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If $A=\{1,2,...,n\}$, $\Omega _A$ is the set of all permutations over $A$, $S_n=(\Omega _A, \circ)$, then for any $\sigma \in \Omega _A$, the order $m$ of $\sigma$ (Smallest $m \in \mathbb{N}$ for which $\sigma^m=1$) equals the least common multiple of the cycle lengths in $\sigma$'s cycle decomposition.

I'm trying to prove it now myself..

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Hint Say $A=\{1,...,5\}$ and $\sigma = (1 2)(3 4 5)$. We know $(1 2)^2 = 1$ and $(3 4 5)^3 = 1$. Then $\sigma^k =1 \Longleftrightarrow \left((1 2)(3 4 5)\right)^k = 1 \Longleftrightarrow (1 2)^k (3 4 5)^k = 1$. So you must have that $(1 2)^k = 1$ and $(3 4 5)^k = 1$. Hence $2 | k$ and $3|k$. The smallest such number is 6.

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  • $\begingroup$ This is quite a Hint :) $\endgroup$ – user76568 Jan 30 '14 at 17:38
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    $\begingroup$ Many people do better with specific examples. Still some work to generalize. $\endgroup$ – John Habert Jan 30 '14 at 17:40
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    $\begingroup$ The key point is that disjoint cycles commute. $\endgroup$ – lhf Jan 31 '14 at 16:19

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