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!!

$|A_4|=O(G)=12$
By Sylow subgroups
$O(G)=12=2^2×3^1$
$A_4$ has a $3$-Sylow subgroup of order $3$ and $A_4$ has $2$-Sylow subgroup of order $4$
$A_4$ has $3$-Sylow subgroup of order $3$
n=1+3r. (n|O(G))
r=0 , n=1
r=1 , n=4
A4 has 2-Sylow subgroup of order 4
n=1+4r. (n|O(G))
r=0 , n=1
2-Sylow subgroup of order 4 is unique, it is normal,
but 3-Sylow subgroup of order 3 is not unique, hence it is not normal
Therefore, A4 has normal subgroup of order 4 and not of order 3
