Thread: Theorem 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?

Originally Posted by dwsmith

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?

If then .

Now, we have that and . So, .

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).