$SL(2, \mathbb F_3)$ 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 $12$?

Outline for a proof:
With the characteristic polynomial, you can see that $A^3 = I$ in $G = \operatorname{SL}(2,3)$ if and only if $tr(A) = -1$. Count that there are $8$ elements of order $3$ in $G$.
A subgroup $H$ of order $12$ would have to contain every element of order $3$. Conclude $H \cong A_4$.
On the other hand $H$ must contain $-I$, so $H$ has a subgroup of order $6$. This is a contradiction since $A_4$ has no subgroup of order $6$.
