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: Up to changing name to the symbols, the fact that $(234)^{-1} (123) (234) = (134)$ is enough, since it proves that any element of order $3$ is conjugated to another element which is not one of its powers.
Here is a non-elementary (but more "adaptable") proof: since the Sylow $3$-subgroups are pairwise conjugated, if there was a normal subgroup of order $3$ it would be the unique Sylow $3$-subgroup, i.e. the unique subgroup of order $3$, and this is false because $A_4$ has more than one subgroup of order $3$.
A: 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 $A_4$.
A: $|A_4|=O(G)=12$
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
