Showing that $V \unlhd S_4$, so that $S_4$ is not simple.

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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?

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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.

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let $S_4$ act on the variables $x_1,x_2,x_3,x_4$ by permutation of suffixes in the obvious way.

now define the three expressions: $$ A =(x_1x_2-x_3x_4)^2 \\ B =(x_1x_3-x_2x_4)^2 \\ C =(x_1x_4-x_2x_3)^2 $$ given any $\sigma \in S_4$ the action just defined induces a permutation of $A,B$ and $C$.

can you (a) check that this construction gives a homomorphism $S_4 \to S_3$. (b) show that the kernel of this homomorphism is your group $V$?

if so, then you have shown that $V \triangleleft S_4$