Special linear group $SL(2,\mathbb Z_3)$ doesn't have subgroups of order $12$ Prove that $SL(2,\mathbb Z_3)$ doesn't have subgroups of order $12$.
I am pretty lost with this problem. I've tried to think of a morphism $$f:SL(2,\mathbb Z_3) \to Aut(\mathbb {Z_3}^2)$$ $$A \to A.k^t \space \forall k \in \mathbb {Z_3}^2$$ and try to play with the orders of subgroups in $\mathbb {Z_3}^2$ and the special linear group but I couldn't arrive to anything. Any suggestions would be greatly appreciated.
 A: Series of hints:
1) Every subgroup of index $2$ is normal.
2) Any normal subgroup is a kernel of some homomorphism.
3) Any homomorphism to an Abelian codomain factors through abelianization of the domain.
A: Suppose that $SL(2,3)$ does contain a subgroup $H$ of order $12$. Denote by $a$ the unique element of order $2$ in $SL(2,3)$. Since $H$ has even order, $H$ must contain $a$. And because the index of $H$ in $SL(2,3)$ is equal to $2$, it must contain the square of every element in $SL(2,3)$. Now consider any element $A$ of order $3$ in $SL(2,3)$. Then $A=A^4=(A^2)^2$ is a square. Hence $H$ must contain all $8$ elements of order $3$. We can multiply these $8$ elements by $a$ to obtain another $8$ elements of order $6$. So $H$ has at least $17$ elements, because $1\in H$. This is a contradiction.
A: I think the following works.
There are four 1-dimensional subspaces of the plabe $\Bbb{Z}_3^2$, namely the spans of $(1,0)$, $(1,1)$, $(1,2)$ and $(0,1)$ repsectivelt. The group $G=SL_2(\Bbb{Z}_3)$ permutes them naturally giving us a homomorphism $\phi: G\to S_4$. It is easy to see that the kernel of this action consists of the scalar matrices, in other words $\ker\phi=\langle -I_2\rangle$.
This implies that $\phi(G)\cong A_4$.
Assume that $G$ has a subgroup $H$ of order $12$. If $H\cap\ker\phi$ is trivial, then $H\cong A_4$. But $H\unlhd G$, so this implies that $G\cong A_4\times \ker\phi$. This is a contradiction because the latter group does not contain elements of order four, but $G$ does have them.
Therefore $\ker\phi\le H$. This means that $\phi(H)$ is an order six subgroup of $A_4$. But such subgroups don't exist. Search this site for proofs.
