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Prove that $A_4$ has no normal subgroup of order $3.$ This is how I started: Assume that v has a normal subgroup of order $3$ for e.g. $K.$ 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! –  Martino Oct 25 '12 at 8:29

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