TITLE: Prove that there exist no element of order 18 in $S_9$
QUESTION [5 upvotes]: Prove that there exist no element of order 18 in $S_9$.
How do I prove this ? I think the idea is that elements of the form: $(123456)(789)$ have order 6 as $\text{lcm}(6,3)=6$. Elements of the form $(123456789)(10\,11)$ surely don't exist in $S_9$. And I don't see any other ways to get to order 18. How do I prove this rigoursly ?

REPLY [7 votes]: It seems you already know that the order is the least common multiple of the cycle lengths. For the least common multiple of a set of numbers to be $2\cdot3^2$, at least one of the numbers must include a factor $3^2$ – but the only partition of $9$ with that property is the one with a single part $9$, and in this case the order is $9$, not $18$.