Smallest positive integer $n$ such that $S_n$ has an element of order $2n$

The Problem:

• Find the smallest positive integer $n$ such that $S_n$ has an element of order greater than $2n$

• Let $n$ be an even positive integer. Prove that $A_n$ had an element of order greater than $2n$ if and only if $n\ge 14$

I know that a permutation can be written as a product of disjoint cycles and the order of the LCM of the lengths of these cycles is the order of that permutation. Also if you write a permutation as a product of its disjoint cycles considering the invariant elements to be $1$-cycles, then the lengths of the disjoint cycles, $\{n_1,n_2,\cdots,n_k\}$, constitute a partition of the integer $n$. Then we need to find out the least integer $n$ such that the LCM of $n_1,n_2,\cdots,n_k$ for some partition of $n$ is $\ge 2n$.

• Yeah, thanks. But it's hard to be satisfied with hit and trial. How should I go about it rigorously? – Abishanka Saha Dec 5 '17 at 20:03
• $(1,2,3,4)(5,6,7,8,9)$ for $n=10$? – user491874 Dec 5 '17 at 20:04
• And there user87.... proves me wrong: $\;n\;$ can be even! – DonAntonio Dec 5 '17 at 20:06
• In the first part, do you mean exactly $2 n$ or at least $2 n$? – Qudit Dec 5 '17 at 20:53
• At least $2n$. I am editing the question likewise. – Abishanka Saha Dec 5 '17 at 20:54