Smallest Normal Subgroup of $S_4$ which contains $\langle(1,3,2,4)\rangle$

Question

We are asked to find smallest normal subgroup of $S_4$ which contains $\langle(1,3,2,4)\rangle = H$.

I know that a subgroup $G$ is normal if: $$\forall x \in S_4, xH = Hx$$

I know that $H$ contains at least $4$ elements generated by $\langle(1,3,2,4)\rangle$. I don't know, however, how should I know which elements should be added from $S_4$.

Answer 1

If there is a $4$-cycle and the group is normal, then it needs to contain any other $4$-cycle. (To see this, consider an element $g$ that maps elements of your $4$-cycle $c$ into $x_1, x_2, x_3, x_4$. Then $gcg^{-1}$ is the $4$-cycle $(x_1, x_2, x_3, x_4)$.)

You can easily show that $4$-cycles generate all of $S_4$.

(Note that $(1,2,3,4)\,(1,3,2,4)\,(1,2,3,4)$ is the $2$-cycle $(2,3)$ so any $2$-cycle must be included.)

Answer 2

We already know the nontrivial proper normal subgroups of $S_4$, namely $A_4$ and $V_4$. Since $V_4$ has only elements of order $2$ and $1$, and $(1324)$ is not in $A_4$, the smallest normal subgroup containing $H$ must be $S_4$.

Reference:

A question on identifying normal subgroups of $S_4$

how do I prove that $S_4$ has no normal subgroup of order 6