular, from the more general Kurosh Subgroup Theorem. See en.wikipedia.org/wiki/Kurosh_subgroup_theorem $\endgroup$ – Autumn Kent Jun 7 '10 at 0:11
  • $\begingroup$ This is a very clean answer (at least, modulo the proof that $SL_2(\mathbb{Z}) \cong Z_4 \ast_{Z_2} Z_6$, which I knew already). Also, the fact that you mention feels like it should be true, and it's nice to know that it is. $\endgroup$ – Peter Samuelson Jun 8 '10 at 13:59

Each finite subgroup $G$ of $GL_2(\mathbb{Z})$ is a finite subgroup of $GL_2(\mathbb{R})$. Taking a positive definite form on $\mathbb{R}^2$ and averaging by the action of $G$ gives a positive definite form invariant under $G$. This implies that $G$ is conjugate in $GL_2(\mathbb{Z})$ to a subgroup of $O(2)$. All such groups are cyclic or dihedral.

A finite subgroup of $GL_2(\mathbb{Z})$ corresponds to a finite subgroup of $O(2)$ leaving a lattice invariant. With respect to generators of this lattice the matrix of a group element has integer coefficients. As it is a rotation or reflection matrix, its trace is $2\cos\theta$ (where $\theta$ is the angle of rotation) or zero (if it is a reflection). The trace must be an integer so it is $2$, $1$, $0$, $-1$ or $-2$, corresponding to rotations of order $1$, $6$, $4$, $3$ or $2$. These are the only possible orders of rotation possible (the crystallographic restriction) and this leads to the classification you cite.

See http://en.wikipedia.org/wiki/Crystallographic_restriction_theorem .

  • $\begingroup$ My argument also - but cos not sin? $\endgroup$ – Charles Matthews Jun 6 '10 at 18:20
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    $\begingroup$ But as Torsten pointed out, this doesn't establish $\mathit{conjugacy\ in}\ SL_2(\mathbb{Z}).$ $\endgroup$ – Victor Protsak Jun 7 '10 at 1:19
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    $\begingroup$ One can get the conjugacy from the geometry of the standard fundamental region of $SL_2(\mathbb{Z})$ on the upper half plane: the only elliptic points are $i$ and $(\pm1+i\sqrt 3)/2$. As a consequence the nontrivial finite subgroups are conjugate to the stabilizers if these points: see texts on modular forms. $\endgroup$ – Robin Chapman Jun 7 '10 at 6:06
  • $\begingroup$ Well, that fuses your argument with Allen Knutson's argument, then. I don't want to appear overly critical: it's just surprising how nontrivial it is to completely prove this "well known fact". $\endgroup$ – Victor Protsak Jun 7 '10 at 13:48

A slightly different argument showing that every finite subgroup of $SL_2(\mathbb Z)$ is of cardinality a divisor of $24$ goes as follows: Consider such a finite subgroup $H$. Since the coefficients of all elements of $H$ involve only a finite number of prime divisors, the obvious group homomorphism from $SL_2(\mathbb Z)$ into $SL_2(\mathbb F_p)$ where $\mathbb F_p$ is the finite field with cardinality a prime number $p$ is injective for almost all primes. Since $SL_2(\mathbb F_p)$ has $p(p^2-1)$ elements, the cardinality $h$ of the finite group $H$ divides $p(p^2-1)$ for almost all prime numbers. This implies that $h$ divides $24$. Indeed, quadratic reciprocity shows that $2$ and $3$ are the only possible prime divisors of $h$ and gives upper bounds on the maximal exponents $\alpha,\beta$ such that $2^\alpha\cdot 3^\beta$ divides $p(p^2-1)$ almost all primes.

