(No-)Generators of $S_4$ Let $S_4$ be the symmetric group on 4 elements and let $x=(1,2)(3,4)$ be a permutation of $S_4$. I try to proof that can be no element $y \in S_4$ such that $<x,y>$ is the whole group $S_4$.
I notice that $x \in K$ where $K$ is the Klein group. Now, I know $S_4 /K$ is isomorphic to $S_3$. How can I use these informations in order to show the target?

You have all the ingredients for the solution. The crucial thing is that
$S_3$ is not cyclic. Let $\pi:S_4\to S_4/K$ be the projection map.
If $H=\left<x,y\right>$, then $\pi(H)=\left<\pi(y)\right>\ne S_4/K$
since $S_4/K$ is not cyclic. Therefore $H\ne S_4$.
