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Hi I am trying to solve this question I don't know where to begin but I have an idea so we have

$$Z(G) = \{x \in G \space |\space xy = yx \space\forall y \in G\}$$

We must show that for all $\sigma$ $\in S_n$ such that $\sigma \neq (1)$ Then there exists y $\in S_n$ such that $\sigma * y$ $\neq$ $y * \sigma$. However I don't know how to proceed further.

I was thinking I could use conjugation somehow but I don't know what would I do with them.

graydad
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Illustionist
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  • Cheat : $Z(G)$ is normal in $G$ for any group $G$. Thus, via simplicity and nonabelianity of $A_n$, $Z(S_n)$ is forced to be trivial. – Balarka Sen Mar 30 '15 at 19:33
  • @BalarkaSen I don't understand the last part via simplicity and nonabelianity of $A_n$ , Z($S_n$) is forced to be trivial can you explain more please? – Illustionist Mar 30 '15 at 20:00
  • $Z(S_n)$ is normal in $S_n$, thus it must be either $A_n$ or trivial via simplicity of $A_n$. But $Z(S_n)$ cannot be $A_n$ since it's not abelian, hence $Z(S_n) = {e}$. – Balarka Sen Mar 31 '15 at 14:49
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    @BalarkaSen; $A_n$ is the only non-trivial subgroup of $S_n$ for $n\ge 5$ ;You need to consider the case for $ n=4$ – Learnmore Sep 24 '17 at 15:50

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