I intended to post the following as a comment but it is too long:

A very easy argument showing that every prime $p>n+1$ works for the injectivity of the reduction modulo $p$ of a finite subgroup $H$ of $SL_n(\mathbb Z)$ is as follows: Since every element of $H$ is finite, its characteristic polynomial is a product of cyclotomic polynomials. Reductions of cyclotomic polynomials modulo $p$ with order (defined as the order of an underlying root of unity in the multiplicative group of invertible elements in $\mathbb C$) prime to $p$ are never congruent to a power of $(1-x)$. Cyclotomic polynomials of order divisible by $p$ are of degree at least $p-1$. Since a non-trivial element $h\in H$ is diagonalisable, its characteristic polynomial is not a power of $(1-x)$. The reduction of this characteristic polynomial modulo $p$ is thus also distinct from a power of $(1-x)$. This implies that $h$ modulo $p$ is non-trivial in $SL_n(\mathbb F_p)$.

  • $\begingroup$ Does this generalize to $SL_n(\mathbb{Z})$? $\endgroup$ – Robin Chapman Jun 7 '10 at 8:17
  • $\begingroup$ No idea. I remember having seen a proof in a lecture by Serre giving an upper bound for cardinalities of finite subgroups in $SL_n(\mathbb Z)$ based on the fact that the canonical morphism into $SL_n(\mathbb F_p)$ is in fact injective when restricted to finite subgroups even for quite small primes (I believe even $p=3$ works.) The argument (which I forgot) was very elementary. $\endgroup$ – Roland Bacher Jun 7 '10 at 8:32
  • $\begingroup$ Indeed, for $p>2$ the reduction is injective and for $p=2$ the kernel consists of elements of order $2$ for a general $\mathrm{GL}_n(\mathbb Z)$. Hence, in the case of $\mathrm{SL}_2(\mathbb Z)$ we get that the kernel has at most order $2$ and order of a finite subgroup is a divisor of $2\cdot|\mathrm{SL}_2(\mathbb Z)/2|=12$. If we restrict ourselves to commutative subgroups we get a divisor of $4$ or $6$ (as the order of elements of $\mathrm{SL}_2(\mathbb Z)/2$ are $1$, $2$ or $3$) but that also follows from the fact that these are the only $d$ with $\varphi(d)\leq 2$. $\endgroup$ – Torsten Ekedahl Jun 7 '10 at 9:31
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    $\begingroup$ @Bacher: Serre considers the finite group G as a subgroup of $SL(n,Z_p)$ where $Z_p$ is the ring of $p$-adic integers. Now, it is easy to see that the kernel of the reduction map from $SL(n, Z_p)$ to $SL(n, F_p)$ is torsion-free. This implies that $G$ maps injectively into $SL(n, F_p)$ $\endgroup$ – Keivan Karai Jun 7 '10 at 11:15
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    $\begingroup$ For those who are interested: I wrote up the arguments from Serre's Lie Algebras and Lie Groups on finite groups of $\operatorname{GL}_n(\mathbb{Q})$ here: math.uga.edu/~pete/8410Chapter9.pdf $\endgroup$ – Pete L. Clark Jun 7 '10 at 13:25

I interpret your statement as being concerned with conjugation in $\mathrm{SL}_2(\mathbb Z)$. In that case I think that the arguments given only give that the groups are cyclic of order $1$, $2$, $3$, $4$ or $6$ not that they are unique up to conjugacy. For this latter fact one need only consider order $1$, $3$ or $4$ as the others are obtained by multplying a generator by $-E$. The case of order $1$ is trivial and that of order $3$ or $4$ gives a module of rank $1$ over the ring of $3$'rd and $4$'th roots of unity and then the statement is equivalent to these rings having class number $1$.

Addendum: I was a little it sketchy as Victor pointed out. In the present case all the relevant representations of $\mathbb Z[\mathbb Z/p]$factor through $\mathbb Z[\zeta_p]$ which is seen by looking at the characteristic polynomial and hence one doesn't have to look at the more general representations. In higher ranks I certainly agree with Victor about the need for the full ring. As for $\mathrm{GL}_2(\mathbb Z)$ versus $\mathrm{SL}_2(\mathbb Z)$-conjugacy I think that is taken care of by the fact that complex conjugation acts trivially on the class groups (which are trivial) and hence there is an automorphism of determinant $-1$ of the modules in question.

