I'm practicing for my upcoming Abstract Algebra midterm, and am trying an exercise in the Dummit and Foote Abstract Algebra book. I would like to verify my solution to the following exercise in Section 1.3 on Symmetric Groups. Did I do this right? If not, is there any way I can improve my answer?
Prove that the order of an element in $S_n$ equals the least common multiple of the lengths of the cycles in its cycle decomposition.
Here is my solution:
Proof: Suppose $\sigma\in S_n.$ Then $\sigma$ can be written as a cycle decomposition of $r\leq n-1$ disjoint cycles $\gamma_i$, so that $\sigma = \gamma_1\gamma_2\dots\gamma_r$. The order of each $\gamma_i$ is its length $m_i,$ so that $\gamma_i^{m_i}=e,$ the identity permutation. Let $L=\text{lcm}(m_1,m_2,\dots,m_r).$ We claim $|\sigma|=L.$ Indeed, since disjoint cycles commute, we have that $\sigma^L=(\gamma_1\gamma_2\dots\gamma_r)^L=\gamma_1^L\gamma_2^L\dots\gamma_r^L.$ Since each $m_i$ divides $L$, we have that $\sigma^L=ee\dots e=e.$ Thus, $|\sigma|$ divides $L.$ Furthermore, if it were true that $|\sigma|<L,$ then one of the $m_i$ would not divide $|\sigma|,$ so that $\gamma_i^{|\sigma|}\neq e.$ This forces $|\sigma|=L,$ meaning the order of $\sigma$ is the least common multiple of the lengths of the cycles in its cycle decomposition. $\blacksquare$
Thank you for your help!