# Elements of finite order in $SL_2(\mathbb{Z})$

Assume $A \in SL_2(\mathbb{Z})$ has finite order, and let $N_0 \in \mathbb{N}$ be the smallest natural number such that $A^{N_0} = I$. I want to show that the only possible values of $N_0$ are $1,2,3,4$ or $6$. I have a proof, but I worry it is incomplete, so my (two part) question is: $$\text{Is the following proof correct? If not, where does the proof break down?}$$$$\text{Does there exist a proof using only first principles? }$$ Basically, my proof follows from Corollary 2.4 here, which says "Any homomorphism $SL_2(\mathbb{Z}) \to \mathbb{C}^\times$ has image in the 12th roots of unity" and Lagrange's theorem. Namely, let $N_0$ be as above, and let $\phi: SL_2(\mathbb{Z}) \to \mathbb{C}^\times$ be a homomorphism. Then, since $\phi(I) = 1$, $$\phi(I)=\phi(A^{N_0}) = \phi(A)^{N_0} = 1.$$ By Lagrange's theorem, and Corollary 2.4, $N_0 | 12$, so $N_0=1,2,3,4$ or $6$. \qed

My trouble is that although $N_0$ is the order of $A$ in $SL_2(\mathbb{Z})$, it does \emph{not} need to be the order of $\phi(A)$ in the 12th roots of unity, in general, but I think I used this implicitly when I used Lagrange's theorem. If $\phi(A)^{N_0} = 1$ then all we can conclude is that the order of $\phi(A)$ divides $N_0$, not that it equals $N_0$. If we were still in $SL_2(\mathbb{Z})$, then this implies it does equal $N_0$ by assumption of minimality, but like I said, I don't see why this needs to hold in $\mathbb{C}^\times$. For exmaple, what's stopping $N_0 = 8$ while $\phi(A)$ has order 4 in the 12th roots of unity? This doesn't seem to yield an immediate contradiction.

Whether or not this proof is correct, I'd like to see how (if) my troubles can be resolved, and if a proof in this way can work. I would also like to know if there is a proof using only the definition of $SL_2(\mathbb{Z})$, and perhaps some ingenuity, since I wouldn't consider Corollary 2.4 a standard fact (to, say, a beginning graduate student).

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This proof certainly can't be correct. You take a single homomorphism $SL_2(\mathbb{Z}) \rightarrow \mathbb{C}$. Well, what if, for example, $\phi = 1$ (the constant homomorphism)? The image of $A$ is then 1, and that tells you nothing about $A$. Or better, it tells you that 1 divides the order of $A$, because it works in this direction: the order of $\phi(A)$ divides the order of $A$. –  nik Jun 30 '12 at 17:03
As for actually proving the statement, do you know basic linear algebra, about diagonalization of matrices and such? If so, you know that $X^N-1$ annihilates $A$, what can you conclude from that? –  nik Jun 30 '12 at 17:05
@nik Actually, I don't know much matrix algebra. If $X^N - 1$ annihilates $A$ then $A$ has order dividing $N$, right? Where do I go from here? –  Derek Allums Jun 30 '12 at 17:11

An element of finite order in $SL_2(\mathbb C)$ has roots of unity as eigenvalues. Thus an element in $SL_2(\mathbb Z)$ of finite order has as eigenvalues a pair of roots of unity as eigenvalues, say $\zeta$ and $\zeta'$. Since the determinant of the element $= 1$, we have $\zeta \zeta' = 1,$ and so $\zeta' = \overline{\zeta}$. Since our matrix is in $SL_2(\mathbb Z)$, it has integral trace, and so $2 \Re(\zeta)$ is an integer.

