Let $V = \{id, (12)(34),(13)(24),(14)(23)\}$. I am trying to show that $V \unlhd S_4$, so that $S_4$ is not simple.
As I understand it, to show that $V \unlhd S_4$ $-$ i.e. that $V$ is a normal subgroup of $S_4$ $-$ I need to show that $\forall x, \forall y \in V (xy^{-1} \in V)$ and that $\forall s \in S_4$, ( $sVs^{-1} = V)$.
Is this correct? If so, how many checks do I need to make?
Since all elements in $V$ are involutions, it's easy to check $V$ is a subgroup. Note that any conjugation will not change the order of an element. If $\exists s,v$ such that $svs^{-1}\notin V$, $svs^{-1}$ must be of the form $(i\ j)$. Then by symmetry, all elements of form $(i\ j)$ will be in the conjugacy class containing $V$. So the conjugacy class containing $V$ will consist of $6+3+1=10$ elements, which doesn't divide $|S_4|=24$. Contradiction.
P.S. To show that $A_4$ is normal is much easier.