# Thread: Theorem Question

1. ## Theorem Question

The order of an element in $S_n$ 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?

2. ## Re: Theorem Question

Originally Posted by dwsmith
The order of an element in $S_n$ 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 $a=(1 5 6)(2 3)(4)$ then $ord(a) = LCM (ord(156), ord(23), ord(4))$.

Now, we have that $ord(156)=3, ord(23)=2$ and $ord(4)=1$. So, $ord(a) =LCM(3,2,1) = 6$.

3. ## Re: 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).