# Order of element in symmetric group with order greater than $2 n$

What is the smallest positive integer $n$ s.t $S_{n}$ has an element of order greater than $2 n$. By brute force, I have figured out than for $n = 9$, this condition is true. Is there any other way to figure this out?

First, every element in $S_n$ is a product of disjoint cycles. If $a \in S_n$ is an element, $a= c_1 \cdots c_m$ where the $c_1, \ldots, c_m$ are disjoint cycles of length $l_i$ for $1 \leq i \leq m$, then $\mathrm{ord}(a)= lcm(l_i)$.
For obvious reasons, given an $a \in S_n$ as above, we have $\sum_{i=1}^m l_i \leq n$. This now becomes a combinatorical problem:
In $S_2$ we find maximum order $2$, in $S_3$ we find maximum order $3$, in $S_4$ we find maximum order $4$, in $S_5$ we find maximum order $2 \cdot 3=6$, in $S_6$ we find maximum order $6$, in $S_7$ we find maximum order $3 \cdot 4 = 12$, in $S_8$ we find maximum order $5\cdot 3 = 15$, in $S_9$ we find maximum order $4 \cdot 5 = 20 > 18 = 2 \cdot 9$.
An element of order $20$ in $S_9$ is given for example by $(1234)(56789)$.
• And $S_{10}$ would have been the first time where looking at three factors kicks in: $(1\,2)(3\,4\,5)(6\,7\,8\,9\,10)$ has order $30$, which could not be achieved with two factors. – Hagen von Eitzen Jun 7 '18 at 13:49