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$\require{extpfeil}\Newextarrow{\xRightarrow}{5,5}{0x21D2}$ Prove that if $\pi\in S_n,n\ge3$ and $\forall\sigma\in S_n, \pi\sigma=\sigma\pi$, then $\pi=e$.

How I did it:

Let $\pi\ne e\Rightarrow\exists i_1,i_2\in\overline{1,n},i_1\ne i_2$ s.t. $\pi(i_1)=i_2,\pi(i_2)=a\in\overline{1,n}$

Let $\sigma\in S_n$ s.t. $\sigma(i_1)=i_2, \sigma(i_2)=b\in\overline{1,n}$ with the condition that $a\ne b$.

$\pi\sigma(i_1)=a,\sigma\pi(i_1)=b\xRightarrow{a\ne b}\pi\sigma(i_1)\ne\sigma\pi(i_1)\Rightarrow\pi\sigma\ne\sigma\pi$

We proved that for $n\ge 3,\pi\ne e\implies \exists\sigma\in S_n$ s.t. $\pi\sigma\ne\sigma\pi$. The proposition is equivalent to $\forall\sigma\in S_n,\pi\sigma=\sigma\pi\implies\pi=e$ which is what we have to prove.

Now, is there an easier and more intuitive solution?

Shaun
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Neox
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  • This is a popular question here. See for example this post for a standard solution. Such a $\pi$ is in the center, and this is trivial. So $\pi=e$. – Dietrich Burde Dec 12 '21 at 14:48
  • @DietrichBurde Uff, I don't know groups, I've just started learning permutations. – Neox Dec 12 '21 at 14:48
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    No problem, the only group you need to know is $S_n$, and you have written it yourself. It is the "symmetric group". You can use the standard proof using permutations, see here for example. – Dietrich Burde Dec 12 '21 at 14:49
  • Well, I guess I'll come back to this problem after learning group theory and stuff. Thanks for help. – Neox Dec 12 '21 at 14:51
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    No need to learn it all now. Start with the group axioms and how $S_n$ with composition of bijections forms a group. – Dietrich Burde Dec 12 '21 at 14:51

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