Let $p \in\mathbb{N}$ be a prime number and $\alpha \in S_n$. If ${\alpha}^p=1$ then $\alpha$ has three disjoint options:
$\ \bullet \alpha = 1$
$\ \bullet \alpha \text{ is a } p-cycle$
$\ \bullet \alpha \text{ is a product of } p-cycles$
Can someone give me some hints to do this? The only thing I know is that I can express $\alpha$ as a product of disjoint cycles, but I don't know how that will be relate to ${\alpha}^p$. Thanks!
Multiplying permutation by itself prime times to get identity
permutations
Best Answer
The order of a product of disjoint cycles is the lowest common multiple of the lengths of its component cycles. If this order is a prime $p$, this implies that all the component cycles (of order greater than 1) have lengths at most $p$ (to keep the LCM at $p$) and also have lengths that are powers of $p$ (to prevent prime factors other than $p$ from creeping into the LCM). Thus those cycles must be of length exactly $p$, which accounts for the last two cases in the question.
Either the permutation $\alpha$ is non-trivial, which leads to the analysis above, or it is trivial and $\alpha^p=1$ still. Thus we have shown that if $\alpha^p=1$ then $\alpha$ is one of the three types of permutations described in the question.