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I think I remember my abstract algebra professor mentioning in class that if $\sigma$ is any permutation belonging to the symmetric group $S_n,$ then $\sigma^n = \iota,$ the identity permutation. Is this true, or can we just say that $\sigma^{n!} = \iota$?

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  • $\begingroup$ This is not true. $\endgroup$ – darij grinberg Oct 2 '14 at 5:12
  • $\begingroup$ Counter example? $\endgroup$ – graydad Oct 2 '14 at 5:12
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    $\begingroup$ Think about $(1\ 2)(3\ 4\ 5) \in S_{5}$. $\endgroup$ – Alastair Litterick Oct 2 '14 at 5:14
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    $\begingroup$ $(12)\in S_3$. Hell, you can use Cauchy's theorem to show it's extremely false. $\endgroup$ – Adam Hughes Oct 2 '14 at 5:16
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This is incorrect, as Alastair's counterexample shows. Two related statements that are correct are:

  • If $\sigma$ is an $n$-cycle (more generally, an $m$-cycle for $m | n$) then $\sigma^n = \iota$.
  • For any finite group $G$, any element $g \in G$ satisfies $g^{\# G} = \iota$, and so for $\sigma \in S_n$, $\sigma^{n!} = 1$.
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  • $\begingroup$ That makes sense, thank you for explaining. How would you prove the second statement? $\endgroup$ – justin Oct 2 '14 at 5:28
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    $\begingroup$ It's a consequence of Lagrange's Theorem, which says that if $H$ is a subgroup of $G$ then $\# H | \# G$, which one can prove by counting cosets of $H$ in $G$. $\endgroup$ – Travis Oct 2 '14 at 5:35
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    $\begingroup$ (And you're welcome, I'm glad you found my answer useful.) $\endgroup$ – Travis Oct 2 '14 at 5:36
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you rememmber wrong, consider $S_4$ and consider $\sigma =(1,2,3)$, then $\sigma^4=(1,2,3)=\sigma$. I think, what your professor said was if $\sigma$ a $n$-cycle, then $\sigma^4=e$

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