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 $\ge 2n$.

I don't know what to do next? Please help.

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

To answer the first question: (1,2,3,4)(5,6,7,8,9) for n=9 appears to be the smallest n.

A counter example to the second statement is provided by the even permutation (1,2,3)(4,5,6,7,8,9,10) for n=10.

  • $\begingroup$ So there is no way other than hit and trial $\endgroup$ – Abishanka Saha Dec 6 '17 at 4:19

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