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If I write (1 3 4 6)(2 3 4)(4 6 1) as a product of disjoint cycles, I get (164)(23), is it true that order of (1 3 4 6)(2 3 4)(4 6 1) is lcm(2,3)=6 (just the order of its disjoint cycles)?

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Write out the definition of what it means for a permutation to have order n and note that disjoint cycles commute. –  Qiaochu Yuan Mar 9 '11 at 12:16

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Yes, it's true that the order of a product of disjoint cycles is the least common multiple of the orders of the disjoint cycles in the product.

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In general, the order of commuting elements in a group is the lcm of their orders. –  lhf Oct 2 '13 at 11:53

Yes.

Let $\pi \in S$ be a permutation; then write $\pi = S_1 \cdots S_k$ where $S_1, \ldots, S_k \in S$ are disjoint cycles. Define $n_i = \#S_i$ for $i \in \{1,\ldots,k\}$. Now $\pi^{n_i}$ fixes the elements of $S_i$. So the smallest number which fixes all elements of $\pi$ is $\#\pi = \mathcal{lcm} \{n_1,\ldots,n_k\}$.

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fixates -> fixes –  user02138 Mar 9 '11 at 22:32
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Oh thx, I haz fixed –  ZulfiqarIII Mar 9 '11 at 22:48

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