  • $\begingroup$ While that surely plays a role, still more work is needed: (1) classification of integral representations of $\mathbb{Z}/p\mathbb{Z}$ does not coincide with the classification of $\mathbb{Z}[\zeta_p]$-modules, cf Curtis and Reiner, (74.3); (2) conjugation by $SL_2(\mathbb{Z})$ is more restrictive than conjugation by $GL_2(\mathbb{Z})$, which is counterbalanced by the fact that cyclic subgroups $\langle g\rangle$ and $\langle h\rangle$ may be conjugate without their generators $g$ and $h$ being conjugate. $\endgroup$ – Victor Protsak Jun 7 '10 at 0:56
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    $\begingroup$ Example of (2): matrices $g=\begin{bmatrix}1 & 1\\ 0 & 1\end{bmatrix}$ and $h=\begin{bmatrix}1 & 0\\ 1 & 1\end{bmatrix}$ aren't even conjugate in $SL_2(\mathbb{R})$, but the cyclic subgroups they generate are conjugate by $\begin{bmatrix}0 & 1\\ -1 & 0\end{bmatrix}$, which maps $g$ into $h^{-1}.$ $\endgroup$ – Victor Protsak Jun 7 '10 at 1:14

Perhaps the following explicit calculations clarify Ekedahl's argument: consider, for example, the case where the group is cyclic of order 3. Let $w$ be a generator of the group. The characteristic polynomial of w is then necessarily $X^2 + X + 1$. Thus the action of $w$ on $M = \mathbf{Z}\oplus\mathbf{Z}$ makes $M$ into a $\mathbf{Z}[X]/(X^2 + X + 1)$-module. This module is torsion free (since it is torsion-free as a $\mathbf{Z}$-module). Consequently, it is locally free (necessarily of rank $1$) and hence free since $\mathbf{Z}[X]/(X^2+X+1)$ is a PID.

The freeness means that one can find a module generator $m\in M$. Since $(-X,1)$ is a $\mathbf{Z}$-module basis for $\mathbf{Z}[X]/(X^2+X+1)$, one finds that that $(-wm,m)$ is a $\mathbf{Z}$-module basis for $M$.

If the basis $(-wm,m)$ has the same orientation as the standard basis for $M$, then the change-of-basis matrix to $(-wm,m)$ is in ${\rm SL}_2(\mathbf{Z})$. Since the matrix of $w$ with respect to $(-wm,m)$ is the standard matrix of the original problem statement, this construction finishes the argument in this case.

If, on the other hand, the basis $(-wm,m)$ has the opposite orientation from the standard basis for $M$, then $(-w^2 m, m)$ has the same orientation as the standard basis of $M$ (algebra omitted). The change-of-basis matrix to $(-w^2 m, m)$ is thus in ${\rm SL}_2(\mathbf{Z})$. The matrix of $w^2$ with respect to $(-w^2 m, m)$ is the standard matrix of the original problem statement, which finishes the argument in this case.

Let me state this last part of the calculation abstractly: since the question involves ${\rm SL}_2(\mathbf{Z})$ conjugacy, one should consider locally free $\mathbf{Z}[X]/(X^2+X+1)$-modules of rank $1$ equipped with an orientation. We use the orientation $(-X,1)$ on $\mathbf{Z}[X]/(X^2+X+1)$. Given such a module $N$, we can obtain a new conjugate module $N'$ by changing the $\mathbf{Z}[X]/(X^2+X+1)$ action by the (orientation-reversing) automorphism $X\mapsto X^2$. For any oriented $N$, either $N$ or $N'$ is module-isomorphic to $\mathbf{Z}[X]/(X^2+X+1)$ (with orientation). Corresponding to these two cases, one can conjugate (in ${\rm SL}_2(\mathbf{Z})$) either $w$ or $w^2$ to the desired standard form.

Note that it is straightforward to find a module generator $m$ (and hence to find a standardizing basis for $M$): the function $$ q(m):m\mapsto \det(-wm \; m) $$ is a quadratic form on $\mathbf{Z}\oplus\mathbf{Z}$. It is either positive definite or negative definite. (One can see that easily in the abstract using a module generator for $M$.) An $m\in M$ is a module generator precisely when $q(m) = \pm 1$. We know that such an $m$ exists, and since $q$ is definite, it is easy to search for one.


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