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..
Best Answer
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.