$SL(2, \mathbb F_3)$ does not have a subgroup of order $12$ Using the characteristic polynomial I can prove that $SL(2, \mathbb F_3)$ does not has an element of order $12$, but how can I prove that $SL(2, \mathbb F_3)$ does not has a subgroup of order $12$?

The sizes of conjugacy classes are $1,1,4,4,4,4,6$ (see http://groupprops.subwiki.org/wiki/Special_linear_group:SL(2,3)).
Now the two elements with conjugacy class of size $1$ are $I, 2I$. These have order $1,2$ respectively.
Any other element lies in a conjugacy class of size $4$ or $6$, so has centralizer group of order $6$ or $4$. In particular an element lies in its own centralizer so the order of any such element must divide $4$ or $6$ so cannot be $12$.
