Orders of the Normal Subgroups of $A_4$ 
Prove that $A_4$ has no normal subgroup of order $3.$

This is how I started:
Assume that $A_4$ has a normal subgroup of order $3$, for example $K$. 
I take the Quotient Group $A_4/K$ with $4$ distinct cosets, each of order $3$.
But I want to prove that these distinct cosets will not contain $(12)(34),(13)(24)$ and $(14)(23)$> Therefore a contradiction.
Please help, I'm really stuck!!

In fact, one can show that  all normal subgroups of $A_4$ are $1$, $K_4$ (Klein four group) and $A_4$. Note that  two permutations of $S_n$ are conjugate iff they have the same type. So we can write down (by some easy calculations) all conjugate classes of $A_4$  are  the following $4 $ classes:


*

*type $1^4$: {(1)}

*type $2^2$: {(12)(34),(13)(24),(14)(23)}

*type $3^1$:{(123),(142),(134),(243)} and {(132),(124),(143),(234)} 


So all normal subgroups of $A_4$ are $1$, $K_4$ and $A_4$.
