When is D_{n} abelian? What's wrong with the proof?

[AdamsJoK]

I agree that this could have been done more simply(i'm not looking for an alternative proof), but I don't understand how it is wrong, any insight?

Since D_n is an dihedral group, we know its elements are symmetries, D_n = (R₁,R₂,R₃...R_i) and since R is a symmetry, we know it's a permutation, so, each R_i can be written as a product of disjoint cycles (w₁,w₂...w_k), now since each element in w represents the vertices's of D_n, which has n vertices's, it follows that w has a total length of n, therefore by problem 4,which states, two non equal cycles of length 2 commute if and only if they are disjoint and also that this isn't the case when the length is larger than 2, we know that D_n is commutative when n <= 2 and therefore abelian when n <= 2.

I'm not sure if this site has latex, thanks.

 

[fresh_42]

... two non equal cycles of length 2 commute if and only if they are disjoint ...

So far so good.

... and also that this isn't the case when the length is larger than 2 ...

What do you mean by this? Why shouldn't, e.g. $(1,2,3)(4,5,6,7)(8,9,10,11,12)$ don't be commutative?

The dihedral groups contain rotations and reflections. Wouldn't it be a lot easier to use the reflections as an argument, or if you will two-cycles?

 

[Martin Rattigan]

$e,\ (ab),\ (cd)$ and $(ab)(cd)$ gives an Abelian group that contains two-cycles from more than two elements and that are disjoint.