Orders of the Normal Subgroups of $A_4$

Question

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!!

Answer 1

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 $3$.

Answer 2

$|A_4|=O(G)=12$

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