Let \(S_9\) be the symmetric group on nine letters.
- Exhibit an element of \(S_9\) of order \(20\).
- Prove that no element of \(S_9\) has order \(18\).
Solution
1. An element of order \(20\) is\[ \boxed{\sigma=(12345)(6789)}. \]
The two displayed cycles are disjoint and have lengths \(5\) and \(4\). Therefore \[ |\sigma| = \operatorname{lcm}(5,4) = 20. \]
2. No element of \(S_9\) has order \(18\).
Suppose that \(\tau\in S_9\) had order \(18\). Write \(\tau\) as a product of disjoint cycles, with cycle lengths \(d_1,\ldots,d_r\). The order of \(\tau\) is \[ \operatorname{lcm}(d_1,\ldots,d_r)=18. \] Since \(9\mid18\), at least one cycle length \(d_i\) must be divisible by \(9\). Every cycle in \(S_9\) has length at most \(9\), so \(d_i=9\). A \(9\)-cycle uses all nine letters, leaving only fixed points. Hence \(\tau\) itself has order \(9\), contradicting the assumption that its order is \(18\).
3. Q.E.D.
Step 1 proves part (1), and step 2 proves part (2).