Which one of the below options is correct?

Question

I think the option $(Q)$ is true since $O(Q/\{-1,1\})= 8/2 = 4 = 2^2$.
Since order is $p^2$ thus $(Q)$ option is true.
Can anyone suggest about option $(P)$?
Thanks

Answer 1

$(P)$ is wrong: there is only one normal subgroup $H$ of order $4$ in $S_4$, and $S_4/H\simeq S_3$. See Show that $S_4/K \cong S_3$.

Answer 2

You are absolutely correct about (Q).

As to (P), you should know there is a unique normal subgroup $V$ of order $4$ in $S_{4}$. This is a transitive subgroup, so $S_{4} = V T$, where $T$ is the stabilizer of $4$, say. But then $T \cong S_{3}$.