Does there exists a non trivial homomorphism $S_3\rightarrow Sl_2(\mathbb{Z})$?

Question

I've been trying to find a homomorphism from $S_3$ to $Sl_2(\mathbb{Z})$, I tried to consider the rotation matrix but I didn't get anything, could someone give me any idea?

Answer 1

Map $\sigma$ to $\begin{pmatrix} \text{sign}(\sigma) & 0 \\ 0 & \text{sign}(\sigma) \end{pmatrix}$.

Answer 2

$Sl_2(\mathbb{Z}) $ is the amalgamated product of $C_4$ and $C_6$. Let $\phi$ be a homomorphism of $S_3$ into it. The image is either trivial (you do not want it) or of order 2 or of order 6 because $S_3$ has three normal subgroups. It cannot be of order $6$ because a finite group acting on the Bass-Serre tree of the amalgamated product must fix a vertex and so has to be conjugate to a subgroup of $C_4$ or $C_6$, hence must be abelian (even cyclic). If the image is of order 2, then again it must be conjugate to a subgroup of order 2 in $C_4$ or $C_6$ which is the intersection of these subgroups in $Sl_2(\mathbb{Z}) $, which is the center of that group, generared by the matrix $diag(-1,-1)$. Thus there exists exactly one nontrivial homomorphism: the one that takes every involution to $diag(-1,-1)$ and every $3$-cycle to 1.