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 $A_4$?
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 $A_4$?
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.