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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!!

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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 $3$.

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    Yes, of course!2012-10-25
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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 $A_4$.