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

Cam someone please help me? I don't know how if I am doing this fine. And I am stuck. Thank you for any help.

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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.

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  • $\begingroup$ thank you , your explanation is very nice. $\endgroup$ – user84028 Sep 3 '15 at 16:39

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