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The statement is relatively simple but the proof is giving me some trouble. Any help will be very much appreciated:

Let $a,b \in S_n$. Assuming $<a> = <b>$ show that $b$ is a conjugate of $a$.

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You are aware that two elements of $S_n$ are conjugate if and only if they have the same cycle structure? That might help. – Gerry Myerson Nov 11 '12 at 23:25
up vote 2 down vote accepted


  1. If $h=(1,2,3,..,r)$ is a cycle, then $ghg^{-1}=(g(1),g(2),..,g(r))$ is a similar cycle of the same lenght, so Gerry Myerson's comment will easily follow.
  2. Assume that $a$ written as the product of disjoint cycles of length $r_1,r_2,..r_k\ $ ($a=c_1\cdot ..\cdot c_k$). Then $\langle a\rangle$ is a cyclic group of order $o(a)={\rm lcm}(r_1,r_2,..,r_k)$.
  3. As $\langle b\rangle=\langle a\rangle$, we have $b=a^s$ for some $s\in\Bbb N$. Show that $s$ has to be relatively prime to all $r_k$'s, and thus $b={c_1}^s\cdot ..\cdot {c_k}^s$ has the same cycle structure.

For this last step expanding $b=a^s$ you also need that disjoint cycles commute.

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