# Prove that there exist no element of order 18 in $S_9$

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 ?

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An idea would be to first try to find a general statement of the special cases you mentioned. Ie, given a decomposition into disjoint cycles, what is the order of a permutation? – Tobias Kildetoft Mar 20 '13 at 13:33

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 $18$.
An element $\sigma$ of $\mathfrak{S}_9$ gives a partition of $9$, and $\text{ord}(\sigma)$ is the lcm of the elements of the partition. If $\text{ord}(\sigma)=18$, the partition contains a multiple of $2$ and of $3$; you can check that the lcm associated to such a partition is never $18$. (There are only five partitions satisfying this condition.)