It is not hard to check that the only roots of unity satisfying $\zeta + \overline{\zeta}$ being an integer are $\zeta = \pm 1, \pm i, (\pm 1 \pm \sqrt{-3})/2$. (We are looking for an angle $\theta = 2\pi/n$ such that $\cos \theta$ is an integer or a half-integer, and the only possibilities are thus $\cos \theta = 0, \pm (1/2), \pm 1$, giving the listed roots of unity.) The order of such an element (which corresponds to the order of its eigenvalues, in the group of all roots of unity) is thus $1,$ $2,$ $3$, $4$, or $6$.

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Thanks for the answer, Matt. Could you provide a source for your first sentence? This may be standard, but I have not heard it before, and I'd like to see a proof, in particular. –  Derek Allums Jun 30 '12 at 19:14
It's simple if you know Jordan's theorem. Just consider the matrix in Jordan normal form. If it wasn't diagonal, it would be not be of finite order. If it is, it must be that the eigenvalues are roots of unity. I think it is true for invertible matrices in any integral domain (except the matrix need not be diagonal in Jordan form for positive characteristic). –  tomasz Jun 30 '12 at 19:56
@unit3000-21 If a matrix $A$ has an eigenvalue $\lambda$ then $\lambda^n$ is an eigenvalue of $A^n$. Thus if $A \in SL_2(\mathbb C)$ has finite order $n$, then if $\lambda$ is an eigenvalue of $A$, $\lambda^n$ is an eigenvalue of $A^n$, but $A^n = I$ so $\lambda^n = 1$. –  Adrián Barquero Jun 30 '12 at 22:01
@Adrian Right, that much is clear. But this seems to work for any matrix, not just for those with determinant one right? I guess the hypothesis that we're in $SL_2(\mathbb{Z})$ threw me off: I was looking for a more complicated reason that wasn't there. –  Derek Allums Jun 30 '12 at 23:10
@unit3000-21 That's right, the argument works for any matrix. To be honest, I didn't know that fact about the eigenvalues being roots of unity either, but after thinking about it for a while I came up with this argument. –  Adrián Barquero Jun 30 '12 at 23:25

You are absolutely correct that the order of $A$ need not agree with the order of $\phi(A)$. Try thinking about what the characteristic polynomial of $A$ can be.

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Thanks, Qiaochu. I guess where you're going with this is the heart of Matt E's answer above: that the eigenvalues of $A \in SL_2(\mathbb{Z})$ of finite order are roots of unity? If so, I'm going to see if I can prove this directly. –  Derek Allums Jun 30 '12 at 19:18

If you know a little about free products, since $\,SL(2,\mathbb Z)\cong C_4*_{C_2} C_6\,$ , then an element in this group has finite order iff it is conjugate to some element in one of the free factors, and we're done.

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Could you expand this a little bit? I know the definition of a free group, but not much more (the relevant section in Hungerford). –  Derek Allums Jun 30 '12 at 17:55
@unit3000-21: "free groups" should probably be "free products". –  t.b. Jun 30 '12 at 18:02
I should have added that the claimed isomorphism is incorrect. It is true that $\operatorname{PSL}(2,\mathbb{Z}) \cong C_2 \ast C_3$, but $\operatorname{SL}(2,\mathbb{Z})$ is the amalgamated product $C_4 \mathbin{\ast_{C_2}} C_6$ (explained on the same Wikipedia page). Don't you need something like the Kurosh subgroup theorem in order to make this into an argument? Is the cyclic case of that result much easier to prove? –  t.b. Jun 30 '12 at 19:09
Thank you @t.b.: "free group = free product" and amalgamated over a group of order 2 was missing. This has been corrected in the answer. Now, Kurosh subgroup theorem gives us at once the order of finite groups and, thus, also of the finite elements in the free amalgamated group, so I guess one can use it though I'm not sure one has to. –  DonAntonio Jun 30 '12 at 23:49
Kurosh's theorem is a couple of orders of magnitud more complicated than the computation of the elements of fintie order in this group, though :) –  Mariano Suárez-Alvarez Jul 1 '12 at 0:00