I was reading online about the topic and this theorem came out, but I haven't found any proof for it
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in fact the center is trivial – Asinomás Aug 12 '16 at 05:15
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4I'm surprised this got put on hold, especially as "unclear" - it's perfectly clear. I've voted to reopen. – Noah Schweber Aug 12 '16 at 06:25
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because $(1,2)(2,3)=(2,3,1)$ and $(2,3)(1,2)=(1,3,2)$
Asinomás
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1+1. Note, though, that this is only half the answer; the other piece is the fact that for $m<n$, $S_m$ can be injected into $S_n$ (it's tempting to say that $S_m\subset S_n$, but this isn't quite true). This means that if $S_m$ is nonabelian, then so is $S_n$ for every $n>m$: if a group has a nonabelian subgroup, it is nonabelian too. So once we know $S_3$ is nonabelian, we know all higher $S_n$s are also nonabelian. This does not address the issue of proving that the center of $S_n$ is trivial for $n>2$, but that's a different problem. – Noah Schweber Aug 12 '16 at 06:27
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4I don't hink we need to inject. those are elements of $S_n$ for $n>2$ – Asinomás Aug 12 '16 at 06:30
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These proofs and this question show that the center of a symmetric group is trivial, as in it only contains the identity element. It follows that the group will always be nonabelian since the center will never equal the group (which is required for abelian groups) ...except, of course, $S_2$. Every element in that group commutes with every other element.
