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Prove that the order of an element in $S_n$ equals the least common multiple of the lengths of the cycles in its cycle decomposition.

Proof: Let $\sigma \in S_n$. Then $\sigma = (a_1a_2...a_{m_1})(a_{m_1+1}a_{m_1+2}...a_{m_2})....(a_{m_{k-1}+1}a_{m_{k-1}+2}...a_{m_{k}})$ represents the cycle decomposition of $\sigma$. Suppose $|\sigma| = n$ is the order of an element in $S_n$. So $\sigma^n = 1$.

Can someone please help me? I don't know how if I am doing this fine. And I am stuck.

Thank you for any help.

Shaun
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user84028
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1 Answers1

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I don't think you should name $\color{red}n$ the order of an element in $S_{\color{red}n}$.

If $\sigma = \gamma_1\gamma_2\dotsm\gamma_k$ is a decomposition of the permutation $\sigma$ as a product of disjoint cycles, these cycles commute with each other. Hence $\sigma^r=\gamma_1^r\gamma_2^r\dotsm\gamma_k^r$. and $$\sigma^r=\gamma_1^r\gamma_2^r\dotsm\gamma_k^r=e\iff \gamma_1^r=\gamma_2^r=\dotsm=\gamma_k^r=e$$ whence $r$ is a common multiple of $o(\gamma_1), o(\gamma_2),\dots,o(\gamma_k)$. The smallest of these $r$s is by definition the least common multiple of the orders of the cycles.

On the other hand, the order of a cycle is no other than its length.

Bernard
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  • thank you , your explanation is very nice. – user84028 Sep 03 '15 at 16:39
  • Waaaait ... $\gamma_1^r\dots \gamma_k^r = e$ doesn't obviously entail $\gamma_1^r=\cdots =\gamma_k^r=e$. – Addem Aug 19 '20 at 18:58
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    It entails so because the cycles are disjoint. – Bernard Aug 19 '20 at 19:11
  • $\gamma_1^r \gamma_2^r = e \Leftrightarrow \gamma_1^r = \gamma_2^r =e$ this statement intuitively makes sense as the cycles are disjoint. But I can't seem to form a proper reasoning for that. So the cycles are disjoint, but why would that follow? Any help? – William Nov 16 '21 at 14:53
  • @William: That's because $\gamma_1^r\gamma_2^r$ is the decomposition into disjoint cycles of $(\gamma_1\gamma_2)^r$, and the decomposition is unique, up to the order of the factors. Further, $e$ is its own decomposition. – Bernard Nov 16 '21 at 15:02
  • @Bernard Alternatively, would it be correct to say that if $\gamma_{1}^r ≠ e ≠ \gamma_{2}^r$ then $\gamma_{1}^r, \gamma_{2}^r$ are cycles of length $≥2$ and since their product is identity they are inverses. But a cycle and its inverse have the same symbols (just written in reverse) so that cannot be the case as they're disjoint. (This only works for product of two disjoint cycles whereas your explanation works for multiple disjoint cycles but I guess you can extend it from two to multiple since $\text{lcm}(a,b,c)= \text{lcm}(a,\text{lcm}(b,c))$ Is this rigorous enough or are there flaws? – William Nov 19 '21 at 21:57
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    @William: For me, it' s quite correct – I might have explained it the same way. – Bernard Nov 19 '21 at 22:03
  • Thanks @Bernard I also understood your explanation but sometimes it's just difficult to "convince" my dumb brain. – William Nov 19 '21 at 22:08
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    @William: You know, it's often a matter of habit of some types of arguments. I also often give only the main lines, leaving some details to be filled, as it helps making progress. Of course, if I'm asked for more details, I explain more. – Bernard Nov 19 '21 at 22:46