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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?

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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 $H\ne S_4$.

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  • $\begingroup$ Then you can say $H \neq S_4$ because of Correspondence theorem, right? $\endgroup$ – Davide Motta Jun 22 '18 at 6:04
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    $\begingroup$ @DavideMotta It's easier than that, simply that $\pi(S_4)=S_4/K$. $\endgroup$ – Angina Seng Jun 22 '18 at 6:07
  • $\begingroup$ Oh great! you're right! thank you so much. Easier then it looks. $\endgroup$ – Davide Motta Jun 22 '18 at 6:13
  • $\begingroup$ @DavideMotta Replying to your comment on the deleted answer: Yes, it does seem that "we" should learn more about presentations of groups. That presentation says (in part) that there exist generators $a,b$ such that $a^2=b^4=(ab)^3=1$. It does not say that any two generators satisfy those relations. So the answer to the (rhetorical) question "Given that $a=(1,2)(3,4)$ and $S_4$ is generated by $a$ and $b$ why does it follow that $a^2=b^4=(ab)^3=1$?" is "it doesn't follow". There's a reason Dietrich deleted his answer... $\endgroup$ – David C. Ullrich Jun 22 '18 at 12:38
  • $\begingroup$ @David C. Ullrich I probably misunderstood the answer . I only wanted to underline my ignorance about presentations of groups. $\endgroup$ – Davide Motta Jun 22 '18 at 17:56

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