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

Question

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

Answer 1

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 $\sigma \in S_n$, $\sigma^{n!} = 1$.

Answer 2

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 $\sigma^4=e$