TITLE: Element of order $2n$ in symmetric group $S_n$
QUESTION [6 upvotes]: I've been recently reading some articles about orders of elements in $S_n$ and I know that in order to find max order in $S_n$ we can use Landau function though I think that for small $n$ it is better to do it "manually".
My question is: For what $n$ can $S_n$ contain an element of order $2n$?
Could you tell me how to answer that question.
It's not possible for $S_3$ to have such element, because here $3-$cycle has order $3 \neq 6$.
Similarly for $S_4$ (max ord = $4$), 
for $S_5$ (max ord = $6$ for cycles  $(ab)(cde)$),
and for  $S_6$ - here we have max ord = $6$. 
for $S_7 $ max ord = $12$ for cycles $(abc)(defg)$. $14=2/cdot7$, but $2+7=9$
Then for $S_8$ we have max ord = $15$ for $(abc)(defgh)$, 
for $S_9$ we have max ord =$20$ for $(abcd)(efghi)$ but we won't get $18$ because $18$ cannot be written as a product of two coprime numbers $\in \{1,2,...,8\}$
It seems that it can occur if we can split permutation into disjoint cycles which lengths are coprime. But this is not sufficient.
Could you help me with that?
Thank you.

REPLY [4 votes]: Suppose that $n$ is not a power of a prime. Then we can write $n=ab$ where $a$ and $b$ are coprime and both at least $2$. Moreover we may assume that we shoved all the factors of $2$ into $a$, so that $b$ is odd. The plan is to prove that $2a+b\leq n$, so we can take the disjoint product of a $2a$-cycle and a $b$-cycle to get an element of order $2n$.
When $a$ and $b$ are both at least $3$ then $a+b \leq n/3 + 3$ (maximize $x+y$ subject to $x,y\geq 3$ and $xy=n$), so $2a+b \leq 2n/3 + 3 \leq n$ since $n\geq 9$. Otherwise we may assume that $a=2$ and $b=n/2$, in which case $2a+b = n/2 + 4 \leq n$ unless $n<8$.
It remains only to convince yourself that $S_n$ does not contain an element of order $2n$ when $n$ is $6$ or a power of a prime.