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).