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

1. Feb 14, 2017

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 Dn is an dihedral group, we know its elements are symmetries, Dn = (R1,R2,R3...Ri) and since R is a symmetry, we know it's a permutation, so, each Ri can be written as a product of disjoint cycles (w1,w2...wk), now since each element in w represents the vertices's of Dn, 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 Dn is commutative when n <= 2 and therefore abelian when n <= 2.

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

2. Feb 15, 2017

### Staff: Mentor

So far so good.
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?

3. Feb 18, 2017

### 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.