The order of an element in equals the LCM of the lengths of the cycles in its cycle decomposition.

Suppose our cycle is (1 5 6)(2 3)(4). How can I use this to find the order of each element?

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- October 4th 2011, 06:13 AMdwsmithTheorem Question
The order of an element in equals the LCM of the lengths of the cycles in its cycle decomposition.

Suppose our cycle is (1 5 6)(2 3)(4). How can I use this to find the order of each element? - October 4th 2011, 01:12 PMzoekRe: Theorem Question
- October 4th 2011, 11:40 PMDevenoRe: Theorem Question
you can actually see this in action, and it makes lots more sense:

all 1-cycles are the identity, so we can disregard them (the map that takes 4-->4, and n-->n for all n ≠ 4, is the map that takes n-->n for all n).

so (1 5 6)(2 3)(4) = (1 5 6)(2 3).

[(1 5 6)(2 3)]^2 = (1 6 5) <--- here the 2-cycle factor "hits" the identity ( )

[(1 5 6)(2 3)]^3 = (2 3) <--- here the 3-cycle factor is the identity

[(1 5 6)(2 3)]^4 = (1 5 6) <--- 2-cycle is the identity

[(1 5 6)(2 3)]^5 = (1 6 5)(2 3)

[(1 5 6)(2 3)]^6 = ( ) <--- both factors are the identity.

powers of disjoint cycles are easy to compute, because they (the disjoint cycles) commute with each other,

(1 5 6)(2 3) = (2 3)(1 5 6).