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Let $a$ be an element of $S_n$ ,the permutation group of order n. $a^k$ is a cycle of length n. Then what is the order of $a$?

If $n$ is prime then a should be a cycle of length $n$.But if $n$ is not prime then how to find the order of $a$? Let $a$ can be expressed as the product of some disjoint cycles. Then the order of $a$ is lcm of the length of the disjoint cycles and $n$ should divide the order of $a$. After that, I'm not getting anything.

Peter Taylor
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Unknown
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  • Hint: Disjoint cycle composition. Can $a$ have fixpoint? – Berci Apr 16 '19 at 10:14
  • If $f$ and $g$ are two disjoint cycles then $f$ fixes all elements of $g$ and $g$ fixes all elements of $f$.What do u mean by 'fixpoint'?I didn't get. – Unknown Apr 16 '19 at 10:26

2 Answers2

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Let $l:=|a|$ be the order of $a$. Hence $n=|a^k|=\frac{l}{(l,k)}$. Therefore, the order of $a$ is a solution $l$ of the integral equation $$l=n(l,k),$$ $(l,k)=\textrm{gcd}\{l,k\}$.

Bruno Rodrigues dos Santos
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Hint: let $f$ and $g$ be disjoint cycles. What can you say about $(fg)^k$?

Peter Taylor
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