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This is meant to be a counter example to $A_4$ being simple, since $\{ (1),(12)(34),(13)(24),(14)(23)\}$ is normal.

But, how can you check it's normal, is there a quick way or do you need to calculate all of $A_4$?

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    To add to the great answers so far, in general, a subgroup is normal if and only if it is the union of conjugacy classes [complete sets of conjugates]. This subgroup is the union of the conjugacy class: $\{(1)\}$ and the class of pairs of disjoint transpositions: $\{(12)(34),(13)(24),(14)(23)\}$.2012-01-23

1 Answers 1

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The simple way is to remember that if $\tau$ is any permutation and $(a_1,\ldots,a_k)$ is a cycle, then $\tau\circ(a_1,\ldots,a_k)\circ \tau^{-1} = (\tau(a_1),\tau(a_2),\ldots,\tau(a_k));$ in particular, conjugation respects the cycle structure.

Since the subgroup you have contains all elements that are the product of two disjoint transpositions plus the identity, and the conjugate of the product of two disjoint transpositions is a product of two disjoint transpositions, it follows that the subgroup is indeed normal.

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    Tha$n$ks a lot for that. This was i$n$ the notes I'm using and forgot to use that fact. I can see how you show it's normal now.2012-01-23