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

Question

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).

Answer 1

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 $6$.

Answer 2

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.

Answer 3

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.