# Is it true that if $\sigma \in S_n,$ then $\sigma^n = \iota$?

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 $\sigma^{n!} = \iota$?

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

• 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 $\sigma \in S_n$, $\sigma^{n!} = 1$.
• 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 $G$. – Travis Oct 2 '14 at 5:35
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 $\sigma^4=e$