3

Prove that if $\sigma=\sigma_{1}\sigma_{2}\cdots\sigma_{m}$ is a product of disjoint cycles, then $\sigma^{lcm(length(\sigma _{1})\cdots length(\sigma_{m}))}=id=e$.

I am having trouble trying to prove this. I tried to first state the fact that the product of disjoint cycles commute. But that is all I've got so far. I am not sure how to connect this fact to the $lcm$.

Did
  • 279,727
  • 1
    So, you are after $\sigma^k=\sigma_{1}^k\sigma_{2}^k\cdots\sigma_{m}^k$. What would be some condition on $k$ ensuring that $\sigma_{1}^k=e$? Ensuring that $\sigma_{1}^k=\sigma_{2}^k=\cdots=\sigma_{m}^k=e$? Thus... – Did Feb 05 '18 at 07:16

1 Answers1

6

As the cycles are disjoint they commute: $\sigma_i\sigma_j=\sigma_j\sigma_i$ for all $i$ and $j$. This means that the power law holds: $\sigma^k=(\sigma_1\sigma_2\cdots\sigma_m)^k=\sigma_1^k\sigma_2^k\cdots\sigma_m^k$. So if $\sigma_i^k$ is the identity for all $i$, so is $\sigma^k$. But this will be the case if $k$ is a multiple of the lengths of all the cycles, so a multiple of their lcm.

Angina Seng
  • 158,341